Practice questions
Lecture 2: Chemistry of Life
1. A linear monosaccharide has its carbonyl group located at carbon 1. How should it be classified?
- An aldose
- A polysaccharide
- A ketose
- A disaccharide
Show answer
An aldose
2. What best explains why sodium and chloride ions can dissolve in water?
- Water forms covalent bonds between sodium and chloride
- Water converts the ions into non-polar molecules
- The ions become hydrophobic when surrounded by water
- Oppositely charged regions of water molecules interact with the ions
Show answer
Oppositely charged regions of water molecules interact with the ions
3. Which combination of properties makes carbon particularly suitable as the backbone of biological molecules?
- It cannot bond with other carbon atoms
- It is always positively charged and strongly hydrophilic
- It forms only ionic bonds and readily dissolves in water
- It forms stable covalent bonds with itself and many other elements
Show answer
It forms stable covalent bonds with itself and many other elements
4. A plant produces a lipid coating that reduces water loss from the surface of its leaves. Which class of lipid most directly performs this function?
- Triacylglycerols
- Phospholipids
- Steroids
- Waxes
Show answer
Waxes
5. Which statement correctly compares O₂ and H₂O?
- Neither is a molecule because each contains fewer than three atoms
- Both are compounds, but only O₂ is a molecule
- O₂ is a compound, whereas H₂O is only a molecule
- Both are molecules, but only H₂O is a compound
Show answer
Both are molecules, but only H₂O is a compound
6. A person who produces insufficient lactase will have difficulty performing which process?
- Converting fatty acids into carbohydrates
- Joining glucose molecules to form glycogen
- Producing lactose by photosynthesis
- Hydrolysing lactose into monosaccharides
Show answer
Hydrolysing lactose into monosaccharides
7. A freshwater organism maintains a stable internal salt concentration even when the surrounding water changes. Which property of life does this demonstrate?
- Homeostasis
- Adaptation
- Evolution
- Reproduction
Show answer
Homeostasis
8. Two monosaccharides join while a molecule of water is released. Which process and product are involved?
- Dehydration synthesis forming a hydrogen bond
- Dehydration synthesis forming a glycosidic bond
- Hydrolysis forming a glycosidic bond
- Hydrolysis forming a peptide bond
Show answer
Dehydration synthesis forming a glycosidic bond
9. Which function demonstrates that nucleotides have roles beyond forming DNA and RNA?
- Waterproofing plant surfaces
- Storing cellular energy as ATP
- Forming bacterial cell walls
- Forming polysaccharide fibres
Show answer
Storing cellular energy as ATP
10. An antibiotic prevents bacteria from constructing a major structural component of their cell walls. Which molecule is the most likely target?
- Amylopectin
- Glycogen
- Chitin
- Peptidoglycan
Show answer
Peptidoglycan
11. How are phospholipids arranged within a membrane separating two aqueous environments?
- Both heads and tails face the water equally
- Hydrophobic tails face the water while hydrophilic heads cluster internally
- Phospholipids remain randomly distributed because they are entirely non-polar
- Hydrophilic heads face the water while hydrophobic tails cluster internally
Show answer
Hydrophilic heads face the water while hydrophobic tails cluster internally
12. A molecule contains a long carbon–hydrogen chain attached to a polar head group. How should this molecule be classified?
- Amphipathic
- Hydrophobic
- Hydrophilic
- Ionic
Show answer
Amphipathic
13. Which structural feature distinguishes the sugar in RNA from the sugar in DNA?
- RNA contains a phosphate group where DNA contains a nucleobase
- RNA contains a hydroxyl group where DNA contains hydrogen at the 2′ carbon
- RNA contains six carbon atoms while DNA contains five
- RNA contains hydrogen where DNA contains a hydroxyl group at the 2′ carbon
Show answer
RNA contains a hydroxyl group where DNA contains hydrogen at the 2′ carbon
14. Which combination correctly represents the amino and carboxyl groups of an amino acid in an aqueous solution at approximately pH 7?
- –NH₃⁻ and –COO⁺
- –NH₃⁺ and –COO⁻
- –NH₂⁻ and –COOH⁺
- –NH₂ and –COOH
Show answer
–NH₃⁺ and –COO⁻
15. The 20 common amino acids share the same basic alpha structure but have different chemical properties. Which part of their structure is primarily responsible for these differences?
- The alpha carbon
- The R group
- The carboxyl group
- The amino group
Show answer
The R group
16. A large body of water warms slowly during the day despite absorbing substantial heat. Which property of water best explains this observation?
- High specific heat
- Versatility as a solvent
- Expansion upon freezing
- High surface tension
Show answer
High specific heat
17. Which comparison between glycogen and plant starch is most accurate?
- Glycogen is structural, whereas starch stores genetic information
- Glycogen is unbranched, whereas all forms of starch are branched
- Glycogen is generally more extensively branched than starch
- Glycogen occurs in plants, whereas starch occurs in animals
Show answer
Glycogen is generally more extensively branched than starch
EASY · 3 MARKS
1. Define monosaccharides, disaccharides and polysaccharides.
Show marking guide
1 mark for identifying monosaccharides as the simplest form of carbohydrate or single sugars.
1 mark for identifying disaccharides as two monosaccharides joined together.
1 mark for identifying polysaccharides as carbohydrate macromolecules composed of many sugars.
EASY · 3 MARKS
2. Describe the carbohydrate storage molecules found in plants and animals.
Show marking guide
1 mark for identifying starch as a carbohydrate storage molecule in plants.
1 mark for stating that starch contains unbranched amylose and branched amylopectin.
1 mark for identifying glycogen as the carbohydrate storage molecule in animals and noting that it is extensively branched.
EASY · 3 MARKS
3. Identify three properties that characterise living organisms.
Show marking guide
Award 1 mark for each correctly identified property, up to 3 marks. Accept any three of:
Order
Energy processing
Sensitivity or response to stimuli
Reproduction
Growth and development
Regulation or homeostasis
Adaptation
Evolution
EASY · 3 MARKS
4. Name three of the four elements most commonly found in living organisms.
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Award 1 mark for each correctly named element, up to 3 marks:
Carbon
Hydrogen
Nitrogen
Oxygen
INTERMEDIATE · 5 MARKS
5. Compare starch and glycogen as carbohydrate storage molecules.
Show marking guide
1 mark for identifying starch as a carbohydrate storage molecule in plants.
1 mark for identifying glycogen as a carbohydrate storage molecule in animals.
1 mark for recognising that both are polysaccharides that can be hydrolysed to provide monosaccharides.
1 mark for stating that starch includes amylose and amylopectin.
1 mark for explaining that glycogen is more extensively branched than starch.
INTERMEDIATE · 5 MARKS
6. Compare the basic composition of a nucleotide and a nucleoside, and describe one biological role of nucleotides other than forming DNA or RNA.
Show marking guide
1 mark for identifying a nucleotide as containing a sugar.
1 mark for identifying a nucleotide as containing a nucleobase.
1 mark for identifying a nucleotide as containing one or more phosphate groups.
1 mark for explaining that a nucleoside contains a sugar and nucleobase but no phosphate group.
1 mark for identifying an appropriate additional role of nucleotides, such as energy storage, metabolic regulation, signal transduction or immune function.
INTERMEDIATE · 5 MARKS
7. Explain how the amphipathic properties of phospholipids allow them to form a lipid bilayer in an aqueous environment.
Show marking guide
1 mark for identifying phospholipids as amphipathic molecules.
1 mark for identifying their polar head region as hydrophilic.
1 mark for identifying their hydrocarbon tail regions as hydrophobic.
1 mark for explaining that the hydrophilic heads interact with the surrounding aqueous environments.
1 mark for explaining that the hydrophobic tails cluster internally away from water, producing a bilayer that separates aqueous environments.
INTERMEDIATE · 5 MARKS
8. Explain why carbon is particularly well suited to forming the molecular backbone of living organisms.
Show marking guide
1 mark for identifying carbon as the major molecular backbone of biological molecules.
1 mark for explaining that carbon can bond with itself.
1 mark for explaining that carbon can also bond with many other elements.
1 mark for recognising that this bonding versatility allows a diverse range of biological molecules to form.
1 mark for explaining that carbon forms covalent bonds with an appropriate level of stability for biological molecules.
HARD · 8 MARKS
9. A hypothetical change causes water molecules to become non-polar and prevents hydrogen bonding between neighbouring water molecules. Predict four consequences this would have for properties of water discussed in the lecture, and explain why each would be affected.
Show marking guide
Award up to 2 marks for each of four appropriate consequences, consisting of 1 mark for identifying the affected property and 1 mark for an appropriate explanation. Accept:
Reduced cohesion or surface tension because hydrogen bonding between water molecules would be lost.
Reduced capillary action because cohesion between water molecules would be reduced.
Reduced ability to moderate temperature because the interactions responsible for water's high specific heat and heat of vaporisation would be disrupted.
Altered freezing behaviour because hydrogen bonds normally hold molecules apart in ice.
Ice would no longer necessarily be less dense than liquid water because the hydrogen-bonded arrangement responsible for expansion on freezing would be lost.
Reduced ability to dissolve polar or charged substances because water would no longer be polar.
Reduced effectiveness as a medium for transporting dissolved nutrients, wastes or chemical messages because solvent properties would be impaired.
Maximum 8 marks.
HARD · 8 MARKS
10. A newly discovered molecule contains a long carbon–hydrogen chain attached to a polar region containing oxygen and phosphorus. Predict how this molecule would interact with water and explain how its structure could allow it to contribute to a biological membrane.
Show marking guide
1 mark for identifying the carbon–hydrogen chain as non-polar.
1 mark for identifying the carbon–hydrogen chain as hydrophobic.
1 mark for identifying the oxygen- and phosphorus-containing region as polar or hydrophilic.
1 mark for classifying the overall molecule as amphipathic.
1 mark for explaining that the hydrophilic region would interact with the aqueous environment.
1 mark for explaining that the hydrophobic region would tend to avoid or be shielded from water.
1 mark for explaining that amphipathic molecules can arrange with hydrophilic regions facing water and hydrophobic regions clustered internally.
1 mark for linking this arrangement to formation of a bilayer that separates aqueous environments.
HARD · 8 MARKS
11. A person cannot efficiently hydrolyse lactose after consuming dairy products. Use the chemistry of carbohydrates described in the lecture to explain the molecular basis of this problem.
Show marking guide
1 mark for identifying lactose as a disaccharide.
1 mark for stating that a disaccharide is composed of two monosaccharides.
1 mark for identifying the covalent linkage between the monosaccharides as a glycosidic bond.
1 mark for stating that glycosidic bonds are formed by dehydration synthesis.
1 mark for explaining that digestion requires the disaccharide to be broken down by hydrolysis.
1 mark for recognising that enzymes facilitate this hydrolysis.
1 mark for identifying lactase as the enzyme required for lactose breakdown.
1 mark for explaining that insufficient lactase prevents efficient hydrolysis of lactose into its component monosaccharides.
HARD · 8 MARKS
12. At approximately pH 7, an α-amino acid is placed in aqueous solution. Describe its expected chemical structure and explain how amino acids can share a common basic structure while having different chemical properties.
Show marking guide
1 mark for identifying the central α-carbon.
1 mark for identifying the amino group.
1 mark for identifying the carboxyl group.
1 mark for identifying the variable R group or side chain.
1 mark for stating that the amino group is normally ionised as –NH₃⁺ at approximately pH 7.
1 mark for stating that the carboxyl group is normally ionised as –COO⁻ at approximately pH 7.
1 mark for recognising that the basic α-amino acid structure is shared by the 20 commonly encoded amino acids.
1 mark for explaining that differences in the R group produce differences in size and chemical properties that are important for protein structure and function.
Lecture 3: Biopolymers
1. A purified nucleic acid sample has an A260:A230 ratio of 1.3. What does this most strongly suggest?
- The sample may contain organic compound contamination
- The sample contains unusually high G-C content
- The nucleic acid has completely denatured
- The sample contains only pure DNA
Show answer
The sample may contain organic compound contamination
2. A mutation changes a polar amino acid sidechain to a nonpolar sidechain. Which property of the resulting protein could be directly affected?
- The identity of the peptide bond itself
- The interactions made by that sidechain with its environment
- The presence of the N-terminus
- The repeating N-C-C arrangement of the backbone
Show answer
The interactions made by that sidechain with its environment
3. Which combination correctly matches an RNA type with its function?
- tRNA — acts as an adaptor that carries amino acids during protein synthesis
- mRNA — structural component of ribosomes
- tRNA — stores the cell's long-term genetic information
- rRNA — carries amino acids to the ribosome
Show answer
tRNA — acts as an adaptor that carries amino acids during protein synthesis
4. A single RNA molecule folds so that two complementary regions pair with one another. Which statement best explains this structure?
- RNA can only base pair with a separate RNA molecule
- RNA must first be converted into DNA
- Complementary base pairing can occur between different regions of the same RNA strand
- Peptide bonds form between complementary RNA bases
Show answer
Complementary base pairing can occur between different regions of the same RNA strand
5. Which experimental observation is most directly explained by the negative charge of the nucleic acid backbone?
- DNA migrates in an electric field during electrophoresis
- Proteins absorb strongly near 280 nm
- RNA can fold through complementary base pairing
- Peptide bonds exhibit resonance
Show answer
DNA migrates in an electric field during electrophoresis
6. A purified sample has an A260:A280 ratio of 1.5. What is the most appropriate interpretation based on the lecture?
- The sample must contain no nucleic acid
- The sample likely contains excess RNA only
- The sample likely contains protein contamination
- The sample is pure DNA
Show answer
The sample likely contains protein contamination
7. Which feature of G-C base pairing best explains its greater contribution to DNA stability compared with A-T pairing?
- G-C pairs contain more phosphate groups
- G-C pairs form stronger phosphodiester bonds
- G-C pairs contain larger sugars
- G-C pairs form three hydrogen bonds rather than two
Show answer
G-C pairs form three hydrogen bonds rather than two
8. A student examines a nucleic acid molecule and finds that its sugar contains an additional hydroxyl group compared with deoxyribose. Which conclusion is most appropriate?
- The molecule is DNA and is generally more susceptible to degradation
- The molecule must be double-stranded DNA
- The molecule is RNA and is generally less chemically stable than DNA
- The molecule must contain thymine rather than uracil
Show answer
The molecule is RNA and is generally less chemically stable than DNA
9. A DNA nucleotide is incorporated into a growing nucleic acid strand. Which statement best describes how it becomes part of the backbone?
- Its phosphate participates in a phosphodiester linkage between neighbouring sugars
- Its sugar forms a hydrogen bond with the previous phosphate
- Its base forms a phosphodiester bond with the next base
- Its base forms a peptide bond with the previous nucleotide
Show answer
Its phosphate participates in a phosphodiester linkage between neighbouring sugars
10. Why does RNA generally not form the same B-type helical structure as DNA?
- RNA contains peptide bonds in its backbone
- RNA has no phosphates
- The additional hydroxyl group in ribose affects its structural properties
- RNA lacks nucleobases
Show answer
The additional hydroxyl group in ribose affects its structural properties
11. A researcher heats double-stranded DNA above its melting temperature. What is the most direct expected outcome?
- The phosphodiester backbone is converted into peptide bonds
- The DNA becomes RNA
- The bases are removed from the sugars
- The two DNA strands separate from one another
Show answer
The two DNA strands separate from one another
12. A nucleic acid sequence is written as 5′-AGTCGA-3′. Which sequence represents the complementary DNA strand aligned antiparallel to it?
- 5′-TCAGCT-3′
- 3′-UCAGCU-5′
- 3′-TCAGCT-5′
- 5′-AGTCGA-3′
Show answer
3′-TCAGCT-5′
13. During ethanol precipitation, why is salt added to a nucleic acid sample?
- To hydrolyse the nucleic acid backbone
- To help neutralise the negative charge of the nucleic acid
- To increase the number of phosphate groups
- To create peptide bonds between nucleotides
Show answer
To help neutralise the negative charge of the nucleic acid
14. Which statement best explains why DNA and proteins should not be considered chemically equivalent even though both can form helices?
- Only proteins contain covalent bonds
- DNA is built from nucleotides, whereas proteins are built from amino acids
- DNA and proteins contain identical backbones but different side groups
- Only DNA has directionality
Show answer
DNA is built from nucleotides, whereas proteins are built from amino acids
15. Cytosine spontaneously deaminates to uracil in an RNA molecule. Why is this change more difficult to identify as an error than the same change in DNA?
- Cytosine cannot be repaired in any nucleic acid
- RNA contains no cytosine
- Uracil is already a normal component of RNA
- RNA lacks a sugar-phosphate backbone
Show answer
Uracil is already a normal component of RNA
EASY · 3 MARKS
1. Describe how nucleotides are linked together within a nucleic acid strand.
Show marking guide
1 mark for stating that nucleotides are joined through the sugar-phosphate backbone.
1 mark for identifying the bond as a phosphodiester bond.
1 mark for stating that the phosphate links the 3′ carbon of one sugar to the 5′ carbon of the next sugar.
EASY · 3 MARKS
2. Describe the directionality convention used for nucleic acid sequences.
Show marking guide
1 mark for stating that a nucleic acid strand has a 5′ end and a 3′ end.
1 mark for stating that nucleic acids are written from 5′ to 3′.
1 mark for stating that nucleic acids are synthesised in one direction only.
EASY · 3 MARKS
3. Describe the three main components of a nucleotide.
Show marking guide
1 mark for identifying a phosphate group.
1 mark for identifying a pentose sugar.
1 mark for identifying a nucleobase.
EASY · 3 MARKS
4. Explain why DNA with a higher G-C content generally has a higher melting temperature.
Show marking guide
1 mark for stating that G pairs with C.
1 mark for stating that a G-C base pair forms three hydrogen bonds compared with two between A and T.
1 mark for explaining that the greater strength of G-C pairing increases DNA stability and therefore increases melting temperature.
INTERMEDIATE · 5 MARKS
5. Cytosine can spontaneously deaminate to form uracil. Compare the consequences of this reaction occurring in DNA and RNA.
Show marking guide
1 mark for stating that cytosine can spontaneously deaminate to form uracil.
1 mark for recognising that uracil is not normally expected as a base in DNA.
1 mark for stating that uracil appearing in DNA can therefore be recognised as incorrect and repaired.
1 mark for recognising that uracil is a normal component of RNA.
1 mark for explaining that the same change is therefore not recognised and repaired in RNA, contributing to RNA being more prone to mistakes.
INTERMEDIATE · 5 MARKS
6. One strand of a double-stranded DNA molecule has the sequence 5′-ATGCCGTA-3′. Write the sequence and direction of the complementary strand and explain the principles used to determine it.
Show marking guide
1 mark for identifying that A pairs with T.
1 mark for identifying that G pairs with C.
1 mark for stating that the two DNA strands are antiparallel.
1 mark for giving the correct complementary sequence: TACGGCAT.
1 mark for correctly writing its direction as 3′-TACGGCAT-5′ when aligned with the supplied strand.
INTERMEDIATE · 5 MARKS
7. Three purified nucleic acid samples have A260:A280 ratios of 1.9, 1.6 and 2.2. Interpret these results and explain the basis of the A260:A280 purity measurement.
Show marking guide
1 mark for identifying the sample with a ratio of 1.9 as consistent with pure DNA.
1 mark for identifying the sample with a ratio of 1.6 as showing likely protein contamination.
1 mark for identifying the sample with a ratio of 2.2 as showing likely RNA contamination.
1 mark for stating that nucleobases have characteristic absorbance around 260 nm.
1 mark for stating that proteins have characteristic absorbance around 280 nm, allowing the ratio of these measurements to provide information about nucleic acid purity.
INTERMEDIATE · 5 MARKS
8. Two double-stranded DNA molecules are the same length, but one contains a much higher proportion of G-C base pairs. Predict which molecule will have the higher melting temperature and explain why.
Show marking guide
1 mark for identifying the DNA molecule with the higher G-C content as having the higher melting temperature.
1 mark for stating that G pairs with C.
1 mark for stating that G-C base pairs form three hydrogen bonds.
1 mark for stating that A-T base pairs form two hydrogen bonds.
1 mark for explaining that the greater strength associated with G-C pairing means more heat is required to separate the two DNA strands.
HARD · 8 MARKS
9. A researcher compares two double-stranded DNA fragments of equal length. Fragment A contains 70% G-C base pairs, whereas Fragment B contains 30% G-C base pairs. Predict how the fragments will differ when progressively heated and explain the molecular basis for your prediction.
Show marking guide
1 mark for predicting that Fragment A will have the higher melting temperature.
1 mark for linking the higher melting temperature to the greater G-C content of Fragment A.
1 mark for stating that G pairs with C in complementary base pairing.
1 mark for stating that G-C base pairs form three hydrogen bonds.
1 mark for stating that A-T base pairs form two hydrogen bonds.
1 mark for explaining that G-C pairing is therefore stronger than A-T pairing.
1 mark for explaining that more thermal energy is required to separate the strands of Fragment A.
1 mark for recognising that heating sufficiently will ultimately separate both double-stranded DNA molecules into single strands.
HARD · 8 MARKS
10. A researcher isolates both DNA and RNA from cells and then stores the purified molecules under identical conditions. Explain why the RNA would generally be expected to be less stable than the DNA, integrating structural and chemical differences between the two molecules.
Show marking guide
1 mark for stating that DNA contains deoxyribose whereas RNA contains ribose.
1 mark for identifying the additional -OH group present in ribose.
1 mark for explaining that this additional -OH group makes the RNA backbone more susceptible to degradation or breakdown.
1 mark for stating that the DNA backbone is comparatively stable.
1 mark for recognising that RNA generally exists as single strands rather than forming a B-type DNA helix.
1 mark for stating that RNase enzymes can hydrolyse RNA.
1 mark for stating that DNase enzymes can accelerate hydrolysis of DNA.
1 mark for concluding that the structural properties of RNA make it intrinsically less stable even though enzymes can promote degradation of both nucleic acids.
HARD · 8 MARKS
11. Cytosine spontaneously deaminates to form uracil in both a DNA molecule and an RNA molecule. Analyse why the same chemical change has different implications for the two nucleic acids.
Show marking guide
1 mark for stating that cytosine can spontaneously deaminate to form uracil.
1 mark for recognising that thymine, rather than uracil, is normally present in DNA.
1 mark for explaining that uracil appearing in DNA can therefore be recognised as incorrect.
1 mark for stating that the incorrect uracil in DNA can be repaired.
1 mark for recognising that uracil is a normal nucleobase in RNA.
1 mark for explaining that uracil produced by cytosine deamination in RNA cannot be distinguished simply as an abnormal base.
1 mark for explaining that this contributes to RNA being more prone to mistakes.
1 mark for relating the greater tolerance of such mistakes to the lecture's distinction that RNA is not the genetic material.
Lecture 4: Central Dogma
1. The codons CAU and CAC both encode histidine. Which property of the genetic code does this demonstrate?
- The code is organism-specific
- The code is overlapping
- The code is redundant
- The code is composed of doublets
Show answer
The code is redundant
2. An AUG codon occurs in the middle of an open reading frame after translation has already begun. How should this codon be interpreted?
- It causes translation to restart from that position
- It cannot be translated because AUG only functions as the first codon
- It can encode methionine within the protein
- It must terminate translation
Show answer
It can encode methionine within the protein
3. An mRNA sequence contains the codon UGA within its open reading frame. What will occur when the ribosome reaches this codon?
- Tryptophan will be incorporated
- The reading frame will shift
- Methionine will be incorporated
- Protein synthesis will terminate
Show answer
Protein synthesis will terminate
4. Two different cell types in the same organism contain the same genome but produce different sets of proteins. Which explanation best accounts for this difference?
- Each cell type contains different nucleotide bases
- Each cell type uses a different genetic code
- Each cell type expresses a different subset of genes
- Each cell type possesses a different number of codons
Show answer
Each cell type expresses a different subset of genes
5. Consider the mRNA sequence 5′-CCCAUGCUAUUUUAA-3′. If translation begins at the AUG codon, which sequence represents the open reading frame?
- AUG-CUA-UUU
- CCA-UGC-UAU-UUU
- CCC-AUG-CUA-UUU
- AUG-CUA-UUU-UAA
Show answer
AUG-CUA-UUU-UAA
6. A point mutation changes one nucleotide in a codon but the resulting codon still specifies the same amino acid. Which feature of the genetic code makes this possible?
- Every amino acid is encoded by AUG
- Each nucleotide belongs to multiple overlapping codons
- Multiple codons can specify the same amino acid
- Codons can be read in either the 5′→3′ or 3′→5′ direction
Show answer
Multiple codons can specify the same amino acid
7. An mRNA sequence begins 5′-AUGCAUUGG-3′. How should this sequence be divided into codons when translation begins at the first nucleotide?
- AUG-CAU-UGG
- A-UGC-AUU-GG
- AU-GCA-UUG-G
- AUGC-AUUG-G
Show answer
AUG-CAU-UGG
8. A point mutation changes one codon so that a different amino acid is incorporated into the polypeptide. Which type of protein-level mutation has occurred?
- Nonsense
- Silent
- Frameshift
- Missense
Show answer
Missense
9. A single nucleotide is inserted into the coding region of an mRNA. What is the most likely consequence described in the lecture?
- The transcriptome becomes identical in every cell
- Only the inserted nucleotide is ignored during translation
- The reading frame changes, altering downstream codons
- The entire genome is duplicated
Show answer
The reading frame changes, altering downstream codons
10. A molecule begins reading a typical circular bacterial chromosome and continues around the chromosome without stopping. What could theoretically occur?
- It could eventually return to its original starting position
- It would reach the physical end of the chromosome
- It would convert the chromosome into a linear molecule
- It would eventually encounter a eukaryotic chromosome
Show answer
It could eventually return to its original starting position
11. A newly studied organism has substantially more chromosomes than humans. What can be concluded about its biological complexity from this information alone?
- It must have a larger proteome than humans
- It must be more complex than humans
- It must contain more protein-coding genes than humans
- Chromosome number alone does not reliably indicate organismal complexity
Show answer
Chromosome number alone does not reliably indicate organismal complexity
12. A point mutation changes an mRNA codon from CUU to CUC. What is the expected effect on the encoded protein?
- Leucine is replaced with isoleucine
- The codon still encodes leucine, so the amino acid sequence is unchanged
- A premature stop codon is introduced
- The reading frame changes from the mutation onwards
Show answer
The codon still encodes leucine, so the amino acid sequence is unchanged
13. What peptide would be produced from the mRNA sequence 5′-AUG CAU UGG UAA-3′ before translation terminates?
- Met-His-Trp-Stop
- Met-His-Trp
- Met-Trp-His
- His-Trp-Met
Show answer
Met-His-Trp
14. Why is identifying the start codon important when determining an open reading frame?
- The start codon prevents mutations in downstream codons
- The start codon identifies the final amino acid of the protein
- The start codon determines which RNA molecule is transcribed
- The start codon establishes the reading frame used for translation
Show answer
The start codon establishes the reading frame used for translation
15. A liver cell changes which genes it expresses after receiving a cellular stimulus. Which cellular feature would be expected to change as a result?
- The universal genetic code
- The nucleotide sequence of its entire genome
- Its transcriptome and potentially its proteome
- The total number of chromosomes
Show answer
Its transcriptome and potentially its proteome
16. Which combination correctly compares typical eukaryotic and prokaryotic chromosomes?
- Eukaryotic chromosomes are circular, whereas prokaryotic chromosomes are linear
- Both are usually linear, but only bacterial chromosomes contain genes
- Both are usually circular and wrapped around histones
- Eukaryotic chromosomes are typically linear, whereas most prokaryotic chromosomes are circular
Show answer
Eukaryotic chromosomes are typically linear, whereas most prokaryotic chromosomes are circular
17. A mature red blood cell lacks a nucleus. How does this distinguish it from most other cells in the same organism?
- It contains RNA but cannot contain proteins
- It uses a different genetic code to other cells
- It has a larger genome than nucleated cells
- It does not contain a nuclear copy of the organism's genome
Show answer
It does not contain a nuclear copy of the organism's genome
EASY · 3 MARKS
1. Define the genome, transcriptome and proteome.
Show marking guide
1 mark for defining the genome as the complete genetic information/DNA of an organism.
1 mark for defining the transcriptome as all RNA present in a given cell.
1 mark for defining the proteome as all proteins expressed in a given cell.
EASY · 3 MARKS
2. Describe three characteristics of typical prokaryotic genomes discussed in the lecture.
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Award 1 mark for each of the following, to a maximum of 3 marks:
Most prokaryotes have circular chromosomes.
Their genomes are generally smaller than eukaryotic genomes.
Most free-living bacterial chromosomes range from approximately 150 thousand to 15 million base pairs.
A molecule moving around a circular chromosome could theoretically return to its point of origin.
EASY · 3 MARKS
3. Describe three important features of the genetic code.
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Award 1 mark for each of the following, to a maximum of 3 marks:
The genetic code is a triplet code, with three nucleotides forming each codon.
The genetic code is non-overlapping, so each nucleotide is part of only one codon within a reading frame.
The genetic code is universal, meaning it is shared by almost all living organisms.
The genetic code is redundant/degenerate, meaning some amino acids are specified by more than one codon.
EASY · 3 MARKS
4. Explain what a frameshift mutation is and identify two types of DNA or RNA mutation that can cause one.
Show marking guide
1 mark for explaining that a frameshift changes the way nucleotides are grouped into codons/the reading frame.
1 mark for identifying an insertion as a mutation that can cause a frameshift.
1 mark for identifying a deletion as a mutation that can cause a frameshift.
EASY · 3 MARKS
5. What is an open reading frame, and how are its beginning and end identified?
Show marking guide
1 mark for defining an open reading frame as the region of a gene/mRNA that encodes a protein or peptide.
1 mark for identifying a start codon as the beginning of the open reading frame and determinant of the reading frame.
1 mark for identifying a stop codon as the end of the open reading frame.
INTERMEDIATE · 5 MARKS
6. A cell increases production of a particular protein in response to a change in its environment. Explain how this response could alter the cell's transcriptome and proteome without changing its genome.
Show marking guide
1 mark for stating that the genome contains the cell's genetic information and does not need to change for gene expression to change.
1 mark for explaining that the cell can increase expression of a particular gene.
1 mark for linking increased gene expression with increased production of the relevant RNA.
1 mark for explaining that this changes the composition or abundance of molecules within the transcriptome.
1 mark for linking increased protein synthesis with a corresponding change in the proteome.
INTERMEDIATE · 5 MARKS
7. A point mutation changes an mRNA codon from UAU to UAA within an open reading frame. Explain how this mutation would affect translation and classify its protein-level outcome.
Show marking guide
1 mark for identifying UAU as a codon for tyrosine.
1 mark for identifying UAA as a stop codon.
1 mark for explaining that the mutation introduces a stop signal earlier than in the original sequence.
1 mark for stating that translation would terminate prematurely.
1 mark for classifying the protein-level outcome as a nonsense mutation.
INTERMEDIATE · 5 MARKS
8. The mRNA sequence 5′-CUGAAGCAUGCUAUUUCUCCCCUAACGCGG-3′ has a number of possible reading frames. Explain how many possible reading frames there are, and why the reading frame containing AUG followed later by UAA is the most likely open reading frame.
Show marking guide
1 mark for recognising that an mRNA sequence can be divided into three possible reading frames.
1 mark for explaining that the grouping of nucleotides into triplet codons differs between reading frames.
1 mark for identifying AUG as a start codon within the relevant reading frame.
1 mark for identifying UAA as a downstream stop codon within the same reading frame.
1 mark for explaining that the region from the start codon to the stop codon represents an open reading frame that can encode a peptide/protein.
INTERMEDIATE · 5 MARKS
9. The mRNA sequences 5′-AUG CAU-3′ and 5′-AUG CAC-3′ encode the same peptide. Explain how this is possible and identify the peptide produced.
Show marking guide
1 mark for translating AUG as methionine (Met).
1 mark for recognising AUG as the start codon in these sequences.
1 mark for translating both CAU and CAC as histidine (His).
1 mark for identifying the peptide as Met-His.
1 mark for explaining that the same peptide can be encoded because the genetic code is redundant/degenerate, allowing more than one codon to specify the same amino acid.
HARD · 8 MARKS
10. A cell experiences an environmental change and increases production of one protein while decreasing production of another. Analyse how this observation relates to the central dogma and to the distinction between the genome, transcriptome and proteome.
Show marking guide
1 mark for stating the central dogma as DNA → RNA → protein.
1 mark for identifying transcription as the production of RNA from DNA.
1 mark for identifying translation as the production of a polypeptide/protein using RNA information.
1 mark for explaining that the cell's genome does not need to change for protein production to change.
1 mark for explaining that expression of different genes can be increased or decreased according to cellular needs.
1 mark for linking these changes in gene expression to changes in the amount or composition of RNA in the transcriptome.
1 mark for linking altered protein synthesis to changes in the proteome.
1 mark for recognising that both transcriptomes and proteomes can therefore vary with stimuli, environment or time.
HARD · 8 MARKS
11. A mutation duplicates part of a coding sequence. Explain why the effect of this duplication on the final protein cannot be predicted simply from knowing that a duplication occurred, and compare its possible effects with those of a single-nucleotide insertion or deletion.
Show marking guide
1 mark for defining a duplication as a mutation in which a segment of DNA or RNA is copied more than once.
1 mark for recognising that the protein-level effect of a duplication depends on the length of the duplicated sequence.
1 mark for recognising that the effect also depends on the position of the duplication.
1 mark for explaining that a duplication may alter the amino acid sequence without necessarily producing the same outcome in every case.
1 mark for explaining that some duplications can alter the reading frame.
1 mark for explaining that a single-nucleotide insertion can cause a frameshift by changing downstream codon groupings.
1 mark for explaining that a single-nucleotide deletion can likewise cause a frameshift.
1 mark for concluding that frameshifts can alter many downstream amino acids and potentially change where a stop codon is encountered.
HARD · 8 MARKS
12. A newly identified organism has 200 chromosomes and a genome containing 40 billion base pairs. A researcher concludes that it must be more biologically complex than humans. Evaluate this conclusion using the principles discussed in the lecture and compare the likely genome organisation of a eukaryote with that of a typical prokaryote.
