AP · thinka-original Practice Paper

2023 AP AP Biology Practice Paper with Answers

Thinka May 2023 AP-Style Mock — AP Biology

34 marks90 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the May 2023 AP AP Biology paper. Not affiliated with or reproduced from AP.

Section II: Long Free-Response Questions

Answer Questions 1 and 2 in paragraph form. Recommended time: approximately 25 minutes per question. Outlines or bulleted lists alone are not acceptable.
2 Question · 18 marks
Question 1 · free-response
9 marks
In eukaryotic microorganisms, cellular zinc homeostasis is maintained through the regulated expression of specific transport proteins. When extracellular zinc (\(\text{Zn}^{2+}\)) concentrations are low, an intracellular signaling pathway activates the transcription factor ZapA. Activation occurs when the protein kinase KinZ phosphorylates ZapA, inducing a conformational change that allows it to bind to the promoter region of the ZUP1 gene and stimulate transcription. ZUP1 encodes a high-affinity plasma membrane zinc transporter, Zup1. When extracellular \(\text{Zn}^{2+}\) is abundant, zinc directly binds to ZapA, promoting its interaction with a co-repressor complex comprising Cbp1 and Cbp2, which prevents ZapA from activating transcription of ZUP1.

To investigate the individual contributions of these regulatory components, researchers generated two mutant strains from a wild-type yeast strain: a mutant lacking functional KinZ (\(kinZ^{mt}\)) and a mutant lacking functional Cbp1 (\(cbp1^{mt}\)). Cells of the wild-type and mutant strains were cultured in media containing either high or low levels of \(\text{Zn}^{2+}\). The researchers measured Zup1 transport activity and quantified relative ZUP1 mRNA levels relative to wild-type cells grown in high \(\text{Zn}^{2+}\) (Table 1).

**TABLE 1. Zup1 TRANSPORT ACTIVITY AND RELATIVE ZUP1 mRNA LEVELS IN WILD-TYPE AND MUTANT YEAST STRAINS UNDER HIGH- AND LOW-ZINC CONDITIONS**

| Yeast Strain | Mutation | Zup1 Activity in High-\(\text{Zn}^{2+}\) (\(\text{nmol}/\text{min}/\text{mg}\) protein \(\pm 2\text{SE}_{\bar{x}}\)) | Zup1 Activity in Low-\(\text{Zn}^{2+}\) (\(\text{nmol}/\text{min}/\text{mg}\) protein \(\pm 2\text{SE}_{\bar{x}}\)) | Relative ZUP1 mRNA in High-\(\text{Zn}^{2+}\) (\(\pm 2\text{SE}_{\bar{x}}\)) | Relative ZUP1 mRNA in Low-\(\text{Zn}^{2+}\) (\(\pm 2\text{SE}_{\bar{x}}\)) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Wild-type | None | \(1.2 \pm 0.2\) | \(25.2 \pm 1.4\) | \(1.0 \pm 0.1\) | \(18.0 \pm 1.5\) |
| \(kinZ^{mt}\) | Nonfunctional KinZ kinase | \(1.1 \pm 0.2\) | \(1.4 \pm 0.3\) | \(0.9 \pm 0.2\) | \(1.2 \pm 0.3\) |
| \(cbp1^{mt}\) | Nonfunctional Cbp1 co-repressor | \(23.8 \pm 1.6\) | \(24.9 \pm 1.5\) | \(17.5 \pm 1.3\) | \(18.2 \pm 1.4\) |

(a) Describe the mechanism by which the addition of a negatively charged phosphate group by a kinase can alter the activity of a transcription factor such as ZapA. Explain how a signaling cascade can amplify a cellular signal during signal transduction.

(b) Based on Table 1, identify ONE dependent variable measured in the researchers' experiment. Justify the researchers' decision to use the wild-type yeast strain as the genetic background for engineering the mutant strains. Justify the researchers' testing of single-gene mutant strains rather than strains with multiple simultaneous mutations.

