Question 1
Paper 2Easy4 marksBefore a root-tip cell divides, it copies its DNA.
(a)[2]
Outline the role of helicase in DNA replication.
(b)[2]
State the role of DNA polymerase in DNA replication.
IB Biology SL Theme D: Continuity and change questions
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Question 2
Paper 2Medium4 marksFigure 1 shows a stage in the elongation of a polypeptide during protein synthesis.
(a)[1]
State the next mRNA codon to be translated as elongation continues.
(b)[3]
Explain how the ribosome adds the next amino acid and continues elongation from the stage shown in Figure 1.
Question 3
Paper 2Medium16 marksPhenylketonuria (PKU) is caused by the recessive mutated allele p. The allele P is dominant, and only individuals with genotype pp have PKU. Two unaffected parents both have genotype Pp.
Their daughter also has genotype Pp. As an adult, she develops a skin cancer. A mutation, T, in a gene involved in regulating cell division is detected in her tumour cells, but not in her non-tumour skin cells, blood cells or gametes.
Non-tumour skin cells and tumour cells from the daughter were cultured under the same conditions. The table shows the mean percentage of cells that underwent division in 48 hours, from three replicate cultures.
| Cell sample | Mean percentage of cells undergoing division in 48 hours |
|---|---|
| Non-tumour skin cells | 8% |
| Tumour cells | 35% |
(a)[4]
Determine the expected genotype ratio among the parents’ offspring and the probability that an offspring will have PKU.
(b)[3]
Explain why the daughter with genotype Pp does not have PKU, and how she could pass allele p to a child who has PKU.
(c)[2]
Describe the difference in cell division between the two cell samples in the table.
(d)[5]
Using the information about mutation T, explain how it could contribute to cancer and whether it could be inherited by the daughter’s offspring.
(e)[2]
Give one limitation of the cell-culture results in deciding whether mutation T caused the higher rate of cell division.
Question 4
Paper 2Medium6 marksA nursery propagates a plant cultivar from stem cuttings, which reproduce asexually, and from seeds produced by sexual reproduction. Groups of 40 plants from each method were grown under two sets of conditions. The number of plants that met the nursery’s marketability standard is shown.
| Growing conditions | Cuttings (asexual), out of 40 | Seedlings (sexual), out of 40 |
|---|---|---|
| Stable | 36 | 28 |
| Hotter and drier | 8 | 25 |
(a)[1]
Calculate the difference between the proportions of marketable cuttings and seedlings in the hotter and drier conditions. Give your answer as a proportion or in percentage points.
(b)[2]
If conditions remain stable and the cultivar is already suited to them, recommend a propagation method. Explain why it may be advantageous.
(c)[3]
Explain how sexual reproduction could produce offspring better suited to hotter and drier conditions.
Question 5
Paper 2Hard16 marksA family has several members diagnosed with phenylketonuria (PKU). Figure 1 shows their pedigree. No other family members are shown.
Use P for the allele not associated with PKU and p for the PKU-associated allele.
(a)[2]
Using Figure 1, state two observations that support PKU having an autosomal recessive inheritance pattern in this family.
(b)[3]
Deduce the genotypes of I-1, I-2, II-2 and II-3.
(c)[3]
Show the genetic cross for a child of II-2 and II-3, and state the probability that this child has PKU.
(d)[3]
III-2 does not have PKU. Determine the probability that III-2 is a carrier of the p allele. Show your reasoning.
(e)[2]
Suppose III-2 has a child with a person known to be a Pp carrier. Determine the probability that their child will have PKU. Use your answer to part (d).
(f)[3]
Explain how inheritance of the p allele can result in PKU and affect the conversion of phenylalanine to tyrosine.
Question 6
Paper 2Hard16 marksA healthy adult volunteer completed two separate 40-minute exposures in a climate chamber, one at 10°C and one at 34°C. Clothing and activity (seated) were the same in both exposures. Measurements and observations at the end of each exposure are shown.
| Air temperature / °C | Core temperature / °C | Skin blood-flow index / arbitrary units | Sweat production / g h⁻¹ | Shivering |
|---|---|---|---|---|
| 10 | 37.0 | 24 | 0 | observed |
| 34 | 37.1 | 72 | 360 | not observed |
A higher skin blood-flow index indicates greater blood flow near the skin.
(a)[3]
Calculate the percentage increase in skin blood-flow index from 10°C to 34°C. State how core temperature changed.
(b)[5]
Explain how vasoconstriction and shivering could help the volunteer at 10°C. State how hair erection could contribute to thermoregulation in humans.
(c)[8]
In a separate 40-minute period at each temperature, the volunteer was free to act. At 10°C, the volunteer put on an insulated coat and moved into shelter. At 34°C, the volunteer moved into shade, removed an outer layer of clothing and rested. Core temperature at the end was 37.0°C at 10°C and 37.1°C at 34°C.