Show marking guide
1 mark for stating that chromosome number does not strongly correlate with perceived organismal complexity.
1 mark for stating that total genome size/DNA length does not strongly correlate with perceived organismal complexity.
1 mark for concluding that the available information is therefore insufficient to establish that the organism is more complex than humans.
1 mark for recognising that eukaryotic genomes can vary greatly in size.
1 mark for stating that eukaryotic chromosomes are typically linear.
1 mark for explaining that eukaryotic DNA is condensed into chromatin and associated with histone proteins.
1 mark for stating that most prokaryotic chromosomes are circular.
1 mark for recognising that typical prokaryotic genomes are generally much smaller than eukaryotic genomes.
HARD · 8 MARKS
13. Consider the mRNA sequence 5′-CUGAAGCAUGCUAUUUCUCCCCUAACGCGG-3′. Identify the open reading frame shown in the lecture, translate the encoded peptide, and explain why the other possible reading frames would produce different interpretations of the same nucleotide sequence.
Show marking guide
1 mark for recognising that there are three possible reading frames when the sequence is read 5′→3′.
1 mark for identifying the relevant reading frame as the one containing AUG as a start codon.
1 mark for grouping the open reading frame correctly as AUG|CUA|UUU|CUC|CCC|UAA.
1 mark for translating AUG as Met.
1 mark for translating CUA, UUU, CUC and CCC as Leu, Phe, Leu and Pro respectively.
1 mark for identifying UAA as a stop codon that does not encode an amino acid.
1 mark for giving the peptide as Met-Leu-Phe-Leu-Pro.
1 mark for explaining that shifting the starting position changes how nucleotides are grouped into non-overlapping triplet codons and therefore changes the amino acids or stop codons encountered.
HARD · 8 MARKS
14. The normal mRNA sequence 5′-AUG CUA UUU CUC CCC UAA-3′ is altered by deletion of one nucleotide within the coding region. Explain why this mutation may have a much greater effect on the amino acid sequence than a silent point mutation.
Show marking guide
1 mark for identifying removal of a nucleotide as a deletion.
1 mark for explaining that the genetic code is read in triplets/codons.
1 mark for explaining that codons are non-overlapping.
1 mark for stating that deletion of a single nucleotide changes the grouping of subsequent nucleotides into codons.
1 mark for classifying this outcome as a frameshift.
1 mark for explaining that many or all downstream amino acids may therefore change.
1 mark for explaining that the altered frame may also encounter a different stop codon and alter the length of the polypeptide.
1 mark for contrasting this with a silent point mutation, in which a changed codon still encodes the same amino acid because of redundancy in the genetic code.
Lecture 5: Copying DNA and RNA
1. Which feature distinguishes RNA transcription from DNA replication?
- Transcription generally copies one DNA strand for a gene rather than both strands of the genome
- Transcription produces nucleic acid in the 3′→5′ direction
- Transcription requires complementary base pairing but replication does not
- Transcription requires multiple RNA primers
Show answer
Transcription generally copies one DNA strand for a gene rather than both strands of the genome
2. Why can a cell produce different amounts of RNA from different genes?
- RNA polymerase permanently removes genes that are not required
- Every gene must be transcribed at the same rate
- DNA replication determines how many RNA molecules each gene produces
- Transcription frequency can be regulated by factors such as promoter strength, repressors and activators
Show answer
Transcription frequency can be regulated by factors such as promoter strength, repressors and activators
3. During transcription elongation, what happens to the DNA after RNA polymerase has moved through a region?
- The template DNA is degraded
- The two DNA strands reanneal
- The DNA is converted into RNA
- Both DNA strands are permanently separated
Show answer
The two DNA strands reanneal
4. Why are regions rich in A/T bases commonly associated with origins of replication?
- A/T-rich regions contain RNA primers before replication begins
- A/T bases are only present in origins of replication
- DNA polymerase specifically recognises A/T-rich sequences
- A/T base pairs are easier to separate than G/C base pairs
Show answer
A/T base pairs are easier to separate than G/C base pairs
5. During rho-independent termination, a G/C-rich region of the RNA forms a hairpin. What is the consequence of this structure?
- It recruits DNA polymerase III
- It causes RNA polymerase to produce an RNA primer
- It begins DNA replication
- It pauses transcription and contributes to RNA polymerase dissociation
Show answer
It pauses transcription and contributes to RNA polymerase dissociation
6. Why are transcription factors important during bacterial transcription initiation?
- They replace RNA primers with DNA
- They join newly synthesised RNA fragments
- They prevent the DNA strands from reassociating throughout the chromosome
- They recognise specific promoter sequences and facilitate RNA polymerase binding
Show answer
They recognise specific promoter sequences and facilitate RNA polymerase binding
7. Which sequence of events correctly describes synthesis of the lagging strand?
- Primase makes RNA primers → DNA polymerase III produces DNA fragments → DNA polymerase I replaces primers → ligase joins fragments
- DNA polymerase III makes primers → primase extends them → DNA polymerase I joins them
- Helicase synthesises RNA primers → DNA polymerase III removes them → ligase copies the DNA
- DNA ligase makes a primer → DNA polymerase I synthesises fragments → primase joins them
Show answer
Primase makes RNA primers → DNA polymerase III produces DNA fragments → DNA polymerase I replaces primers → ligase joins fragments
8. Which comparison between DNA replication and RNA transcription is correct?
- Replication and transcription both require DNA ligase to join fragments
- Replication requires a primer, whereas transcription does not
- Replication copies approximately one gene, whereas transcription copies the whole genome
- Replication uses U instead of T during complementary base pairing
Show answer
Replication requires a primer, whereas transcription does not
9. A DNA replication fork is moving along a chromosome. Why must the leading and lagging strands be synthesised differently?
- DNA polymerase can only synthesise DNA in the 5′→3′ direction while the template strands are antiparallel
- Only the lagging strand requires complementary base pairing
- DNA polymerase can synthesise DNA in either direction, but only one strand contains a promoter
- The leading strand contains RNA whereas the lagging strand contains DNA
Show answer
DNA polymerase can only synthesise DNA in the 5′→3′ direction while the template strands are antiparallel
10. A bacterial promoter contains regions located approximately at −35 and −10 relative to the transcription start site. What is the main significance of these regions?
- They are origins from which DNA replication begins
- They act as termination signals
- They encode the final nucleotides of the RNA transcript
- They contribute to recognition of the promoter during transcription initiation
Show answer
They contribute to recognition of the promoter during transcription initiation
11. Why does lagging-strand synthesis require multiple RNA primers?
- The lagging strand is synthesised discontinuously as the replication fork progresses
- Each nucleotide requires a separate primer
- DNA polymerase changes direction after every few nucleotides
- RNA primers are used to join completed DNA molecules together
Show answer
The lagging strand is synthesised discontinuously as the replication fork progresses
12. Why do eukaryotic chromosomes generally contain multiple origins of replication?
- Each origin replicates only the lagging strand
- Their large linear genomes would be inefficient to replicate from only one starting location
- Each gene must have its own origin of replication
- Eukaryotic DNA polymerases synthesise DNA more slowly than bacterial polymerases
Show answer
Their large linear genomes would be inefficient to replicate from only one starting location
13. A protein blocks RNA polymerase from binding to a gene's promoter. How would this protein affect gene expression?
- It would increase DNA replication at the gene
- It would act as a repressor and decrease transcription
- It would increase transcription of the gene
- It would convert the promoter into a terminator
Show answer
It would act as a repressor and decrease transcription
14. What would most likely happen if single-stranded binding proteins were unable to function during DNA replication?
- Okazaki fragments would become longer
- DNA replication would proceed without a template
- The separated DNA strands would be more likely to reassociate
- DNA polymerase would begin synthesising RNA
Show answer
The separated DNA strands would be more likely to reassociate
15. A researcher inhibits DNA ligase during replication. Which structure would be expected to remain unjoined?
- The two parental DNA strands
- RNA nucleotides within a primer
- The two replication forks
- Okazaki fragments on the lagging strand
Show answer
Okazaki fragments on the lagging strand
EASY · 3 MARKS
1. Describe how Okazaki fragments are processed to produce a continuous DNA strand.
Show marking guide
Award 1 mark for each of the following:
DNA polymerase III synthesises the DNA within the Okazaki fragments.
DNA polymerase I replaces the RNA primers with DNA.
DNA ligase joins the DNA fragments together.
EASY · 3 MARKS
2. Describe three general features of DNA replication.
Show marking guide
Award 1 mark for each of the following:
DNA replication is semi-conservative.
DNA replication is bi-directional.
DNA replication is semi-discontinuous.
EASY · 3 MARKS
3. Outline the three main stages of RNA transcription.
Show marking guide
Award 1 mark for each of the following:
Initiation: RNA polymerase is recruited to/binds the promoter and transcription begins.
Elongation: RNA polymerase moves along the DNA template and synthesises the RNA strand.
Termination: transcription stops at a termination signal and the RNA transcript is released.
EASY · 3 MARKS
4. State three differences between DNA replication and RNA transcription.
Show marking guide
Award 1 mark for each valid difference, to a maximum of 3 marks. Acceptable differences include:
DNA replication uses DNA polymerases and several other replication enzymes, whereas transcription mainly uses RNA polymerase with transcription factors.
Replication copies both DNA strands, whereas transcription uses one template strand for a particular gene.
Replication requires an RNA primer, whereas transcription does not require a primer.
Replication uses A–T and G–C base pairing, whereas transcription uses A–U and G–C.
Replication copies the whole genome, whereas transcription generally produces a transcript approximately the length of one gene.
Both processes synthesise in the 5′→3′ direction, so this is not a difference.
INTERMEDIATE · 5 MARKS
5. Compare rho-independent and rho-dependent termination of bacterial transcription.
Show marking guide
Award 1 mark for each of the following:
Both mechanisms result in termination of transcription and release of the RNA transcript.
Rho-independent termination uses a signal encoded within the transcribed RNA.
A G/C-rich region can form an RNA hairpin that pauses transcription and promotes RNA polymerase dissociation.
Rho-dependent termination involves Rho binding to a C-rich rut site on ribosome-free RNA.
Rho reaches RNA polymerase and promotes release of the transcription components through its helicase activity.
INTERMEDIATE · 5 MARKS
6. Compare the organisation of origins of replication in typical prokaryotic and eukaryotic genomes and explain why they differ.
Show marking guide
Award 1 mark for each of the following:
Prokaryotic genomes are generally smaller and circular.
They generally contain a single origin of replication.
Eukaryotic chromosomes are linear and much larger.
Eukaryotic chromosomes contain multiple origins of replication.
Multiple origins allow large eukaryotic chromosomes to be replicated more efficiently than replication from only one starting location.
INTERMEDIATE · 5 MARKS
7. Explain why DNA replication is described as semi-conservative, bi-directional and semi-discontinuous.
Show marking guide
Award 1 mark for each of the following:
Semi-conservative replication produces DNA molecules containing one parental strand and one newly synthesised strand.
Replication begins at an origin and proceeds in two directions.
This produces two replication forks moving away from the origin.
The leading strand is synthesised continuously.
The lagging strand is synthesised discontinuously as Okazaki fragments.
INTERMEDIATE · 5 MARKS
8. Explain why synthesis of the lagging strand requires more processing than synthesis of the leading strand.
Show marking guide
Award 1 mark for each of the following:
DNA polymerase can synthesise new DNA only in the 5′→3′ direction.
The two parental DNA strands are antiparallel.
The leading strand can therefore be synthesised continuously as the replication fork progresses.
The lagging strand must be synthesised discontinuously as multiple Okazaki fragments initiated from multiple RNA primers.
DNA polymerase I replaces the RNA primers with DNA and DNA ligase joins the resulting fragments.
HARD · 8 MARKS
9. A bacterial RNA transcript cannot form the hairpin required for rho-independent termination, but it contains a functional C-rich rut site. Explain how transcription could still terminate.
Show marking guide
Award 1 mark for each of the following:
Rho-independent termination normally involves a signal encoded in the RNA.
A G/C-rich sequence can normally form a hairpin structure.
The hairpin pauses transcription and contributes to RNA polymerase dissociation.
In this scenario, that termination mechanism would be impaired because the hairpin cannot form.
The C-rich rut site provides a binding site for Rho protein.
Rho binds to ribosome-free RNA at the rut site.
Rho can reach RNA polymerase and use helicase activity to induce release of the transcription components.
Transcription could therefore terminate through the rho-dependent pathway despite failure of rho-independent termination.
HARD · 8 MARKS
10. A bacterial gene has an intact coding region, but mutations alter both its −10 and −35 promoter regions so that the sigma factor no longer recognises them effectively. Predict the effect on transcription and explain why the RNA sequence itself may remain unchanged in any transcripts that are produced.
Show marking guide
Award 1 mark for each of the following:
The −10 and −35 regions are components of the bacterial promoter.
The sigma factor is a transcription factor that recognises specific promoter sequences.
Sigma factor binding facilitates recruitment of RNA polymerase.
Reduced promoter recognition would therefore reduce effective RNA polymerase binding/initiation.
Transcription of the gene would consequently occur less frequently.
The promoter lies upstream of the transcription start site.
The RNA sequence is determined by complementary copying of the DNA template downstream from the promoter.
Therefore, any transcripts that are successfully initiated could retain their normal sequence if the transcribed DNA sequence itself is unchanged.
HARD · 8 MARKS
11. A eukaryotic chromosome is experimentally modified so that nearly all of its origins of replication are inactive. Explain why this would reduce the efficiency of genome replication compared with a normal chromosome.
Show marking guide
Award 1 mark for each of the following:
Eukaryotic chromosomes are large and linear.
They normally contain multiple origins of replication.
Replication begins at an origin by locally opening the DNA.
Each activated origin produces two replication forks.
Replication proceeds bi-directionally away from each origin.
Multiple origins therefore allow many regions of a chromosome to be copied simultaneously.
With very few functional origins, much larger regions would need to be replicated from each remaining origin.
Replication of the chromosome would therefore take longer or be substantially less efficient.
HARD · 8 MARKS
12. A student claims that DNA replication and RNA transcription are essentially identical because both copy information from DNA and synthesise nucleic acids in the 5′→3′ direction. Evaluate this claim by identifying important similarities and differences between the two processes.
Show marking guide
Award 1 mark for each of the following:
Both processes use DNA as a template.
Both depend on complementary base pairing.
Both synthesise the new nucleic acid strand in the 5′→3′ direction.
Replication copies both parental DNA strands, whereas transcription uses one template strand for a particular gene.
Replication requires a primer, whereas transcription does not.
Replication uses DNA polymerases and several additional replisome proteins, whereas transcription mainly uses RNA polymerase with transcription factors.
Replication produces DNA using A–T and G–C pairing, whereas transcription produces RNA using A–U and G–C pairing.
Replication copies the whole genome, whereas transcription generally produces a much shorter RNA transcript, so the processes are related but not identical.
Lecture 6: Making Proteins
1. A newly synthesised polypeptide consists of a particular sequence of amino acids linked by peptide bonds. Which level of protein structure has been produced directly by translation?
- Primary structure
- Secondary structure
- Tertiary structure
- Quaternary structure
Show answer
Primary structure
2. Why is tRNA described as an adapter molecule during translation?
- It forms the ribosome binding site on mRNA
- It links mRNA codons with their corresponding amino acids
- It directly catalyses peptide bond formation
- It converts proteins back into RNA sequences
Show answer
It links mRNA codons with their corresponding amino acids
3. Two regions of the same polypeptide that are more than 100 amino acid residues apart contribute to the same β-sheet. Which feature of β-sheets does this demonstrate?
- β-sheets can form only between neighbouring amino acids
- β-sheets require covalent bonding between every participating residue
- β-sheets cannot occur within a single polypeptide
- Hydrogen bonding can occur between segments that are distant in the primary sequence
Show answer
Hydrogen bonding can occur between segments that are distant in the primary sequence
4. A protein functions as a single folded polypeptide and does not associate with other polypeptide subunits. Which statement is most appropriate?
- The protein cannot form a functional three-dimensional shape
- The protein has quaternary structure because every protein must contain four structural levels
- The protein can have primary, secondary and tertiary structure without having quaternary structure
- The protein cannot contain α-helices or β-sheets
Show answer
The protein can have primary, secondary and tertiary structure without having quaternary structure
5. Haemoglobin consists of four individual polypeptide subunits assembled into a functional protein. Which level of structure is specifically illustrated by this assembly?
- Quaternary structure
- Tertiary structure
- Primary structure
- Secondary structure
Show answer
Quaternary structure
6. Which observation best explains why transcription and translation can be coupled in prokaryotic cells?
- Prokaryotic mRNA does not need to be transported out of a nucleus before translation
- Prokaryotes use the same enzyme for transcription and translation
- Prokaryotic ribosomes are located inside the nucleus
- Prokaryotic translation does not require mRNA
Show answer
Prokaryotic mRNA does not need to be transported out of a nucleus before translation
7. A newly charged tRNA has entered the A site of a ribosome and its anticodon correctly matches the mRNA codon. What is expected to occur next?
- The ribosome immediately disassembles
- The mRNA moves into the nucleus
- Peptide bond formation occurs between the positioned amino acids
- The tRNA exits directly through the E site
Show answer
Peptide bond formation occurs between the positioned amino acids
8. Which description best distinguishes the stabilisation of secondary structure from tertiary structure?
- Secondary structure requires multiple polypeptides, whereas tertiary structure occurs only in monomers
- Secondary structure involves hydrogen bonding between backbone groups, whereas tertiary structure involves interactions between amino acid R groups
- Secondary structure is stabilised by R-group interactions, whereas tertiary structure involves only backbone hydrogen bonds
- Secondary structure depends on peptide bonds, whereas tertiary structure depends only on phosphodiester bonds
Show answer
Secondary structure involves hydrogen bonding between backbone groups, whereas tertiary structure involves interactions between amino acid R groups
9. During termination, peptidyl transferase contributes to release of the completed polypeptide by:
- charging the final tRNA with ATP
- adding water rather than another amino acid
- breaking down the mRNA template
- attaching another amino acid to the chain
Show answer
adding water rather than another amino acid
10. A protein contains several α-helices and β-sheets that fold together into an overall three-dimensional shape. Which level of structure describes this overall spatial arrangement?
- Secondary structure
- Primary structure
- Tertiary structure
- Quaternary structure
Show answer
Tertiary structure
11. An α-helix is disrupted when hydrogen bonding between peptide backbone groups is disturbed. Which level of protein structure is most directly affected?
- Primary structure
- Tertiary structure only
- Secondary structure
- Quaternary structure
Show answer
Secondary structure
12. A protein is exposed to conditions that cause denaturation but do not chemically break its peptide bonds. Which feature would be expected to remain intact?
- Its original three-dimensional fold
- Its complete tertiary structure
- Its primary amino acid sequence
- Its biological activity
Show answer
Its primary amino acid sequence
13. Which feature distinguishes initiation of protein synthesis in eukaryotes from that in prokaryotes as described in the lecture?
- Only eukaryotes use ribosomes
- Eukaryotes do not require initiation factors
- Eukaryotes begin translation with normal methionine rather than formylated methionine
- Eukaryotes translate mRNA in the 3′ to 5′ direction
Show answer
Eukaryotes begin translation with normal methionine rather than formylated methionine
14. After peptide bond formation and movement of the ribosome by three bases, what happens to the tRNA that has donated its amino acid?
- It remains permanently in the P site
- It moves to the E site and exits the ribosome
- It moves back to the A site to collect another amino acid
- It binds a stop codon
Show answer
It moves to the E site and exits the ribosome
15. Why does charging a tRNA contribute to the energy requirements of protein synthesis?
- ATP is required to form the mRNA codons
- ATP is used to move the ribosome along the mRNA
- ATP is used by aminoacyl-tRNA synthetase when attaching an amino acid to its tRNA
- ATP is incorporated directly into the growing polypeptide
Show answer
ATP is used by aminoacyl-tRNA synthetase when attaching an amino acid to its tRNA
16. A tRNA has successfully paired its anticodon with an mRNA codon during elongation. Which feature ensures that the amino acid carried by this tRNA is appropriate for that codon?
- A release factor attaches the correct amino acid to the tRNA
- An aminoacyl-tRNA synthetase charges the tRNA with its specific amino acid
- Peptidyl transferase determines which amino acid the tRNA should carry
- The ribosome directly selects the amino acid before tRNA binding
Show answer
An aminoacyl-tRNA synthetase charges the tRNA with its specific amino acid
EASY · 3 MARKS
1. Describe primary, secondary and tertiary protein structure.
Show marking guide
1 mark for describing primary structure as the amino acid sequence of a polypeptide linked by peptide bonds.
1 mark for describing secondary structure as local structures such as α-helices and β-sheets stabilised by hydrogen bonding.
1 mark for describing tertiary structure as the overall three-dimensional arrangement of a protein involving interactions between amino acid R groups.
EASY · 3 MARKS
2. Describe the functions of the A, P and E sites of the ribosome during translation.
Show marking guide
1 mark for identifying the A site as the site where incoming charged tRNAs bind.
1 mark for identifying the P site as the site associated with the growing peptide chain/peptide elongation.
1 mark for identifying the E site as the site from which uncharged tRNA exits the ribosome.
EASY · 3 MARKS
3. Describe the roles of mRNA, tRNA and rRNA during protein synthesis.
Show marking guide
1 mark for stating that mRNA contains the template/information specifying which amino acids are added during protein synthesis.
1 mark for stating that tRNA matches amino acids to mRNA codons and carries the appropriate amino acid.
1 mark for stating that rRNA combines with proteins to form the ribosome and/or contributes to catalysing peptide bond formation.
EASY · 3 MARKS
4. Describe three key events that occur during initiation of translation in bacteria.
Show marking guide
1 mark for stating that the ribosome recognises/binds the ribosome binding site or Shine-Dalgarno sequence on the mRNA.
1 mark for stating that the small and large ribosomal subunits assemble, assisted by initiation factors.
1 mark for stating that the initiating tRNA carries formylated methionine (fMet) and is positioned at the start codon.
EASY · 3 MARKS
5. Describe three key events that occur during the elongation stage of translation.
Show marking guide
1 mark for stating that a new charged tRNA enters the A site according to codon-anticodon matching.
1 mark for stating that peptidyl transferase catalyses peptide bond formation between the positioned amino acids.
1 mark for stating that the ribosome moves forward by three bases, allowing tRNAs to progress through the ribosomal sites.
INTERMEDIATE · 5 MARKS
6. Compare translation in prokaryotic and eukaryotic cells using features described in the lecture.
Show marking guide
1 mark for recognising that translation follows the same general stages of initiation, elongation and termination in both cell types.
1 mark for stating that eukaryotes use a different mechanism from the bacterial Shine-Dalgarno sequence to identify the start codon.
1 mark for stating that prokaryotic initiation uses fMet whereas eukaryotic initiation uses normal methionine.
1 mark for stating that eukaryotes use a single release factor to recognise all three stop codons whereas prokaryotes use multiple release factors.
1 mark for explaining that transcription and translation can be coupled in prokaryotes because mRNA does not need to be transported out of a nucleus.
INTERMEDIATE · 5 MARKS
7. Compare translation initiation and termination in bacteria.
Show marking guide
1 mark for stating that initiation begins when the ribosome recognises the ribosome binding site/Shine-Dalgarno sequence.
1 mark for stating that the ribosomal subunits assemble with the assistance of initiation factors.
1 mark for stating that the initiating tRNA carries formylated methionine (fMet) and is positioned at the start codon.
1 mark for stating that termination begins when the ribosome reaches a stop codon, which has no matching tRNA.
1 mark for stating that a release factor binds and promotes release of the polypeptide followed by disassembly of the translation machinery.
INTERMEDIATE · 5 MARKS
8. Explain how protein denaturation can affect protein structure and biological activity without changing the amino acid sequence.
Show marking guide
1 mark for defining denaturation as a loss of protein structural integrity.
1 mark for stating that denaturation disrupts higher-order protein structure.
1 mark for stating that primary structure generally remains intact because peptide bonds are not normally disrupted by denaturation alone.
1 mark for explaining that loss of the normal three-dimensional structure results in loss of biological activity.
1 mark for recognising that denaturation may be irreversible for some proteins but reversible, through renaturation, for others.
INTERMEDIATE · 5 MARKS
9. Explain how tRNA charging and codon-anticodon recognition work together to ensure that the correct amino acid is added during translation.
Show marking guide
1 mark for stating that each tRNA contains an anticodon that recognises a complementary codon on the mRNA.
1 mark for stating that an aminoacyl-tRNA synthetase recognises a specific tRNA and its corresponding amino acid.
1 mark for stating that the synthetase uses ATP to attach the amino acid to the tRNA.
1 mark for stating that the resulting charged tRNA carries the amino acid to the ribosome.
1 mark for explaining that codon-anticodon matching positions the charged tRNA so that the appropriate amino acid can be incorporated into the growing polypeptide.
INTERMEDIATE · 5 MARKS
10. Explain what happens to tRNAs and the growing polypeptide during one cycle of translation elongation.
Show marking guide
1 mark for stating that a charged tRNA with a complementary anticodon enters the A site.
1 mark for stating that the amino acids associated with the A-site and P-site tRNAs are positioned next to one another.
1 mark for stating that peptidyl transferase catalyses peptide bond formation, transferring/extending the growing peptide chain.
1 mark for stating that the ribosome moves forward by three bases along the mRNA.
1 mark for stating that the uncharged tRNA moves to the E site and exits while the tRNA carrying the peptide moves to the P site.
HARD · 8 MARKS
11. A mutation changes a stop codon in a bacterial mRNA into a codon that is recognised by a charged tRNA. Predict what would happen during translation and explain the molecular basis for your prediction.
Show marking guide
1 mark for stating that normal stop codons do not have matching tRNAs.
1 mark for stating that a release/termination factor normally binds when a stop codon enters the ribosome.
1 mark for stating that this normally causes peptidyl transferase to add water rather than another amino acid.
1 mark for stating that this releases the completed polypeptide.
1 mark for recognising that the mutated codon can instead be recognised by a complementary charged tRNA.
1 mark for predicting that an amino acid would therefore be incorporated at this position.
1 mark for predicting that elongation could continue beyond the original termination position.
1 mark for concluding that translation would continue until another functional stop codon is encountered, potentially producing a longer polypeptide.
HARD · 8 MARKS
12. An aminoacyl-tRNA synthetase incorrectly attaches the wrong amino acid to a tRNA, but the tRNA anticodon remains unchanged. Explain how this error could affect the primary structure and function of the resulting protein.
Show marking guide
1 mark for stating that aminoacyl-tRNA synthetases normally charge tRNAs with the appropriate amino acid.
1 mark for stating that the tRNA anticodon determines which complementary mRNA codon the tRNA recognises.
1 mark for recognising that the incorrectly charged tRNA could still bind its normal codon because its anticodon is unchanged.
1 mark for explaining that the wrong amino acid could therefore be incorporated into the growing polypeptide.
1 mark for stating that this would alter the primary amino acid sequence of the protein.
1 mark for explaining that protein sequence influences protein folding.
1 mark for explaining that altered folding can change the three-dimensional structure of the protein.
1 mark for concluding that altered three-dimensional structure can change or disrupt protein function.
HARD · 8 MARKS
13. Compare the consequences of disrupting hydrogen bonds that stabilise an α-helix with breaking peptide bonds in the same region of a protein.
Show marking guide
1 mark for identifying the α-helix as secondary structure.
1 mark for stating that α-helices are stabilised by hydrogen bonds between backbone C=O and N-H groups.
1 mark for recognising that disrupting these hydrogen bonds can destabilise or unfold the α-helix.
1 mark for stating that disruption of these hydrogen bonds does not necessarily change the amino acid sequence.
1 mark for identifying peptide bonds as covalent bonds linking amino acids in the primary structure.
1 mark for explaining that breaking peptide bonds would disrupt the primary structure/polypeptide chain.
1 mark for explaining that disruption of either level of structure can affect higher-order folding.
1 mark for linking altered higher-order structure to possible loss or alteration of protein function.
HARD · 8 MARKS
14. Haemoglobin is composed of four folded polypeptide subunits. Suppose each individual subunit retains its normal primary, secondary and tertiary structure, but the subunits can no longer associate with one another. Explain which level of protein structure is directly affected and how this illustrates the relationship between the four levels of protein structure.
Show marking guide
1 mark for defining primary structure as the amino acid sequence of each polypeptide.
1 mark for identifying α-helices and β-sheets as examples of secondary structure.
1 mark for defining tertiary structure as the overall three-dimensional arrangement of an individual folded polypeptide.
1 mark for defining quaternary structure as the association of multiple polypeptide subunits into a functional complex.
1 mark for identifying haemoglobin as a tetramer containing four polypeptide subunits.
1 mark for recognising that the primary, secondary and tertiary structures of the individual subunits could remain intact in the scenario.
1 mark for identifying quaternary structure as the level directly disrupted by failure of the subunits to associate.
1 mark for recognising that not all proteins require quaternary structure, but proteins such as haemoglobin depend on assembly of multiple subunits to form their normal multimeric structure.
Lecture 7: Enzymes and Thermodynamics
1. An acid phosphatase has maximum activity at approximately pH 5. What would most likely happen if a strong base shifted the reaction mixture to a very high pH?
- The enzyme concentration would automatically increase
- Enzyme activity would fall substantially or stop
- The equilibrium constant would increase
- The reaction would become more exergonic
Show answer
Enzyme activity would fall substantially or stop
2. A reaction begins with mostly reactant and has a strongly negative ΔG. As product accumulates, what is expected to happen to ΔG?
- It remains unchanged throughout the reaction
- It rises towards zero
- It becomes increasingly negative
- It immediately becomes positive
Show answer
It rises towards zero
3. Sucrose hydrolysis is exergonic but can proceed extremely slowly in sterile water. What does this demonstrate?
- All reactions with negative ΔG occur rapidly
- Activation energy determines whether ΔG is positive or negative
- Thermodynamic favourability does not determine reaction speed
- Exergonic reactions cannot occur spontaneously
Show answer
Thermodynamic favourability does not determine reaction speed
4. Two theoretical enzyme reactions are performed under the same experimental conditions with infinite substrate concentration. The second reaction contains twice the enzyme concentration. What is expected for the initial reaction rate?
- It will remain unchanged
- It will approximately double
- It will become zero
- It will be half as fast
Show answer
It will approximately double
5. A biochemical reaction has a positive ΔG and therefore cannot proceed favourably on its own. How can cells drive such a reaction?
- Increase its activation energy
- Allow entropy within the universe to decrease
- Couple it to an exergonic reaction that provides energy
- Remove all enzymes from the pathway
Show answer
Couple it to an exergonic reaction that provides energy
6. In a reversible reaction, the forward and reverse reactions continue to occur at equal rates. What does this indicate?
- The system is at dynamic equilibrium
- The reaction has stopped
- The reaction is necessarily exergonic
- All reactants have been converted into products
Show answer
The system is at dynamic equilibrium
7. Which change would most directly increase the rate of a thermodynamically favourable reaction without changing its overall ΔG?
- Making ΔG more negative
- Changing the equilibrium constant
- Increasing the Gibbs free energy of the products
- Lowering the activation energy
Show answer
Lowering the activation energy
8. During an energy transformation, some energy becomes dispersed as heat. Which principle best explains this observation?
- Energy transformations are inefficient and entropy increases
- Gibbs free energy always increases
- Energy is created during cellular reactions
- Equilibrium prevents energy loss
Show answer
Energy transformations are inefficient and entropy increases
9. A contracting muscle converts stored chemical energy into movement. Which energy transformation is occurring?
- Kinetic energy is being converted into chemical energy
- Potential energy is being converted into kinetic energy
- Thermal energy is being converted into potential energy
- Entropy is being converted into chemical energy
Show answer
Potential energy is being converted into kinetic energy
10. Why can an enzyme be used repeatedly to catalyse the same type of reaction?
- It changes into a new enzyme after each reaction
- It permanently binds to each product
- It increases its own concentration during the reaction
- It is not consumed in the overall reaction
Show answer
It is not consumed in the overall reaction
11. An enzyme normally functions within a specific cellular compartment but is detected at an unusual location. Why might this be biologically important?
- Enzyme location determines the overall ΔG of its reactions
- Enzyme compartmentalisation can contribute to regulation, and abnormal location may indicate a problem
- Enzymes function identically regardless of location
- Enzymes can only function outside cells
Show answer
Enzyme compartmentalisation can contribute to regulation, and abnormal location may indicate a problem
12. A mutation causes an enzyme within a metabolic pathway to lose activity. What is a possible consequence described in the lecture?
- The enzyme automatically changes into a different enzyme
- The equilibrium constant of all cellular reactions changes
- An alternative pathway may partly compensate because of biological redundancy
- Every metabolic pathway in the cell must stop
Show answer
An alternative pathway may partly compensate because of biological redundancy
13. ATP hydrolysis has a ΔG of approximately −30.5 kJ/mol. Why is this useful for cellular energy coupling?
- ATP hydrolysis prevents any energy from being lost as heat
- ATP hydrolysis is exergonic and can provide energy to drive endergonic processes
- ATP hydrolysis changes the equilibrium constant of all coupled reactions
- ATP hydrolysis is endergonic and absorbs energy
Show answer
ATP hydrolysis is exergonic and can provide energy to drive endergonic processes
14. Two identical reactions are performed with and without an enzyme. Which feature should remain unchanged between the reactions?
- The reaction rate
- The time required to reach equilibrium
- The activation energy
- The overall ΔG
Show answer
The overall ΔG
15. An enzyme is added to a reaction that would otherwise proceed very slowly. What is the expected effect?
- The enzyme supplies energy to make the reaction favourable
- The reaction reaches equilibrium more quickly
- The equilibrium position shifts towards products
- The overall ΔG becomes more negative
Show answer
The reaction reaches equilibrium more quickly
16. An enzyme specifically binds a substrate because its active site interacts appropriately with the substrate's transition-state form. Which property of enzymes does this demonstrate?