(c) Based on the data in Table 1, identify the yeast strain and growth condition that resulted in the lowest relative ZUP1 mRNA level. Calculate the percent change in Zup1 transport activity in wild-type yeast cells grown in low-\(\text{Zn}^{2+}\) compared to wild-type yeast cells grown in high-\(\text{Zn}^{2+}\).

(d) In a subsequent experiment, researchers generated a yeast strain with a loss-of-function mutation in the gene encoding Cbp2. Predict the effect of this mutation on ZUP1 mRNA expression when the mutant yeast is grown in a high-\(\text{Zn}^{2+}\) environment. Provide reasoning to justify your prediction.
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Worked solution

(a)
- Description: The addition of a negatively charged phosphate group introduces electrostatic interactions (such as repulsion or attraction with neighboring amino acid residues) that alter the tertiary/three-dimensional conformation of the transcription factor, exposing its DNA-binding domain or altering its affinity for target DNA sequences.
- Explanation: Signal amplification occurs because a single activated enzyme (such as an upstream protein kinase) can catalyze the modification/activation of many downstream target proteins or secondary messengers, multiplying the signal at each successive step in the pathway.

(b)
- Identification: Either "Zup1 transport activity" or "relative ZUP1 mRNA levels / expression".
- Justification (wild-type control): Using the isogenic wild-type background ensures that all strains are genetically identical except for the specific targeted mutation, guaranteeing that any observed phenotypic differences are directly attributable to the introduced mutation rather than background genetic variation.
- Justification (single mutations): Mutating only a single gene at a time allows researchers to isolate and determine the specific, independent function of each individual component in the regulatory pathway.

(c)
- Identification: \(kinZ^{mt}\) in a high-\(\text{Zn}^{2+}\) environment (relative mRNA level of \(0.9 \pm 0.2\)).
- Calculation:
\[\text{Percent Change} = \frac{\text{Value}_{\text{low}} - \text{Value}_{\text{high}}}{\text{Value}_{\text{high}}} \times 100\%\]
\[\text{Percent Change} = \frac{25.2 - 1.2}{1.2} \times 100\% = \frac{24.0}{1.2} \times 100\% = 2,000\%\]
*(An increase of \(2,000\%\))*

(d)
- Prediction: ZUP1 mRNA expression will be high/constitutively expressed (elevated compared to wild-type under high-\(\text{Zn}^{2+}\) conditions).
- Justification: Cbp2 is an essential subunit of the co-repressor complex that normally functions with Cbp1 to inhibit ZapA under high-\(\text{Zn}^{2+}\) conditions. Without functional Cbp2, the co-repressor complex cannot form or inhibit ZapA, leaving ZapA active and free to promote ZUP1 transcription even in the presence of high zinc.

Marking scheme

Part (a): 2 points maximum
- 1 point for describing that phosphorylation changes the conformation/tertiary shape/charge distribution of the protein.
- 1 point for explaining that one activated enzyme/kinase can catalyze reactions on multiple downstream protein molecules, thereby multiplying the molecular signal.

Part (b): 3 points maximum
- 1 point for identifying a dependent variable:
- Accept one of: Zup1 (transport) activity OR (relative) ZUP1 mRNA levels.
- 1 point for justifying using the wild-type strain:
- Accept one of: Ensures differences are solely due to the introduced mutation / controls for background genetic differences.
- 1 point for justifying using single-gene mutations:
- Accept one of: Allows the effect of each individual component/gene to be evaluated independently / determines the specific role of each pathway component.

Part (c): 2 points maximum
- 1 point for identifying \(kinZ^{mt}\) yeast in a high-\(\text{Zn}^{2+}\) environment.
- 1 point for correctly calculating the percent change:
- Accept: \(2,000\%\) (or \(+2,000\%\)) based on \(\frac{25.2 - 1.2}{1.2} \times 100\%\) [also accept \(-95.2\%\) if comparing high relative to low: \(\frac{1.2 - 25.2}{25.2} \times 100\%\)].

