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
Multiple choiceMedium1 markPCR primers flank a 300 bp region of a DNA sample. After repeated temperature cycles, the products and a 600 bp DNA fragment are loaded into separate wells of the same gel. Which statement correctly links DNA synthesis in PCR to the relative movement of the fragments?
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Answer: C. Taq polymerase extends primers by adding nucleotides, making copies of the region between them. The 300 bp product is shorter than the 600 bp fragment, so it moves farther through the gel. A: assigns DNA synthesis to helicase and reverses the gel-size relationship; helicase unwinds strands, while the shorter fragment travels farther. B: correctly identifies polymerase synthesis but reverses the size-dependent migration. D: assigns DNA synthesis to helicase, which separates strands rather than making DNA copies. Remember: PCR copies DNA with polymerase; gel electrophoresis separates fragments by size.
Question 3
Multiple choiceEasy1 markA particular gene is inactive in a non-dividing somatic cell. Which description is consistent with the gene having been switched off at the first stage of gene expression?
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Answer: A. The first stage of gene expression is transcription. When transcription is switched off, no new RNA is transcribed from the gene; using DNA as a template does not change its base sequence.
B: assumes that switching off transcription alters the DNA sequence.
C: confuses switching off transcription with continued transcription while a later stage is inactive.
D: assumes the DNA template is consumed or destroyed rather than used without changing its sequence.
Remember: transcription changes gene expression, not the DNA base sequence.
Question 4
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 5
Multiple choiceMedium1 markTwo versions of a gene occur in a population. In version X, one nucleotide is replaced by another; in version Y, one nucleotide is inserted within the coding sequence, before its final codon. No other sequence changes are present. Which statements are correct?
I. Version X is a single-nucleotide polymorphism (SNP) resulting from a base substitution.
II. The replacement in X must change the amino acid encoded at that position.
III. The insertion in Y shifts the reading frame of subsequent codons, making it likely that the polypeptide ceases to function.
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Answer: B. A base substitution can give rise to an SNP, but degeneracy of the genetic code means it may or may not change the amino acid. A one-nucleotide insertion within a coding sequence shifts the reading frame of subsequent codons, making loss of polypeptide function likely. Thus I and III are correct; II is not.
A: accepts the misconception in II that every substitution changes an amino acid, and omits the consequence of the insertion.
C: accepts II and fails to identify the base substitution as an SNP.
D: includes II; degeneracy means the replacement need not change the amino acid.
Remember: substitutions may be silent, while a one-nucleotide insertion can shift the reading frame.
Question 6
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 7
Multiple choiceEasy1 markDuring budding in yeast, a small bud receives much less cytoplasm than the larger cell. Which distribution of mitochondria is expected when the cells separate?
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Answer: D. Cytokinesis can divide cytoplasm unequally, but each daughter cell must receive at least one mitochondrion. The larger cell may receive more mitochondria, but neither cell can receive none. A: assigns all mitochondria to the larger cell, leaving the bud without one. B: reverses that allocation, leaving the larger cell without one. C: assumes mitochondria can be made anew rather than inherited from pre-existing mitochondria. Remember: Unequal cytokinesis does not mean either daughter can lack mitochondria.
Question 8
Multiple choiceEasy1 markThe figure compares equal-volume samples; each dot represents the same dissolved solute. Which statements are correct?
I. A plant cell in medium P gains water and develops turgor pressure.
II. In medium Q, the plasma membrane pulls away from the cell wall because water enters the plant cell.
III. IV solution R is isotonic to blood-cell contents and is suitable because it causes no net water movement across cell membranes.
Which statements are correct?
Show answer and explanation
Answer: B. The equal-volume samples show that P has a lower solute concentration than the plant-cell contents, so water enters and turgor pressure develops. Q has a higher solute concentration, so water leaves and the plasma membrane pulls away from the cell wall (plasmolysis). R has the same solute concentration as the blood-cell contents, so it is isotonic and causes no net water movement.
A: omits III; equal concentrations in R indicate an isotonic solution suitable for intravenous use.
