IB Biology SL Theme D: Continuity and change questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceMedium1 mark

PCR 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?

Show answer and explanation

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 2

Paper 2Medium4 marks

Figure 1 shows a stage in the elongation of a polypeptide during protein synthesis.

ribosome5′3′AUGGCUUUUCAAgrowing polypeptide chain3′-CGA-5′
(a)

State the next mRNA codon to be translated as elongation continues.

[1]
(b)

Explain how the ribosome adds the next amino acid and continues elongation from the stage shown in Figure 1.

[3]

Question 3

Multiple choiceMedium1 mark

Two 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.

Show answer and explanation

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 4

Paper 2Medium16 marks

Phenylketonuria (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 sampleMean percentage of cells undergoing division in 48 hours
Non-tumour skin cells8%
Tumour cells35%
(a)

Determine the expected genotype ratio among the parents’ offspring and the probability that an offspring will have PKU.

[4]
(b)

Explain why the daughter with genotype Pp does not have PKU, and how she could pass allele p to a child who has PKU.

[3]
(c)

Describe the difference in cell division between the two cell samples in the table.

[2]
(d)

Using the information about mutation T, explain how it could contribute to cancer and whether it could be inherited by the daughter’s offspring.

[5]
(e)

Give one limitation of the cell-culture results in deciding whether mutation T caused the higher rate of cell division.

[2]

Question 5

Paper 2Medium6 marks

A 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 conditionsCuttings (asexual), out of 40Seedlings (sexual), out of 40
Stable3628
Hotter and drier825
(a)

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.

[1]
(b)

If conditions remain stable and the cultivar is already suited to them, recommend a propagation method. Explain why it may be advantageous.

[2]
(c)

Explain how sexual reproduction could produce offspring better suited to hotter and drier conditions.

[3]

Question 6

Multiple choiceMedium1 mark

A 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 7

Multiple choiceMedium1 mark

After 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?

Show answer and explanation

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 8

Paper 2Medium16 marks

In 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 colourShell patternNumber at startNumber alive after two weeks
DarkPlain4024
DarkSpeckled4032
PalePlain4032
PaleSpeckled4024

Assume shell pattern is inherited as described and does not change with soil colour. Surviving beetles are equally likely to reproduce.

(a)

Explain how the mutation described could provide heritable variation in the beetle population.

[4]
(b)

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.

[5]
(c)

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.

[7]

Question 9

Multiple choiceMedium1 mark

A 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?

234500.10.20.30.40.50.60.70.80.9Trophic levelMean concentration in tissue / mg kg-1DDTmercury
Show answer and explanation

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 10

Multiple choiceMedium1 mark

A 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?

Show answer and explanation

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 11

Paper 2Medium16 marks

Three 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.

05101501020304050Years since restoration began / yearsMean net increase in carbon stored / t C ha⁻¹○ Non-native tree plantation□ Native forest regeneration△ Restored peat-forming wetland

Table 1

Restoration approachMean number of native plant species per 20 quadrats after 15 years
Non-native tree plantation7
Native forest regeneration19
Restored peat-forming wetland15
(a)

Identify the restoration approach with the greatest increase in carbon stored after 10 years.

[1]
(b)

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.

[2]
(c)

Describe and compare the changes in carbon stored in the three restoration approaches over the 15-year period.

[3]
(d)

Explain how tree-based restoration and restoration of a peat-forming wetland can each increase carbon storage.

[4]
(e)

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.

[4]
(f)

Suggest one limitation of the investigation’s design and explain how it could affect the comparison of restoration approaches.

[2]

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