Part (d): 2 points maximum
- 1 point for predicting that ZUP1 will be expressed / transcribed / mRNA levels will be high/elevated in the mutant strain in a high-\(\text{Zn}^{2+}\) environment.
- 1 point for providing reasoning that without functional Cbp2, the co-repressor complex fails to assemble/inhibit ZapA, allowing ZapA to bind the promoter and activate transcription.
Question 2 · long_free_response
9 marks
Light availability is a critical factor influencing the rate of photosynthesis and primary productivity in marine ecosystems. Researchers studying photoadaptation examined several species of marine macroalgae to evaluate how changes in irradiance affect cellular photosynthetic capacity. The researchers measured the rate of oxygen evolution (a proxy for the rate of photosynthesis) in six different algal species when cultured under both low-light (​50 \(\mu\text{mol}\cdot\text{m}^{-2}\cdot\text{s}^{-1}\)) and high-light (​400 \(\mu\text{mol}\cdot\text{m}^{-2}\cdot\text{s}^{-1}\)) conditions (Table 1).

TABLE 1. AVERAGE RATE OF OXYGEN EVOLUTION IN SIX MACROALGAL SPECIES EXPOSED TO LOW-LIGHT AND HIGH-LIGHT CONDITIONS

| Species | Oxygen Evolution at Low Light (\(\mu\text{mol } \text{O}_2\cdot\text{g}^{-1}\cdot\text{h}^{-1}\)) \(\pm 2\text{SE}_{\bar{x}}\) | Oxygen Evolution at High Light (\(\mu\text{mol } \text{O}_2\cdot\text{g}^{-1}\cdot\text{h}^{-1}\)) \(\pm 2\text{SE}_{\bar{x}}\) |
| :--- | :--- | :--- |
| 1 | \(4.2 \pm 0.4\) | \(8.6 \pm 0.5\) |
| 2 | \(2.1 \pm 0.3\) | \(4.8 \pm 0.4\) |
| 3 | \(3.0 \pm 0.5\) | \(6.5 \pm 0.6\) |
| 4 | \(1.5 \pm 0.2\) | \(3.2 \pm 0.3\) |
| 5 | \(5.0 \pm 0.6\) | \(5.4 \pm 0.7\) |
| 6 | \(2.8 \pm 0.4\) | \(7.1 \pm 0.5\) |

(a) Describe the role of the thylakoid membrane in the light-dependent reactions of photosynthesis.

(b) Using the grid provided, construct an appropriately labeled graph that represents the data in Table 1. Determine which species show(s) a statistically significant difference in the rate of oxygen evolution between low-light and high-light conditions.

(c) Based on the data in Table 1, describe the relationship between light irradiance and the rate of oxygen evolution in species 1 through 4 and species 6.

(d) In a particular species of macroalga, individual isolates typically display dark green pigmentation, but researchers isolated a pale green mutant strain with severely impaired photosynthetic electron transport. Genetic analysis indicates that this pale green phenotype results from a mutation in chloroplast DNA (cpDNA). In this algal species, chloroplasts are exclusively contributed by the maternal gamete (female strain) during sexual reproduction. Researchers crossed a female pale green mutant with a male wild-type (dark green) strain. Predict the phenotype(s) of the offspring produced from this cross. Provide reasoning to justify your prediction. Explain why genetically identical algal clones grown in environments with different light intensities can exhibit differences in photosynthetic enzyme concentration.
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Worked solution

(a) The thylakoid membrane is the site of the light-dependent reactions of photosynthesis. It embeds Photosystems II and I, electron transport chain complexes, and ATP synthase. It acts as an impermeable barrier that allows protons pumped into the thylakoid lumen during electron transport to establish a proton motive force/electrochemical gradient, which powers ATP synthesis via chemiosmosis, while also reducing \(\text{NADP}^+\) to \(\text{NADPH}\).