- Enzymes alter equilibrium constants
- Enzymes make all reactions spontaneous
- Enzymes are highly specific
- Enzymes are consumed during reactions
Show answer
Enzymes are highly specific
17. A cell uses ATP to move ions against their concentration gradient. Which cellular process is being powered by ATP?
- Active transport
- Thermoregulation
- Signal transduction
- Equilibrium
Show answer
Active transport
18. A chemical reaction has a negative ΔG but occurs extremely slowly at room temperature. Which explanation is most appropriate?
- The reaction must actually have a positive ΔG
- The reaction cannot occur without changing its equilibrium constant
- The reaction is thermodynamically favourable but has a high activation energy
- The reaction has already reached equilibrium
Show answer
The reaction is thermodynamically favourable but has a high activation energy
EASY · 3 MARKS
1. Define activation energy and explain its relationship to reaction rate.
Show marking guide
1 mark for defining activation energy as the initial energy investment required to start a reaction.
1 mark for recognising activation energy as an energy barrier that must be overcome for the reaction to proceed.
1 mark for explaining that lowering the activation energy allows the reaction to proceed more rapidly.
EASY · 3 MARKS
2. Distinguish between kinetic energy and potential energy.
Show marking guide
1 mark for describing kinetic energy as energy associated with motion.
1 mark for describing potential energy as stored energy associated with the location or structure of matter.
1 mark for providing an appropriate example of either form, such as muscle contraction for kinetic energy or chemical bonds for potential energy.
EASY · 3 MARKS
3. Explain the basic concept of energy coupling in cells.
Show marking guide
1 mark for defining energy coupling as using energy from an exergonic process to drive an endergonic reaction.
1 mark for identifying ATP as a major mediator of energy coupling in cells.
1 mark for recognising ATP hydrolysis as an exergonic process that releases energy that can support endergonic cellular processes.
EASY · 3 MARKS
4. Explain the difference between thermodynamics and kinetics when considering a chemical reaction.
Show marking guide
1 mark for stating that thermodynamics determines whether a reaction is energetically favourable.
1 mark for relating thermodynamic favourability to ΔG.
1 mark for stating that kinetics describes how quickly a reaction proceeds towards equilibrium.
INTERMEDIATE · 5 MARKS
5. A cellular reaction has a positive ΔG and therefore does not proceed favourably on its own. Explain how energy coupling involving ATP can allow the reaction to occur.
Show marking guide
1 mark for recognising that a positive ΔG indicates an endergonic and thermodynamically unfavourable reaction.
1 mark for defining energy coupling as using energy from an exergonic process to drive an endergonic process.
1 mark for identifying ATP as a major mediator of energy coupling in cells.
1 mark for recognising ATP hydrolysis as an exergonic reaction that releases usable energy.
1 mark for explaining that this released energy can be coupled to the endergonic cellular reaction.
INTERMEDIATE · 5 MARKS
6. Explain how catabolic and anabolic processes are linked through ATP in cells.
Show marking guide
1 mark for describing catabolism as the breakdown of fuel molecules.
1 mark for recognising that catabolic processes can be exergonic and release energy.
1 mark for explaining that released energy can be used to synthesise ATP from ADP and inorganic phosphate.
1 mark for recognising that ATP hydrolysis can release energy for cellular work.
1 mark for explaining that this energy can support endergonic anabolic processes such as protein synthesis.
INTERMEDIATE · 5 MARKS
7. Two enzyme reactions are performed under identical experimental conditions with a high substrate concentration. The second reaction contains twice the enzyme concentration of the first. Predict how their initial reaction rates should compare and explain how the rates would be measured.
Show marking guide
1 mark for predicting that the reaction with twice the enzyme concentration should have approximately twice the initial reaction rate.
1 mark for recognising that the comparison assumes the same experimental conditions.
1 mark for recognising that substrate concentration is sufficiently high in this example.
1 mark for explaining that reaction rate can be measured from the change in product concentration over time.
1 mark for stating that the slope of the initial linear portion of the product-versus-time curve is used to calculate the reaction rate.
HARD · 8 MARKS
8. A cellular reaction has ΔG = +20 kJ/mol. The cell couples this reaction to ATP hydrolysis, which has ΔG = −30.5 kJ/mol. Explain how energy coupling can allow the cellular reaction to proceed and determine the overall ΔG of the coupled processes.
Show marking guide
1 mark for identifying the cellular reaction as endergonic because its ΔG is positive.
1 mark for recognising that the reaction is thermodynamically unfavourable on its own.
1 mark for identifying ATP hydrolysis as an exergonic process.
1 mark for recognising that ATP hydrolysis releases energy that can be coupled to an endergonic cellular process.
1 mark for defining energy coupling as using an exergonic process to drive an endergonic process.
1 mark for correctly combining the ΔG values: +20 + (−30.5) = −10.5 kJ/mol.
1 mark for recognising that the overall coupled process therefore has a negative ΔG.
1 mark for concluding that coupling to ATP hydrolysis can make the combined process thermodynamically favourable.
HARD · 8 MARKS
9. A reversible reaction begins with a high concentration of reactant and very little product. Its initial ΔG is strongly negative. Explain how the reaction and its ΔG are expected to change as the system progresses towards equilibrium.
Show marking guide
1 mark for recognising that the initial negative ΔG provides a thermodynamic drive towards product formation.
1 mark for stating that the forward reaction is initially favoured.
1 mark for recognising that reactant concentration decreases as the reaction proceeds.
1 mark for recognising that product concentration increases.
1 mark for explaining that ΔG depends on the reactant-to-product ratio and is therefore not fixed throughout the reaction.
1 mark for stating that ΔG becomes less negative and approaches zero as product accumulates.
1 mark for stating that ΔG = 0 at equilibrium.
1 mark for explaining that equilibrium is dynamic, with equal forward and reverse reaction rates and no net change in reactant or product concentrations.
HARD · 8 MARKS
10. An enzyme normally functions within a particular cellular compartment and is initially produced as an inactive zymogen. A mutation causes the enzyme to become active before it reaches its normal location. Explain why both compartmentalisation and zymogen formation can regulate enzyme activity, and predict why premature activation could be problematic.
Show marking guide
1 mark for recognising that compartmentalisation restricts enzymes to particular locations inside or outside cells.
1 mark for explaining that localisation helps control where an enzyme can interact with its substrates.
1 mark for recognising that enzyme location is therefore one mechanism of regulating activity.
1 mark for defining a zymogen as an inactive form of an enzyme.
1 mark for explaining that zymogen formation allows enzyme activity to be controlled through activation.
1 mark for recognising that activation can provide regulation when localisation alone is insufficient.
1 mark for predicting that premature activation could allow the enzyme to act in an inappropriate location or at an inappropriate time.
1 mark for recognising that enzymes in the wrong place can cause or indicate biological problems.
Lecture 9: Photosynthesis I
1. What is the functional significance of water oxidation at photosystem II?
- It replenishes electrons lost from the reaction centre while releasing oxygen and protons
- It generates the final sugar products of the Calvin cycle
- It consumes oxygen to create glucose in the thylakoid lumen
- It converts ATP directly into NADP⁺
Show answer
It replenishes electrons lost from the reaction centre while releasing oxygen and protons
2. Why did oxygenic photosynthesis profoundly alter early Earth?
- It stopped all organisms from using redox chemistry
- Water-splitting photosynthesis released oxygen that accumulated and changed atmospheric chemistry
- It made biological carbon fixation independent of light or chemical energy
- It converted atmospheric oxygen permanently into nitrogen
Show answer
Water-splitting photosynthesis released oxygen that accumulated and changed atmospheric chemistry
3. Why are two photon-absorbing photosystems involved in the non-cyclic light reactions?
- Both photosystems are used exclusively to export ATP from the chloroplast
- Light-driven excitation at successive reaction centres supports electron flow from water towards NADP⁺
- One photosystem absorbs carbon dioxide while the other directly synthesises glucose
- Two photosystems cancel the need for a redox reaction
Show answer
Light-driven excitation at successive reaction centres supports electron flow from water towards NADP⁺
4. Why is photosystem II called “II” even though it operates before photosystem I?
- Photosystem II absorbs no light until after photosystem I has ceased working
- The Roman numeral refers to the number of membranes surrounding each photosystem
- The photosystems were named in discovery order rather than functional order
- Photosystem I belongs to respiration whereas photosystem II belongs to photosynthesis
Show answer
The photosystems were named in discovery order rather than functional order
5. Which event initiates charge separation within a photosystem?
- Carbon dioxide reacts with a proton to form NADPH
- An antenna pigment hydrolyses ATP to produce a photon
- An excited reaction-centre chlorophyll transfers an electron to an acceptor
- Oxygen accepts ATP from the thylakoid lumen
Show answer
An excited reaction-centre chlorophyll transfers an electron to an acceptor
6. Why does excitation energy move through an antenna complex towards the reaction centre?
- The same electron travels sequentially through every antenna pigment to the stroma
- Energy is transferred between pigments before the first actual electron-transfer event at the centre
- Antenna pigments release carbon dioxide to transport excitation
- ATP synthase transports the excited electron through chlorophyll molecules
Show answer
Energy is transferred between pigments before the first actual electron-transfer event at the centre
7. Which set of products correctly links the light reactions to carbon fixation?
- Glucose provides the energy and carbon dioxide supplies the electron acceptor for ATP synthase
- ADP provides reducing power and oxygen provides phosphate
- ATP provides usable chemical energy and NADPH provides reducing power
- Water provides all carbon atoms and oxygen directly reduces NADP⁺
Show answer
ATP provides usable chemical energy and NADPH provides reducing power
8. What does the label P680 signify in photosystem II?
- A protein consisting of 680 amino acids in every photosystem
- The number of ATP molecules produced by a chloroplast
- A reaction-centre chlorophyll with strongest absorption near 680 nm
- The wavelength of oxygen released by water oxidation
Show answer
A reaction-centre chlorophyll with strongest absorption near 680 nm
9. Which sequence correctly links the two principal stages of photosynthesis?
- The Calvin cycle oxidises water, then the antenna fixes carbon dioxide
- Light reactions directly assemble glucose without a subsequent carbon-fixation stage
- Light reactions generate ATP and NADPH, which can support carbon fixation in the Calvin cycle
- The Calvin cycle generates light, which is absorbed by the photosystems
Show answer
Light reactions generate ATP and NADPH, which can support carbon fixation in the Calvin cycle
10. Which direction best describes proton movement through thylakoid ATP synthase during ATP formation?
- From the chloroplast into the cytosol through the nuclear envelope
- From the stroma into the mitochondrial matrix
- From the stroma towards the lumen through ATP synthase
- From the lumen towards the stroma
Show answer
From the lumen towards the stroma
11. Why must cells continuously regenerate ATP?
- Cells cannot couple energy-releasing and energy-requiring reactions
- Energy-releasing reactions always require a new chlorophyll molecule
- ATP is rapidly recycled as an energy carrier rather than maintained as a large long-term reserve
- ATP is biologically stable only in the complete absence of water
Show answer
ATP is rapidly recycled as an energy carrier rather than maintained as a large long-term reserve
12. Which comparison of NADH and NADPH reflects their roles in the lecture?
- Both are structural pigments in the chloroplast antenna
- NADPH is the principal electron sink of aerobic respiration and NADH fixes carbon
- NADH is prominent in catabolic energy capture, whereas NADPH supplies reducing power for biosynthesis
- Neither can donate electrons after being reduced
Show answer
NADH is prominent in catabolic energy capture, whereas NADPH supplies reducing power for biosynthesis
13. Why are electron carriers often absent from the overall equations for respiration or photosynthesis?
- They move energy only as heat rather than through reactions
- They are used and regenerated across individual steps and cancel in the net equation
- They do not participate in either pathway
- They are always consumed completely before products appear
Show answer
They are used and regenerated across individual steps and cancel in the net equation
14. Which observation would identify glucose synthesis from carbon dioxide as anabolic rather than catabolic?
- It produces ATP through hydrolysis of ADP
- It releases stored chemical energy by oxidising glucose
- It constructs an energy-rich organic molecule and requires an energy input
- It transfers electrons to oxygen during glucose breakdown
Show answer
It constructs an energy-rich organic molecule and requires an energy input
15. During aerobic glucose breakdown, which change best identifies the oxidation of glucose?
- All glucose carbon atoms become part of ATP
- Glucose acquires electrons from NADH without being degraded
- Electrons are removed from glucose-derived intermediates and transferred to carriers
- Electrons are created from chemical energy rather than transferred
Show answer
Electrons are removed from glucose-derived intermediates and transferred to carriers
EASY · 3 MARKS
1. Describe the role of water in the light reactions.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for water supplies electrons that replace those lost from PSII.
1 mark for water oxidation releases protons.
1 mark for water oxidation releases molecular oxygen.
EASY · 3 MARKS
2. Distinguish heterotrophy from autotrophy as introduced in the lecture.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for heterotrophs acquire pre-formed organic compounds as food.
1 mark for heterotrophs extract energy from those organic compounds.
1 mark for autotrophs synthesise organic compounds from inorganic carbon using an energy source such as sunlight.
EASY · 3 MARKS
3. Explain the function of a photosystem antenna complex.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for antenna complexes contain light-absorbing pigments.
1 mark for excitation energy is transferred between pigments towards the reaction centre.
1 mark for reaction-centre excitation can initiate actual electron transfer to an acceptor.
EASY · 3 MARKS
4. Identify the order and principal reaction-centre labels of the two photosystems.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify photosystem II as first in non-cyclic transport.
1 mark for identify PSII reaction centre as P680.
1 mark for identify PSI reaction centre as P700.
INTERMEDIATE · 5 MARKS
5. Explain how PSII supplies electrons for non-cyclic photosynthetic electron transport.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify PSII as operating before PSI.
1 mark for describe light-induced electron loss from the P680 reaction centre.
1 mark for state that water provides replacement electrons.
1 mark for explain that water oxidation releases oxygen.
1 mark for explain how replacement supports continued electron flow towards PSI/NADP⁺.
INTERMEDIATE · 5 MARKS
6. Explain how an apparently simple overall equation can conceal the mechanisms of metabolism.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify the overall equation as a net balance of reactants and products.
1 mark for describe metabolic transformations as occurring via multiple steps.
1 mark for explain that intermediate redox reactions transfer electrons.
1 mark for identify electron carriers such as NAD(H) or NADP(H) as participating in those steps.
1 mark for explain why carriers may cancel from the summed net equation.
INTERMEDIATE · 5 MARKS
7. Explain the importance of chloroplast compartmentalisation for ATP production.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify lumen and stroma as distinct compartments.
1 mark for locate the photosynthetic electron transport chain in the thylakoid membrane.
1 mark for explain proton accumulation in the lumen.
1 mark for describe the resulting electrochemical gradient.
1 mark for explain the use of proton flow via ATP synthase to produce ATP.
INTERMEDIATE · 5 MARKS
8. Explain why photosynthesis is an anabolic process and how it differs energetically from the breakdown of glucose.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for describe photosynthesis as building energy-rich organic compounds from smaller/inorganic precursors.
1 mark for explain that synthesis requires an energy input.
1 mark for identify glucose breakdown as catabolic.
1 mark for explain that catabolism releases energy stored in organic molecules.
1 mark for connect catabolic energy capture to cellular work via ATP or electron carriers.
INTERMEDIATE · 5 MARKS
9. Trace the transfer of energy from sunlight to reaction-centre electron transfer in a photosystem.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify chlorophyll/pigments as photon absorbers.
1 mark for describe excitation within an antenna pigment.
1 mark for explain transfer of excitation energy among antenna pigments.
1 mark for distinguish movement of excitation energy from physical electron movement through the antenna.
1 mark for describe reaction-centre electron transfer to an acceptor as charge separation.
HARD · 8 MARKS
10. A chloroplast mutant has normal light absorption and electron transport but a thylakoid membrane that leaks protons. Predict its effects on the proton gradient, ATP formation and carbon fixation, explaining each link.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify that the H⁺ gradient across the thylakoid membrane will dissipate.
1 mark for explain that the electrochemical driving force across the membrane is diminished.
1 mark for explain that proton passage through ATP synthase is less effectively coupled to gradient energy.
1 mark for predict reduced photophosphorylation/ATP production.
1 mark for recognise that light absorption need not itself stop immediately.
1 mark for identify the Calvin cycle as an ATP-requiring stage.
1 mark for predict that carbon-fixation capacity may decrease owing to ATP limitation.
1 mark for distinguish the ATP limitation from any separately possible effects on NADPH supply.
HARD · 8 MARKS
11. A researcher can measure proton concentrations in the thylakoid lumen and stroma but cannot measure ATP directly. Explain how the readings could inform a prediction of ATP formation and identify relevant limitations.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify the lumen and stroma as spaces on opposite sides of the thylakoid membrane.
1 mark for explain how light-driven electron transport accumulates H⁺ in the lumen.
1 mark for identify water oxidation as an additional lumenal H⁺ source.
1 mark for explain that a lumen-to-stroma H⁺ concentration difference represents a chemical gradient.
1 mark for recognise that H⁺ also contributes an electrical component to the electrochemical gradient.
1 mark for explain that a stronger driving gradient can support H⁺ passage through ATP synthase.
1 mark for predict ATP synthesis only if ATP synthase and substrates are functional/available.
1 mark for explain why proton concentration alone does not quantify ATP synthesis without flux and other conditions.
HARD · 8 MARKS
12. An investigator finds that a chloroplast continues receiving photons but the Calvin cycle has very little ATP or NADPH available. Using the lecture mechanisms, propose how failures at different locations could explain the observation.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify the light reactions as the normal source of ATP.
1 mark for identify the light reactions as the normal source of NADPH.
1 mark for explain how an antenna/reaction-centre defect could reduce excitation-driven electron transfer.
1 mark for explain how impaired PSII water oxidation could restrict replacement electrons.
1 mark for explain how disrupted electron transport could reduce gradient generation.
1 mark for explain how a leaky thylakoid membrane could dissipate the H⁺ gradient.
1 mark for explain how inactive ATP synthase could specifically impede ATP generation.
1 mark for explain how limited NADP⁺ reduction could independently restrict NADPH available for carbon fixation.
HARD · 8 MARKS
13. Design the key labels and arrows for a simplified chloroplast schematic showing non-cyclic photophosphorylation and justify the direction of each major flow.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for draw or identify the thylakoid membrane separating lumen and stroma.
1 mark for place PSII within the membrane.
1 mark for place PSI downstream of PSII within the membrane.
1 mark for identify water as an electron source to PSII.
1 mark for indicate electron flow from PSII towards PSI and ultimately NADP⁺/NADPH.
1 mark for indicate H⁺ accumulation in the lumen.
1 mark for place ATP synthase spanning the membrane with H⁺ flowing lumen to stroma.
1 mark for identify ATP production in association with H⁺ passage through ATP synthase.
Lecture 10: Photosynthesis II
1. What is the immediate fate of the unstable six-carbon intermediate made during Rubisco carboxylation?
- It is oxidised to molecular oxygen by Rubisco
- It is exported from the leaf as sucrose without further reactions
- It becomes PEP and then returns to the mesophyll as pyruvate
- It splits into two three-carbon 3-phosphoglycerate molecules
Show answer
It splits into two three-carbon 3-phosphoglycerate molecules
2. Which description correctly links 3-phosphoglycerate and G3P in the Calvin cycle?
- G3P is oxidised by photosystem II to release water
- 3-phosphoglycerate is the long-term vacuolar store of night-time carbon
- ATP and NADPH support conversion of 3-phosphoglycerate into G3P
- ATP is unnecessary because the conversion is driven only by CO₂ diffusion
Show answer
ATP and NADPH support conversion of 3-phosphoglycerate into G3P
3. What does the return of pyruvate to mesophyll cells achieve in the depicted C4 pathway?
- It allows regeneration of PEP to sustain the carbon-concentrating cycle
- It replaces all ATP and NADPH in the Calvin cycle
- It permanently removes three carbon atoms from the plant
- It delivers oxygen for the photorespiratory pathway in chloroplasts
Show answer
It allows regeneration of PEP to sustain the carbon-concentrating cycle
4. Why is most Calvin-cycle G3P retained instead of exported for sugar synthesis?
- It is the primary product of photorespiratory oxygenation
- It directly opens stomata by replacing guard cells
- It is needed to regenerate RuBP and maintain continued carbon fixation
- It is required to split water in photosystem II
Show answer
It is needed to regenerate RuBP and maintain continued carbon fixation
5. In the two-stage redox description, what happens to NADPH during Calvin-cycle carbon reduction?
- NADPH becomes the five-carbon acceptor RuBP
- NADPH donates electrons and is oxidised to NADP⁺
- NADPH releases oxygen from water in the stroma
- NADPH accepts electrons from CO₂ and becomes NADP⁺
Show answer
NADPH donates electrons and is oxidised to NADP⁺
6. What is the main biochemical difference between Rubisco carboxylation and oxygenation?
- Carboxylation uses oxygen as substrate whereas oxygenation uses carbon dioxide
- Carboxylation releases all fixed carbon as CO₂ whereas oxygenation makes sugars efficiently
- They compete for Rubisco activity, with oxygenation initiating wasteful photorespiration
- Only oxygenation can occur inside chloroplasts
Show answer
They compete for Rubisco activity, with oxygenation initiating wasteful photorespiration
7. Why is night-time stomatal opening beneficial for CAM plants in dry habitats?
- Cooler, more humid night air reduces water loss per unit CO₂ uptake
- Water vapour can no longer diffuse through stomata at night under any conditions
- It eliminates the need to assimilate carbon dioxide
- Stomata become permanently impermeable to all gases
Show answer
Cooler, more humid night air reduces water loss per unit CO₂ uptake
8. Which sequence correctly represents the general C4 carbon-concentrating pathway described?
- Calvin-cycle fixation in mesophyll, oxygen transfer to bundle sheath, nightly malate storage
- Initial bicarbonate fixation in mesophyll, C4-acid transfer, CO₂ release in bundle sheath, Calvin-cycle fixation
- Stomatal closure by day, vacuolar malate storage by night, all in the same photosynthetic cell
- Carbon dioxide fixation by Rubisco in air, followed by sugar transfer into PEP carboxylase
Show answer
Initial bicarbonate fixation in mesophyll, C4-acid transfer, CO₂ release in bundle sheath, Calvin-cycle fixation
9. Why is C4 photosynthesis effective at suppressing photorespiration?
- It removes the Calvin cycle and therefore removes all RuBP
- It reduces carbon dioxide availability specifically at Rubisco active sites
- It converts all atmospheric oxygen into water in mesophyll stomata
- It raises CO₂ concentration around bundle sheath Rubisco so carboxylation outcompetes oxygenation
Show answer
It raises CO₂ concentration around bundle sheath Rubisco so carboxylation outcompetes oxygenation
10. A leaf has a photosynthetic CO₂ uptake of 17.9 µmol m⁻² s⁻¹ and water loss of 2.4 mmol m⁻² s⁻¹. Which approximate molar water-use efficiency follows?
- 135 mol CO₂ fixed per mol H₂O lost
- 0.75 mol CO₂ fixed per mol H₂O lost
- 7.5 mol CO₂ fixed per mol H₂O lost
- 0.0075 mol CO₂ fixed per mol H₂O lost
Show answer
0.0075 mol CO₂ fixed per mol H₂O lost
11. Why is the Calvin cycle described as light-independent although photosynthesis requires light?
- It can only operate when all chloroplasts are kept in darkness
- It fixes carbon without any demand for chemical energy
- It does not directly absorb photons but uses ATP and NADPH supplied by the light reactions
- It directly oxidises water instead of receiving reduced electron carriers
Show answer
It does not directly absorb photons but uses ATP and NADPH supplied by the light reactions
12. Why does the Calvin cycle require regeneration of RuBP?
- RuBP is needed to open guard cells rather than for carbon fixation
- Every sugar molecule is directly formed by irreversible breakdown of RuBP without replacement
- RuBP must be replenished to continue accepting carbon dioxide in new turns
- RuBP is the sole terminal electron acceptor of the light reactions
Show answer
RuBP must be replenished to continue accepting carbon dioxide in new turns
13. Which comparison correctly distinguishes the chief separation strategies of C4 and CAM?
- Neither pathway uses PEP carboxylase or Rubisco
- C4 separates initial and Calvin-cycle fixation across cell types; CAM separates initial capture and Calvin-cycle use across night and day
- C4 separates the reactions by time only; CAM separates them by different mesophyll and bundle sheath cells
- C4 and CAM both require stomata to remain open throughout midday
Show answer
C4 separates initial and Calvin-cycle fixation across cell types; CAM separates initial capture and Calvin-cycle use across night and day
14. What is the functional role of vacuolar malic-acid storage in CAM plants?
- It permanently replaces NADPH in all biosynthetic reactions
- It allows the Calvin cycle to operate only in bundle sheath cells
- It directly stores the oxygen released by light-driven water splitting
- It holds carbon captured at night until CO₂ can be released for daytime Calvin-cycle fixation
Show answer
It holds carbon captured at night until CO₂ can be released for daytime Calvin-cycle fixation
15. Why did the oxygenase activity of ancient Rubisco become problematic as Earth changed?
- Early Rubisco evolved only after modern oxygen concentrations were reached
- Atmospheric oxygen accumulated after oxygenic photosynthesis evolved, increasing competition with CO₂
- Carbon dioxide no longer existed once oxygenic photosynthesis began
- Photorespiration occurs only if all atmospheric oxygen disappears
Show answer
Atmospheric oxygen accumulated after oxygenic photosynthesis evolved, increasing competition with CO₂
EASY · 3 MARKS
1. Describe the main night-time events of CAM photosynthesis.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for explain that stomata open at night for CO₂ uptake.
1 mark for explain that PEP carboxylase fixes carbon into malic acid.
1 mark for identify the vacuole as the store of malic acid.
EASY · 3 MARKS
2. Explain the basic photosynthesis–transpiration compromise.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for stomata admit CO₂ required for photosynthesis.
1 mark for open stomata also allow water vapour to diffuse out of leaves.
1 mark for closing stomata reduces water loss but restricts CO₂ entry.
EASY · 3 MARKS
3. Explain why the Calvin cycle is called C3 photosynthesis.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify Rubisco-mediated fixation of carbon dioxide.
1 mark for describe formation of an unstable six-carbon intermediate that splits.
1 mark for identify the first stable product as a three-carbon compound (3-phosphoglycerate).
EASY · 3 MARKS
4. State where the Calvin cycle takes place and identify its principal energy and electron supplies.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for locate the Calvin cycle in the chloroplast stroma.
1 mark for identify ATP from the light reactions as its energy input.
1 mark for identify NADPH from the light reactions as its reducing-power input.
INTERMEDIATE · 5 MARKS
5. Compare C3, C4 and CAM photosynthesis in the environments to which the lecture links them.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify C3 as direct Calvin-cycle fixation with relatively greater photorespiration risk under some conditions.
1 mark for identify cool, wet conditions as generally favourable for C3 in the lecture summary.
1 mark for identify C4 as spatial CO₂ concentration and link it to hot, sunny conditions.
1 mark for identify CAM as night/day separation and link it to hot, dry habitats.
1 mark for state that Rubisco and the Calvin cycle remain important in all three types.
INTERMEDIATE · 5 MARKS
6. Explain the physiological link between stomatal opening, CO₂ uptake and water loss.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify stomata as leaf gas-exchange pores.
1 mark for identify guard cells as controlling stomatal aperture.
1 mark for explain that open stomata permit CO₂ diffusion into leaves.
1 mark for explain outward water-vapour diffusion down the humidity gradient.
1 mark for explain why stomatal closure conserves water but limits photosynthesis.
INTERMEDIATE · 5 MARKS
7. Explain the relationship between light reactions and the Calvin cycle, tracing the main energy and electron carriers.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for explain water oxidation and electron flow in the light reactions.
1 mark for explain reduction of NADP⁺ to NADPH.
1 mark for explain generation of ATP in the light reactions.
1 mark for explain that the Calvin cycle consumes NADPH to reduce CO₂.
1 mark for explain that ATP/ADP and NADPH/NADP⁺ cycle between the stages.
INTERMEDIATE · 5 MARKS
8. Explain why lower atmospheric CO₂ could favour the evolution of C4 photosynthesis.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify Rubisco as having competing carboxylase and oxygenase activities.
1 mark for explain that lower CO₂ increases the relative competition from O₂.
1 mark for explain the cost of photorespiration to net carbon fixation.
1 mark for describe the C4 mechanism as concentrating CO₂ near Rubisco.
1 mark for connect improved carbon fixation under low CO₂ to possible selection for the C4 trait.
INTERMEDIATE · 5 MARKS
9. Trace the major movement of carbon through the representative C4 pathway described.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for explain bicarbonate fixation by PEP carboxylase in mesophyll cells.
1 mark for identify formation of a four-carbon acid such as malate.
1 mark for explain transfer of the C4 acid to bundle sheath cells.
1 mark for explain CO₂ release near Rubisco and Calvin-cycle operation there.
1 mark for explain return of a three-carbon compound for regeneration of PEP.
HARD · 8 MARKS
10. A mutation disrupts Calvin-cycle RuBP regeneration while light reactions remain active. Explain the expected metabolic consequences, distinguishing immediate from downstream effects.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify RuBP as the CO₂ acceptor used by Rubisco.
1 mark for explain its consumption during carboxylation.
1 mark for identify the need to regenerate RuBP from cycle intermediates.
1 mark for predict reduced sustained CO₂ fixation as RuBP becomes limiting.
1 mark for predict reduced production/export of carbon-containing G3P over time.
1 mark for recognise that photosystems may initially still generate ATP.
1 mark for recognise that light reactions may initially still produce NADPH.
1 mark for explain why ongoing ATP/NADPH supply alone cannot substitute for the missing carbon acceptor.
HARD · 8 MARKS
11. A researcher measures photosynthesis of 17.9 µmol CO₂ m⁻² s⁻¹ and transpiration of 2.4 mmol H₂O m⁻² s⁻¹ in wheat. Calculate water-use efficiency and explain its physiological significance.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify CO₂ uptake as 17.9 µmol m⁻² s⁻¹.
1 mark for convert 2.4 mmol water loss to 2400 µmol m⁻² s⁻¹.
1 mark for compute 17.9/2400 ≈ 0.00746.
1 mark for state water-use efficiency as ≈0.0075 mol CO₂ per mol H₂O.
1 mark for recognise this is equivalent to ≈134–135 mol H₂O lost per mol CO₂ gained.
1 mark for explain CO₂ diffusion into leaves through open stomata.
1 mark for explain simultaneous outward diffusion of water vapour.
1 mark for connect the trade-off to selection for water-saving adaptations such as CAM.
HARD · 8 MARKS
12. Compare the carbon-concentrating solutions used by C4 and CAM plants, including their roles in conserving carbon and water.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify PEP carboxylase in initial CO₂/bicarbonate fixation in both pathways.
1 mark for explain C4 production of an organic four-carbon acid.
1 mark for identify mesophyll-to-bundle-sheath spatial segregation in C4.
1 mark for explain decarboxylation near Rubisco in C4.
1 mark for identify night-time stomatal opening and organic-acid fixation in CAM.
1 mark for describe vacuolar storage of malic acid until daylight.
1 mark for explain daytime CO₂ release to Rubisco with stomata generally closed.
1 mark for explain reduced photorespiration in both and particularly high water-use efficiency in CAM.
HARD · 8 MARKS
13. Explain how oxygenic photosynthesis ultimately created an evolutionary disadvantage for the ancient enzyme Rubisco, and how later pathways evolved in response.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify Rubisco as an ancient enzyme that originally functioned when atmospheric O₂ was scarce.
1 mark for identify the oxygenase activity of Rubisco.
1 mark for explain that water oxidation by oxygenic photosynthesis released O₂.
1 mark for explain subsequent atmospheric O₂ accumulation.
1 mark for describe increasing competition between O₂ and CO₂ at Rubisco.
1 mark for describe photorespiratory loss of carbon and energy.
1 mark for connect declining CO₂ to evidence for later C4 evolution.
1 mark for explain that C4 reduces oxygenation by concentrating CO₂ around Rubisco.