C: omits I; the lower solute concentration in P causes water to enter the plant cell.
D: reverses water movement in Q; water leaves the cell, rather than entering it.
Remember: hypotonic media cause water entry; hypertonic media cause plasmolysis.
Question 9
Multiple choiceEasy1 markA student traces the route taken by sperm in a male reproductive system diagram. Which structure carries sperm from the epididymis towards the urethra?
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Answer: A. The sperm duct carries sperm from the epididymis towards the urethra.
B: confuses the later passage through which semen leaves the body with the duct that carries sperm from the epididymis.
C: identifies a site of sperm production rather than the structure that transports sperm from the epididymis.
D: confuses a gland that contributes fluid to semen with a sperm-transporting duct.
Remember: sperm travel from the epididymis through the sperm duct towards the urethra.
Question 10
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 11
Multiple choiceMedium1 markA mother with XX sex chromosomes and a father with XY sex chromosomes have a child. The father has allele R on his X chromosome, and R is absent from his Y chromosome. Which statement(s) are correct?
I. A sperm carrying Y results in an XY zygote, associated with certain male-typical physical characteristics.
II. A child with XY sex chromosomes inherits allele R from the father.
III. The X chromosome carries more genes than the Y chromosome.
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Answer: C. The egg contributes X, while a Y-bearing sperm results in an XY zygote, associated with certain male-typical physical characteristics. An XY child receives the father’s Y chromosome, not his X chromosome, so cannot inherit R from him. The X chromosome carries more genes than the Y chromosome.
A: assumes the father passes his X-linked allele R to an XY child.
B: identifies the Y-sperm outcome but omits the true comparison of gene numbers.
D: assumes R can pass from father to an XY child and omits the sperm’s role in determining the zygote’s sex chromosomes.
Remember: an XY child receives its X chromosome from the egg and its Y chromosome from the sperm.
Question 12
Multiple choiceMedium1 markAfter a meal, a patient’s blood glucose remains high even though the pancreas releases a normal amount of insulin. Which option identifies the likely type of diabetes and its relevant physiological change?
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Answer: A. Type 2 diabetes involves reduced responsiveness of body cells to insulin. Insulin is being released, but its effect in lowering blood glucose is reduced, matching the pattern described.
B: identifies type 2 but substitutes insufficient insulin release for reduced cellular responsiveness; the case specifies normal release.
C: identifies reduced responsiveness but assigns it to type 1, which is associated with insufficient insulin production.
D: recalls insufficient insulin production in type 1, but that does not fit the normal insulin release in this case.
Remember: Type 1 involves insufficient insulin production; type 2 involves reduced responsiveness to insulin.
Question 13
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 14
Multiple choiceMedium1 markA survey measured the mean concentrations of DDT and mercury in the tissues of consumers at trophic levels 2–5 in a lake. Which interpretation is supported by the graph?
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Answer: B. Biomagnification is an increase in pollutant concentration in consumer tissues at higher trophic levels. Both plotted concentrations rise from level 2 to level 5; at level 5, DDT is 0.8 mg kg−1 and mercury is 0.4 mg kg−1, so DDT is higher. A: recognizes the rising trends but reverses the comparison at level 5. C: follows the DDT trend but misses that mercury also increases, from 0.1 to 0.4 mg kg−1. D: recognizes both rising trends but reads the two level-5 values as equal. Remember: biomagnification is indicated by pollutant concentration increasing across higher trophic levels.
Question 15
Multiple choiceMedium1 markA shift in ocean currents leaves surface water in a coastal region warmer during the season when nutrients normally rise from deeper water. Which paired change is most likely?
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Answer: D. Warmer surface water can impede nutrient upwelling, so fewer nutrients reach the surface. The reduced nutrient supply lowers primary production, decreasing energy flow through marine food chains.
A: identifies the decline in nutrients and production but reverses the resulting change in energy flow. B: reverses the effect of reduced nutrient availability on primary production. C: reverses the effect of warmer surface water on nutrient upwelling, while keeping the later links consistent.
Remember: Lower primary production means less energy enters marine food chains.
Question 16
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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