(b) Graph Construction:
1. Axis Labels & Units: y-axis labeled as 'Oxygen Evolution (\(\mu\text{mol } \text{O}_2\cdot\text{g}^{-1}\cdot\text{h}^{-1}\))' and x-axis labeled as 'Species' with categories 1 through 6.
2. Graph Type: A grouped bar graph or modified column chart comparing Low Light vs. High Light for each species.
3. Plotting & Error Bars: Correct plotting of all 12 mean values and corresponding \(\pm 2\text{SE}_{\bar{x}}\) error bars.
Determination: Species 1, 2, 3, 4, and 6 show statistically significant differences because the error bars (\(\pm 2\text{SE}_{\bar{x}}\)) for low-light and high-light conditions do not overlap in these species. Species 5 shows no significant difference because its error bars overlap (\(5.0 \pm 0.6\) overlaps with \(5.4 \pm 0.7\)).

(c) There is a positive relationship between light intensity and the rate of oxygen evolution for species 1–4 and 6; exposure to higher light irradiance leads to an increased rate of oxygen production per unit mass.

(d) Prediction: All offspring will exhibit the pale green phenotype.
Justification: In this species, chloroplast DNA is inherited strictly maternally through the female gamete. Because the female parent carries the mutated cpDNA that causes the pale green phenotype, all progeny inherit only the mutant chloroplasts.
Explanation: Phenotypic variation among genetically identical individuals (clones) occurs because environmental factors, such as light irradiance, alter cellular signaling pathways that increase or decrease the transcription and translation (expression) of genes encoding photosynthetic enzymes.

Marking scheme

Part (a) (1 point):
• 1 point for describing that the thylakoid membrane contains photosystems/ETC/ATP synthase for photophosphorylation OR that it maintains a proton concentration gradient between the lumen and stroma.

Part (b) (4 points total):
• 1 point for appropriate axis labels and units (y-axis: Oxygen Evolution with units; x-axis: Species numbered 1–6).
• 1 point for representing data in an appropriate grouped bar graph format.
• 1 point for accurately plotted mean data points and error bars (\(\pm 2\text{SE}_{\bar{x}}\)).
• 1 point for correctly determining that Species 1, 2, 3, 4, and 6 show a significant difference (or that Species 5 does not).

Part (c) (1 point):
• 1 point for describing that higher light irradiance results in a higher rate of oxygen evolution / a positive correlation between irradiance and oxygen evolution.

Part (d) (3 points total):
• 1 point for predicting that all offspring will be pale green / have the mutant phenotype.
• 1 point for justifying that chloroplasts/cpDNA are inherited maternally / through the female parent's ovule/gamete.
• 1 point for explaining that changes in environmental conditions (light intensity) affect the expression of certain genes (transcription/translation).

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Section II: Short Free-Response Questions

Answer Questions 3 through 6 in paragraph form. Recommended time: approximately 10 minutes per question.
4 Question · 16 marks
Question 1 · Scientific Investigation
4 marks
Mayfly nymphs of the genus Baetis are abundant primary consumers in temperate freshwater streams. These aquatic insects graze on benthic periphyton (algae and cyanobacteria) and serve as a vital energy source for secondary consumers such as juvenile trout and predatory stoneflies. Environmental disturbances, such as elevated stream temperatures and nitrate pollution resulting from agricultural runoff, threaten stream macroinvertebrate communities.

Baetis nymphs typically inhabit clear, cool streams where typical baseline conditions are approximately \(12^\circ\text{C}\) with low dissolved nitrate levels (approximately \(0.5\text{ mg/L}\)). To investigate how rising water temperatures and nutrient enrichment affect macroinvertebrate survival, researchers monitored cohorts of Baetis nymphs reared in stream mesocosms under combinations of three temperatures (\(12^\circ\text{C}\), \(16^\circ\text{C}\), and \(20^\circ\text{C}\)) and three dissolved nitrate concentrations (\(0.5\text{ mg/L}\), \(5.0\text{ mg/L}\), and \(15.0\text{ mg/L}\)) over a 60-day period.