Lecture 11: Respiration in Plants
1. Why is the Krebs cycle a cycle rather than a linear pathway?
- It generates new carbon atoms that replace all released carbon dioxide
- The four-carbon acceptor is regenerated so another acetyl group can enter
- All metabolic products leave the pathway without reuse
- Its reactions occur only when light drives a photosystem
Show answer
The four-carbon acceptor is regenerated so another acetyl group can enter
2. How does the respiratory electron transport chain contribute energy to ATP synthesis?
- It directly attaches electrons to ADP to generate ATP without a proton gradient
- Electron transfers release energy used to pump protons and establish a gradient
- It converts oxygen into chlorophyll to absorb light
- It makes the matrix and intermembrane space chemically identical
Show answer
Electron transfers release energy used to pump protons and establish a gradient
3. Which description best captures the electrochemical nature of the mitochondrial proton gradient?
- The only difference is in glucose concentration because protons are uncharged
- The intermembrane space differs from the matrix in both proton concentration and electrical potential
- An electrochemical gradient is the physical movement of ATP between organelles
- The matrix and intermembrane space must have exactly equal proton concentrations
Show answer
The intermembrane space differs from the matrix in both proton concentration and electrical potential
4. Which observation supports thermogenesis by sacred lotus flowers?
- Flower temperatures always track ambient temperatures exactly
- Their heat production is exclusively from sunlight stored in stomata
- Flower receptacles can remain warmer than surrounding air while respiration increases in cooler air
- Lotus flowers release no carbon dioxide or consume no oxygen when warm
Show answer
Flower receptacles can remain warmer than surrounding air while respiration increases in cooler air
5. How do AOX and uncoupling proteins differ mechanistically?
- Both directly fix carbon dioxide in mitochondria
- Both are alternative names for Rubisco
- AOX bypasses proton-pumping electron-transfer steps; uncoupling proteins provide proton leak pathways
- AOX carries protons through ATP synthase; uncoupling proteins directly oxidise pyruvate
Show answer
AOX bypasses proton-pumping electron-transfer steps; uncoupling proteins provide proton leak pathways
6. Which statement best explains why NADH and FADH₂ production in the Krebs cycle is important?
- They are the two principal products of oxygen reduction at complex IV
- They directly fix carbon dioxide into sugars during respiration
- They act as the structural phospholipids of the inner mitochondrial membrane
- They deliver electrons from carbon oxidation to the respiratory electron transport chain
Show answer
They deliver electrons from carbon oxidation to the respiratory electron transport chain
7. Why can mitochondrial uncoupling promote heat production?
- The protein converts all NADH directly into ATP
- An uncoupler increases the ATP yield by preventing proton return
- Protons dissipate gradient energy through a route that bypasses ATP synthase
- Protons are created from heat and stored as oxygen
Show answer
Protons dissipate gradient energy through a route that bypasses ATP synthase
8. Why might AOX or uncoupling pathways benefit a plant when reducing power exceeds ATP demand?
- They make oxygen unnecessary as the terminal acceptor of respiration
- They prevent all redox reactions and all electron transfer indefinitely
- They require the chloroplast Calvin cycle to operate inside mitochondria
- They allow electron flow or gradient dissipation without forcing all available energy into ATP production
Show answer
They allow electron flow or gradient dissipation without forcing all available energy into ATP production
9. What connects glycolysis to the Krebs cycle?
- Fixation of carbon dioxide by Rubisco
- Conversion of pyruvate to acetyl-CoA with CO₂ release and NADH production
- Movement of citrate into the thylakoid lumen
- Conversion of oxygen directly into glucose in the matrix
Show answer
Conversion of pyruvate to acetyl-CoA with CO₂ release and NADH production
10. Which comparison identifies the different terminal electron acceptors in the two organelles?
- Both terminate with Rubisco accepting electrons
- Mitochondrial NADP⁺ is reduced to glucose; chloroplast oxygen is reduced to water
- Mitochondrial oxygen is reduced to water; chloroplast NADP⁺ is reduced to NADPH
- Both terminate with FADH₂ being irreversibly destroyed
Show answer
Mitochondrial oxygen is reduced to water; chloroplast NADP⁺ is reduced to NADPH
11. Why is oxygen indispensable to sustained normal aerobic oxidative phosphorylation?
- It is the source of carbon atoms released from acetyl-CoA
- It directly phosphorylates ADP without ATP synthase
- It accepts electrons at the chain end, allowing upstream electron transfer to continue
- It is reduced to carbon dioxide as the major product
Show answer
It accepts electrons at the chain end, allowing upstream electron transfer to continue
12. Which general carbon progression occurs between glycolysis and the Krebs cycle?
- Oxygen is converted into a two-carbon acetyl group
- Carbon dioxide is converted to glucose, then enters the link reaction
- Glucose is converted to pyruvate, then to an acetyl group that enters the cycle
- Glucose carbon moves directly into the intermembrane space for oxidation
Show answer
Glucose is converted to pyruvate, then to an acetyl group that enters the cycle
13. Which electron-source comparison between mitochondria and chloroplasts is correct?
- Both rely on carbon dioxide as the terminal electron donor
- Mitochondria receive electrons from NADH/FADH₂; oxygenic chloroplasts obtain electrons from water
- Neither system uses reduced electron carriers
- Mitochondria obtain electrons only from sunlight; chloroplasts use NADH as their ultimate donor
Show answer
Mitochondria receive electrons from NADH/FADH₂; oxygenic chloroplasts obtain electrons from water
14. Which proton-flow comparison correctly distinguishes photosynthetic and respiratory ATP synthase?
- Mitochondrial H⁺ returns into the matrix; thylakoid H⁺ returns into the stroma
- In both, ATP synthase eliminates the need for a separating membrane
- Both organelles require protons to return into the thylakoid lumen
- Mitochondrial H⁺ moves into the stroma; chloroplast H⁺ moves into the matrix
Show answer
Mitochondrial H⁺ returns into the matrix; thylakoid H⁺ returns into the stroma
15. Why is coenzyme A released after it delivers an acetyl group to the Krebs cycle?
- It becomes the terminal acceptor of mitochondrial electrons
- It converts NADH into a chlorophyll pigment
- It stores the proton gradient across the inner membrane
- It acts as a reusable carrier rather than the carbon substrate that is oxidised
Show answer
It acts as a reusable carrier rather than the carbon substrate that is oxidised
16. Why does cellular respiration release usable energy more controllably than burning glucose?
- It bypasses electron transfer and releases all energy at once
- It creates energy from ADP without using the stored energy in glucose
- It oxidises substrates through a series of coupled reactions that transfer energy to carriers and ATP
- It converts all heat directly into carbon dioxide
Show answer
It oxidises substrates through a series of coupled reactions that transfer energy to carriers and ATP
17. Why does the folding of the mitochondrial inner membrane support respiration?
- Cristae create carbon atoms needed by the Krebs cycle
- Membrane folds make glycolysis occur inside mitochondria
- Folding eliminates the separation of matrix and intermembrane space
- Cristae increase membrane surface area available for electron-transport proteins and ATP synthase
Show answer
Cristae increase membrane surface area available for electron-transport proteins and ATP synthase
EASY · 3 MARKS
1. Describe how uncoupling proteins can generate heat.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify uncoupling proteins as H⁺ leak routes across the inner mitochondrial membrane.
1 mark for explain that protons bypass ATP synthase.
1 mark for explain that proton-gradient energy is dissipated as heat rather than conserved as ATP.
EASY · 3 MARKS
2. Explain the effect of alternative oxidase on respiration.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify AOX as an alternative electron route towards oxygen.
1 mark for explain that it bypasses some normal proton-pumping steps.
1 mark for explain that this reduces ATP conservation and allows more heat dissipation.
EASY · 3 MARKS
3. Explain the role of oxygen in mitochondrial electron transport.
Show marking guide
Not at
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify oxygen as the terminal electron acceptor.
1 mark for explain that it receives electrons at the end of the chain.
1 mark for explain that oxygen combines with electrons and protons to form water.
EASY · 3 MARKS
4. State three mechanistic similarities between photosynthetic and respiratory ATP production.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for both use an electron transport chain embedded in a membrane.
1 mark for both establish an H⁺ electrochemical gradient.
1 mark for both use ATP synthase to couple proton return to ATP synthesis.
EASY · 3 MARKS
5. State where glycolysis occurs and identify its main products.
Show marking guide
Award up to 3 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify the cytoplasm as the site of glycolysis.
1 mark for identify pyruvate as the major carbon-containing product.
1 mark for identify ATP and NADH as energy/reducing-power products.
INTERMEDIATE · 5 MARKS
6. Compare how proton gradients are generated and used in chloroplasts and mitochondria.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for locate photosynthetic proton pumping across the thylakoid membrane.
1 mark for identify accumulation of H⁺ in the thylakoid lumen.
1 mark for locate respiratory proton pumping across the mitochondrial inner membrane.
1 mark for identify accumulation of H⁺ in the intermembrane space.
1 mark for explain ATP synthase-dependent H⁺ return to stroma or matrix, respectively.
INTERMEDIATE · 5 MARKS
7. Compare the sources and fates of electrons in chloroplast and mitochondrial electron transport.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify water as the ultimate source of electrons in oxygenic photosynthetic electron transport.
1 mark for explain the role of sunlight in exciting photosynthetic electrons.
1 mark for identify NADP⁺ reduction to NADPH as a photosynthetic endpoint.
1 mark for identify NADH/FADH₂ as direct electron donors in mitochondria.
1 mark for identify oxygen reduction to water as the mitochondrial endpoint.
INTERMEDIATE · 5 MARKS
8. Explain the evidence and mechanism for thermogenesis in sacred lotus flowers.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify observation of warm receptacles despite cooler ambient conditions.
1 mark for identify oxygen consumption as a measure of respiratory activity.
1 mark for describe increased respiration at lower ambient temperatures in the reported data.
1 mark for explain AOX-mediated bypass of energy-conserving electron-transport steps.
1 mark for connect the reduced ATP conservation to heat production that warms floral tissue.
INTERMEDIATE · 5 MARKS
9. Explain the relationship between electron transport, oxygen reduction and the mitochondrial proton gradient.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify NADH/FADH₂ as electron donors.
1 mark for describe electron transfer through inner-membrane carrier complexes.
1 mark for explain use of released energy to pump H⁺ into the intermembrane space.
1 mark for identify oxygen as terminal electron acceptor forming water.
1 mark for connect the gradient to ATP formation by chemiosmosis.
INTERMEDIATE · 5 MARKS
10. Explain why oxygen limitation strongly reduces normal aerobic ATP production.
Show marking guide
Award up to 5 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify oxygen as the terminal electron acceptor.
1 mark for explain why chain electron flow slows when terminal acceptance fails.
1 mark for explain the consequent reduction in proton pumping.
1 mark for explain the loss of proton-motive force available for ATP synthase.
1 mark for predict a marked decline in oxidative ATP production.
HARD · 8 MARKS
11. A sacred lotus receptacle remains warm despite cooler surrounding air. Evaluate how measured oxygen uptake and the respiratory mechanism could support thermoregulation.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify the observation that receptacle temperature can exceed ambient temperature.
1 mark for explain oxygen uptake as evidence of respiratory electron flow.
1 mark for describe greater oxygen consumption in cooler conditions in the lecture example.
1 mark for identify AOX as a route to oxygen in plant mitochondria.
1 mark for explain bypass of some proton-pumping complexes.
1 mark for predict reduced ATP formation per unit electron transfer.
1 mark for explain heat release from reduced energy conservation.
1 mark for explain how increased respiratory heat output can maintain elevated flower temperature.
HARD · 8 MARKS
12. Explain the difference between capturing electrons during carbon oxidation and using their energy to synthesise ATP, using the Krebs cycle and ETC.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify oxidation of acetyl-derived carbon as a major Krebs-cycle process.
1 mark for identify carbon release as CO₂.
1 mark for identify NAD⁺ reduction to NADH.
1 mark for identify FAD reduction to FADH₂.
1 mark for explain reduced carriers transferring electrons to the inner-membrane ETC.
1 mark for explain downhill electron flow releasing free energy.
1 mark for explain ETC-dependent H⁺ pumping and gradient storage.
1 mark for explain ATP synthase conversion of proton-gradient energy into ATP.
HARD · 8 MARKS
13. Trace how oxidation of glucose supports mitochondrial ATP synthesis, following carbon, electron carriers, protons and oxygen.
Show marking guide
Award up to 8 marks (one per scientifically correct point; accept equivalent wording and valid alternatives):
1 mark for identify glucose as the starting organic substrate.
1 mark for describe glycolytic formation of pyruvate.
1 mark for describe pyruvate conversion to acetyl-CoA with CO₂ release.
1 mark for explain Krebs-cycle oxidation of acetyl-derived carbon to CO₂.
1 mark for identify NADH/FADH₂ as carriers transporting released electrons.
1 mark for explain electron transfer to oxygen at the ETC terminus with H₂O formation.
1 mark for describe H⁺ pumping from matrix to intermembrane space.
1 mark for explain H⁺ return through ATP synthase driving ATP formation.
Lecture 13: Metabolism
1. Which summary best describes the carbon flow from glucose to fatty acid presented in the lecture?
- Glucose becomes lysine and then glycogen without intermediary reactions
- All glucose carbons become phosphate groups before entering fat
- Glucose is translated directly into fatty acid by ribosomes
- Glucose is converted through pyruvate and acetyl units, which can be used for fatty acid synthesis
Show answer
Glucose is converted through pyruvate and acetyl units, which can be used for fatty acid synthesis
2. Which enzyme activity reverses the addition of a phosphate group to many proteins?
- A protein phosphatase removes the phosphate group
- A glycogen synthase removes the phosphate group
- A DNA polymerase removes the phosphate group
- A ribosome removes the phosphate group
Show answer
A protein phosphatase removes the phosphate group
3. What is the scientific weakness in the film’s claim that a lysine-dependent animal could not survive outside a facility?
- Animals can obtain lysine only by photosynthesis
- Lysine occurs in ordinary foods, so escaped animals could potentially obtain it by eating
- Lysine is non-essential in all animals
- Lysine cannot be present in either plants or animals
Show answer
Lysine occurs in ordinary foods, so escaped animals could potentially obtain it by eating
4. How is ATP hydrolysis often coupled to biosynthetic reactions?
- ATP functions only as a structural lipid
- ATP always releases energy by converting its ribose to carbon dioxide
- ATP supplies all carbon atoms incorporated into new fatty acids
- ATP can phosphorylate a substrate or a protein, helping drive an otherwise unfavourable step
Show answer
ATP can phosphorylate a substrate or a protein, helping drive an otherwise unfavourable step
5. Why is glycogen stored in skeletal muscle not simply equivalent to a liver glucose reserve?
- Liver glycogen is used only for muscle contraction
- Muscle glycogen is assembled from amino acids rather than glucose
- Skeletal muscle cannot break down glycogen
- Muscle glycogen is used predominantly to support local muscle energy needs
Show answer
Muscle glycogen is used predominantly to support local muscle energy needs
6. After a carbohydrate-rich meal, blood glucose rises. Which response supports short-term glucose storage?
- Insulin directly hydrolyses every glycogen bond
- Glucagon converts fatty acids into intact glycogen without intermediates
- Glucagon causes immediate conversion of all glycogen into protein
- Insulin signalling promotes glucose uptake and glycogen synthesis in responsive tissues
Show answer
Insulin signalling promotes glucose uptake and glycogen synthesis in responsive tissues
7. How can a varied plant-based diet supply essential amino acids if one food is relatively low in a particular amino acid?
- Other plant foods can contribute complementary essential amino acids
- The body can manufacture every essential amino acid from water
- Essential amino acids are unnecessary when dietary protein is plant-derived
- All plant proteins contain identical proportions of amino acids
Show answer
Other plant foods can contribute complementary essential amino acids
8. A triglyceride is formed by joining three fatty acids to glycerol. How is this related to fat storage?
- It turns fatty acids into essential amino acids without a nitrogen source
- It hydrolyses the glycerol backbone into glucose immediately
- It builds an energy-rich lipid molecule that can be stored in adipose tissue
- It yields a molecule that is identical to glycogen
Show answer
It builds an energy-rich lipid molecule that can be stored in adipose tissue
9. Which structural feature explains why fatty acid chains tend to be hydrophobic?
- They carry multiple permanent positive charges along the chain
- They are made mainly of charged phosphate groups
- They contain repeated peptide bonds exposed to water
- They contain predominantly non-polar carbon–hydrogen bonds
Show answer
They contain predominantly non-polar carbon–hydrogen bonds
10. How can transfer of phosphate from ATP to a protein change protein function?
- Added negative charge can alter electrostatic interactions and protein conformation
- Phosphorylation permanently destroys every enzyme
- ATP inserts an extra amino acid into the protein backbone
- The phosphate always replaces the entire polypeptide chain
Show answer
Added negative charge can alter electrostatic interactions and protein conformation
11. Two glucose molecules join with the removal of water. How should this reaction be classified?
- A catabolic reaction that releases free amino acids
- An anabolic condensation reaction that produces a glycosidic bond
- An anabolic reaction that removes a fatty acid
- A catabolic hydrolysis reaction that breaks a peptide bond
Show answer
An anabolic condensation reaction that produces a glycosidic bond
12. During a period between meals, which response helps the liver supply glucose?
- Insulin stimulates hepatic glycogen synthesis as the only available process
- Glucagon signalling promotes hepatic glycogen breakdown
- All blood glucose is converted immediately into triglyceride
- Liver cells use skeletal-muscle glycogen directly as circulating glucose
Show answer
Glucagon signalling promotes hepatic glycogen breakdown
13. When comparing 100 g of lean beef with 86 g of roasted soybeans in the lecture table, what is the most defensible interpretation?
- The table proves that no other dietary factors should influence food choice
- The soybeans contain no methionine
- The data show that beef supplies every amino acid in greater amounts
- Both foods supply the listed essential amino acids, though quantities differ among amino acids
Show answer
Both foods supply the listed essential amino acids, though quantities differ among amino acids
14. Why can digested protein from a meal contribute to new human proteins?
- Absorbed amino acids can be assembled into human-specific polypeptide sequences
- Dietary proteins retain their original sequence and function in human cells
- Only plant-derived amino acids can enter human protein synthesis
- The digestive tract transfers intact meat proteins directly into muscle fibres
Show answer
Absorbed amino acids can be assembled into human-specific polypeptide sequences
15. Why is a phosphate group generally hydrophilic?
- It is composed solely of non-polar C–H bonds
- It has the same structure as a fatty acid chain
- It is completely electrically neutral at physiological pH
- Its negatively charged oxygens interact favourably with water
Show answer
Its negatively charged oxygens interact favourably with water
16. Why does ATP commonly link catabolic and anabolic metabolism?
- ATP is a permanent store that never needs replenishment
- ATP replaces all catalysts in synthetic reactions
- ATP converts every absorbed nutrient directly into protein
- Energy released by breakdown pathways can support ATP regeneration, and ATP-dependent reactions can drive biosynthesis
Show answer
Energy released by breakdown pathways can support ATP regeneration, and ATP-dependent reactions can drive biosynthesis
EASY · 3 MARKS
1. Describe glycogen and identify the two principal tissues highlighted for glycogen storage.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for describing glycogen as a carbohydrate polymer composed of glucose units
1 mark for identifying the liver as a major glycogen store
1 mark for identifying skeletal muscle as a major glycogen store
EASY · 3 MARKS
2. Describe how two glucose units can be joined by a glycosidic bond.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying two monosaccharides or glucose units as the reactants
1 mark for describing formation of a covalent glycosidic linkage between sugar units
1 mark for noting that water is released in a condensation reaction
EASY · 3 MARKS
3. Explain the meaning of an essential amino acid using lysine as an example.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for defining an essential amino acid as one that must be obtained in sufficient quantity from the diet
1 mark for identifying lysine as an essential amino acid for humans
1 mark for explaining that appropriate foods, including plant-derived foods, can provide lysine
EASY · 3 MARKS
4. Identify the components of ATP and state one way ATP can support cellular work.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying adenine and ribose as components of adenosine
1 mark for identifying three phosphate groups as part of ATP
1 mark for describing ATP-dependent phosphorylation or another correct ATP-coupled reaction as a mechanism for doing work
INTERMEDIATE · 5 MARKS
5. Compare lean beef and soybeans as sources of essential amino acids, using the lecture table cautiously.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for recognising that both listed portions contain all nine essential amino acids
1 mark for identifying at least one amino acid present in greater mass in the soybean portion, such as leucine
1 mark for identifying methionine as higher in the beef portion in the provided table
1 mark for noting that compared portions are different masses and must not be mistaken for equal-mass comparisons
1 mark for concluding that both can contribute to adequate protein intake and quality depends on overall diet
INTERMEDIATE · 5 MARKS
6. Compare macronutrients and micronutrients in terms of examples and functions.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for explaining macronutrients are needed in relatively larger dietary amounts
1 mark for identifying protein, carbohydrate or lipid as a correct macronutrient
1 mark for explaining micronutrients are required in relatively smaller amounts
1 mark for identifying a vitamin or mineral as a micronutrient
1 mark for explaining one micronutrient function such as supporting enzyme activity or tissue structure
INTERMEDIATE · 5 MARKS
7. Explain how catabolism and anabolism cooperate in continual protein maintenance.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for describing protein damage or normal turnover
1 mark for explaining catabolic degradation of proteins into smaller components
1 mark for identifying the recovery or dietary supply of amino acids
1 mark for explaining anabolic assembly of new polypeptide chains
1 mark for connecting energy requirements for synthesis to ATP supplied by energy metabolism
HARD · 8 MARKS
8. A research proposal states that dietary carbohydrates can replace all dietary essential amino acids because nutrients are interconvertible. Assess the claim.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for recognising that metabolic interconversion exists for some nutrient-derived carbon skeletons
1 mark for describing the demonstrated glucose-to-pyruvate step
1 mark for describing pyruvate-to-acetyl-derived carbon as a connection to fatty acid synthesis
1 mark for distinguishing these carbon conversions from the synthesis of amino acids
1 mark for explaining amino acids also require nitrogen in addition to carbon skeletons
1 mark for defining essential amino acids as those humans cannot make in sufficient quantities
1 mark for giving lysine as an example of a dietary essential amino acid
1 mark for concluding carbohydrate intake does not eliminate the requirement for adequate dietary essential amino acids
HARD · 8 MARKS
9. Design a conceptual explanation of how one dietary protein molecule can contribute both to replacement tissue protein and to general metabolism rather than remaining intact in the body.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for explaining digestive catabolism of the dietary protein
1 mark for identifying amino acids or small peptides as absorption products
1 mark for explaining absorbed amino acids enter intracellular or systemic amino acid pools
1 mark for identifying some amino acids as essential dietary inputs
1 mark for explaining anabolic polymerisation can create proteins with new sequences specified by the recipient cells
1 mark for explaining cellular proteins undergo subsequent damage or turnover
1 mark for connecting nutrient catabolism to ATP production or availability for cellular work
1 mark for recognising that metabolic fates depend on amino acid needs and pathways rather than preserving intact food proteins
HARD · 8 MARKS
10. Evaluate the claim that eating plant protein cannot supply a complete set of essential amino acids.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for defining essential amino acids in terms of the body’s limited capacity to make them
1 mark for identifying lysine as an example of an essential amino acid
1 mark for recognising plants can synthesise amino acids essential to humans
1 mark for explaining dietary protein is digested into absorbable amino acids or small peptides
1 mark for recognising soy as a plant source that contains all the listed essential amino acids
1 mark for noting individual plant foods can differ in relative amounts of particular amino acids
1 mark for explaining complementary plant foods or varied diets can meet essential amino acid needs
1 mark for concluding protein adequacy depends on overall amino acid supply and intake rather than plant-versus-animal labelling alone
HARD · 8 MARKS
11. Explain how glycogen breakdown, ATP recycling and ATP-mediated phosphorylation can be linked when a cell must activate a transport protein.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying glycogen as a polymeric glucose store
1 mark for describing catabolic breakdown of glycogen to supply sugar intermediates
1 mark for explaining further fuel metabolism can contribute to ATP regeneration
1 mark for identifying ATP as adenosine with three phosphate groups
1 mark for describing transfer of a phosphate group from ATP to a protein
1 mark for explaining that phosphorylation changes local charge and may alter protein conformation
1 mark for connecting conformational change to transporter activity
1 mark for explaining a phosphatase can remove the phosphate and alter or reset protein state
HARD · 8 MARKS
12. Use carbon accounting to explain how six-carbon glucose can supply precursors for fatty acid synthesis without directly turning into a lipid polymer.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying glucose as a six-carbon precursor
1 mark for explaining glycolysis produces two three-carbon pyruvates per glucose
1 mark for explaining pyruvate oxidation releases one carbon as carbon dioxide per pyruvate
1 mark for identifying the remaining two-carbon acetyl group attached to coenzyme A
1 mark for explaining a two-carbon acetyl group joins four-carbon oxaloacetate
1 mark for identifying the resulting citrate as a six-carbon molecule
1 mark for explaining citrate export can provide acetyl units outside the mitochondrion
1 mark for connecting assembly of acetyl-derived carbon units to fatty acid formation
Lecture 14: Cell Diversity
1. Which process requires both a flexible membrane and cellular machinery to engulf an external particle?
- Transcription of DNA into RNA
- Passive diffusion of a gas across a bilayer
- Synthesis of peptidoglycan in the nucleolus
- Phagocytic uptake by membrane extension around the particle
Show answer
Phagocytic uptake by membrane extension around the particle
2. Why can plant cell walls contribute to shape support without replacing plasma-membrane transport functions?
- The cell wall acts as a second nucleus
- The cellulose wall provides external support, while the plasma membrane still regulates exchange
- The plant cell wall consists mainly of transmembrane proteins
- Cellulose performs all ATP-dependent membrane transport
Show answer
The cellulose wall provides external support, while the plasma membrane still regulates exchange
3. How can a prokaryotic membrane carry out functions even without internal membrane-bound organelles?
- Prokaryotes rely exclusively on nuclei for energy metabolism
- Membrane-embedded proteins can perform transport and energy-related processes at the cell boundary
- Prokaryotes lack all proteins associated with membranes
- Their cell walls perform all intracellular translation
Show answer
Membrane-embedded proteins can perform transport and energy-related processes at the cell boundary
4. What is the principal contribution of peptidoglycan to many bacterial cells?
- A nuclear envelope that encloses DNA
- A protein complex that translates mRNA
- An internal reserve of triglyceride
- A supportive external cell wall built from sugar chains cross-linked by peptides
Show answer
A supportive external cell wall built from sugar chains cross-linked by peptides
5. A microorganism has no nucleus and its cell wall lacks peptidoglycan. Which group is most consistent with these observations?
- Archaea
- Flowering plants
- Bacteria with a typical peptidoglycan wall
- Multicellular animals
Show answer
Archaea
6. Why are evolutionary models for the origin of the nucleus considered hypotheses rather than direct observations?
- They claim that membranes cannot change shape
- They have been directly photographed in ancient living cells
- They reconstruct ancient events from present-day evidence rather than witnessed historical processes
- They require every cell to be completely identical today
Show answer
They reconstruct ancient events from present-day evidence rather than witnessed historical processes
7. Which feature best supports an endosymbiotic origin for mitochondria?
- Their location exclusively outside living cells
- The absence of any membrane surrounding them
- Their function as the plant cellulose wall
- Their bacterial-like ancestry and retention of a double membrane
Show answer
Their bacterial-like ancestry and retention of a double membrane
8. A scientist observes carbohydrates projecting from the external face of a membrane protein. How should this structure be described?
- A triglyceride converted into a chromosome
- A glycoprotein that contributes to surface carbohydrate display
- A bare phospholipid with no polar head
- A peptidoglycan chain forming the nucleus
Show answer
A glycoprotein that contributes to surface carbohydrate display
9. Why does an apparently simple textbook animal-cell drawing often under-represent actual cell structure?
- It omits the bacterial peptidoglycan layer from all animal cells
- It omits extensive cytoskeletal organisation, membranes and dynamic organelle arrangements
- It incorrectly shows genetic material in a nucleus
- It exaggerates the fact that cells have a plasma membrane
Show answer
It omits extensive cytoskeletal organisation, membranes and dynamic organelle arrangements
10. Why can a phospholipid bilayer act as a selective barrier rather than an unrestricted opening?
- Its hydrophobic interior impedes many ions and polar solutes
- All dissolved molecules diffuse through the interior at identical rates
- The bilayer contains no transport proteins
- The bilayer is made only of solid cellulose
Show answer
Its hydrophobic interior impedes many ions and polar solutes
11. What is the likely contribution of carbohydrates attached to external membrane proteins and lipids?
- They make all membrane proteins hydrophobic throughout
- They contribute to a cell-surface glycocalyx involved in recognition and surface interactions
- They prevent any interaction with extracellular water
- They replace the plasma membrane with a rigid DNA coat
Show answer
They contribute to a cell-surface glycocalyx involved in recognition and surface interactions
12. A cell has a plasma membrane and ribosomes but no membrane-enclosed nucleus. Which classification is best supported?
- Animal cell
- Fungal cell with a nucleus
- Prokaryotic cell
- Plant cell
Show answer
Prokaryotic cell
13. Which comparison of plant and animal cell boundaries is accurate?
- Animal cells have cellulose cell walls while plants have only cholesterol
- Plant cells typically have cellulose cell walls outside the plasma membrane; animal cells lack cellulose walls
- Neither group has a plasma membrane
- Both cell types normally contain bacterial peptidoglycan walls
Show answer
Plant cells typically have cellulose cell walls outside the plasma membrane; animal cells lack cellulose walls
14. A bacterium swims towards a nutrient source. Which cellular feature may enable its directed movement?
- Contraction of multicellular muscle tissue
- Movement of a nucleus along microtubules
- Rotation of flagella powered by a membrane-associated motor
- Cellulose growth without membrane transport
Show answer
Rotation of flagella powered by a membrane-associated motor
15. Why is cholesterol important for animal plasma membranes?
- It helps modulate membrane organisation and fluidity over changing conditions
- It catalyses translation of mRNA
- It replaces all phospholipid molecules
- It forms a peptidoglycan cell wall
Show answer
It helps modulate membrane organisation and fluidity over changing conditions
16. An animal membrane loses its integral transport proteins but retains phospholipids. What would be most directly affected?
- Regulated transport of solutes that require protein-mediated passage
- The presence of DNA in the nucleus
- The existence of a hydrophobic bilayer interior
- The construction of cellulose microfibrils
Show answer
Regulated transport of solutes that require protein-mediated passage
17. A phospholipid in an aqueous environment has its hydrophobic tails directed towards neighbouring tails. What physical principle explains this?
- Non-polar tails minimise contact with water while polar heads remain hydrated
- Phosphate groups spontaneously become proteins
- All tails are permanently positively charged
- Water becomes hydrophobic after touching a membrane
Show answer
Non-polar tails minimise contact with water while polar heads remain hydrated
EASY · 3 MARKS
1. Distinguish prokaryotic and eukaryotic cells using three defining features or examples.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying a membrane-enclosed nucleus as characteristic of eukaryotic cells
1 mark for explaining prokaryotic cells lack a membrane-enclosed nucleus
1 mark for providing a correct example or distinction such as bacteria/archaea versus animals/plants/fungi/protists
EASY · 3 MARKS
2. Outline how information encoded in nuclear DNA leads to protein synthesis in the cytoplasm.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying transcription of DNA to messenger RNA inside the nucleus
1 mark for explaining messenger RNA exits through nuclear pores
1 mark for explaining ribosomes translate messenger RNA into protein
EASY · 3 MARKS
3. State three features distinguishing Bacteria from Archaea and eukaryotic organisms.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for recognising Bacteria and Archaea are prokaryotic groups
1 mark for identifying peptidoglycan as characteristic of many bacterial cell walls
1 mark for identifying that archaea lack true peptidoglycan and eukaryotes have membrane-enclosed nuclei
INTERMEDIATE · 5 MARKS
4. Explain how a cell membrane can act as a barrier while still supporting exchange and signalling.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for explaining the hydrophobic interior restricts many polar solutes and ions
1 mark for identifying membrane transport proteins as selective passage routes
1 mark for identifying receptor proteins as means of signal detection
1 mark for describing the bilayer as laterally fluid or capable of shape changes
1 mark for connecting these combined properties to controlled cellular physiology
INTERMEDIATE · 5 MARKS
5. Explain how glycoproteins and glycolipids contribute to the outer cell surface.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying glycoproteins as proteins carrying carbohydrate chains
1 mark for identifying glycolipids as lipids carrying carbohydrate groups
1 mark for placing these carbohydrate groups on the extracellular membrane face
1 mark for describing their contribution to the glycocalyx
1 mark for explaining a role in recognition, adhesion or hydration of the surface
INTERMEDIATE · 5 MARKS
6. Explain how the structure of a phospholipid produces a stable bilayer in water.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying the phospholipid polar or charged head
1 mark for identifying the non-polar hydrocarbon tails
1 mark for explaining the amphipathic nature of the molecule
1 mark for describing heads orienting towards water on either bilayer surface
1 mark for explaining tails cluster in the interior to minimise contact with water
INTERMEDIATE · 5 MARKS
7. Explain why the endoplasmic reticulum and nuclear envelope illustrate the value of compartmentalisation.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying the nucleus as a compartment containing most eukaryotic chromosomal DNA
1 mark for identifying the nuclear envelope as separating nucleoplasm from cytoplasm
1 mark for explaining nuclear pores permit regulated exchange
1 mark for describing membrane continuity between nuclear envelope and endoplasmic reticulum
1 mark for connecting internal membranes to distinct but coordinated cellular functions
HARD · 8 MARKS
8. A cell contains a bacterial-type chromosome, has motile flagella and lacks a nuclear envelope. Analyse what can and cannot be concluded about its classification.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying absence of a membrane-enclosed nucleus as evidence for prokaryotic organisation
1 mark for explaining a chromosome does not by itself make a cell eukaryotic
1 mark for identifying flagella as compatible with bacterial motility
1 mark for explaining a membrane-associated motor can drive bacterial flagellar movement
1 mark for recognising pili and flagella are not sufficient to determine a taxonomic domain
1 mark for identifying peptidoglycan as a useful feature supporting classification as Bacteria
1 mark for identifying absence of true peptidoglycan as compatible with Archaea
1 mark for concluding additional evidence is needed to distinguish bacterial from archaeal identity
HARD · 8 MARKS
9. A drug removes much of the cholesterol from an animal cell membrane. Predict potential consequences for membrane dynamics and transport.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying cholesterol as an animal membrane component
1 mark for describing its interaction with phospholipid organisation
1 mark for explaining its ability to modulate or buffer membrane fluidity
1 mark for predicting membrane physical properties may become more temperature sensitive
1 mark for recognising altered lipid organisation may affect membrane-protein environments
1 mark for connecting transport or signalling changes to effects on embedded proteins
1 mark for recognising phospholipids still provide the primary bilayer architecture
1 mark for avoiding an unconditional prediction that removing cholesterol instantly destroys every plasma membrane
HARD · 8 MARKS
10. A synthetic cell membrane contains amphipathic lipids but no membrane proteins. Predict its physical properties and functional limitations.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying hydrophilic heads and hydrophobic tails in amphipathic lipids
1 mark for explaining self-assembly into a bilayer in an aqueous environment
1 mark for identifying a hydrophobic interior that impedes many ions and polar solutes
1 mark for recognising some small non-polar molecules can still cross by diffusion
1 mark for explaining protein-mediated selective transport would be reduced or absent
1 mark for explaining receptor-dependent signalling would be limited
1 mark for explaining the lipid bilayer could still form a flexible physical boundary
1 mark for concluding membranes need both lipid organisation and specialised proteins for many physiological functions
HARD · 8 MARKS
11. Critically evaluate membrane infolding and endosymbiosis as explanations for features of early eukaryotic cell evolution.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying internal membrane infolding as one proposed route to a nuclear envelope or endomembrane system
1 mark for explaining that a membrane could be reorganised around genetic material in such a model
1 mark for identifying bacterial endosymbiosis as the widely supported origin of mitochondria
1 mark for recognising mitochondrial double membranes as consistent with this history
1 mark for distinguishing origin of mitochondria from the precise origin of the nucleus
1 mark for recognising alternative membrane-protrusion models have been proposed
1 mark for explaining present-day molecular or structural evidence is used to evaluate these models
1 mark for recognising that detailed ancient sequences cannot be established solely from a modern schematic drawing
Lecture 15: Compartmentalisation
1. Why are intermediate filaments often associated with mechanical resilience?
- They function as open sodium channels
- They are the only molecules capable of generating ATP
- Their rope-like arrangements help distribute mechanical stresses through cells
- They replace plasma membranes with solid walls
Show answer
Their rope-like arrangements help distribute mechanical stresses through cells
2. A ligand opens a membrane channel and sodium ions move down their electrochemical gradient. How should this be classified?