A research team concluded that these behavioural actions alone were sufficient to maintain core temperature, and that physiological responses were not important. Evaluate this conclusion using the table and the follow-up observations. Include physiological and behavioural responses, and one limitation of the evidence.
Question 7
Paper 2Medium16 marksIn a population of ground beetles, shell pattern is determined by one gene with two alleles, P and p. P is dominant: beetles with PP or Pp have plain shells, while beetles with pp have speckled shells. A base substitution in a germ cell changed a P allele to p, and the p allele was inherited by descendants.
Two heterozygous beetles mate. In a separate investigation, equal numbers of plain and speckled beetles were placed on dark or pale soil. The number alive after two weeks was recorded.
| Soil colour | Shell pattern | Number at start | Number alive after two weeks |
|---|---|---|---|
| Dark | Plain | 40 | 24 |
| Dark | Speckled | 40 | 32 |
| Pale | Plain | 40 | 32 |
| Pale | Speckled | 40 | 24 |
Assume shell pattern is inherited as described and does not change with soil colour. Surviving beetles are equally likely to reproduce.
(a)[4]
Explain how the mutation described could provide heritable variation in the beetle population.
(b)[5]
Use a genetic cross to determine the expected genotype and phenotype ratios among the offspring of the two heterozygous beetles. State how sexual reproduction generates variation in this cross.
(c)[7]
Calculate the percentage survival of each shell pattern on each soil colour. Use the results to explain how natural selection may act on variation in this population, distinguishing the role of mutation from the role of natural selection.
Question 8
Paper 2Hard16 marksA research group collected the following illustrative monitoring data for one defined region of the Amazon rainforest. The forest area in 1980 is the original area for this region.
| Year | Area of forest remaining / km² | Mean annual rainfall / mm |
|---|---|---|
| 1980 | 100 000 | 1600 |
| 2000 | 90 000 | 1510 |
| 2020 | 80 000 | 1400 |
The group also investigated ecosystem stability in aquatic mesocosms. Each mesocosm was a sealed glass vessel containing freshwater plants and microorganisms. Four vessels were used for each plant-cover treatment. The initial total abundance index was 100 in all vessels. All other conditions were kept the same. Light was reduced to half its starting level for three days and then restored. The table shows the mean total abundance index.
| Plant cover | Before disturbance | At end of disturbance | 14 days after light was restored |
|---|---|---|---|
| 25% | 100 | 64 | 82 |
| 75% | 100 | 80 | 96 |
In this investigation, greater stability is defined as a smaller decrease in abundance during the disturbance and a value closer to the initial abundance 14 days after the disturbance.
(a)[3]
Calculate the percentage decrease in forest area between 1980 and 2020. Describe the change in mean annual rainfall over the same period. Show your working.
(b)[4]
Explain how continued deforestation could make the Amazon rainforest a possible tipping point in ecosystem stability.
(c)[3]
Using the stated definition of stability, compare the stability of the two mesocosm treatments.
(d)[6]
Assess how far the monitoring data and the mesocosm results support the claim that continued deforestation could destabilize the Amazon rainforest. Use evidence from both data sets, explain one advantage of the mesocosm design and identify one limitation when applying the model to the Amazon.
Question 9
Paper 2Medium16 marksThree temperate sites were managed using different approaches to restoration: planting non-native trees, allowing native forest to regenerate, or restoring a peat-forming wetland by re-establishing waterlogged conditions. Waterlogging at the wetland limits the decomposition of dead plant material.
Each approach was studied at one site. Three plots were sampled at each site. Figure 1 shows the mean net increase in carbon stored in vegetation and soil relative to the start of restoration. Table 1 shows the mean number of native plant species recorded per 20 quadrats after 15 years, averaged across the three plots.
Table 1
| Restoration approach | Mean number of native plant species per 20 quadrats after 15 years |
|---|---|
| Non-native tree plantation | 7 |
| Native forest regeneration | 19 |
| Restored peat-forming wetland | 15 |
(a)[1]
Identify the restoration approach with the greatest increase in carbon stored after 10 years.
(b)[2]
Calculate the mean annual increase in carbon stored in the restored peat-forming wetland between years 5 and 15. Give your answer with a unit.
(c)[3]
Describe and compare the changes in carbon stored in the three restoration approaches over the 15-year period.
(d)[4]
Explain how tree-based restoration and restoration of a peat-forming wetland can each increase carbon storage.
(e)[4]
A project manager argues that non-native tree plantations should be chosen as the best option, considering carbon storage and native plant diversity over the first 15 years. Evaluate this claim using Figure 1 and Table 1.
(f)[2]
Suggest one limitation of the investigation’s design and explain how it could affect the comparison of restoration approaches.
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