(a) Describe the effect of high species diversity among primary consumers on the resilience of an aquatic ecosystem subjected to an environmental disturbance.

(b) Justify the researchers' choice of \(12^\circ\text{C}\) and \(0.5\text{ mg/L}\) dissolved nitrate as the baseline conditions in their experimental design.

(c) State an appropriate null hypothesis for this experiment.

(d) The researchers claim that a substantial decline in the Baetis nymph population will alter the biomass of benthic periphyton in the stream. Provide reasoning to support the researchers' claim.
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Worked solution

(a) High species diversity provides functional redundancy within a trophic level. If one species experiences population decline due to environmental stress, other tolerant primary consumers can maintain grazing pressure and continue transferring energy to higher trophic levels, thereby enhancing the resilience and stability of the ecosystem.

(b) Selecting \(12^\circ\text{C}\) and \(0.5\text{ mg/L}\) dissolved nitrate serves as an essential baseline or negative control representing natural, undisturbed habitat conditions. This allows researchers to isolate and determine the specific effects caused by elevated temperatures and elevated nitrate concentrations relative to standard development.

(c) A properly formulated null hypothesis predicts no significant difference or effect: "Changes in water temperature and dissolved nitrate concentration (or their combination) will have no effect on the survival rate of Baetis nymphs."

(d) Baetis nymphs exert top-down control on primary producers by grazing on benthic periphyton. A decrease in the nymph population relieves this consumption pressure, resulting in an unchecked increase/accumulation in periphyton biomass.

Marking scheme

Part (a) [1 point]:
• Accept one of the following:
- Increased species diversity increases ecosystem resilience/stability because other primary consumers can fill the ecological niche/role if one species declines.
- Ecosystems with higher diversity are more likely to contain species tolerant of the environmental disturbance, ensuring continued primary consumption and energy transfer.

Part (b) [1 point]:
• Accept one of the following:
- These values reflect the natural/ambient stream conditions and provide a control/baseline for comparison with the elevated treatment groups.
- It serves to establish the normal survival rate of the nymphs under typical environmental conditions to determine whether elevated temperature or nitrate has an effect.

Part (c) [1 point]:
• Accept one of the following:
- (Increases in) temperature and/or dissolved nitrate levels will have no effect on the survival (or maturation) rate of Baetis nymphs.
- There is no difference in the survival rates of Baetis nymphs across the different temperature and nitrate treatments.

Part (d) [1 point]:
• Accept one of the following:
- Periphyton biomass will increase because reduced numbers of Baetis nymphs lead to reduced grazing/herbivory pressure (top-down regulation).
- The reduction in primary consumer predation allows primary producers (periphyton) to grow and accumulate biomass without being consumed at normal rates.
Question 2 · Subjective
4 marks
During aerobic cellular respiration, glucose is catabolized through glycolysis in the cytosol, producing pyruvate, ATP, and NADH. In eukaryotic cells, pyruvate is transported into the mitochondrial matrix, where it is converted into acetyl-CoA by the pyruvate dehydrogenase (PDH) complex, releasing \(\text{CO}_2\) and reducing \(\text{NAD}^+\) to NADH. Acetyl-CoA then enters the citric acid cycle.

Activity of the PDH complex is tightly controlled. Pyruvate dehydrogenase kinase (PDK1) inhibits PDH by phosphorylating a specific serine residue on the enzyme. Conversely, pyruvate dehydrogenase phosphatase removes the phosphate group, reactivating PDH. In many proliferating cancer cells, PDK1 is constitutively active, leading to significant metabolic reprogramming.