- Synthesis of sodium ions by osmosis
- Channel-mediated passive transport
- ATP-powered pumping against the gradient
- Exocytosis of sodium vesicles
Show answer
Channel-mediated passive transport
3. Why is a vesicle an effective carrier of protein cargo in an aqueous cytoplasm?
- It works by dissolving proteins into free fatty acids
- It transports cargo only as unprotected nuclear DNA
- Its lipid membrane encloses cargo while separating it from surrounding cytosol
- It lacks any boundary separating it from the cytoplasm
Show answer
Its lipid membrane encloses cargo while separating it from surrounding cytosol
4. Why must descriptions of osmosis consider membrane permeability to solutes?
- An impermeant solute can establish a sustained osmotic difference across the membrane
- Osmosis is the movement of salt alone through protein channels
- Water transport requires a direct ATP pump in all situations
- Every solute moves freely across every biological membrane
Show answer
An impermeant solute can establish a sustained osmotic difference across the membrane
5. Two compartments are separated by a membrane permeable to water but not salt. Which way will water move if salt concentration is higher on the right?
- Only towards the side with more mitochondria
- Net movement towards the left because salt repels all water
- No movement because water is polar
- Net movement towards the right until water potential differences are reduced
Show answer
Net movement towards the right until water potential differences are reduced
6. Why can ATP generation and ATP utilisation be considered coupled cellular processes?
- ATP cannot donate phosphate to substrates or proteins
- Energy-releasing reactions support ATP regeneration, and ATP hydrolysis can drive energy-demanding work
- All ATP production is itself an ATP-dependent protein export event
- ATP synthase consumes DNA to generate ATP
Show answer
Energy-releasing reactions support ATP regeneration, and ATP hydrolysis can drive energy-demanding work
7. Which association of filament, constituent and location is most consistent with the lecture?
- Actin is the major mineral building block of mitochondria
- Microtubules are made of cellulose and found only outside cells
- Actin microfilaments frequently contribute to the cell cortex near the plasma membrane
- Intermediate filaments are composed only of ATP
Show answer
Actin microfilaments frequently contribute to the cell cortex near the plasma membrane
8. A protein is produced in the secretory pathway and released outside the cell. Which final event is required?
- Conversion of protein cargo into peptidoglycan
- Breakdown of the nuclear chromosome
- Fusion of a cargo-containing vesicle with the plasma membrane
- Active transport of the entire ribosome through an ion channel
Show answer
Fusion of a cargo-containing vesicle with the plasma membrane
9. Which set of functions correctly reflects the lecture’s account of mitochondria?
- Oxidative ATP production, participation in apoptosis and steroidogenesis
- Translation of every cellular protein, osmosis and exocytosis
- Protein secretion, cellulose synthesis and bacterial flagellar assembly
- Glycogen storage only, DNA export and peptidoglycan formation
Show answer
Oxidative ATP production, participation in apoptosis and steroidogenesis
10. A microscope shows elongated interconnected mitochondria. What should a student infer?
- The observed structures must be bacterial peptidoglycan
- The sample cannot contain mitochondria because they are not bean-shaped
- Mitochondria are dynamic structures that may form networks
- The mitochondria cannot participate in ATP production
Show answer
Mitochondria are dynamic structures that may form networks
11. A cell internalises extracellular cholesterol within a membrane-bounded compartment. Which process is most likely involved?
- Exocytosis
- Passive oxygen diffusion
- Endocytosis
- Direct translation of cholesterol
Show answer
Endocytosis
12. What is the main difference between an ion channel and a membrane pump?
- Every channel uses ATP to move ions against a gradient
- A channel provides a pathway for downhill movement; a pump can couple energy to uphill transport
- Channels internalise vesicles but pumps secrete proteins
- Pumps are made of lipid alone whereas channels contain only DNA
Show answer
A channel provides a pathway for downhill movement; a pump can couple energy to uphill transport
13. Which situation most clearly requires an energy-coupled transporter?
- Moving an ion from a lower electrochemical potential to a higher one
- Oxygen crossing a lipid bilayer down its concentration gradient
- Water moving passively towards higher effective solute concentration
- A small solute moving through an open channel down its gradient
Show answer
Moving an ion from a lower electrochemical potential to a higher one
14. A cell must both move secretory vesicles and maintain its surface shape. Which pairing is most plausible?
- Triglycerides as tracks and ATP molecules as surface filaments
- Peptidoglycan as the only intracellular cytoskeleton
- Chromosomal DNA as tracks and free calcium as a membrane scaffold
- Microtubules for transport tracks and cortical actin for surface support
Show answer
Microtubules for transport tracks and cortical actin for surface support
15. How can ATP-dependent phosphorylation change the function of a transport protein?
- ATP removes the plasma membrane so no transport protein is needed
- Phosphorylation always means the protein diffuses out of the cell
- ATP replaces the transporter with a cellulose fibre
- Phosphate transfer alters the transporter’s interactions or conformation, allowing a transport step
Show answer
Phosphate transfer alters the transporter’s interactions or conformation, allowing a transport step
16. Why might mitochondria be positioned along microtubules near sites of high cellular demand?
- ATP can only be made inside the nuclear envelope
- Motor-associated redistribution can place ATP-producing organelles where they are useful
- Mitochondria are permanently fixed in one position in every cell
- Microtubules convert mitochondria into plant cell walls
Show answer
Motor-associated redistribution can place ATP-producing organelles where they are useful
17. Why are cristae relevant to mitochondrial ATP production?
- They contain cellulose needed for ATP hydrolysis
- They are extracellular secretory vesicles
- Their folds provide extensive inner-membrane area associated with oxidative phosphorylation
- They act as nuclear pores for importing mRNA
Show answer
Their folds provide extensive inner-membrane area associated with oxidative phosphorylation
EASY · 3 MARKS
1. Define osmosis and explain the direction of water movement between two solutions separated by a suitable membrane.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for defining osmosis as net water movement across a selectively permeable membrane
1 mark for identifying a difference in water potential or effective impermeant solute concentration as the driver
1 mark for explaining net water moves towards lower water potential or higher effective solute concentration
EASY · 3 MARKS
2. Describe channel-mediated transport and distinguish it from active pumping.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying channels as proteins that offer selective paths through a membrane
1 mark for explaining channel-mediated movement generally proceeds down an electrochemical gradient
1 mark for explaining an energy-coupled pump can move solutes against a gradient
EASY · 3 MARKS
3. Distinguish endocytosis from exocytosis.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for defining endocytosis as uptake into the cell through membrane invagination and vesicle formation
1 mark for defining exocytosis as release from the cell by vesicle fusion with the plasma membrane
1 mark for providing one correct example such as uptake of extracellular cargo or secretion of a protein
EASY · 3 MARKS
4. Explain how cytoskeletal microtubules and motor proteins assist intracellular transport.
Show marking guide
Indicative marking guide (3 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying microtubules as tubulin-based cytoskeletal tracks
1 mark for identifying motor proteins as associated with movement of cargo along the tracks
1 mark for explaining vesicles or organelles can be repositioned within the cell
INTERMEDIATE · 5 MARKS
5. Describe how a newly synthesised protein can travel in a vesicle to a different cellular location.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying cargo packaging into a membrane-bound vesicle
1 mark for explaining vesicle budding from a donor compartment
1 mark for describing movement of a vesicle through cytoplasm
1 mark for identifying microtubules and motor proteins as possible transport machinery
1 mark for explaining vesicle fusion with a target membrane delivers its cargo
INTERMEDIATE · 5 MARKS
6. Describe why channel opening need not be equivalent to ATP-driven pumping.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying selective channel proteins as pathways across the membrane
1 mark for recognising opening or closing can be regulated
1 mark for explaining net ion movement through an open channel follows the electrochemical gradient
1 mark for explaining pumping can drive movement against that gradient
1 mark for explaining energy coupling is needed for uphill pumping but not each downhill ion movement through a channel
INTERMEDIATE · 5 MARKS
7. Explain osmosis in a system with two compartments and an impermeant solute concentrated on one side.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for describing the water-permeable and solute-impermeable membrane
1 mark for identifying the unequal effective solute concentrations
1 mark for explaining this creates a difference in water potential
1 mark for predicting net movement of water towards the higher effective solute concentration
1 mark for explaining net movement slows as the water-potential difference decreases or pressure counterbalances it
INTERMEDIATE · 5 MARKS
8. Explain why the lecture challenged the standard bean-shaped depiction of mitochondria.
Show marking guide
Indicative marking guide (5 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying the common oval textbook representation as a simplification
1 mark for explaining mitochondria can form elongated or connected networks
1 mark for recognising mitochondrial morphology can change over time or among cells
1 mark for explaining intracellular positioning can change relative to local demand
1 mark for identifying cytoskeletal interactions as a mechanism contributing to positioning
HARD · 8 MARKS
9. A cell has normal ATP synthesis but defects in both tubulin polymerisation and vesicle motor proteins. Predict consequences for trafficking and organelle distribution.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying tubulin as the building block of microtubules
1 mark for explaining microtubules normally provide intracellular tracks
1 mark for explaining motor proteins transport vesicles or organelles along these tracks
1 mark for predicting decreased efficiency of long-range vesicle movement
1 mark for predicting altered mitochondrial positioning or distribution
1 mark for recognising ATP presence alone is insufficient if mechanical machinery is disrupted
1 mark for distinguishing other cytoskeletal systems that may remain intact
1 mark for linking impaired distribution to potential mislocalisation of cargo or cellular processes
HARD · 8 MARKS
10. A student claims that oxygen, sodium ions and large protein cargo must all use the same membrane transport mechanism. Explain why this is incorrect.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying small non-polar gases such as oxygen as able to diffuse through a lipid bilayer
1 mark for explaining diffusion proceeds down an appropriate concentration gradient
1 mark for recognising sodium is charged and generally needs a selective protein pathway
1 mark for describing channel-mediated passive ion movement down an electrochemical gradient
1 mark for describing energy-coupled ion pumping as a different option for uphill transport
1 mark for recognising a large folded protein cannot simply diffuse through the bilayer core
1 mark for explaining protein cargo can be packaged within a vesicle
1 mark for explaining vesicle fusion or budding supports secretion or uptake of large cargo
HARD · 8 MARKS
11. A tissue undergoes rapid ATP demand near one end of a long cell. Discuss why mitochondrial networks and cytoskeleton-mediated positioning could be advantageous.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying mitochondria as important for oxidative ATP generation
1 mark for identifying localised energy demand as the physiological challenge
1 mark for recognising mitochondria can be elongated or interconnected rather than always isolated
1 mark for explaining network organisation can be dynamic
1 mark for identifying microtubules as cytoskeletal tracks
1 mark for explaining motor-mediated movement can reposition mitochondria
1 mark for relating redistribution to support of local energy-demanding processes
1 mark for recognising morphology and positioning do not imply all ATP is made in a single fixed spot
HARD · 8 MARKS
12. An ATP-driven ion pump creates a gradient that later drives ion entry through a gated channel. Analyse how the two transport mechanisms work together.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying the pump as an energy-coupled membrane protein
1 mark for explaining it transports an ion against an electrochemical gradient
1 mark for explaining ATP hydrolysis or phosphorylation can provide the required energy
1 mark for recognising this builds or maintains a concentration or electrical gradient
1 mark for identifying the channel as a selective, gated pathway
1 mark for explaining ion entry through an open channel can proceed passively down the stored gradient
1 mark for noting that channel opening may be triggered by voltage or a ligand
1 mark for concluding the pump uses energy to establish the gradient while channel flow can exploit it without direct ATP use
HARD · 8 MARKS
13. Analyse why mitochondrial function extends beyond ATP production in a cell responding to stress signals.
Show marking guide
Indicative marking guide (8 marks; accept scientifically correct equivalent wording and examples):
1 mark for identifying oxidative ATP synthesis as a major mitochondrial activity
1 mark for connecting ATP availability to energy-demanding stress responses
1 mark for identifying mitochondrial contributions to steroidogenesis
1 mark for explaining steroid synthesis connects mitochondrial chemistry with hormone-related physiology
1 mark for identifying mitochondrial involvement in apoptotic signalling
1 mark for explaining apoptosis provides a controlled route to remove damaged or unneeded cells
1 mark for recognising mitochondrial structures and activities are dynamically regulated
1 mark for concluding the organelle has integrated metabolic, biosynthetic and regulatory functions
Lecture 16: Cells, Tissues and Communication
1. How does Desmosomes, basal lamina and invasion contribute to the wider biological or ecological process discussed?
- Desmosomes are identical to endocrine hormones
- The basal lamina is the exposed apical surface
- Epithelial cells rely on cell junctions to resist physical stress
- Desmosomes are pores that pump ATP into cells
Show answer
Epithelial cells rely on cell junctions to resist physical stress
2. How does Four animal tissue classes contribute to the wider biological or ecological process discussed?
- All animal tissues are variants of epithelium
- Connective tissue lacks extracellular components
- Muscle tissue is limited to the brain
- Nervous tissue coordinates responses through electrical and chemical signalling
Show answer
Nervous tissue coordinates responses through electrical and chemical signalling
3. An ion channel opens in an excitable neurone and the membrane becomes less negative. Which interpretation best explains this observation?
- Selective ion movement can produce depolarisation and initiate electrical signalling.
- Nervous tissue cannot communicate with muscles
- Neurons depolarise only by synthesising new DNA
- Action potentials travel through extracellular collagen rather than cells
Show answer
Selective ion movement can produce depolarisation and initiate electrical signalling.
4. How does Vectorial epithelial transport contribute to the wider biological or ecological process discussed?
- The apical and basal membranes are a single continuous transport channel
- Passive movement may then occur down a gradient on the opposite side
- Active transport and passive transport are synonyms
- Directional epithelial transport requires both faces to be impermeable
Show answer
Passive movement may then occur down a gradient on the opposite side
5. Adjacent epithelial cells experience repeated mechanical strain without separating. Which interpretation best explains this observation?
- Intercellular adhesive junctions such as desmosomes help maintain the sheet.
- Desmosomes are pores that pump ATP into cells
- Cancer spread is permanently impossible while any cell junction exists
- Desmosomes are identical to endocrine hormones
Show answer
Intercellular adhesive junctions such as desmosomes help maintain the sheet.
6. An epithelial solute is pumped across its apical membrane and later leaves across the basal membrane. Which interpretation best explains this observation?
- Directional epithelial transport requires both faces to be impermeable
- The apical and basal membranes are a single continuous transport channel
- Energy-dependent and gradient-driven steps can cooperate to achieve net transport.
- Solutes must move uphill through both membranes without energy
Show answer
Energy-dependent and gradient-driven steps can cooperate to achieve net transport.
7. Stress activates a sequence involving the brain, pituitary gland and adrenal cortex. Which interpretation best explains this observation?
- This is the coordinated hypothalamic-pituitary-adrenal endocrine axis.
- Cortisol is the neurotransmitter responsible for every immediate nerve impulse
- Pituitary peptide hormones leave cells by simple diffusion through phospholipids
- The adrenal cortex makes CRH to activate the hypothalamus
Show answer
This is the coordinated hypothalamic-pituitary-adrenal endocrine axis.
8. How does Epithelial sheets and polarity contribute to the wider biological or ecological process discussed?
- The apical and basal surfaces are functionally identical
- The basal surface associates with the basal lamina
- Epithelial sheets are always electrically excitable neurons
- Epithelial cells are defined by abundant extracellular matrix between cells
Show answer
The basal surface associates with the basal lamina
9. A tissue sample has relatively dispersed cells surrounded by fibres and a hydrated background material. Which interpretation best explains this observation?
- All connective tissues contain no living cells
- Connective tissue consists exclusively of tightly packed cell membranes
- The combination is characteristic of connective tissue with extracellular matrix.
- Ground substance refers to nuclear DNA
Show answer
The combination is characteristic of connective tissue with extracellular matrix.
10. A sheet of cells has an exposed surface and a second face anchored to supportive material. Which interpretation best explains this observation?
- Epithelial sheets are always electrically excitable neurons
- The basal lamina sits inside the cell nucleus
- This is a polarised epithelium with distinct apical and basal domains.
- The apical and basal surfaces are functionally identical
Show answer
This is a polarised epithelium with distinct apical and basal domains.
11. How does Steroid and protein hormone signalling contribute to the wider biological or ecological process discussed?
- All hormone receptors sit inside the nucleus
- Steroid hormones cannot cross a lipid bilayer under any circumstances
- Protein hormones always enter directly through the membrane's hydrophobic core
- Protein hormones are generally hydrophilic and bind cell-surface receptors
Show answer
Protein hormones are generally hydrophilic and bind cell-surface receptors
12. A section from an intestine contains contractile fibres in the wall but no heart tissue. Which interpretation best explains this observation?
- Smooth muscle is the relevant contractile tissue for that organ.
- Skeletal muscle generates most hormonal signals in blood
- Cardiac muscle is the main contractile tissue in intestinal walls
- All muscle tissue has identical microscopic organisation and control
Show answer
Smooth muscle is the relevant contractile tissue for that organ.
13. A hormone binds a receptor inside the cytoplasm and changes gene transcription. Which interpretation best explains this observation?
- All hormone receptors sit inside the nucleus
- Steroid hormones cannot cross a lipid bilayer under any circumstances
- Cyclic AMP is the same molecule as a steroid receptor
- A lipid-soluble steroid acting via an intracellular receptor is a plausible mechanism.
Show answer
A lipid-soluble steroid acting via an intracellular receptor is a plausible mechanism.
14. A cell responds to a cue within seconds by activating an existing enzyme. Which interpretation best explains this observation?
- Receptors cannot alter cytoplasmic enzyme activity
- Apoptosis always prevents normal tissue development
- Every signal has an identical effect in all cells
- Rapid signalling can change pre-existing protein activity without waiting for new transcription.
Show answer
Rapid signalling can change pre-existing protein activity without waiting for new transcription.
15. How does Nervous tissue and depolarisation contribute to the wider biological or ecological process discussed?
- Neurons depolarise only by synthesising new DNA
- Depolarisation means every ion gradient disappears permanently
- Nervous tissue cannot communicate with muscles
- Opening suitable ion channels can permit depolarisation
Show answer
Opening suitable ion channels can permit depolarisation
EASY · 3 MARKS
1. Describe three important features or roles of epithelial sheets and polarity as covered in this lecture.
Show marking guide
Indicative marking guide (3 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that epithelial cells form closely associated sheets over surfaces and cavities.
1 mark for correctly recognising that the apical surface faces a lumen or external space.
1 mark for correctly recognising that the basal surface associates with the basal lamina.
EASY · 3 MARKS
2. Describe three important features or roles of four animal tissue classes as covered in this lecture.
Show marking guide
Indicative marking guide (3 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that epithelium forms protective and exchange surfaces.
1 mark for correctly recognising that connective tissue provides support through cells embedded in matrix.
1 mark for correctly recognising that nervous tissue coordinates responses through electrical and chemical signalling.
EASY · 3 MARKS
3. Describe three important features or roles of steroid and protein hormone signalling as covered in this lecture.
Show marking guide
Indicative marking guide (3 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that steroid hormones are lipid-soluble molecules derived from cholesterol.
1 mark for correctly recognising that steroid hormones commonly interact with intracellular receptors.
1 mark for correctly recognising that protein hormones are generally hydrophilic and bind cell-surface receptors.
EASY · 3 MARKS
4. Describe three important features or roles of the HPA stress-response axis and secretion as covered in this lecture.
Show marking guide
Indicative marking guide (3 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that the hypothalamus releases corticotropin-releasing hormone or CRH.
1 mark for correctly recognising that cRH stimulates pituitary release of adrenocorticotropic hormone or ACTH.
1 mark for correctly recognising that aCTH acts on the adrenal cortex to stimulate cortisol secretion.
INTERMEDIATE · 5 MARKS
5. Explain cell-to-cell signalling and development using relevant mechanisms or relationships from the lecture. Apply your explanation to this observation: A cell responds to a cue within seconds by activating an existing enzyme.
Show marking guide
Indicative marking guide (5 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that cells respond selectively to signals through receptors.
1 mark for correctly recognising that some signalling changes existing cytoplasmic proteins rapidly.
1 mark for correctly recognising that changes requiring transcription and translation take longer.
1 mark for correctly recognising that development depends on coordinated survival, division, differentiation and apoptosis signals.
1 mark for correctly recognising that rapid signalling can change pre-existing protein activity without waiting for new transcription.
INTERMEDIATE · 5 MARKS
6. Explain desmosomes, basal lamina and invasion using relevant mechanisms or relationships from the lecture. Apply your explanation to this observation: Adjacent epithelial cells experience repeated mechanical strain without separating.
Show marking guide
Indicative marking guide (5 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that desmosomes mechanically connect neighbouring epithelial cells.
1 mark for correctly recognising that the basal lamina supports and anchors the epithelial sheet.
1 mark for correctly recognising that epithelial cells rely on cell junctions to resist physical stress.
1 mark for correctly recognising that breaching the basal lamina can facilitate movement of malignant cells into underlying tissue.
1 mark for correctly recognising that intercellular adhesive junctions such as desmosomes help maintain the sheet.
INTERMEDIATE · 5 MARKS
7. Explain the HPA stress-response axis and secretion using relevant mechanisms or relationships from the lecture. Apply your explanation to this observation: Stress activates a sequence involving the brain, pituitary gland and adrenal cortex.
Show marking guide
Indicative marking guide (5 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that the hypothalamus releases corticotropin-releasing hormone or CRH.
1 mark for correctly recognising that cRH stimulates pituitary release of adrenocorticotropic hormone or ACTH.
1 mark for correctly recognising that aCTH acts on the adrenal cortex to stimulate cortisol secretion.
1 mark for correctly recognising that cortisol helps mobilise energy stores during physiological stress.
1 mark for correctly recognising that this is the coordinated hypothalamic-pituitary-adrenal endocrine axis.
HARD · 8 MARKS
8. Compare and analyse cell-to-cell signalling and development and steroid and protein hormone signalling in the context of this lecture. Use the following cases to explain mechanisms and predict consequences: (i) During digit formation, cells between developing fingers are not removed despite otherwise normal growth. (ii) A hormone binds a receptor inside the cytoplasm and changes gene transcription.
Show marking guide
Indicative marking guide (8 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that cells respond selectively to signals through receptors.
1 mark for correctly recognising that some signalling changes existing cytoplasmic proteins rapidly.
1 mark for correctly recognising that changes requiring transcription and translation take longer.
1 mark for correctly recognising that development depends on coordinated survival, division, differentiation and apoptosis signals.
1 mark for correctly recognising that steroid hormones are lipid-soluble molecules derived from cholesterol.
1 mark for correctly recognising that steroid hormones commonly interact with intracellular receptors.
1 mark for correctly recognising that failure of the normal apoptosis programme could contribute to incomplete separation of digits.
1 mark for correctly recognising that a lipid-soluble steroid acting via an intracellular receptor is a plausible mechanism.
HARD · 8 MARKS
9. Compare and analyse epithelial sheets and polarity and desmosomes, basal lamina and invasion in the context of this lecture. Use the following cases to explain mechanisms and predict consequences: (i) An epithelial layer loses its basal attachment yet remains in contact with a lumen. (ii) Adjacent epithelial cells experience repeated mechanical strain without separating.
Show marking guide
Indicative marking guide (8 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that epithelial cells form closely associated sheets over surfaces and cavities.
1 mark for correctly recognising that the apical surface faces a lumen or external space.
1 mark for correctly recognising that the basal surface associates with the basal lamina.
1 mark for correctly recognising that epithelial polarity permits distinct functions on opposite cell surfaces.
1 mark for correctly recognising that desmosomes mechanically connect neighbouring epithelial cells.
1 mark for correctly recognising that the basal lamina supports and anchors the epithelial sheet.
1 mark for correctly recognising that the disruption compromises its normal anchorage and tissue organisation even if its apical surface remains exposed.
1 mark for correctly recognising that intercellular adhesive junctions such as desmosomes help maintain the sheet.
HARD · 8 MARKS
10. Compare and analyse nervous tissue and depolarisation and connective tissue and extracellular matrix in the context of this lecture. Use the following cases to explain mechanisms and predict consequences: (i) Channel function is blocked while neuronal structure remains intact. (ii) A tissue sample has relatively dispersed cells surrounded by fibres and a hydrated background material.
Show marking guide
Indicative marking guide (8 marks; accept scientifically accurate equivalent wording):
1 mark for correctly recognising that nervous tissue supports rapid coordination and behaviour.
1 mark for correctly recognising that a neuron's membrane potential depends on unequal ion distributions.
1 mark for correctly recognising that opening suitable ion channels can permit depolarisation.
1 mark for correctly recognising that membrane electrical changes can lead to signalling to another cell.
1 mark for correctly recognising that connective tissue contains cells and non-cellular extracellular material.
1 mark for correctly recognising that fibres such as collagen contribute mechanical properties.
1 mark for correctly recognising that rapid electrical communication can fail despite the neuron still being physically present.
1 mark for correctly recognising that the combination is characteristic of connective tissue with extracellular matrix.
Lecture 19: Microbiology and One Health
1. A microscopic predator has a nucleus and feeds on other cells. Which group best fits?
- A filamentous mould
- A bacteriophage
- A photosynthetic bacterium
- A predatory protist
Show answer
A predatory protist
2. Two patients have the same symptoms, but distinct microorganisms are recovered. Which concern does this raise for a simple causal claim?
- Agar culture always produces identical isolates
- Microscopy cannot reveal any microbial differences
- Similar symptoms need not establish the same causative agent
- Every disease must have exactly two causative agents
Show answer
Similar symptoms need not establish the same causative agent
3. A scientist sees several cell types in a mixed broth culture. Why would isolating colonies on agar improve the investigation?
- It removes the need to examine cell characteristics
- It makes every microbe produce identical symptoms
- It guarantees that all microbes will grow
- It helps separate populations for subsequent identification
Show answer
It helps separate populations for subsequent identification
4. A microbe is detected on a person's hands immediately after touching a door handle but is absent after washing. What is the best interpretation?
- It must have reproduced within the skin
- It was probably a transient microbe
- It was necessarily an obligate pathogen
- It was necessarily a permanent member of skin microbiota
Show answer
It was probably a transient microbe
5. A flask remains free of growth when sealed but develops growth after opening. What explanation is most consistent with Pasteur's position?
- Air itself transformed into new living cells
- Sterile broth generated cells once exposed to light
- Microorganisms entered from the surroundings
- The flask walls converted nutrients into microbes
Show answer
Microorganisms entered from the surroundings
6. A person acquires microbes from diet and environmental contact, and the community changes over time. Which property of microbiota does this illustrate?
- Its composition is dynamic and varies among individuals
- Its composition is fixed permanently at birth
- Its members are all obligate pathogens
- Its members are exclusively fungi
Show answer
Its composition is dynamic and varies among individuals
7. A bacterium causes disease mainly in patients whose immune defences are weakened. How is it best described?
- Non-living infectious particle
- Obligate pathogen in every host
- Transient microbe incapable of infection
- Opportunistic pathogen
Show answer
Opportunistic pathogen
8. A microorganism has a nucleus and mitochondria and can occur as a single-celled yeast. Which group best fits?
- Bacteria
- Viruses
- Archaea
- Fungi
Show answer
Fungi
9. A photosynthetic microscopic organism has a nucleus. Which classification is compatible with the lecture?
- An acellular virus
- A eukaryotic alga
- A prokaryotic bacillus
- A bacterial cell lacking organelles
Show answer
A eukaryotic alga
10. A microorganism is recovered from healthy skin over many months and is adapted to that body site. Which category fits best?
- Resident microbiota
- Transient contaminant
- Obligate pathogen
- Acellular virus
Show answer
Resident microbiota
11. A patient and an experimentally infected animal have similar symptoms. What additional evidence is needed for the final Koch postulate?
- Demonstrating that the animal was exposed to air
- Proving that the organism can grow without nutrients
- Recovering the same candidate organism from the animal
- Showing that the organism is found on healthy skin
Show answer
Recovering the same candidate organism from the animal
12. Why does a negative agar-culture result not always show that a sample contains no microorganisms?
- Some microorganisms cannot grow under the chosen culture conditions
- Agar automatically destroys any organism present
- Microscopy cannot detect microorganisms in samples
- All microbes must grow in every agar medium
Show answer
Some microorganisms cannot grow under the chosen culture conditions
13. A normally harmless skin bacterium enters a deep wound and causes infection. Which explanation best fits?
- An obligate pathogen must have been present from birth
- The bacterium has become a virus in the wound
- An opportunistic pathogen has reached an unusual body location
- All normal microbiota always cause disease after contact
Show answer
An opportunistic pathogen has reached an unusual body location
14. An infectious agent contains RNA within a protein coat and cannot replicate without a host cell. How should it be classified?
- Protist
- Virus
- Fungus
- Bacterium
Show answer
Virus
15. Why did sealed-flask experiments matter to debates about spontaneous generation?
- They demonstrated that agar is necessary for growth
- They tested whether new growth required contamination from outside
- They established that microbes arise from sterile nutrients
- They proved that all microbes need oxygen to survive
Show answer
They tested whether new growth required contamination from outside
16. A candidate pathogen is consistently present in sick hosts but cannot be grown in a cell-free medium. Which conclusion is most defensible?
- Its presence in patients alone establishes causation
- Failure to culture it limits application of the classical postulates
- The organism must be a harmless member of the microbiota
- Its inability to grow on agar proves it cannot cause disease
Show answer
Failure to culture it limits application of the classical postulates
17. A disease emerges in people following increased contact with infected wildlife. Which approach best reflects One Health?
- Investigate wildlife exposure alongside human infection patterns
- Study wildlife without considering human transmission
- Examine human cases without recording animal contact
- Restrict the investigation to hospital equipment
Show answer
Investigate wildlife exposure alongside human infection patterns
EASY · 3 MARKS
1. Describe three different bacterial cell shapes using their standard names.
Show marking guide
1 mark for cocci are spherical.
1 mark for bacilli are rod-shaped.
1 mark for vibrios are comma-shaped.
Accept scientifically equivalent wording and well-justified alternatives.
EASY · 3 MARKS
2. Distinguish a transient microorganism from resident microbiota using three relevant points.
Show marking guide
1 mark for transient microbes are present temporarily.
1 mark for resident microbiota occupy body sites more persistently.
1 mark for transient microbes may be acquired through contact with people, animals or surfaces.
Accept scientifically equivalent wording and well-justified alternatives.
EASY · 3 MARKS
3. State three features of the human microbiota.
Show marking guide
1 mark for microbes live in or on the body.
1 mark for communities are specialised for particular body sites.
1 mark for composition varies between people or over time.
Accept scientifically equivalent wording and well-justified alternatives.
EASY · 3 MARKS
4. State three key distinctions between a predatory protist and a virus.
Show marking guide
1 mark for a predatory protist is cellular.
1 mark for a predatory protist is eukaryotic.
1 mark for a virus depends on a host cell for replication.
Accept scientifically equivalent wording and well-justified alternatives.
INTERMEDIATE · 5 MARKS
5. A bacterium causes disease in immunocompromised people but not most healthy carriers. Distinguish opportunistic from obligate pathogenicity using this example.
Show marking guide
1 mark for opportunistic disease depends on conditions.
1 mark for impaired immunity can permit disease.
1 mark for healthy carriage may occur without illness.
1 mark for an obligate pathogen is characterised in the lecture as causing disease.
1 mark for disease outcome depends on both microbial and host factors.
Accept scientifically equivalent wording and well-justified alternatives.
INTERMEDIATE · 5 MARKS
6. A patient develops an infection after a normally harmless skin bacterium enters a deep wound. Explain why this is an opportunistic infection and identify two relevant contributing conditions.
Show marking guide
1 mark for the organism may normally be part of the resident microbiota.
1 mark for it causes disease under particular conditions rather than inevitably.
1 mark for the wound creates access to an unusual or vulnerable location.
1 mark for damaged barriers increase the opportunity for infection.
1 mark for host immune status or high microbial numbers can further influence risk.
Accept scientifically equivalent wording and well-justified alternatives.
INTERMEDIATE · 5 MARKS
7. A suspected pathogen is found in every sick animal sampled. Explain why this observation is not enough to demonstrate causation using Koch’s postulates.
Show marking guide
1 mark for presence establishes an association.
1 mark for the organism should be isolated as a defined culture where possible.
1 mark for an isolate should reproduce disease in a susceptible host.
1 mark for the organism should be recovered again from the experimentally infected host.
1 mark for host or culture limitations can complicate the classical test.
Accept scientifically equivalent wording and well-justified alternatives.
INTERMEDIATE · 5 MARKS
8. A swab from healthy skin and a swab taken after touching a public surface yield different microorganisms. Explain how resident and transient communities could account for the findings.
Show marking guide
1 mark for resident microbes persist at characteristic body sites.
1 mark for contact can introduce transient microbes.
1 mark for transient composition may change quickly.
1 mark for hygiene can remove many recently acquired microbes.
1 mark for detection does not by itself establish pathogenicity.
Accept scientifically equivalent wording and well-justified alternatives.
INTERMEDIATE · 5 MARKS
9. Explain why a photosynthetic microscopic organism should not be classified as a protist solely because it photosynthesises.
Show marking guide
1 mark for photosynthesis is a functional trait.