(a) Describe the role of the citric acid cycle in generating molecules required for the electron transport chain.

(b) Explain why a high ratio of ATP to ADP in the mitochondrial matrix acts as a signal to activate PDK1 in non-cancerous cells.

(c) Researchers treat cancer cells overexpressing PDK1 with dichloroacetate (DCA), a small-molecule inhibitor that specifically blocks PDK1 activity. Predict the effect of DCA treatment on the rate of mitochondrial oxygen (\(\text{O}_2\)) consumption in these cells.

(d) Justify your prediction in part (c).
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Worked solution

(a) The citric acid cycle functions in the mitochondrial matrix to oxidize organic substrates derived from acetyl-CoA, reducing the coenzymes \(\text{NAD}^+\) and \(\text{FAD}\) to \(\text{NADH}\) and \(\text{FADH}_2\). These reduced electron carriers transfer electrons to the electron transport chain to power oxidative phosphorylation.

(b) A elevated \(\text{ATP}:\text{ADP}\) ratio reflects a high cellular energy charge. Activating PDK1 serves as a negative feedback mechanism that inactivates the PDH complex, thereby slowing carbohydrate oxidation when sufficient ATP is already present and sparing substrate for other metabolic pathways.

(c) Treatment with dichloroacetate (DCA) will increase the rate of mitochondrial oxygen (\(\text{O}_2\)) consumption.

(d) By inhibiting PDK1, DCA prevents the inhibitory phosphorylation of PDH, allowing the enzyme complex to remain active. This leads to increased conversion of pyruvate to acetyl-CoA and greater flux through the citric acid cycle. The resulting increase in \(\text{NADH}\) and \(\text{FADH}_2\) delivery to the electron transport chain accelerates electron flow to molecular oxygen (the terminal electron acceptor), thus increasing \(\text{O}_2\) consumption.

Marking scheme

Part (a): 1 point
- Describe that the citric acid cycle reduces electron carriers (\(\text{NAD}^+\) and/or \(\text{FAD}\) to \(\text{NADH}\) and/or \(\text{FADH}_2\)) / provides high-energy electrons to the electron transport chain.

Part (b): 1 point
- Explain that high ATP signifies an abundance of cellular energy/energy demands met, so activating PDK1 (which inhibits PDH) prevents unnecessary catabolism/breakdown of glucose.

Part (c): 1 point
- Predict that the rate of mitochondrial oxygen consumption will increase / be higher.

Part (d): 1 point
- Justify that uninhibited PDH increases acetyl-CoA entry into the citric acid cycle, producing more \(\text{NADH}\)/\(\text{FADH}_2\) to supply electrons to the electron transport chain where \(\text{O}_2\) is reduced / acts as the terminal electron acceptor.
Question 3 · Free-Response
4 marks
Carnivorous plants have independently evolved specialized structures and biochemical adaptations to capture and digest prey, allowing them to survive in nutrient-poor, waterlogged soils. Researchers investigating the evolutionary relationships among five related plant species (Species 1 through 5) within a single family constructed a phylogenetic tree using molecular sequencing data of chloroplast genes (Figure 1). An ancestral non-carnivorous plant, Species 0, was used as the outgroup.

```
Species 0 (Outgroup)
|
+------ Species 1
|
`------+
|------ Species 2
|
`------+
|------ Species 3
|
`------+
|------ Species 4
|
`------ Species 5
```
Figure 1. Molecular cladogram showing the evolutionary relationships among five plant species and an outgroup.

The researchers also cataloged the presence (+) or absence (−) of three key morphological and physiological traits related to carnivory across the five species (Table 1).

TABLE 1. PRESENCE (+) OR ABSENCE (−) OF SPECIALIZED TRAITS IN FIVE PLANT SPECIES
| Trait | Description | Species 1 | Species 2 | Species 3 | Species 4 | Species 5 |
| :--- | :--- | :---: | :---: | :---: | :---: | :---: |
| 1 | Secretion of acid phosphatase and endopeptidase enzymes | + | + | + | + | + |
| 2 | Stalked glandular trichomes producing sticky mucilage | − | + | + | + | + |
| 3 | Rapid leaf-closure movement (active snap-trap) | − | + | − | − | + |

(a) Describe how comparing homologous DNA sequences across different plant species allows scientists to determine their evolutionary relatedness.

(b) Based on the data in Table 1, identify the trait that represents an ancestral characteristic (symplesiomorphy) shared by all carnivorous species in this lineage.

(c) Based on Figure 1 and Table 1, identify the branch or lineage segment on the cladogram where Trait 2 most likely arose.

(d) Based on the cladogram in Figure 1, explain why the presence of Trait 3 in both Species 2 and Species 5 is more parsimoniously explained by convergent evolution rather than inheritance from a shared common ancestor without trait loss.
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Worked solution

(a) Homologous DNA sequences accumulate mutations over time through substitutions, insertions, and deletions. Organisms that diverged more recently have had less time to accumulate mutations and therefore share a higher percentage of sequence identity compared to organisms that diverged further in the evolutionary past.

(b) Trait 1 is present in all five species (Species 1 through 5). Because it is shared across all ingroup members, it represents the ancestral state that evolved prior to the divergence of Species 1 from the remaining species.

(c) Trait 2 is present in Species 2, Species 3, Species 4, and Species 5, but absent in Species 1 (and the outgroup). Under maximum parsimony, Trait 2 evolved once on the lineage immediately following the branch point that separates Species 1 from the common ancestor of Species 2 through 5.

(d) If Trait 3 arose in the most recent common ancestor of Species 2, 3, 4, and 5, it would require one gain and two separate loss events (in the lineage leading to Species 3 and the lineage leading to Species 4), totaling 3 evolutionary transitions. Conversely, independent evolution (convergent evolution) of Trait 3 in Species 2 and Species 5 requires only 2 evolutionary transitions (two independent gains). Since 2 changes is fewer than 3 changes, convergent evolution is the more parsimonious hypothesis.

Marking scheme

Part (a): 1 point
- Accept: A description that species with higher sequence similarity/fewer nucleotide differences share a more recent common ancestor / diverged more recently.
- Do not accept: Simply stating 'DNA is the genetic material' without mentioning similarity or divergence time.

Part (b): 1 point
- Accept: Trait 1 (or 'Secretion of acid phosphatase and endopeptidase enzymes').

Part (c): 1 point
- Accept: On the stem/lineage leading to the common ancestor of Species 2, 3, 4, and 5 (after the divergence of Species 1 / between the node of Species 1 and the node of Species 2).

Part (d): 1 point
- Accept: Reasoning that convergent evolution requires fewer evolutionary steps/events (2 independent gains) than common ancestry with multiple subsequent losses (1 gain and 2 losses / loss in Species 3 and Species 4 lineages).
- Accept: An explanation that Species 3 and Species 4 are more closely related to Species 5 than Species 2 is, meaning sharing by common descent would require trait loss in intervening taxa.
Question 4 · subjective
4 marks
Heat-shock proteins (HSPs) function as molecular chaperones that stabilize and refold denatured proteins during cellular stress. Researchers investigated the regulation of several genes involved in cellular homeostasis during thermal stress in mammalian cells. Cultured cells grown at a baseline temperature of 37°C were either maintained at 37°C or exposed to elevated heat-shock conditions of 42°C.

The researchers quantified mRNA stability by measuring the mRNA half-life (the time required for 50% of the mRNA molecules to be degraded) and determined the relative protein synthesis rate for three heat-shock genes (HSP70, HSP90, and HSPA8) as well as a reference structural gene (ACTB). The results are summarized in Table 1.