1 mark for some eukaryotic protists photosynthesise.
1 mark for the lecture also recognises photosynthetic prokaryotic algae.
1 mark for cellular organisation is needed to classify it.
1 mark for a nucleus or other eukaryotic structures would provide additional evidence.
Accept scientifically equivalent wording and well-justified alternatives.
HARD · 8 MARKS
10. A bacterium is recovered from 90% of sick animals and 40% of healthy animals. Design a cautious investigation to test whether it contributes to disease, and explain why prevalence alone is insufficient.
Show marking guide
1 mark for recognise that association is not proof of causation.
1 mark for account for the organism in healthy hosts.
1 mark for compare exposure or host susceptibility.
1 mark for attempt isolation and reliable identification.
1 mark for use an appropriate susceptible host model where ethical and feasible.
1 mark for assess whether inoculation reproduces disease.
1 mark for attempt recovery of the matching organism after infection.
1 mark for interpret limitations of the classical postulates.
Accept scientifically equivalent wording and well-justified alternatives.
HARD · 8 MARKS
11. A candidate infectious agent is detected by microscopy but cannot grow on nutrient agar and replicates only inside host cells. Evaluate how the classical Koch framework should be applied.
Show marking guide
1 mark for recognise dependence on host cells.
1 mark for explain why nutrient agar is inadequate.
1 mark for distinguish detection from pure isolation.
1 mark for identify the need for a suitable host-cell system.
1 mark for seek reproducible association with illness.
1 mark for consider experimental infection only where appropriate.
1 mark for seek recovery or identification after experimental infection.
1 mark for state that failure of classical culture does not rule out causation.
Accept scientifically equivalent wording and well-justified alternatives.
HARD · 8 MARKS
12. A person carries an opportunistic bacterium without symptoms, then becomes immunosuppressed and develops disease. Explain how host and microbial factors interact, and assess what would be needed to establish causation.
Show marking guide
1 mark for describe asymptomatic carriage.
1 mark for identify changed host immune defence.
1 mark for explain conditional pathogenicity.
1 mark for consider microbial abundance.
1 mark for consider anatomical location.
1 mark for distinguish carriage from infection.
1 mark for identify the need for organism-specific evidence.
1 mark for avoid assuming temporal sequence alone proves cause.
Accept scientifically equivalent wording and well-justified alternatives.
HARD · 8 MARKS
13. A student claims that cloudy broth in an open flask proves spontaneous generation. Construct an experimental comparison that distinguishes this claim from environmental contamination.
Show marking guide
1 mark for prepare comparable initially sterilised broths.
1 mark for include a sealed condition.
1 mark for include an exposed condition.
1 mark for control handling and incubation.
1 mark for record whether growth occurs.
1 mark for test whether exposure permits microbial entry.
1 mark for explain predicted results under contamination.
1 mark for avoid treating cloudiness alone as proof of origin.
Accept scientifically equivalent wording and well-justified alternatives.
Lecture 20: Microbes, Food and Nutrition
1. A cook cuts raw chicken and then slices cooked vegetables on the same unwashed board. Which pathway is the principal concern?
- Fermentation of the raw chicken by yeast
- Microbial decomposition of soil organic matter
- Nitrogen fixation in the cooked vegetables
- Cross-contamination of ready-to-eat food
Show answer
Cross-contamination of ready-to-eat food
2. A food producer wants to slow microbial spoilage during transport without changing the recipe. Which intervention best targets microbial growth rate?
- Package food only after spoilage becomes visible
- Allow raw and cooked products to share utensils
- Maintain appropriate refrigeration throughout transport
- Increase the time food spends at room temperature
Show answer
Maintain appropriate refrigeration throughout transport
3. A plant receives additional phosphorus and water through a fungal network attached to its roots. What does the fungus most likely obtain in return?
- Atmospheric nitrogen fixed by the leaves
- Oxygen released by bacterial fermentation
- Cellulose digested in the animal rumen
- Sugars produced by the plant
Show answer
Sugars produced by the plant
4. Two genetically uniform banana plantations are exposed to the same fungal pathogen. One contains a mixture of resistant varieties, while the other contains one susceptible clone. What is the likely difference?
- Disease may spread less extensively in the genetically mixed plantation
- The susceptible clone must develop resistance after the first infection
- Disease must spread equally because the pathogen is identical
- Genetic diversity prevents fungal spores from reaching either plantation
Show answer
Disease may spread less extensively in the genetically mixed plantation
5. A manufacturer ferments milk with lactic acid bacteria. Which change contributes directly to preservation?
- Oxygen release prevents all bacterial growth
- Acid production lowers the pH and inhibits some spoilage microbes
- Nitrogen fixation removes every food-borne pathogen
- Alcohol production raises the pH and sterilises the milk
Show answer
Acid production lowers the pH and inhibits some spoilage microbes
6. A tick carries Borrelia from an infected animal to a person. What role does the tick play?
- Vector transmitting the pathogen
- Primary source of atmospheric nitrogen
- Host that necessarily develops the same symptoms as the person
- Fermentation organism preserving the pathogen
Show answer
Vector transmitting the pathogen
7. A baker adds yeast to dough and allows it to ferment. Which product chiefly causes the dough to rise?
- Oxygen produced during anaerobic metabolism
- Lactic acid produced by fungal photosynthesis
- Nitrogen gas fixed by yeast cells
- Carbon dioxide released from sugar metabolism
Show answer
Carbon dioxide released from sugar metabolism
8. A person contracts rabies after an infected animal bites them. Which route of zoonotic transmission is illustrated?
- Transfer by a tick carrying a bacterial pathogen
- Transmission only through a food-processing surface
- Exposure to a harmless resident gut microbe
- Direct transmission from an infected animal
Show answer
Direct transmission from an infected animal
9. A chilled meal is left at 25°C for several hours before serving. Why is this a concern?
- The temperature prevents transfer of microorganisms from hands
- The temperature destroys all microbial spores
- The temperature permits growth of many food-associated microorganisms
- The temperature makes any preformed toxin harmless
Show answer
The temperature permits growth of many food-associated microorganisms
10. A company claims its probiotic product improves every digestive disorder because it contains live bacteria. Which appraisal best fits the lecture?
- A probiotic is effective whenever it survives packaging
- All live bacterial strains produce identical clinical effects
- The presence of bacteria alone proves a health benefit
- Benefits depend on the particular strain and outcome, so broad claims need evidence
Show answer
Benefits depend on the particular strain and outcome, so broad claims need evidence
11. A legume grows poorly in nitrogen-poor soil but improves after inoculation with compatible root-nodule bacteria. Which process most directly explains the improvement?
- Transfer of photosynthetic oxygen into bacterial protein
- Release of carbon dioxide as the plant's nitrogen source
- Conversion of atmospheric nitrogen into ammonium
- Conversion of soil phosphate into atmospheric nitrogen
Show answer
Conversion of atmospheric nitrogen into ammonium
12. A stool sample is analysed by sequencing microbial DNA rather than culturing on agar. What advantage does this provide?
- Identification of every organism's contribution to human health
- Measurement of toxin concentration without further testing
- Proof that every detected microbe is alive and active
- Detection of taxa that do not grow under the selected culture conditions
Show answer
Detection of taxa that do not grow under the selected culture conditions
13. A person develops gastroenteritis after eating contaminated chicken; bacteria multiply in the intestine. Which mechanism best describes the illness?
- Non-infectious food spoilage alone
- Food-borne infection
- Food-borne intoxication from preformed toxin
- Allergic response to bacterial fermentation
Show answer
Food-borne infection
14. A food sample looks and smells normal, but a pathogen is detected. Which conclusion is best supported?
- Normal appearance proves that microbial toxins are absent
- Food can be unsafe without obvious spoilage
- Only spoiled food can contain infectious organisms
- Food is safe whenever no mould is visible
Show answer
Food can be unsafe without obvious spoilage
15. Several diners develop symptoms soon after eating a dish in which bacteria had produced a toxin before serving. Which mechanism is most consistent with this observation?
- Food-borne intoxication
- Spoilage that cannot cause illness
- A beneficial fermentation with no health risk
- Food-borne infection requiring intestinal multiplication
Show answer
Food-borne intoxication
EASY · 3 MARKS
1. Give three contrasting ways microorganisms affect food production, preparation or consumption.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that microbes affect food from farms to the consumer
1 mark for recognising that some soil and plant microbes improve production
1 mark for recognising that some microbes cause spoilage or food-borne illness
EASY · 3 MARKS
2. Identify three ways diet or environmental factors can influence gut microbial communities.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that the gut contains complex microbial communities detectable by DNA sequencing
1 mark for recognising that dietary fibre can support microbes producing short-chain fatty acids
1 mark for recognising that antibiotic courses can temporarily disrupt gut microbial diversity
EASY · 3 MARKS
3. State three key characteristics or consequences of foot-and-mouth disease in livestock systems.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that foot-and-mouth disease is caused by a virus affecting several cloven-hoofed livestock species
1 mark for recognising that major outbreaks can impose severe production and trade costs
1 mark for recognising that imported infected animal products can introduce disease risk
INTERMEDIATE · 5 MARKS
4. Explain how a bacterium can be part of the normal microbial flora of one animal yet cause disease in a person consuming contaminated food.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that a zoonosis is an infection transmitted from an animal source to humans
1 mark for recognising that rabies is presented as pathogenic in both animal and human hosts
1 mark for recognising that salmonella can be normal flora in chickens yet cause human disease
1 mark for recognising that some pathogens use an animal vector such as a tick to reach humans
1 mark for recognising that differences in host context can make an animal-associated organism pathogenic in people
INTERMEDIATE · 5 MARKS
5. Explain why a cow can use cellulose-rich forage even though host-produced cellulase is inadequate, identifying the roles of microbes and fermentation products.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that plant cellulose is difficult for ruminants to digest without microbial assistance
1 mark for recognising that rumen microbes produce cellulolytic enzymes
1 mark for recognising that microbial breakdown produces organic acids and other products usable by animals
1 mark for recognising that rumen metabolism also generates methane with climate implications
1 mark for recognising that symbiotic rumen microorganisms explain conversion of cellulose into accessible nutrients
INTERMEDIATE · 5 MARKS
6. Explain why contaminated animal products arriving from an affected region may justify biosecurity action before any local animal shows symptoms.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that foot-and-mouth disease is caused by a virus affecting several cloven-hoofed livestock species
1 mark for recognising that major outbreaks can impose severe production and trade costs
1 mark for recognising that imported infected animal products can introduce disease risk
1 mark for recognising that border biosecurity helps reduce introduction into livestock systems
1 mark for recognising that preventing their entry is a plausible measure to protect susceptible livestock
HARD · 8 MARKS
7. A treatment suppresses rumen microbial activity without changing feed consumption. Predict nutritional consequences and discuss the trade-off between rumen fermentation and methane emissions.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying the rumen microbial community as crucial for cellulose breakdown
1 mark for explaining why ruminants rely on microbes to access energy from fibre
1 mark for predicting reduced digestion or energy extraction with suppressed fermentation
1 mark for recognising that unchanged food intake need not imply unchanged nutrient absorption
1 mark for identifying microbial fermentation as a contributor to methane generation
1 mark for predicting a possible reduction in methane production under suppression
1 mark for explaining the trade-off between climate effects and animal nutrition
1 mark for proposing evaluation of both digestive efficiency and methane outcomes
HARD · 8 MARKS
8. Design a microbial food-safety strategy that extends from farm soil to consumers. Evaluate beneficial microbial activities alongside the risks that would be overlooked by focusing only on packaged foods.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for recognising that food-related microbiology begins before packaging
1 mark for identifying beneficial soil or plant-associated microbes in production
1 mark for identifying beneficial fermentation or preservation activities
1 mark for identifying a microbial pathogen or spoilage risk along the chain
1 mark for recognising potential contamination during processing, storage or handling
1 mark for explaining that consumer food safety also depends on preparation and storage
1 mark for proposing prevention or monitoring at multiple stages
1 mark for justifying protection of beneficial processes while controlling harmful organisms
HARD · 8 MARKS
9. Explain why detection of foot-and-mouth disease in livestock after a suspected importation could affect animal health, agricultural production and international trade. Propose proportionate border precautions.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying foot-and-mouth disease as a contagious viral livestock disease
1 mark for recognising susceptibility of cloven-hoofed livestock
1 mark for explaining potential animal-health and productivity effects
1 mark for recognising movement or imports as potential transmission routes
1 mark for explaining trade or quarantine consequences for an exporting country
1 mark for proposing appropriate detection, reporting or surveillance at borders
1 mark for proposing biosecurity or movement restrictions proportionate to risk
1 mark for justifying a preventative strategy to limit animal and economic damage
HARD · 8 MARKS
10. Two patients have similar symptoms after a shared meal, but one ingested a pre-formed toxin and the other ingested viable pathogens. Compare the biological mechanisms and identify implications for prevention and interpretation of food appearance.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying food intoxication as ingestion of pre-formed toxin
1 mark for identifying food infection as ingestion of viable pathogens
1 mark for explaining that pre-formed toxin need not require growth within the patient
1 mark for explaining why viable pathogens can multiply or cause disease after ingestion
1 mark for recognising that similar symptoms do not identify the same mechanism
1 mark for identifying handling or storage conditions that allow microbial hazards to develop
1 mark for explaining why food can look normal yet remain hazardous
1 mark for proposing prevention appropriate to both living pathogens and toxins
Lecture 21: Microbes and Ecosystems
1. Antibiotic runoff enters a river microbial community. What selective effect is possible?
- Selection for antimicrobial resistance
- Prevention of gene exchange among environmental bacteria
- Conversion of antibiotics into photosynthetic pigments
- Guaranteed elimination of all resistant bacteria
Show answer
Selection for antimicrobial resistance
2. A marine bloom converts dissolved carbon dioxide into biomass. Which process is occurring?
- Production of methane from carbon dioxide
- Release of carbon dioxide by decomposition
- Photosynthetic carbon fixation
- Oxidation of methane to carbon dioxide
Show answer
Photosynthetic carbon fixation
3. A lichen grows on exposed rock despite intermittent drying. Which fungal function is important?
- Production of all photosynthetic sugars
- Structure and water retention
- Methane oxidation as the sole carbon source
- Production of oxygen by photosynthesis
Show answer
Structure and water retention
4. A cyanobacterium fixes carbon dioxide in illuminated water. Which energy source drives fixation?
- Light
- Oxidation of organic detritus
- Methane oxidation
- Hydrogen oxidation without light
Show answer
Light
5. Lichens decline near a source of sulphur dioxide pollution. Which characteristic explains their use in monitoring?
- Obligate growth in heavily polluted air
- Ability to remove every pollutant without damage
- Dependence on petroleum hydrocarbons
- Sensitivity to atmospheric pollutants
Show answer
Sensitivity to atmospheric pollutants
6. A microbial cell builds organic material using dissolved inorganic carbon. Which carbon strategy does it use?
- Carbon dioxide
- Dead plant material
- Preformed organic molecules
- Methane as an obligatory organic nutrient
Show answer
Carbon dioxide
7. A microbe fixes carbon dioxide in a dark environment. What can power this metabolism?
- Light captured by chlorophyll
- Energy from inorganic chemical reactions
- Heat alone without a chemical energy source
- Digestion of preformed organic carbon as its sole carbon source
Show answer
Energy from inorganic chemical reactions
8. A soil sample yields few colonies but contains many microbial cells. Which method best reveals uncultured diversity?
- Measuring colony diameter on agar
- Environmental DNA sequencing
- Counting colonies on one nutrient agar
- Recording only visible organisms in soil
Show answer
Environmental DNA sequencing
9. Methanotrophs can also transform some industrial solvents. Which enzyme supports this activity?
- Rubisco
- Nitrogenase
- DNA ligase
- Methane monooxygenase
Show answer
Methane monooxygenase
10. A survey identifies microbial bacteria, archaea and unicellular algae. Which domain-level classification fits?
- Bacteria, Archaea and Eukaryota
- Bacteria, fungi and protists as three domains
- Archaea, Plantae and Animalia
- Bacteria, Archaea and viruses
Show answer
Bacteria, Archaea and Eukaryota
11. Dead organic matter accumulates when microbial decomposition is inhibited. Which process has slowed?
- Preventing organic matter from entering food webs
- Converting all soil nitrogen directly into atmospheric oxygen
- Recycling nutrients from dead organic material
- Fixing carbon dioxide using light
Show answer
Recycling nutrients from dead organic material
12. An anaerobic digester contains archaea using carbon dioxide and hydrogen. Which gas is produced?
- Methane oxidation to carbon dioxide
- Oxygenic photosynthesis
- Methane production
- Aerobic decomposition requiring oxygen
Show answer
Methane production
13. Coral turns white after a marine heatwave. Which change explains the loss of colour?
- Loss of symbiotic algae under heat stress
- Enhanced photosynthesis within the coral skeleton
- Replacement of coral polyps by lichens
- Increased nitrogen fixation by coral tissue
Show answer
Loss of symbiotic algae under heat stress
EASY · 3 MARKS
1. Define bioremediation and describe two microbial capabilities used to transform pollutants.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that bioremediation exploits microbial activities to remove or transform pollutants
1 mark for recognising that hydrocarbon-degrading bacteria can use petroleum-derived carbon sources
1 mark for recognising that some microbial enzymes act on industrial solvents
EASY · 3 MARKS
2. Describe three components of the exchange between corals and their photosynthetic symbionts.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that photosynthetic algal symbionts supply much of a coral host's energy
1 mark for recognising that the coral provides a protected environment and nutrients to the algae
1 mark for recognising that coral-associated algae contribute to the coral's colour
EASY · 3 MARKS
3. Give three features of methanotrophs and their role in methane cycling.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that methanotrophs use methane as carbon and energy
1 mark for recognising that methanotrophic bacteria oxidise methane towards carbon dioxide
1 mark for recognising that some methane is consumed in soils before reaching the atmosphere
EASY · 3 MARKS
4. State three ways ecosystem conditions can influence human or animal health.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that human wellbeing depends on functioning environmental systems
1 mark for recognising that animal, human and ecosystem health are linked in One Health
1 mark for recognising that ecological change can alter the distribution of infectious disease
INTERMEDIATE · 5 MARKS
5. Explain how heterotrophic fungi and other decomposers return resources from leaf litter to a forest ecosystem.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that heterotrophs obtain carbon and energy from organic matter
1 mark for recognising that decomposers recycle nutrients by breaking down dead material
1 mark for recognising that soil fungi are important terrestrial decomposers
1 mark for recognising that microbial predators can regulate the abundance of other microbes
1 mark for recognising that heterotrophic decomposition contributes to nutrient cycling and plant growth
INTERMEDIATE · 5 MARKS
6. Explain how the location of methanotrophs within an oxygen gradient can influence methane emissions from a wetland.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that methanotrophs use methane as carbon and energy
1 mark for recognising that methanotrophic bacteria oxidise methane towards carbon dioxide
1 mark for recognising that some methane is consumed in soils before reaching the atmosphere
1 mark for recognising that methane oxidation can moderate net greenhouse-gas emissions
1 mark for recognising that methanotrophs in the surface layer can consume some methane before its release
INTERMEDIATE · 5 MARKS
7. Explain why methane can be released from an oxygen-poor wetland and identify the relevant microbial metabolism.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that methanogens are archaeal microorganisms
1 mark for recognising that they can use carbon dioxide and hydrogen to produce methane
1 mark for recognising that methanogens thrive under oxygen-free conditions such as wetlands and digesters
1 mark for recognising that methane production links microbial metabolism to greenhouse-gas dynamics
1 mark for recognising that anaerobic archaeal methanogenesis is a plausible source of the gas
INTERMEDIATE · 5 MARKS
8. Explain why molecular sequencing may detect many more soil microorganisms than traditional plate counts.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that much of Earth's biological diversity is microbial
1 mark for recognising that microbial lineages occur across Bacteria, Archaea and Eukaryotes
1 mark for recognising that many environmental microbes do not grow using routine laboratory culture methods
1 mark for recognising that DNA sequencing reveals diversity missed by cultivation and microscopy alone
1 mark for recognising that culture-independent sequencing detects organisms not represented among cultured isolates
HARD · 8 MARKS
9. A contaminant inhibits methane-oxidising bacteria at the wetland surface, while anaerobic methane production is unchanged. Predict consequences for net atmospheric release and describe the microbial processes involved.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying methanogenesis as a source of methane in anaerobic wetland zones
1 mark for identifying methanotrophs as methane-consuming microbes
1 mark for recognising methane oxidation at or near oxygenated interfaces
1 mark for explaining that suppressed oxidation does not mean production has fallen
1 mark for predicting more methane surviving transport towards the surface
1 mark for predicting a possible rise in net atmospheric methane emission
1 mark for distinguishing gross production from net emission
1 mark for qualifying the outcome by transport and other environmental conditions
HARD · 8 MARKS
10. Assess the feasibility of using naturally occurring oil-degrading microbes in a coastal clean-up. Identify environmental constraints and explain how targeted microbial enzymes might complement the strategy.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying microbial metabolism as a driver of oil breakdown
1 mark for recognising petroleum hydrocarbons as potential microbial substrates
1 mark for explaining that appropriate organisms or enzymes must be present
1 mark for identifying oxygen or nutrient availability as a possible limitation
1 mark for explaining that temperature or other conditions can influence degradation
1 mark for predicting incomplete removal when suitable conditions are absent
1 mark for describing how targeted enzymes may transform a further pollutant class
1 mark for assessing effectiveness using monitored contaminant change rather than assumed disappearance
HARD · 8 MARKS
11. During a marine heatwave, a reef experiences widespread bleaching. Trace how disruption of the coral–algal mutualism could affect energy supply, colony survival and reef-dependent organisms.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying coral–algal association as a mutualism
1 mark for explaining that photosynthetic algae provide organic carbon to coral hosts
1 mark for recognising bleaching as loss or impairment of the algal partnership
1 mark for linking reduced photosynthate supply to decreased coral energy availability
1 mark for predicting poorer growth, reproduction or survival under sustained stress
1 mark for explaining potential loss of habitat for reef-associated organisms
1 mark for recognising that recovery depends on stress duration or re-establishment of symbiosis
1 mark for tracing the indirect consequences without assuming every bleached coral dies
HARD · 8 MARKS
12. Predict ecological consequences when forest-soil microbial decomposition remains suppressed for several years. Trace potential effects on organic carbon, mineral nutrients and primary production.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying decomposers as processors of dead organic material
1 mark for explaining the role of microbes in carbon transformations
1 mark for predicting accumulation of undecomposed material
1 mark for predicting reduced mineralisation or nutrient availability
1 mark for recognising that nutrients must cycle rather than be permanently locked in biomass
1 mark for explaining how nutrient limitation can constrain plant production
1 mark for predicting possible downstream effects on consumers
1 mark for qualifying the prediction according to other decomposition pathways or conditions
Lecture 22: Cell Factories and Biotechnology
1. A plasmid must be cut at a chosen DNA sequence. Which enzyme type is used?
- Select transformed cells by antibiotic resistance
- Express an inserted gene as protein
- Join DNA fragments through peptide bonds
- Cut DNA at particular recognition sequences
Show answer
Cut DNA at particular recognition sequences
2. A yeast expression system produces a hepatitis B surface component. Which antigen is made?
- HIV envelope glycoprotein gp120
- Hepatitis B viral DNA polymerase
- Influenza haemagglutinin
- HBsAg, the hepatitis B surface antigen
Show answer
HBsAg, the hepatitis B surface antigen
3. A researcher wants to deliver a target gene into a host cell. What does the vector do?
- Carries DNA of interest into a host
- Replaces the need for a host cell
- Destroys the target gene before insertion
- Translates proteins outside living cells
Show answer
Carries DNA of interest into a host
4. A food manufacturer uses living microbes to ferment a substrate. Which type of biotechnology is involved?
- Studying DNA sequences without organisms
- Using whole living cells in production
- Using only synthetic DNA outside cells
- Producing materials without biological activity
Show answer
Using whole living cells in production
5. Only some cells take up the plasmid. Which feature helps identify carriers?
- Converts foreign DNA into a protein without expression
- Prevents plasmid replication in every host
- Allows identification or selection of plasmid-containing cells
- Replaces the gene of interest during cloning
Show answer
Allows identification or selection of plasmid-containing cells
6. A plasmid has been joined to a foreign gene. What is the resulting DNA called?
- A host cell lacking the cloning vector
- A purified protein without any DNA
- A vector joined to an inserted DNA sequence
- An unmodified plasmid lacking an insert
Show answer
A vector joined to an inserted DNA sequence
7. A host cell uses an introduced gene to make its encoded protein. Which stage is this?
- Isolation of a host chromosome without transcription
- Production of RNA and protein from introduced genetic information
- Purification of a protein before it has been made
- Insertion of DNA into a plasmid without cellular activity
Show answer
Production of RNA and protein from introduced genetic information
8. A manufacturer needs many batches of a recombinant protein. Which property makes microbes attractive?
- Inability to express foreign genes
- Guaranteed absence of all production contaminants
- Rapid growth and scalable cultivation
- Requirement for harvesting organs from animals
Show answer
Rapid growth and scalable cultivation
9. A vector and insert have compatible ends. Which enzyme joins them?
- Breaks down a purified vaccine antigen
- Translates mRNA into a protein
- Cuts DNA at specific recognition sites
- Joins DNA fragments together
Show answer
Joins DNA fragments together
10. Recombinant yeast produces the intended surface protein. What must be done to obtain the vaccine antigen?
- Isolation and purification of the antigen
- Removal of the antigen-coding DNA before expression
- Deliberate infection of recipients with hepatitis B virus
- Re-insertion of the antigen into the viral genome
Show answer
Isolation and purification of the antigen
11. A vaccine contains a viral surface protein recognised by the immune system. What is the protein's role?
- Acts only as an antibiotic against bacteria
- Stimulates a specific immune response
- Replaces all immune memory with immediate digestion
- Replicates as a complete infectious virus
Show answer
Stimulates a specific immune response
12. A company uses living cells to manufacture a useful compound. Which field does this illustrate?
- Studying organisms without producing a product
- Manufacturing only with non-biological raw materials
- Using biological systems to produce useful products
- Using only inorganic catalysts without cells
Show answer
Using biological systems to produce useful products
13. A recombinant plasmid must be maintained in growing host cells. Which feature allows copying?
- Determines the amino-acid sequence of the antigen
- Purifies the expressed protein
- Enables plasmid DNA replication in a compatible host
- Cuts the inserted gene into fragments
Show answer
Enables plasmid DNA replication in a compatible host
14. A scientist places a chosen gene into a plasmid. What is the chosen gene called?
- The DNA sequence of interest
- The antibiotic used for selection
- The host cell receiving the plasmid
- The entire host chromosome
Show answer
The DNA sequence of interest
15. A bacterium is made to take up plasmid DNA. Which step is this?
- Copying a protein sequence into DNA
- Introduction of DNA into a host cell
- Purifying a protein after expression
- Joining two DNA fragments with ligase
Show answer
Introduction of DNA into a host cell
16. A bacterium takes up a plasmid carrying an inserted gene. How is the cell described?
- A purified antigen outside the host
- A cell with no introduced DNA
- An isolated gene before transformation
- A host cell containing recombinant DNA
Show answer
A host cell containing recombinant DNA
EASY · 3 MARKS
1. Identify three defining features or approaches used in biotechnology.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that biotechnology uses biological systems or components to make useful products
1 mark for recognising that cellular biotechnology can use whole living organisms
1 mark for recognising that molecular biotechnology exploits DNA and cellular machinery
EASY · 3 MARKS
2. Identify three steps or components in the preparation of recombinant DNA.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that molecular cloning begins by identifying a gene of interest
1 mark for recognising that restriction enzymes can cut DNA at suitable sites
1 mark for recognising that DNA ligase joins DNA fragments to create recombinant molecules
EASY · 3 MARKS
3. Identify three steps or considerations linking a recombinant gene to a usable biotechnology product.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that gene expression is used to produce a protein encoded by an inserted sequence
1 mark for recognising that some expressed proteins are final medicines whereas others are enzymes for subsequent production
1 mark for recognising that useful products may need isolation and purification from cells or medium
EASY · 3 MARKS
4. Name three functional components commonly used in plasmid vector design.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that an origin of replication supports plasmid maintenance in a suitable host
1 mark for recognising that a cloning site permits insertion of a chosen DNA sequence
1 mark for recognising that a selectable marker helps identify or enrich cells retaining a vector
INTERMEDIATE · 5 MARKS
5. Describe how restriction enzymes and DNA ligase can be used to insert a gene into a plasmid vector.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that molecular cloning begins by identifying a gene of interest
1 mark for recognising that restriction enzymes can cut DNA at suitable sites
1 mark for recognising that DNA ligase joins DNA fragments to create recombinant molecules
1 mark for recognising that preparing DNA precedes transformation and expression in the production sequence
1 mark for recognising that cutting and ligation are steps used to assemble recombinant DNA
INTERMEDIATE · 5 MARKS
6. Distinguish successful gene delivery from successful expression, and explain why a manufactured protein commonly requires purification.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that gene expression is used to produce a protein encoded by an inserted sequence
1 mark for recognising that some expressed proteins are final medicines whereas others are enzymes for subsequent production
1 mark for recognising that useful products may need isolation and purification from cells or medium
1 mark for recognising that gMO risks must be assessed in relation to the inserted traits and potential gene transfer
1 mark for recognising that successful DNA introduction does not by itself establish adequate gene expression
INTERMEDIATE · 5 MARKS
7. Explain how acquisition of a plasmid could give a bacterium a new resistance trait, distinguishing horizontal transfer from vertical inheritance.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that plasmids are often circular DNA molecules that can replicate separately from a chromosome
1 mark for recognising that plasmids can carry genes useful in particular environments
1 mark for recognising that horizontal gene transfer can move genetic information between microbes
1 mark for recognising that modified plasmids can serve as recombinant DNA vectors
1 mark for recognising that a plasmid and horizontal gene transfer can explain acquisition of the trait
INTERMEDIATE · 5 MARKS
8. Explain how expression of hepatitis B surface antigen in yeast can provide a component for a vaccine.
Show marking guide
Marking guide (accept scientifically equivalent explanations):
1 mark for recognising that vaccination trains adaptive immune responses to recognise pathogen-associated antigens
1 mark for recognising that some vaccines are manufactured using recombinant biotechnology
1 mark for recognising that the hepatitis B example expresses a viral surface antigen in yeast
1 mark for recognising that purified antigen can be used as the vaccine product without injecting the whole infectious virus
1 mark for recognising that recombinant expression and purification can supply an antigen for vaccination
HARD · 8 MARKS
9. A manufacturer compares bacteria, yeast and algae for several production goals. Evaluate the trade-offs among rapid growth, protein processing and light-driven carbon fixation before recommending appropriate hosts.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying rapid growth or ease of genetic manipulation as a bacterial advantage
1 mark for recognising limits of bacterial protein processing for some eukaryotic products
1 mark for identifying yeast as a eukaryotic host able to perform some relevant processing
1 mark for recognising that algal hosts can exploit light and inorganic carbon
1 mark for explaining that culture conditions and yields differ across hosts
1 mark for matching at least one production goal to a justified host
1 mark for matching another contrasting goal to a justified host
1 mark for evaluating trade-offs rather than declaring one organism universally superior
HARD · 8 MARKS
10. A vaccine developer aims to stimulate immunity against a virus while avoiding administration of the whole infectious agent. Evaluate recombinant antigen production, purification and immune recognition as steps towards this goal.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying an antigen as a target recognised by the immune system
1 mark for describing gene-based production of a viral antigen in a suitable host
1 mark for explaining that antigen production need not use an infectious whole virus
1 mark for recognising purification or quality control as part of vaccine manufacture
1 mark for explaining that exposure to antigen can stimulate an immune response
1 mark for distinguishing immunogenicity from active viral infection
1 mark for identifying the need to test safety and immune response
1 mark for drawing a qualified conclusion about recombinant subunit vaccine design
HARD · 8 MARKS
11. Explain how a manufacturer could improve the consistency of mixed-culture fermentation by selecting production organisms and controlling growth conditions. Identify limitations that would still require monitoring.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for identifying mixed-culture composition as a source of process variability
1 mark for recognising selection of a defined production strain as one control
1 mark for explaining why nutrients or substrate availability influence product formation
1 mark for explaining why temperature or growth conditions influence microbial metabolism
1 mark for recognising that fermentation time or conditions need monitoring
1 mark for predicting more repeatable products under controlled conditions
1 mark for recognising that contamination can still disrupt production
1 mark for proposing a practical measurement of output quality or consistency
HARD · 8 MARKS
12. Recommend how to establish stable recombinant production when a gene-delivery procedure transfers DNA into microbial cells but yields no stable modified population. Consider vector–host compatibility and verification.
Show marking guide
Marking guide (8 marks; award one mark for each distinct criterion; accept scientifically equivalent answers):
1 mark for recognising DNA entry as distinct from stable maintenance
1 mark for identifying vector–host compatibility as a requirement
1 mark for explaining loss of plasmids that cannot replicate in the host
1 mark for recognising that selection markers do not guarantee long-term maintenance
1 mark for proposing a compatible origin or alternative host
1 mark for proposing appropriate selection and verification measures
1 mark for distinguishing stable transformants from transient DNA uptake
1 mark for specifying a test across growth or passages to check maintenance
Lecture 25: Individuals, Behaviour and Environment
1. A bird selects a food item yielding more energy but requiring substantially longer handling. What must be compared to evaluate its choice under optimal foraging theory?
- Its reproductive age relative to other birds
- Net energy gained relative to the time spent obtaining it
- The number of other species living near the food item
- The food’s colour relative to available alternatives
Show answer
Net energy gained relative to the time spent obtaining it
2. A lizard alternately lifts its feet from very hot sand. Which explanation best connects its action with environmental conditions?
- The behaviour allows the lizard to absorb more nutrients
- The behaviour helps manage heat gained from the substrate
- The lizard is displaying to potential mates
- The lizard is increasing exposure to a predator
Show answer
The behaviour helps manage heat gained from the substrate
3. A small mammal feeds less in an exposed meadow despite abundant grass. Which interpretation best reflects the lecture?
- It is necessarily incapable of digesting meadow grasses
- It may sacrifice food intake to reduce exposure to predators
- Its diet breadth must have narrowed to a single plant species
- Its reduced feeding proves that it has no energy requirements
Show answer
It may sacrifice food intake to reduce exposure to predators
4. Why might a specialist forager be more constrained in its choice of feeding locations than a generalist?
- A specialist can use a narrower range of acceptable foods
- A specialist must forage alone to locate food
- A specialist always moves more slowly than a generalist
- A specialist necessarily has a higher metabolic rate than a generalist
Show answer
A specialist can use a narrower range of acceptable foods
5. What makes food mixed into an inedible matrix useful for a giving-up-density experiment?
- Finding the remaining food becomes harder as food is removed, creating diminishing returns
- The matrix supplies extra energy each time food becomes scarce
- The matrix keeps the nutritional quality of all food items constant by changing it over time
- The matrix ensures the forager must consume every food item before leaving
Show answer
Finding the remaining food becomes harder as food is removed, creating diminishing returns
6. Why does an ambush predator benefit from camouflage?
- It can remain less detectable while prey approaches
- It can increase its maximum pursuit speed
- It can consume a wider range of plant tissues
- It can remove all energetic costs of predation
Show answer
It can remain less detectable while prey approaches
7. Which observation most directly connects individual behaviour with community-level ecology?
- A desert animal raising its feet to cool down
- A bird increasing its time spent resting
- A predator changing its feeding behaviour and thereby altering prey interactions
- A plant root extending towards soil nutrients
Show answer
A predator changing its feeding behaviour and thereby altering prey interactions
8. When might a forager leave a profitable-looking patch even though edible items remain?
- When the current harvest rate has declined relative to alternatives
- When the forager can no longer perceive any predators
- Only when no edible items are left anywhere
- Only after all food items become nutritionally worthless
Show answer
When the current harvest rate has declined relative to alternatives
9. The lecture reports fewer pademelon scats farther from vegetation. What does this evidence most directly indicate?
- Predators capture every pademelon that crosses an exposed site
- Open sites contain no suitable food
- Pademelons use locations closer to cover more frequently under the measured conditions
- Pademelons are physiologically unable to enter open habitat
Show answer
Pademelons use locations closer to cover more frequently under the measured conditions
10. Which statement most accurately links parental care to natural selection?
- Care must increase the number of breeding attempts to be beneficial
- Care always evolves only in species with two caregiving parents
- Care improves parental survival but does not influence reproductive fitness
- Care can be favoured when increased offspring success compensates for parental investment costs
Show answer
Care can be favoured when increased offspring success compensates for parental investment costs
11. Why might grouping reduce an individual’s risk of predation?
- Group members may detect threats and dilute an individual’s chance of being targeted
- Grouping removes all costs associated with competition for food
- Predators always avoid groups regardless of group size or behaviour
- Group members no longer need to respond to potential threats
Show answer
Group members may detect threats and dilute an individual’s chance of being targeted
12. A student claims all adaptive behaviours are beneficial in every respect. Which response best reflects the lecture?
- Adaptations arise only when there are no energetic costs
- All behaviours that increase survival necessarily maximise reproduction
- Behaviour cannot be selected because it changes with context
- A behaviour may have costs but still be favoured if its fitness benefits exceed them
Show answer
A behaviour may have costs but still be favoured if its fitness benefits exceed them
13. Why can an increase in vigilance reduce food intake?
- Vigilance makes all food items harder to digest
- Vigilance necessarily causes prey to move farther from shelter
- Time spent scanning for predators cannot simultaneously be spent collecting food
- Vigilance removes the need for nutritional intake
Show answer
Time spent scanning for predators cannot simultaneously be spent collecting food
14. Which comparison correctly distinguishes acting costly from being costly to a predator?
- Acting costly requires a physical toxin; being costly requires mimicry only
- Acting costly reduces prey energy gain; being costly reduces predator food supply
- Acting costly means hiding; being costly means escaping by running
- Acting costly can involve mimicking an unpalatable organism; being costly can involve actual toxins or spines
Show answer
Acting costly can involve mimicking an unpalatable organism; being costly can involve actual toxins or spines
15. Male redback spiders that somersault into a female’s mouth can fertilise more eggs. Why can this behaviour be favoured by selection despite the male being eaten?
- The behaviour increases the male’s survival through repeated mating
- The female becomes unable to produce offspring after consuming the male
- The behaviour removes the need for any inherited variation
- Extended mating and reduced subsequent mating by the female can increase the male’s paternity
Show answer
Extended mating and reduced subsequent mating by the female can increase the male’s paternity
16. An animal forages in a patch where food becomes increasingly difficult to find as items are removed. Which model directly addresses when it should leave?
- The marginal value theorem
- The model of reproductive investment in parental care
- The model of mimicry in predator avoidance
- The model of sexual selection by mate choice
Show answer
The marginal value theorem
17. Why might a conspicuous peacock tail persist despite costs of production and predation exposure?
- Sexual selection acts only on traits that improve survival
- The tail is maintained because all males with shorter tails are infertile
- Predator avoidance must always outweigh any reproductive disadvantage
- Greater mating success can outweigh the survival and energetic costs
Show answer
Greater mating success can outweigh the survival and energetic costs
EASY · 3 MARKS
1. Define behavioural ecology and identify the two broad kinds of pressure that shape behaviour.
Show marking guide
1 mark for defining behavioural ecology as study of organismal actions or decisions in ecological and evolutionary context.
1 mark for identifying ecological pressures such as food availability or predators.
1 mark for identifying evolutionary pressures such as selection associated with reproductive success.
EASY · 3 MARKS
2. Define fitness and explain why behaviour can be acted on by natural selection.
Show marking guide
1 mark for defining fitness as relative genetic contribution to subsequent generations.
1 mark for stating that behaviour can affect reproductive success or survival that influences reproduction.
1 mark for identifying heritable variation/genetic contribution to behaviour as permitting selection.
EASY · 3 MARKS
3. Describe three ways foraging strategies can differ among organisms.
Show marking guide
1 mark for explaining diet breadth as the specialist-to-generalist spectrum.
1 mark for identifying differences in food type such as nectar, insects or meat.
1 mark for identifying differences in how food is obtained such as ambush versus active pursuit.
EASY · 3 MARKS
4. State the central prediction of optimal foraging theory and the environmental assumption used in the lecture.
Show marking guide
1 mark for identifying a non-depleting environment in the lecture’s basic model.
1 mark for stating that foragers choose food or foraging strategies non-randomly.
1 mark for stating that net energy intake per unit time should be maximised.
INTERMEDIATE · 5 MARKS
5. A prey animal forages less after predators become active in its habitat. Explain the benefits and costs of this behavioural change.
Show marking guide
1 mark for identifying reduced predation risk as a potential benefit.
1 mark for describing an anti-predator behaviour such as vigilance or feeding near shelter.
1 mark for identifying reduced foraging time or missed food as a cost.
1 mark for relating food acquisition to energy available for survival or reproduction.
1 mark for explaining that the behaviour can be favoured when its fitness benefits outweigh costs.
INTERMEDIATE · 5 MARKS
6. Explain how hiding, mimicry and actual defences differ as ways to avoid predation.
Show marking guide
1 mark for explaining concealment or camouflage as reducing detection.
1 mark for identifying mimicry as resembling something predators avoid.
1 mark for explaining that mimicry can make harmless prey appear costly to attack.
1 mark for describing a genuine defence such as spines or toxic compounds.
1 mark for linking these strategies to lowered predation risk while recognising that defence strategies can carry costs.
INTERMEDIATE · 5 MARKS
7. Explain how male redback spider somersaulting may affect fitness, despite the risk of being consumed.
Show marking guide
1 mark for identifying the male’s post-copulatory somersault towards the female’s mouth.
1 mark for linking the behaviour to longer copulation.
1 mark for linking longer copulation to more eggs fertilised.
1 mark for noting lower female acceptance of subsequent mates.
1 mark for concluding that increased paternity can outweigh the cost to the male; recognise the lecture reports a genetic basis.
INTERMEDIATE · 5 MARKS
8. Explain why a peacock’s tail is an example of sexual selection rather than a trait that must maximise survival.
Show marking guide
1 mark for identifying the conspicuous tail as costly in energy or predation exposure.
1 mark for identifying mate attraction as a potential reproductive advantage.
1 mark for describing intersexual mate choice as the relevant selective process.
1 mark for explaining that courtship display behaviour contributes alongside tail morphology.
1 mark for concluding that net reproductive success can favour a costly trait despite survival disadvantages.
HARD · 8 MARKS
9. A group-living prey species detects predators sooner than solitary individuals but competes more intensely for food. Analyse why group living could evolve in some conditions but not others.
Show marking guide
1 mark for identifying vigilance or earlier predator detection as a group benefit.
1 mark for identifying dilution or confusion of predator attacks as another potential benefit.
1 mark for noting the potential improvement to survival and thus reproductive success.
1 mark for identifying increased within-group food competition.
1 mark for identifying another possible cost such as aggression or reduced access to food.
1 mark for explaining the resulting trade-off between avoiding predators and acquiring resources.
1 mark for predicting that increased predator pressure could favour grouping, other factors equal.
1 mark for explaining why selection depends on net fitness benefits exceeding costs under local conditions.
HARD · 8 MARKS
10. A rodent’s mean food intake rate is initially high in a patch but declines with time. A nearby patch is replenished regularly and requires travel. Explain how you would test the marginal value theorem’s departure prediction.
Show marking guide
1 mark for identifying the current patch as a depleting food source.
1 mark for describing collection of within-patch harvest rate over time.
1 mark for describing estimation of average intake rate available across patches including travel.
1 mark for stating the predicted departure when the marginal patch rate reaches the relevant average.
1 mark for identifying travel time as a factor influencing the comparison.
1 mark for proposing repeated observations of when animals actually leave patches.
1 mark for explaining how observed departure times could be compared against model predictions.
1 mark for recognising that predator exposure or other costs may help explain departures inconsistent with an energy-only prediction.
HARD · 8 MARKS
11. A small mammal forages in two patches containing equal amounts of food. One is open and exposed to predators; the other is sheltered. Design a giving-up-density comparison and predict how predation risk could affect the results.
Show marking guide
1 mark for setting out the two patch types as the main contrast.
1 mark for controlling the starting amount and type of food.
1 mark for using comparable inedible matrices or patch conditions.
1 mark for measuring food remaining after the animals leave each patch.
1 mark for defining the measured remainder as the giving-up density.
1 mark for predicting a higher GUD in the more exposed patch under greater perceived risk.
1 mark for explaining that earlier departure or lower foraging investment reduces predator exposure.
1 mark for identifying a relevant alternative influence or limitation, such as travel distance, food accessibility or individual differences, that should be controlled or considered.
HARD · 8 MARKS
12. A species can either provide prolonged parental care to its current offspring or attempt another breeding event. Explain how ecological conditions could change which strategy is favoured.
Show marking guide
1 mark for identifying increased survival of current offspring as a benefit of care.
1 mark for identifying improved offspring growth or future reproduction as another benefit.
1 mark for identifying time or energy devoted to care as a parental cost.
1 mark for identifying missed opportunities for additional reproduction.
1 mark for predicting stronger benefits of care when offspring survival otherwise would be low.
1 mark for predicting greater opportunity costs when additional breeding opportunities are abundant.
1 mark for recognising that parents may be male, female or both and that strategies can vary.
1 mark for integrating expected contributions to future generations to justify context-dependent selection.
HARD · 8 MARKS
13. In a feeding experiment, small mammals leave exposed patches with more food remaining than sheltered patches. Analyse what can be concluded and how the experimental design could be strengthened.
Show marking guide
1 mark for defining GUD as the amount of food remaining when a patch is abandoned.
1 mark for identifying higher GUD in exposed sites as evidence of reduced foraging investment.
1 mark for linking this pattern to a possible increase in perceived predation risk.
1 mark for explaining the food-versus-safety trade-off.
1 mark for identifying initial food quantity as a variable to standardise.
1 mark for identifying substrate, food accessibility or patch quality as a variable to control.
1 mark for proposing replication across sites, trials or individuals.
1 mark for explaining that differences in cover alone do not prove predation is the causal mechanism without appropriate controls.
Lecture 26: Groups and Populations
1. Under the logistic model, what happens to the growth factor (1 − N/K) as N approaches K from below?
- It approaches one
- It approaches zero
- It stays constant while population size increases
- It becomes progressively larger than one
Show answer
It approaches zero
2. A population has N = 200 and K = 400. What is the logistic density-limitation factor (1 − N/K)?
- 2.00
- 0.50
- 0.25
- 1.50
Show answer
0.50
3. Which information is represented by a life table?
- The geographic position of every individual
- The daily movement route of each disperser
- The abundance of all species in a community
- Survivorship probabilities across age classes
Show answer
Survivorship probabilities across age classes
4. A researcher tags reef fish that remain close to one coral patch and later checks for survivors. Why is this useful?
- Limited movement guarantees that tagged fish never die
- Limited movement removes variation in individual survival
- Limited movement ensures all new fish are detected
- Limited movement makes disappearance easier to interpret
Show answer
Limited movement makes disappearance easier to interpret
5. Why might a seasonal breeding population show a saw-tooth trajectory?
- All individuals survive until the following breeding season
- Breeding pulses increase numbers between periods of mortality
- Immigration remains constant between breeding events
- Carrying capacity increases after each breeding event
Show answer
Breeding pulses increase numbers between periods of mortality
6. Why are mortality rates often harder to estimate than birth rates in wild animals?
- Mortality only occurs during the breeding season
- Dead animals are consistently easier to count than young
- Every birth necessarily occurs outside the study area
- Disappearing individuals may have died or moved elsewhere
Show answer
Disappearing individuals may have died or moved elsewhere
7. An open population starts with 80 individuals; 12 are born, 7 die, 6 immigrate and 4 emigrate. What is the final size?
- 101 individuals
- 91 individuals
- 83 individuals
- 87 individuals
Show answer
87 individuals
8. How can dispersal influence a metapopulation after a local patch becomes empty?
- Individuals from occupied patches can recolonise it
- Dispersal prevents extinction in every occupied patch
- All local populations become closed after extinction
- The empty patch must remain permanently unoccupied
Show answer
Individuals from occupied patches can recolonise it
9. In mark–release–recapture, 30 animals are initially marked, and 10 of 50 animals in the second sample bear marks. What population is estimated?
- 60 individuals
- 250 individuals
- 150 individuals
- 90 individuals
Show answer
150 individuals
10. Why does the proportion of reproductive adults matter when forecasting population growth?
- Individuals of different ages differ in fecundity
- Total population size alone determines every birth rate
- Age structure influences movement but not births
- All individuals contribute equally to births at every age
Show answer
Individuals of different ages differ in fecundity
11. A population grows by the same percentage each year. How will the number of individuals added annually change?
- It will increase as the population becomes larger
- It will decrease as the population becomes larger
- It will alternate between gains and losses each year
- It will remain the same because the percentage is fixed
Show answer
It will increase as the population becomes larger
12. Why can a population of sessile trees still show movement across generations?
- Mature trees routinely migrate between habitat patches
- Individual trunks relocate when resources are depleted
- Sessile populations cannot change their spatial distribution
- Seeds can disperse before establishing as sessile plants
Show answer
Seeds can disperse before establishing as sessile plants
13. Which parameter set would be most relevant when building a PVA model?
- Predator diet, colour patterns and soil texture alone
- Species richness, trophic diversity and leaf morphology
- Individual behaviour, food preference and habitat colour alone
- Fecundity, age-specific mortality and environmental variation
Show answer
Fecundity, age-specific mortality and environmental variation
14. Which change would most directly violate a closed-population assumption during mark–release–recapture?
- Retention of marks on recaptured individuals
- Thorough mixing of animals after release
- Substantial immigration between the two sampling events
- Equal likelihood of capture among marked animals
Show answer
Substantial immigration between the two sampling events
15. An isolated population begins with 120 animals, has 18 births and 11 deaths. What is its new size?
- 131 animals
- 113 animals
- 149 animals
- 127 animals
Show answer
127 animals
16. A habitat has K = 600 individuals. Which population size is expected to show zero growth under the logistic model with r > 0?
- 450 individuals
- 600 individuals
- 150 individuals
- 300 individuals
Show answer
600 individuals
17. Two bird flocks occur together because both use a temporary feeding site, without social attraction. What type of grouping is this?
- A single-species population defined by reproduction
- Indirect grouping around a shared resource
- Positive grouping through social attraction
- Accidental grouping unrelated to resources
Show answer
Indirect grouping around a shared resource
EASY · 3 MARKS
1. Define a population in ecology and identify two properties used to describe it.
Show marking guide
1 mark for defines a population as individuals of the same species within a defined geographical area.
1 mark for identifies population size or abundance.
1 mark for identifies a second property such as age structure, sex ratio or occupied area.
EASY · 3 MARKS
2. Describe three ways organisms can come together in groups.
Show marking guide
1 mark for identifies positive or social attraction.
1 mark for identifies indirect grouping around a common resource.
1 mark for identifies accidental grouping by chance.
EASY · 3 MARKS
3. Distinguish discrete from continuous population growth.
Show marking guide
1 mark for discrete growth involves periodic breeding or reproduction.
1 mark for continuous growth involves births throughout the year.
1 mark for recognises that both can show exponential population growth.
EASY · 3 MARKS
4. Explain what is meant by positive, zero and negative population growth.
Show marking guide
1 mark for positive growth means population size increases.
1 mark for zero growth means population size stays stable.
1 mark for negative growth means population size decreases.
EASY · 3 MARKS
5. Give three examples of biological structures used to estimate age.
Show marking guide
1 mark for identifies tree rings in woody trees.
1 mark for identifies otoliths in fish.
1 mark for identifies tooth cementum rings or tooth wear in mammals.
INTERMEDIATE · 5 MARKS
6. Explain how PVA can inform management of a threatened mammal population.
Show marking guide
1 mark for describes modelling population trajectories over time.
1 mark for identifies inputs such as size, fecundity and mortality.
1 mark for includes variation in environmental or demographic conditions.
1 mark for describes comparing management scenarios such as habitat protection.
1 mark for relates outputs to persistence probability or extinction risk.
INTERMEDIATE · 5 MARKS
7. Explain how a group of organisms can differ from an ecological population, using an example of organisms gathering at a water source.
Show marking guide
1 mark for defines a population using a single species and a specified area.
1 mark for explains a group may include multiple species.
1 mark for explains groups may be temporary.
1 mark for identifies a common water source as an indirect grouping mechanism.
1 mark for explains that sharing a resource does not itself establish a single population.
INTERMEDIATE · 5 MARKS
8. For a closed population initially containing 180 animals, with 28 births and 16 deaths, calculate the final size and explain the model.
Show marking guide
1 mark for states the closed-population equation N(t+1) = N(t) + births − deaths.
1 mark for substitutes the starting size of 180.
1 mark for adds 28 births.
1 mark for subtracts 16 deaths.
1 mark for gives the final size of 192 animals.
INTERMEDIATE · 5 MARKS
9. In a mark–release–recapture study, 24 animals are marked initially, then 30 are caught of which 6 are marked. Estimate population size and state the relevant assumptions.
Show marking guide
1 mark for identifies initial marked sample M = 24.
1 mark for identifies second sample n = 30 and recaptures R = 6.
1 mark for applies N = Mn/R.
1 mark for calculates N = 120 individuals.
1 mark for states a relevant assumption such as closure, stable marks or equal catchability.
INTERMEDIATE · 5 MARKS
10. Outline how birth and death rates can be estimated in wild bird or fish populations and why mortality is especially challenging.
Show marking guide
1 mark for describes counting newborns, eggs, fertilised gametes or active nests.
1 mark for describes tagging identifiable individuals.
1 mark for describes repeat surveys or resightings to assess survival.
1 mark for explains disappearances do not always provide evidence of death.
1 mark for notes movement or incomplete detection can confound mortality estimates.
HARD · 8 MARKS
11. A population follows logistic growth with r = 0.2 per year, N = 200 and K = 500. Calculate its instantaneous growth and predict the consequence if habitat loss reduces K to 150, with r and N unchanged.
Show marking guide
1 mark for states dN/dt = rN(1 − N/K).
1 mark for calculates the initial N/K = 0.4.
1 mark for calculates the initial density factor = 0.6.
1 mark for calculates initial dN/dt = 24 individuals per year.
1 mark for identifies reduced K = 150.
1 mark for calculates the new factor 1 − 200/150 = −1/3.
1 mark for calculates new dN/dt = −13.33 individuals per year approximately.
1 mark for explains that N exceeding K predicts decline under the model.
HARD · 8 MARKS
12. A team marks 40 animals and later captures 50, including 10 marked animals. Estimate population size, then analyse how trap-shyness and mark loss may affect the estimate.
Show marking guide
1 mark for identifies M = 40.
1 mark for identifies n = 50.
1 mark for identifies R = 10.
1 mark for states or applies N = Mn/R.
1 mark for calculates N = 200 individuals.
1 mark for explains trap-shyness reduces the marked recapture fraction.
1 mark for explains mark loss also reduces observed marked recaptures.
1 mark for predicts either effect tends to inflate the simple estimate.
HARD · 8 MARKS
13. Two bird populations each contain 500 individuals. One contains mostly juveniles, while the other contains mostly breeding adults. Analyse why equal population size may conceal different conservation prospects and propose how to improve predictions.
Show marking guide
1 mark for identifies total population size as insufficient for demographic forecasts.
1 mark for distinguishes the two populations’ age structures.
1 mark for explains age-specific fecundity.
1 mark for explains age-specific survival.
1 mark for predicts potential differences in short-term births.
1 mark for considers future recruitment as juveniles mature.
1 mark for proposes measuring age-specific birth or survival rates.
1 mark for explains that a stage-structured forecast or PVA improves assessment.
Lecture 27: Species
1. A park survey is conducted only in dry weather. Which limitation is most likely?
- All recorded species will be misidentified
- Species active after heavy rain may be overlooked
- A greater proportion of wet-weather specialists will be recorded
- Richness cannot be compared with any other site
Show answer
Species active after heavy rain may be overlooked
2. A biodiversity map shows many records around Sydney and very few in remote bushland. Which inference is safest?
- Observer effort may be greater around populated areas
- Remote habitats contain fewer species of every group
- Species are intrinsically restricted to city centres
- All remote areas have been sampled comparably
Show answer
Observer effort may be greater around populated areas
3. A survey of a rare, burrowing species produces few records. Which strategy best addresses potential under-detection?
- Replace all sampling with historical records
- Treat missing observations as confirmed extinction
- Include only more conspicuous species
- Repeat surveys using methods matched to its ecology
Show answer
Repeat surveys using methods matched to its ecology
4. Researchers compare plant species counts in 0.1-hectare plots across habitats. What is the principal benefit?
- Identical plot size removes all detectability differences
- A consistent area makes richness comparisons more meaningful
- It allows direct comparison of animal abundance
- All species in each habitat will be recorded
Show answer
A consistent area makes richness comparisons more meaningful
5. Two samples each contain ten species and 100 individuals. Site A has one dominant species; site C has equal abundances. Which site is more diverse?
- Site A, because rare species contribute more individuals
- Site C, because individuals are distributed more evenly
- Both, because species richness is the same
- Site A, because a dominant species increases richness
Show answer
Site C, because individuals are distributed more evenly
6. Two equally sampled habitats each have six species, but only two are shared. What does this indicate?
- The habitats contain no locally restricted species
- Identical species composition despite equal richness
- Substantial species turnover between habitats
- Low gamma diversity despite high richness
Show answer
Substantial species turnover between habitats
7. For counting canopy insects rather than ground insects, what is the strongest survey-design choice?
- Adapt the sampling method to reach the canopy
- Use only soil cores for both habitats
- Use identical pitfall traps at ground level
- Use only night-time bird observations
Show answer
Adapt the sampling method to reach the canopy
8. A camera survey records mammals that a daytime visual survey missed. What is the main explanation?
- Visual surveys always underestimate species richness by half
- Camera surveys detect only nocturnal species
- Detection probability depends on method and animal behaviour
- Cameras always detect every species present
Show answer
Detection probability depends on method and animal behaviour
9. Two taxonomists disagree about whether genetically distinct dingo populations merit separate species status. What does this illustrate?
- Species classifications are independent of conservation policy
- Reproductive compatibility always settles classification
- All genetically distinct populations must belong to separate species
- Species boundaries can involve judgement when evidence differs in interpretation
Show answer
Species boundaries can involve judgement when evidence differs in interpretation
10. The cumulative species curve has nearly levelled off. What does this suggest?
- Each new sample adds many common species
- Sampling effort can no longer affect detection
- Further sampling is finding relatively few new species
- The habitat contains no undetected species
Show answer
Further sampling is finding relatively few new species
11. A species-rich habitat contains mostly individuals from one species. Which statement is accurate?
- Richness must therefore be low
- Its gamma diversity must equal its abundance
- All habitats with that richness are equally diverse
- Its diversity may be low because evenness is low
Show answer
Its diversity may be low because evenness is low
12. An eDNA survey fails to detect an elusive fish in a stream. Which conclusion is most justified?
- The species has become locally extinct
- The fish is absent throughout the catchment
- The species is present in another stream
- Non-detection alone does not demonstrate absence
Show answer
Non-detection alone does not demonstrate absence
13. A botanist identifies two visually distinct garden plants that readily hybridise. What challenge does this pose?
- The biological concept may not give a clear species boundary
- The phylogenetic concept requires identical DNA
- The morphological concept cannot compare them
- The ecological concept forbids hybridisation
Show answer
The biological concept may not give a clear species boundary
14. Two cat populations look similar but have consistently different DNA profiles. Which additional species concept could help distinguish them?
- Ecological, based on resource use
- Phylogenetic, based on genetic differentiation
- Morphological, based on habitat choice
- Biological, based only on appearance
Show answer
Phylogenetic, based on genetic differentiation
15. A landscape contains three habitats with overlapping species lists. Which count represents gamma diversity?
- The mean number of species per habitat
- The number of unique species across all three habitats
- The number exclusive to one habitat
- The sum of individuals across all habitats
Show answer
The number of unique species across all three habitats
EASY · 3 MARKS
1. Describe three approaches for estimating the total number of undescribed species.
Show marking guide
1 mark for extrapolating from rates of new species discoveries.
1 mark for extrapolating from well-surveyed to poorly surveyed regions.
1 mark for using expert estimates or knowledge of species associations.
EASY · 3 MARKS
2. Explain the meanings of alpha, beta and gamma diversity.
Show marking guide
1 mark for alpha as local species richness within a habitat.
1 mark for beta as species turnover or compositional differences between habitats.
1 mark for gamma as total species richness across the combined region.
EASY · 3 MARKS
3. Explain why a biodiversity observation map may contain many records near cities.
Show marking guide
1 mark for identifying greater observer or sampling effort near cities.
1 mark for recognising that fewer records elsewhere need not mean species absence.
1 mark for suggesting more systematic surveys in poorly sampled areas.
EASY · 3 MARKS
4. Give three reasons why species classification matters in ecology and conservation.
Show marking guide
1 mark for identifying a consistent means of naming or communicating about organisms.
1 mark for identifying the importance of species units in monitoring biodiversity.
1 mark for explaining their use in conservation planning or threatened-species protection.
INTERMEDIATE · 5 MARKS
5. Assess the strengths and limitations of eDNA when testing whether an invasive fish occurs in a stream.
Show marking guide
1 mark for explaining collection of environmental samples such as water.
1 mark for identifying species DNA as the detection target.
1 mark for explaining the usefulness of a positive detection.
1 mark for recognising that failure to detect DNA does not prove absence.
1 mark for recognising transport, assay specificity or complementary field surveys as a consideration.
INTERMEDIATE · 5 MARKS
6. Design a bird survey in a local park that accounts for variation in detectability.
Show marking guide
1 mark for defining the area and target bird species or assemblage.
1 mark for proposing visual or acoustic observations as appropriate.
1 mark for sampling at suitable times of day.
1 mark for including seasonal or repeated sampling.
1 mark for recognising the limitation of any single detection method.
INTERMEDIATE · 5 MARKS
7. Explain why species richness comparisons require attention to sampling area, effort and season.
Show marking guide
1 mark for noting that larger areas can encompass more habitats or species.
1 mark for noting that greater sampling effort can detect more taxa.
1 mark for explaining that seasonal activity affects detectability.
1 mark for identifying the need to standardise or account for these factors.
1 mark for describing how inconsistent sampling can mislead comparisons.
INTERMEDIATE · 5 MARKS
8. Explain why the dingo presents a challenge for species delimitation and describe one conservation implication.
Show marking guide
1 mark for noting that dingoes can interbreed with domestic dogs.
1 mark for noting differences in morphology, ecology, behaviour or genetics.
1 mark for explaining why distinct species concepts may lead to different classifications.
1 mark for linking species status to management or conservation decisions.
1 mark for explaining the need to consider multiple kinds of evidence.
HARD · 8 MARKS
9. A dingo management policy depends on treating dingoes as either a separate species or domestic dogs. Critically evaluate why classification can be contested and how managers should use the evidence.
Show marking guide
1 mark for identifying reproductive compatibility with domestic dogs.
1 mark for explaining what that implies under the biological species concept.
1 mark for discussing morphological differentiation.
1 mark for discussing ecological or behavioural differentiation.
1 mark for discussing genetic differentiation.
1 mark for explaining the absence of a universally sharp species boundary.
1 mark for identifying conservation or management consequences.
1 mark for recommending transparent use of multiple evidence types.
HARD · 8 MARKS
10. A field team aims to count all insect species in a bushland reserve but has one week of sampling. Propose a survey strategy and critically assess its likely biases.
Show marking guide
1 mark for defining spatial extent and taxonomic scope.
1 mark for proposing methods for ground-dwelling insects.
1 mark for proposing methods for flying or canopy insects.
1 mark for proposing methods for soil-dwelling or cryptic taxa.
1 mark for considering weather, time of day or season.
1 mark for specifying replicates or standardised effort.
1 mark for explaining likely under-detection of rare or seasonal species.
1 mark for suggesting repeat sampling or qualified interpretation.
HARD · 8 MARKS
11. Habitats A and B each contain six species. They share only two species, and habitat A is dominated by one species while B is relatively even. Explain what richness, evenness, beta diversity and gamma diversity reveal.
Show marking guide
1 mark for identifying alpha richness of six in each habitat.
1 mark for recognising that equal richness does not mean equal diversity.
1 mark for identifying lower evenness in A.
1 mark for predicting higher diversity in B if using a comparable appropriate index.
1 mark for identifying the two shared species.
1 mark for identifying four unique species in each habitat.
1 mark for calculating combined gamma richness of ten.
1 mark for explaining high turnover and the conservation value of both habitats.
HARD · 8 MARKS
12. Two taxonomists disagree over whether a cryptic glider complex is one species or two. Develop a defensible assessment using the species concepts from the lecture and explain the conservation implications.
Show marking guide
1 mark for accurately outlining the biological species concept.
1 mark for discussing difficulty testing interbreeding in the relevant populations.
1 mark for identifying morphological comparisons.
1 mark for identifying phylogenetic or DNA comparisons.
1 mark for identifying ecological differentiation.
1 mark for explaining how evidence may support competing classifications.
1 mark for discussing consequences for biodiversity counts or management.
1 mark for proposing an evidence-based, transparent classification decision.
Lecture 28: Trophic Ecology
1. Ants protect sap-feeding insects and consume their honeydew. What explains the mutual benefit?
- The ants and insects eat each other
- Protection is exchanged for a food reward
- Both species are unaffected by the association
- The insects gain light and ants lose food
Show answer
Protection is exchanged for a food reward
2. In the tree-hollow experiment, what was measured as a response to productivity?
- The mass of surrounding trees
- The number of trophic links
- The temperature of incoming rain
- The annual rainfall at each site
Show answer
The number of trophic links
3. A large animal accidentally crushes a small plant without experiencing an effect. What sign combination applies?
- (0, −)
- (+, 0)
- (+, −)
- (−, −)
Show answer
(0, −)
4. A seal eats penguins that feed on krill, which consume phytoplankton. What is the seal's role in that chain?
- Primary consumer
- Secondary consumer
- Primary producer
- Tertiary consumer
Show answer
Tertiary consumer
5. A mite eats both eucalypt leaves and springtails. How should its trophic position be described?
- It is always a primary consumer
- It must be a tertiary consumer
- It depends on which food is consumed
- It is always a secondary consumer
Show answer
It depends on which food is consumed
6. A harmless mite lives on a mammal, gaining shelter without affecting its host. What is the interaction?
- Competition
- Amensalism
- Commensalism
- Predation
Show answer
Commensalism
7. In the moose–birch study, which measure was reduced under heavy browsing even though hatching numbers were not significantly different?
- Moose reproductive rate
- Great tit fledgling numbers
- Tree-hollow rainfall
- Pollinator visits
Show answer
Great tit fledgling numbers
8. Two mutualist species gain benefits together but can each survive separately. What type is this?
- Facultative mutualism
- Obligate parasitism
- Obligate mutualism
- Interspecific competition
Show answer
Facultative mutualism
9. A drought causes the longest food chains in a pond system to shorten. Which hypothesis does this most directly support?
- Ecological species concept
- Obligate mutualism hypothesis
- Dynamic stability hypothesis
- Energy hypothesis only
Show answer
Dynamic stability hypothesis
10. An ecosystem supports numerous producers but few high-level predators. Which explanation is most consistent with energy transfer?
- Predators convert all prey biomass to growth
- Energy increases with consumer body size
- Producers lose all energy before being eaten
- Less energy remains at successive levels
Show answer
Less energy remains at successive levels
11. A predator suppresses herbivores, allowing plants to increase. What kind of ecological effect does the predator have on plants?