TABLE 1. mRNA HALF-LIFE AND RELATIVE PROTEIN SYNTHESIS AT 37°C AND 42°C

| Gene | mRNA Half-Life at 37°C (hours) $\pm 2\text{SE}_\bar{x}$ | mRNA Half-Life at 42°C (hours) $\pm 2\text{SE}_\bar{x}$ | Relative Protein Synthesis at 37°C (arbitrary units) $\pm 2\text{SE}_\bar{x}$ | Relative Protein Synthesis at 42°C (arbitrary units) $\pm 2\text{SE}_\bar{x}$ |
| :--- | :---: | :---: | :---: | :---: |
| ACTB | $12.0 \pm 0.8$ | $3.2 \pm 0.4$ | $100 \pm 5$ | $22 \pm 4$ |
| HSP70 | $1.5 \pm 0.2$ | $8.5 \pm 0.6$ | $15 \pm 2$ | $240 \pm 12$ |
| HSP90 | $4.0 \pm 0.5$ | $4.2 \pm 0.4$ | $50 \pm 6$ | $55 \pm 5$ |
| HSPA8 | $6.0 \pm 0.5$ | $11.0 \pm 0.7$ | $40 \pm 4$ | $130 \pm 8$ |

(a) Based on the data in Table 1, identify the gene whose mRNA stability shows the greatest increase in response to thermal stress (42°C).

(b) Based on the data in Table 1, describe the relationship between mRNA half-life and relative protein synthesis for HSP70 when cells are transferred from 37°C to 42°C.

(c) The researchers claim that the post-transcriptional regulation of ACTB mRNA contributes to shifting the cell's translational machinery toward protective chaperone production during thermal stress. Use the data in Table 1 to support the researchers' claim.

(d) Explain how an increase in mRNA half-life allows a cell to increase the accumulation of a specific protein without altering the rate of transcription of that gene.
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Worked solution

(a) HSP70 exhibits the greatest relative increase in mRNA stability (half-life increases from $1.5\text{ hours}$ at 37°C to $8.5\text{ hours}$ at 42°C, which is a greater than 5.6-fold increase; HSPA8 exhibits a $5.0\text{ hour}$ absolute increase, so either HSP70 or HSPA8 is acceptable with appropriate reasoning).

(b) For HSP70, there is a positive/direct correlation: as the mRNA half-life increases (from $1.5\text{ hours}$ to $8.5\text{ hours}$), the relative protein synthesis rate markedly increases (from $15$ to $240\text{ arbitrary units}$).

(c) The data show that under heat shock (42°C), ACTB mRNA half-life decreases substantially (from $12.0\text{ h}$ to $3.2\text{ h}$, with non-overlapping standard error ranges), accompanied by a steep decline in ACTB protein synthesis (from $100$ to $22\text{ units}$). This rapid degradation of non-essential housekeeping mRNA reduces competition for translational machinery, supporting the synthesis of essential chaperone proteins like HSP70 and HSPA8.

(d) An increased mRNA half-life means that individual mRNA transcripts are protected from degradation and persist in the cytoplasm for a longer duration. Consequently, ribosomes can engage in repeated rounds of translation on the same existing mRNA molecules, producing more polypeptide products per transcript over time without requiring higher rates of de novo transcription.

Marking scheme

Part (a): 1 point
- Identification of HSP70 (or HSPA8).

Part (b): 1 point
- Description that as the mRNA half-life increases, the rate of protein synthesis increases / there is a positive (direct) relationship between mRNA stability and protein synthesis for HSP70.

Part (c): 1 point
- Support connects the substantial decrease in ACTB mRNA half-life / protein synthesis at 42°C to the reduction of non-heat-shock protein production, allowing resources/ribosomes to be redirected toward chaperone synthesis.

Part (d): 1 point
- Explanation that longer-lived mRNA transcripts remain available in the cytoplasm to be translated multiple times / repeatedly by ribosomes, yielding more protein molecules per transcript without changing transcription rate.

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