- Direct consumption
- Direct competition
- Indirect positive effect
- Indirect parasitism
Show answer
Indirect positive effect
12. Moose prefer willow and aspen over spruce. What community response is consistent with this preference?
- Spruce may increase relative to browsed trees
- Aspen gains an advantage over spruce
- Willow must increase because it is eaten
- All tree species must decline equally
Show answer
Spruce may increase relative to browsed trees
13. In a food web, a fox eats a rodent that feeds on seeds. Which link shows a direct energy transfer to the fox?
- Rodent → fox
- Fox → seeds
- Fox → rodent
- Seeds → fox
Show answer
Rodent → fox
14. Two wallaby species use the same limited pasture. What is the most appropriate interaction classification?
- Interspecific competition
- Obligate commensalism
- Facultative parasitism
- Intraspecific competition
Show answer
Interspecific competition
15. Why can two explanations for short food chains both receive experimental support?
- One hypothesis must be mathematically wrong
- Energy and environmental stability can both matter
- Chain length is independent of conditions
- Only the most recent hypothesis can apply
Show answer
Energy and environmental stability can both matter
EASY · 3 MARKS
1. Describe how energy moves through a simple food chain consisting of a plant, a herbivore and a predator.
Show marking guide
1 mark for identifying the plant as the primary producer.
1 mark for identifying the herbivore as the primary consumer and predator as the next consumer.
1 mark for explaining that energy transfers from food to consumer in that direction.
EASY · 3 MARKS
2. Distinguish an autotroph from a heterotroph and give one example of each.
Show marking guide
1 mark for explaining that autotrophs synthesise organic compounds from inorganic sources using light or chemical energy.
1 mark for explaining that heterotrophs obtain organic energy from other organisms or their material.
1 mark for giving one accurate example of each category.
EASY · 3 MARKS
3. Distinguish interspecific and intraspecific competition and state the sign of their effects.
Show marking guide
1 mark for identifying interspecific competition as between different species.
1 mark for identifying intraspecific competition as within the same species.
1 mark for explaining the negative-negative classification when organisms compete for limited resources.
EASY · 3 MARKS
4. Explain why a food web provides a more realistic representation of feeding interactions than a single food chain.
Show marking guide
1 mark for describing a food chain as a linear feeding pathway.
1 mark for describing a food web as interconnected feeding pathways.
1 mark for explaining that many organisms consume or are consumed by multiple species.
EASY · 3 MARKS
5. Explain why herbivory is considered a form of predation and identify one plant defence.
Show marking guide
1 mark for stating that herbivory involves consumption of plant material.
1 mark for identifying a benefit to the herbivore and cost to the plant.
1 mark for identifying a valid defence such as toxic compounds or structural barriers.
INTERMEDIATE · 5 MARKS
6. Ants defend sap-feeding scale insects and collect their honeydew. Explain the ecological classification and when it would be facultative.
Show marking guide
1 mark for identifying the ants' food benefit.
1 mark for identifying protection for the scale insects.
1 mark for classifying the net interaction as mutualism.
1 mark for explaining that a close persistent association may also be symbiotic.
1 mark for stating that it is facultative if partners can survive without the association.
INTERMEDIATE · 5 MARKS
7. Compare the energy and dynamic stability hypotheses for why food chains are generally short.
Show marking guide
1 mark for stating that energy is progressively lost between trophic levels.
1 mark for predicting longer chains when productivity is higher.
1 mark for explaining that disturbances can disproportionately affect higher levels.
1 mark for predicting longer chains in more stable environments.
1 mark for noting that the hypotheses can act together.
INTERMEDIATE · 5 MARKS
8. Describe how moose feeding preferences could benefit spruce while harming willow and aspen.
Show marking guide
1 mark for identifying stronger browsing of willow.
1 mark for identifying stronger browsing of aspen.
1 mark for recognising spruce as less palatable or less browsed.
1 mark for explaining reduced competition from browsed trees.
1 mark for identifying spruce's gain as an indirect effect.
INTERMEDIATE · 5 MARKS
9. In a woodland, moose browse birch leaves and insectivorous birds feed on arthropods associated with birch. Explain a plausible indirect consequence of increased browsing.
Show marking guide
1 mark for describing increased herbivory as a direct negative effect on birch.
1 mark for predicting reduced birch canopy.
1 mark for linking canopy reduction to reduced arthropod biomass.
1 mark for predicting reduced food for insectivorous birds.
1 mark for identifying the effect on birds as indirect rather than direct predation.
INTERMEDIATE · 5 MARKS
10. Two species share a limited food resource, while another pair exchanges pollination services. Compare the interactions and their fitness effects.
Show marking guide
1 mark for identifying the food-sharing pair as competitors.
1 mark for explaining the negative-negative effect of competition.
1 mark for linking competition to a limited shared resource.
1 mark for identifying the pollination pair as mutualists.
1 mark for explaining the positive-positive effects of reward and pollen transfer.
HARD · 8 MARKS
11. A fish feeds on algae and herbivorous invertebrates, and is itself eaten by a bird. Construct the relevant feeding pathways and explain why a fixed trophic label for the fish is misleading.
Show marking guide
1 mark for identifying algae as producers.
1 mark for identifying invertebrates eating algae as primary consumers.
1 mark for tracing algae to fish to bird.
1 mark for identifying fish as a primary consumer on that route.
1 mark for tracing algae to invertebrate to fish to bird.
1 mark for identifying fish as a secondary consumer on the second route.
1 mark for explaining the arrow direction as energy flow.
1 mark for concluding that food webs reveal context-dependent trophic positions.
HARD · 8 MARKS
12. Darwin noted that villages with many cats also had many flowers. Develop a plausible indirect trophic explanation involving cats, small mammals, pollinators and plants, and identify how it could be tested.
Show marking guide
1 mark for proposing cats as predators of small mammals.
1 mark for proposing small mammals as consumers of pollinators or their nests.
1 mark for predicting suppression of small mammals where cats are abundant.
1 mark for predicting greater pollinator activity as a consequence.
1 mark for linking greater pollination to flowering plant reproductive success.
1 mark for labelling the cat-to-plant influence as indirect.
1 mark for recognising the chain as a hypothesis rather than a demonstrated explanation from the lecture.
1 mark for suggesting comparisons or monitoring of the intervening populations and pollination.
HARD · 8 MARKS
13. In an experiment, heavily browsed birch stands have lower canopy cover and fewer great tit fledglings than lightly browsed controls, but similar hatchling numbers. Explain the ecological mechanism and interpret this result.
Show marking guide
1 mark for identifying the direct negative effect of moose on birch.
1 mark for identifying reduced birch canopy in browsed stands.
1 mark for linking reduced vegetation to lower arthropod resources.
1 mark for linking arthropod food to developing great tit chicks.
1 mark for identifying similar hatching numbers across treatments.
1 mark for identifying reduced fledgling numbers in browsed stands.
1 mark for inferring an effect on post-hatching survival or development rather than necessarily on hatching.
1 mark for identifying the effect on great tits as indirect.
Lecture 29: Assemblages and Ecosystems
1. A forest canopy tree falls, increasing light at ground level. Which change is expected first?
- Establishment of light-demanding colonisers
- Permanent cessation of plant succession
- Immediate dominance by shade-tolerant hardwoods
- Loss of all seed dispersal from the opening
Show answer
Establishment of light-demanding colonisers
2. Why might moderate patchy fires maintain more species than complete fire exclusion?
- Every patch is repeatedly reset to bare soil
- Different successional stages can coexist
- Pioneer species are permanently removed
- The whole area remains at climax stage
Show answer
Different successional stages can coexist
3. What happens during denitrification?
- Organic carbon is converted to methane
- Nitrate is returned to gaseous nitrogen
- Nitrogen gas is converted to plant protein
- Phosphate is released from rock surfaces
Show answer
Nitrate is returned to gaseous nitrogen
4. What is a likely outcome of fires repeated before plants can replace their seeds?
- Some resprouting and reseeding species decline
- Recovery becomes independent of fire frequency
- All successional stages become equally common
- More late-successional trees establish
Show answer
Some resprouting and reseeding species decline
5. A soil nutrient is recycled locally with no large atmospheric reservoir. Which cycle is most consistent?
- Water
- Carbon
- Nitrogen
- Phosphorus
Show answer
Phosphorus
6. A trap captures several insect species at one site, but their interactions are unknown. What term best describes this sample?
- Population
- Ecosystem
- Assemblage
- Community
Show answer
Assemblage
7. Why may vegetation maps be imperfect substitutes for maps of all animal communities?
- Abiotic conditions affect only animal species
- Animal distributions need not match vegetation types
- Vegetation communities never change over time
- Vegetation has no links to other taxa
Show answer
Animal distributions need not match vegetation types
8. A bare quarry wall has little soil and almost no organic matter. What challenge particularly affects early colonists?
- Excessive canopy shade
- Limited nutrients and rooting sites
- A surplus of soil nitrogen
- High competition from hardwoods
Show answer
Limited nutrients and rooting sites
9. Which observation most strongly suggests facilitation during primary succession?
- Pioneers and newcomers never influence each other
- Pioneers build soil that supports later colonists
- Early shrubs exclude all arriving seedlings
- Late trees prevent soil from accumulating
Show answer
Pioneers build soil that supports later colonists
10. Why can open oceans have low NPP per unit area despite covering large areas?
- Oceanic producers never photosynthesise
- Ocean water cannot support autotrophs
- All ocean energy is stored in consumers
- Nutrient supply can limit producer growth
Show answer
Nutrient supply can limit producer growth
11. What happens to kelp when sea otter abundance rises and sea urchin grazing falls?
- Kelp stops taking up carbon dioxide
- Kelp carbon storage must disappear
- Kelp biomass and carbon storage can increase
- Kelp biomass becomes independent of grazing
Show answer
Kelp biomass and carbon storage can increase
12. How can mycorrhizal fungi benefit plants growing in low-phosphorus soils?
- They remove the need for all root uptake
- They transform atmospheric CO2 into nitrate
- They convert plant sugars directly into rainfall
- They enhance phosphorus acquisition
Show answer
They enhance phosphorus acquisition
13. Dense introduced shrubs prevent native seedlings establishing beneath them. Which model is illustrated?
- Intermediate disturbance
- Facilitation
- Inhibition
- Tolerance
Show answer
Inhibition
14. Why is the water cycle global even though water may be scarce within an ecosystem?
- Atmospheric movement transfers water between regions
- Rainfall permanently traps water within regions
- Soil water is always recycled in place
- Only oceans participate in the water cycle
Show answer
Atmospheric movement transfers water between regions
EASY · 3 MARKS
1. Define resilience in the context of ecological disturbance and state one difficulty in measuring it.
Show marking guide
1 mark for describing recovery towards a reference or pre-disturbance community state.
1 mark for recognising that recovery may take time after disturbance.
1 mark for identifying ambiguity about the reference state or equilibrium.
EASY · 3 MARKS
2. Describe ecological succession and name two stages in the classical abandoned-field sequence.
Show marking guide
1 mark for describing succession as change in species composition over time.
1 mark for identifying an early stage such as annual plants or perennial grasses.
1 mark for identifying a later stage such as shrubs, softwoods or hardwoods.
EASY · 3 MARKS
3. Explain why nitrogen fixation matters for plant productivity.
Show marking guide
1 mark for noting that plants generally cannot use atmospheric nitrogen gas directly.
1 mark for identifying microbial fixation as making nitrogen biologically available.
1 mark for linking usable nitrogen to plant growth and productivity.
EASY · 3 MARKS
4. Identify three typical characteristics of pioneer plant species.
Show marking guide
1 mark for identifying rapid growth or a short generation time.
1 mark for identifying good dispersal or small seeds.
1 mark for identifying high-light tolerance, nitrogen fixation or weak competition.
EASY · 3 MARKS
5. State the intermediate disturbance hypothesis and describe its predicted pattern of richness.
Show marking guide
1 mark for identifying disturbance frequency or intensity as the predictor.
1 mark for predicting highest richness at intermediate disturbance.
1 mark for recognising lower richness at low and high disturbance levels.
INTERMEDIATE · 5 MARKS
6. A researcher records plant species in several traps but does not know whether they interact. Explain whether assemblage or community is the more defensible term, and how interactions could change that interpretation.
Show marking guide
1 mark for defining assemblage as co-occurring species without an interaction requirement.
1 mark for defining community as co-occurring interacting species.
1 mark for choosing assemblage for the evidence given.
1 mark for explaining that co-occurrence alone does not demonstrate interaction.
1 mark for identifying an observation or experiment that could reveal ecological interactions.
INTERMEDIATE · 5 MARKS
7. Compare facilitation, tolerance and inhibition as mechanisms underlying succession.
Show marking guide
1 mark for explaining facilitation as a positive influence of early species.
1 mark for explaining tolerance as a neutral influence.
1 mark for explaining inhibition as a negative influence.
1 mark for giving a defensible mechanism or example for any two models.
1 mark for recognising that which mechanism dominates depends on the ecosystem.
INTERMEDIATE · 5 MARKS
8. Explain how sea otters can affect carbon storage without directly photosynthesising.
Show marking guide
1 mark for identifying otter predation on sea urchins.
1 mark for identifying sea urchin grazing on kelp.
1 mark for explaining reduced grazing when otters are present.
1 mark for linking greater kelp biomass to carbon fixed by photosynthesis.
1 mark for recognising the otter effect on carbon storage as an indirect trophic cascade.
INTERMEDIATE · 5 MARKS
9. Explain why a landscape with patchy intermediate disturbance can support more species than landscapes with very little or very frequent disturbance.
Show marking guide
1 mark for describing competitive dominance in low-disturbance habitats.
1 mark for identifying a mosaic of succession stages under intermediate disturbance.
1 mark for linking this mosaic to coexistence of different species.
1 mark for explaining repeated removal or interrupted recovery under high disturbance.
1 mark for connecting these processes to a peaked richness–disturbance relationship.
HARD · 8 MARKS
10. An area mapped as one vegetation community contains distinct fungi and animal species on opposite slopes. Explain why vegetation-based community classification may be insufficient and design a more robust survey.
Show marking guide
1 mark for identifying vegetation mapping as a proxy for wider community types.
1 mark for recognising that non-plant taxa need not match vegetation patterns.
1 mark for considering slope, water, light or soil variation.
1 mark for defining appropriate sampling units across both slopes.
1 mark for including surveys of animal taxa.
1 mark for including surveys of fungal or other non-plant taxa.
1 mark for proposing comparisons of composition across locations.
1 mark for explaining how added observations improve conservation decisions.
HARD · 8 MARKS
11. An ecologist proposes that all ecological communities naturally converge to a permanent hardwood climax. Critically evaluate this claim using succession, disturbance and time.
Show marking guide
1 mark for describing the classical progression towards later-successional trees.
1 mark for identifying traits favouring long-lived late species.
1 mark for explaining that disturbances can reset stages.
1 mark for noting that mosaics of stages can coexist spatially.
1 mark for identifying local colonisation and extinction over time.
1 mark for noting that community composition is not permanently fixed.
1 mark for recognising variation among ecosystems including Australian examples.
1 mark for concluding that permanent universal climax is not supported by the lecture.
HARD · 8 MARKS
12. In a woodland model, every patch advances towards mature trees unless a disturbance resets it. Explain how varying disturbance frequency and spatial extent could generate a non-linear relationship with species richness.
Show marking guide
1 mark for identifying stage progression in the absence of disturbance.
1 mark for explaining low stage diversity under very little disturbance.
1 mark for identifying disturbance as a reset mechanism.
1 mark for explaining spatial mosaics under moderate disturbance.
1 mark for linking mosaics to richer species assemblages.
1 mark for explaining loss of late stages under frequent disturbance.
1 mark for considering effects of large disturbance footprint on patch survival.
1 mark for predicting an intermediate peak while recognising the model is simplified.
Lecture 30: Human Footprint
1. Researchers find persistent industrial chemicals in remote Arctic wildlife. Which explanation is supported by the lecture?
- Pollutants are produced naturally by sea-ice formation
- Predators remove all chemicals from their prey
- Only local Arctic manufacturing could explain detection
- Atmospheric transport followed by cold-region deposition
Show answer
Atmospheric transport followed by cold-region deposition
2. Raptor breeding success falls despite few adult deaths after pesticide use. Which mechanism best matches the lecture example?
- Eggshell thinning lowers successful recruitment
- Habitat corridors prevent new nest formation
- Food chain shortening raises the adult death rate
- Adult birds increase their investment in eggs
Show answer
Eggshell thinning lowers successful recruitment
3. A bird refuses to cross farmland separating two forest patches. What landscape feature is limiting connectivity?
- A short food chain
- A high predator abundance within both patches
- An inhospitable matrix
- A large interior habitat area
Show answer
An inhospitable matrix
4. Why may a national ban fail to eliminate exposure to a volatile persistent pollutant?
- All contaminants are generated by natural local sources
- Ocean currents prevent cross-border contaminant transport
- National bans immediately remove stored chemicals globally
- Emissions elsewhere can travel through the atmosphere
Show answer
Emissions elsewhere can travel through the atmosphere
5. A forest survey records temperature changes and earlier insect emergence. Which conclusion is most justified?
- A single correlation excludes all other ecological influences
- The biological shift is consistent with warming but needs wider context
- The insect shift proves temperature is its sole cause
- The temperature data cannot inform ecological interpretation
Show answer
The biological shift is consistent with warming but needs wider context
6. What distinguishes biomass collapse from the predator-loss cascade in the lecture case studies?
- Both arise solely from reduced sunlight reaching the canopy
- One requires climate warming; the other requires ocean acidification
- One centres on edge-driven tree loss; the other on disrupted trophic control
- Both are direct effects of heavy-metal pollution
Show answer
One centres on edge-driven tree loss; the other on disrupted trophic control
7. Predatory fish contain higher concentrations of a persistent toxin than their prey. Which concept is illustrated?
- Succession after habitat disturbance
- Atmospheric transport across latitudes
- Recolonisation through a habitat corridor
- Biomagnification across trophic levels
Show answer
Biomagnification across trophic levels
8. A narrow native-vegetation strip connects two reserves. What is its most likely advantage and limitation?
- It supports only interior specialists and excludes mobile taxa
- It eliminates isolation and recreates extensive core forest
- It prevents dispersal while removing surrounding disturbances
- It permits movement but may consist mostly of edge habitat
Show answer
It permits movement but may consist mostly of edge habitat
9. Policy changes reduce native-vegetation clearing. What ecological benefit is most directly expected?
- Complete prevention of climate-related range shifts
- Instant elimination of all existing edge effects
- Immediate recovery of all locally extinct species
- Less habitat loss and fragmentation for native species
Show answer
Less habitat loss and fragmentation for native species
10. After an oil spill, dispersant application reduces visible oil but harms marine larvae. What does this show?
- Visible shoreline condition measures all biological effects
- Removing surface oil always restores all affected species
- Dispersants prevent pollutants entering food chains
- Clean-up interventions can create additional ecological impacts
Show answer
Clean-up interventions can create additional ecological impacts
11. A chemical is detected above background levels but no harmful effect is established. Which description is most appropriate?
- It is a contaminant; its pollutant status is unresolved
- It is a biomagnified toxin because it is widespread
- It is harmless because it has not caused death
- It is necessarily a pollutant because it is detectable
Show answer
It is a contaminant; its pollutant status is unresolved
12. Why are small fragments particularly vulnerable to edge effects?
- Their interiors are always wetter than continuous forests
- They contain a higher total area than larger fragments
- They receive no sunlight from outside the fragment
- A greater proportion of their area lies near boundaries
Show answer
A greater proportion of their area lies near boundaries
13. Why can long-lived marine mammals be significant sources of dietary pollutant exposure?
- Persistent substances can build up in their tissues
- Cold seawater makes all pollutants biologically inactive
- Long lifespans prevent their tissues absorbing toxins
- Their high trophic level destroys chemical pollutants
Show answer
Persistent substances can build up in their tissues
14. Why are expected climate-driven range shifts often towards higher latitudes or altitudes?
- Species are compelled to follow national boundaries
- Higher latitudes always contain more habitat area
- Organisms may track suitable temperature conditions
- Lower altitudes always experience less competition
Show answer
Organisms may track suitable temperature conditions
15. Two reserves have equal total habitat area: one large square and four small squares. Which likely has more core habitat?
- Neither can contain any edge habitat
- The four small squares
- Both have identical perimeter relative to area
- The single large square
Show answer
The single large square
EASY · 3 MARKS
1. Define biomagnification and explain why top predators may have high contaminant concentrations.
Show marking guide
1 mark for biomagnification is an increase in tissue concentration at successive trophic levels.
1 mark for predators consume multiple contaminated prey.
1 mark for persistent substances can accumulate at high trophic levels.
EASY · 3 MARKS
2. Describe one mechanism by which rodenticides can poison powerful owls indirectly.
Show marking guide
1 mark for rodenticide bait is consumed by non-target prey such as possums.
1 mark for a powerful owl consumes the contaminated prey.
1 mark for the owl is exposed to rodenticide and may suffer secondary poisoning.
EASY · 3 MARKS
3. Describe the basic predator–herbivore–plant cascade expected after predators disappear.
Show marking guide
1 mark for predator removal reduces top-down control of herbivores.
1 mark for herbivore abundance or feeding pressure can increase.
1 mark for plant abundance or seedling recruitment can decline.
EASY · 3 MARKS
4. Describe what an edge effect is and give two changes that can occur at a forest edge.
Show marking guide
1 mark for an edge effect is an ecological change near the boundary of a habitat patch.
1 mark for one relevant change, such as altered wind exposure or light.
1 mark for a second change, such as altered hydrology, vine growth or tree mortality.
EASY · 3 MARKS
5. Identify three landscape properties that affect the consequences of habitat fragmentation.
Show marking guide
1 mark for patch size or extent of remaining habitat.
1 mark for patch shape or edge-to-core ratio.
1 mark for isolation, connectivity or permeability of the surrounding matrix.
INTERMEDIATE · 5 MARKS
6. Describe how researchers could investigate whether tree biomass declines towards forest edges after fragmentation.
Show marking guide
1 mark for compare sites at different measured distances from the edge.
1 mark for include suitable forest-interior or continuous-forest controls.
1 mark for measure tree diameter or comparable data for biomass estimation.
1 mark for follow plots through time after fragmentation.
1 mark for test whether biomass change is more negative near boundaries.
INTERMEDIATE · 5 MARKS
7. Explain how pesticides reduced bald eagle populations without necessarily causing widespread adult mortality.
Show marking guide
1 mark for persistent pesticides entered food webs.
1 mark for exposure affected birds at higher trophic levels.
1 mark for toxins interfered with eggshell formation or integrity.
1 mark for eggs were more likely to break during incubation.
1 mark for lower hatching success reduced recruitment and population growth.
INTERMEDIATE · 5 MARKS
8. Explain how the loss of predators from newly isolated islands can eventually affect forest regeneration.
Show marking guide
1 mark for isolation can lead to loss of large predators.
1 mark for predator release can increase the numbers of some herbivores or seed predators.
1 mark for greater feeding pressure affects seeds, seedlings or saplings.
1 mark for fewer young trees survive to replace older canopy trees.
1 mark for long-term changes in forest structure can produce ecological meltdown.
INTERMEDIATE · 5 MARKS
9. Explain why the lecture treats the evidence for ecological climate change as stronger across many systems than in a single case.
Show marking guide
1 mark for natural variation can produce changes in individual systems.
1 mark for long-term observations provide directional trends.
1 mark for climate warming predicts particular types of biological and physical responses.
1 mark for many independent systems exhibit changes in those predicted directions.
1 mark for consistent cross-system patterns strengthen inference without proving every individual cause.
HARD · 8 MARKS
10. A chemical manufacturer argues that a national PCB ban makes further national and international action unnecessary. Evaluate this claim using the Arctic case study.
Show marking guide
1 mark for identify the historical persistence of PCBs in environmental compartments.
1 mark for explain volatilisation in warmer locations.
1 mark for explain atmospheric transport across borders.
1 mark for explain deposition in colder environments.
1 mark for identify slow degradation or retention under cold conditions.
1 mark for explain accumulation in marine animal tissues and dietary transfer.
1 mark for argue why coordinated international controls are needed.
1 mark for explain why monitoring and enforcement remain necessary after restrictions.
HARD · 8 MARKS
11. A local government can protect either a large circular forest patch or an equal area distributed among four thin strips. Evaluate the options for forest specialists and landscape connectivity.
Show marking guide
1 mark for recognise that the alternatives protect equal total areas.
1 mark for compare perimeter-to-area or edge-to-core relationships.
1 mark for explain the larger circular patch generally retains more core habitat.
1 mark for link edges to altered wind, light or hydrology.
1 mark for explain the resulting potential for tree mortality or recruitment changes.
1 mark for identify how strips could offer connectivity or dispersal pathways.
1 mark for note limitations for edge-sensitive or interior-dependent species.
1 mark for give a conditional recommendation reflecting species needs and matrix permeability.
HARD · 8 MARKS
12. Researchers attribute forest biomass loss after clearing solely to reduction in patch area. Critique their inference and propose a stronger study.
Show marking guide
1 mark for identify patch area as a potentially relevant factor.
1 mark for explain that smaller patches also often have a larger proportion of edge habitat.
1 mark for identify distance from edge as a potential confounder.
1 mark for include plots near edges and far into interiors.
1 mark for include large fragments and continuous-forest controls.
1 mark for measure tree size or diameter using consistent methods.
1 mark for compare biomass change through time rather than a single snapshot.
1 mark for consider microclimatic mechanisms and recruitment or mortality rates.
Lecture 31: Conservation
1. Fox populations expand following rabbit establishment. What ecological process may help explain this?
- Two invaders must occupy different trophic levels
- One invader facilitates another through food supply
- Fox spread is unrelated to prey availability
- Rabbits directly suppress fox reproduction
Show answer
One invader facilitates another through food supply
2. A species has not been assessed for the IUCN Red List. What follows from that fact?
- Its extinction risk has not been formally assessed
- It is necessarily extinct in the wild
- It has already recovered from previous decline
- It must be classified as Least Concern
Show answer
Its extinction risk has not been formally assessed
3. Cane toads are advancing through a dry landscape where they depend on waterholes. Which intervention follows the lecture's reasoning?
- Provide more permanent water points ahead of the front
- Release more toads to increase intraspecific competition
- Remove native predators while keeping water available
- Restrict access to strategic artificial water points
Show answer
Restrict access to strategic artificial water points
4. Which observation would support top-down effects of dingoes across a landscape?
- No variation in herbivore abundance across the fence
- Identical fox densities regardless of dingoes
- Fewer foxes where dingoes persist
- More foxes where dingoes persist
Show answer
Fewer foxes where dingoes persist
5. After dingoes decline, foxes become more abundant. What is this pattern called?
- Mesopredator release
- Nutrient immobilisation
- Competitive exclusion by dingoes
- Primary succession
Show answer
Mesopredator release
6. Of 1,000 species deliberately introduced, how many would the classic tens rule predict become invasive?
- Approximately 500
- Approximately 10
- Approximately 100
- Approximately 1
Show answer
Approximately 1
7. A host-specific parasite disappears following the extinction of its sole host. Which mechanism applies?
- Co-extinction
- Facilitation
- Competitive release
- Naturalisation
Show answer
Co-extinction
8. A formerly unrecorded mammal is found in a single surviving population. Which action best addresses the risk of one local disaster?
- Establish additional secure populations
- Assume rediscovery removes the extinction risk
- Keep all animals in the original habitat
- Delay management until a decline is measured
Show answer
Establish additional secure populations
9. A conservation team documents a decline, estimates its drivers, and tests responses. What broader sequence does this follow?
- Choose solutions first and ignore causal evidence
- Increase harvesting, record declines, remove controls
- List names, assign habitats, disregard causes
- Understand threats, predict effects, develop solutions
Show answer
Understand threats, predict effects, develop solutions
10. A revegetated woodland has few native ant species but high rates of seed removal. What does this demonstrate?
- All original ant species have necessarily returned
- An ecological function can recover before species composition
- Seed removal cannot occur without a remnant community
- Vegetation structure is unrelated to any ecosystem process
Show answer
An ecological function can recover before species composition
11. Why should a restoration project define a reference condition and measurable goal?
- To guarantee recovery of every historic species
- To make all restored habitats structurally identical
- To avoid collecting data after planting
- To judge ecological outcomes rather than actions alone
Show answer
To judge ecological outcomes rather than actions alone
12. A newly introduced herbivore reproduces quickly and survives on many plant species. Which feature makes establishment more likely?
- Very narrow habitat tolerance
- Inability to disperse between sites
- Complete reliance on one host plant
- Broad ecological flexibility
Show answer
Broad ecological flexibility
13. Which example best shows why conservation biology requires social as well as biological evidence?
- A model estimates survival of young mammals
- A survey identifies an animal using genetic markers
- A sound recovery plan lacks political support for implementation
- A field study records the number of native plants
Show answer
A sound recovery plan lacks political support for implementation
14. A port receives cargo containing live insects. Which pathway of introduction is most directly implicated?
- Primary ecological succession
- Natural background extinction
- Human trade and transport
- Local nutrient recycling
Show answer
Human trade and transport
EASY · 3 MARKS
1. Define co-extinction and give an example.
Show marking guide
1 mark for one species disappears following the loss of another it depends upon.
1 mark for the dependence may involve a host, pollinator or prey species.
1 mark for an example is a specialist parasite lost with its sole host.
EASY · 3 MARKS
2. Explain the three stages described in the traditional tens rule.
Show marking guide
1 mark for a proportion of imported species escapes into the wild.
1 mark for a proportion of escaped species becomes naturalised.
1 mark for a proportion of naturalised species becomes invasive.
EASY · 3 MARKS
3. Give three indicators of success in an ecological restoration project.
Show marking guide
1 mark for recovery of vegetation structure or habitat quality.
1 mark for return of target species or native assemblages.
1 mark for recovery of an ecological function such as pollination or seed dispersal.
EASY · 3 MARKS
4. List three categories of threat included in the Evil Quartet.
Show marking guide
1 mark for alien or introduced species.
1 mark for overhunting or overexploitation.
1 mark for habitat loss or co-extinction as a third distinct threat.
INTERMEDIATE · 5 MARKS
5. Compare the Evil Quartet and HIPPO as frameworks for conservation threats.
Show marking guide
1 mark for both include habitat destruction or loss.
1 mark for both include alien or invasive species.
1 mark for both recognise overhunting or overharvesting.
1 mark for the Evil Quartet explicitly includes co-extinction.
1 mark for hIPPO also highlights pollution and human overpopulation.
INTERMEDIATE · 5 MARKS
6. Explain how creating an insurance population can reduce extinction risk for a rare mammal.
Show marking guide
1 mark for a single small population is vulnerable to local disasters.
1 mark for translocating some individuals creates another population.
1 mark for the new site should be secure and suitable for persistence.
1 mark for separate populations reduce the risk of one event affecting every individual.
1 mark for insurance populations may help recolonise an affected site later.
INTERMEDIATE · 5 MARKS
7. Explain why controlling introduced foxes may require monitoring another predator.
Show marking guide
1 mark for foxes prey upon many small native mammals.
1 mark for control may reduce fox density and direct predation.
1 mark for foxes can also suppress other predators such as cats.
1 mark for cats might increase if fox suppression is removed.
1 mark for monitoring prey and both predator populations supports integrated management.
INTERMEDIATE · 5 MARKS
8. Explain why restoration outcomes should be assessed using both species and ecological functions.
Show marking guide
1 mark for vegetation structure can recover without all species returning.
1 mark for assemblage composition indicates which species have recolonised.
1 mark for functional measures assess processes such as seed dispersal or pollination.
1 mark for different organisms may provide similar functions after restoration.
1 mark for both measures reveal successes and limitations relative to stated goals.
HARD · 8 MARKS
9. A council claims a revegetation project is “nature positive” because it planted thousands of trees. Propose an evidence-based evaluation.
Show marking guide
1 mark for clarify whether the target is to halt or reverse biodiversity decline.
1 mark for establish baseline or reference conditions and relevant time frames.
1 mark for measure vegetation survival, structure and habitat quality rather than planting effort alone.
1 mark for survey relevant native taxa using appropriate methods.
1 mark for assess ecological functions such as pollination and seed dispersal.
1 mark for use comparable sites and repeated monitoring to identify trends.
1 mark for account for incomplete species recovery and possible introduced-species substitution.
1 mark for judge success against explicit ecological outcomes and revise actions as required.
HARD · 8 MARKS
10. A threatened marsupial exists in one fire-prone reserve. Design a conservation strategy that addresses immediate and longer-term extinction risks.
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1 mark for identify the vulnerability of a single geographically concentrated population.
1 mark for assess fire, introduced predators and other local threats.
1 mark for protect or improve the existing habitat and manage those threats.
1 mark for select one or more ecologically suitable independent sites.
1 mark for establish insurance populations through carefully managed translocation.
1 mark for monitor reproduction, survival and genetic or demographic viability.
1 mark for plan responses to future fires or local population losses.
1 mark for use monitoring outcomes to revise the recovery strategy.
HARD · 8 MARKS
11. In a landscape, rabbits support foxes and dingoes suppress foxes. Predict consequences of dingo removal and design a management comparison.
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1 mark for explain that rabbits can subsidise fox populations.
1 mark for explain that dingoes can limit fox abundance.
1 mark for predict increased foxes following removal of dingo suppression.
1 mark for identify a possible rise in predation on small native mammals.
1 mark for consider changes in herbivore pressure and vegetation through trophic interactions.
1 mark for compare replicated areas or periods with contrasting dingo presence.
1 mark for measure foxes, rabbits, native prey and vegetation, controlling relevant differences.
1 mark for use results to plan integrated rather than single-species management.
HARD · 8 MARKS
12. Restored woodland has few native ant species but high seed removal, while native pollinators remain scarce. Evaluate whether restoration has succeeded.
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1 mark for define the restoration goals and appropriate reference sites.
1 mark for identify that ant species richness remains below remnant values.
1 mark for recognise that seed-removal function has recovered despite species differences.
1 mark for explain that key functional groups can provide services without identical assemblages.
1 mark for identify reduced native pollinator recovery as a separate limitation.
1 mark for consider whether introduced pollinators provide similar but different services.
1 mark for recommend monitoring species composition and multiple ecosystem functions.
1 mark for conclude that success is partial and depends on explicit objectives.