IB Biology HL 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.

Type
Multiple choiceWritten
Difficulty
EasyMediumHard

Question 1

Multiple choiceMedium1 mark

At a replication fork moving to the right, the upper parental template runs 3′→5′ from left to right, and the lower template runs 5′→3′. Which description of new-strand synthesis is correct?

Show answer and explanation

Answer: D. DNA polymerase extends a new strand at its 3′ end, so synthesis proceeds 5′→3′. The upper template allows continuous synthesis toward the fork from one primer. The lower template is copied away from the fork in Okazaki fragments, requiring repeated primers. A: reverses which template supports leading-strand synthesis toward the fork. B: reverses the direction of DNA synthesis; new DNA is made 5′→3′. C: reverses primer frequency: continuous synthesis uses one primer, whereas discontinuous synthesis requires repeated primers. Remember: the lagging strand is made discontinuously as Okazaki fragments.

Question 2

Paper 2Medium16 marks

A researcher follows one replication fork in Escherichia coli. Figure 1 shows the two parental template strands and newly synthesized strands A and B. The arrow indicates the direction in which the replication fork moves.

upper parental templatelower parental template3′5′5′3′5′3′5′3′A3′5′3′5′3′5′Bdirection of fork movementshaded sections = RNA primers; unshaded sections = newly synthesized DNA
(a)

Using Figure 1, state which of A and B is synthesized continuously and which is synthesized discontinuously. Name the short DNA pieces on B.

[3]
(b)

Explain, using the strand orientations in Figure 1, why A requires one RNA primer whereas B requires repeated RNA primers.

[3]
(c)

A newly synthesized lagging-strand region is 2.4 kb long. The mean length of an Okazaki fragment is 200 nucleotides. For this calculation, use 1 kb = 1000 nucleotides. Estimate the number of Okazaki fragments and the number of RNA primers required.

[2]
(d)

In four separate bacterial preparations, the replication fork has already formed and exactly one enzyme is selectively inactivated. All other replication components remain active.

PreparationObservation
PThe existing leading strand continues to extend, but no new RNA primers are detected on the lagging-strand template.
QRNA primers are present, but no DNA is added to their 3′ ends.
RDNA has been extended from the primers, but RNA primer sections remain in the daughter strands.
SRNA primers have been removed and replaced with DNA, but nicks remain between adjacent lagging-strand fragments.

For each preparation, identify the inactive enzyme and explain how its normal function accounts for the observation.

[8]

Question 3

Multiple choiceEasy1 mark

A cell contains gene X, which codes for an enzyme. Transcription of gene X is switched off. What is directly prevented?

Show answer and explanation

Answer: A. Transcription uses a gene’s DNA sequence to synthesize mRNA. If transcription of gene X is switched off, new mRNA from that gene is not produced; translation uses mRNA to synthesize a polypeptide.
B: confuses transcription with translation by using DNA directly as the template for polypeptide synthesis.
C: confuses gene expression with DNA replication; switching off transcription does not itself prevent DNA copying.
D: confuses transcriptional control with translational control; existing mRNA can still be translated.
Remember: transcription makes mRNA; translation makes a polypeptide from mRNA.

Question 4

Paper 2Easy5 marks

Researchers examine mature mRNA produced from the same gene in two cell types. The exons present in each mRNA are shown.

Cell typeExons in mature mRNA
X1, 2, 4
Y1, 3, 4
(a)

Identify the exon present in the mRNA from cell type X but absent from the mRNA from cell type Y.

[1]
(b)

State the two exons present in both mRNAs.

[2]
(c)

Using the table, explain how this gene can code for different polypeptides.

[2]

Question 5

Multiple choiceMedium1 mark

The diagram shows the base sequences in two cell lineages. Which interpretation correctly identifies the offspring that inherited the germ-cell mutation and the fate of the somatic-cell mutation?

C–G–A–TReference sequenceC–G–G–Tchanged baseMutated sequenceGerm-cell lineageParent germ cell/ C–G–G–TGamete/ C–G–G–TOffspring P: body cells/ C–G–G–TSomatic-cell lineageParent somatic cell/ C–G–G–TTumour cells/ C–G–G–TParent germ cell/ C–G–A–TGamete/ C–G–A–TOffspring Q: body cells/ C–G–A–T
Show answer and explanation

Answer: D. A mutation in a germ cell can pass through a gamete to the offspring, as shown by the mutated sequence in offspring P. The somatic-cell mutation is present in the tumour cells, while the separate germ-cell lineage and offspring Q retain the reference sequence.

A: swaps the offspring labels and incorrectly assigns the somatic mutation to offspring P.
B: selects Q rather than P as the offspring with the inherited germ-cell mutation.
C: reverses the somatic-cell pathway, placing its mutation in offspring Q rather than the tumour cells.
Remember: mutations in germ cells can be inherited; mutations in somatic cells can be associated with cancer but are not passed to offspring.

Question 6

Paper 2Medium16 marks

A gene in a population encodes part of a polypeptide. The table shows the coding-strand DNA sequence from a reference individual and three variants found in other individuals. The mRNA codons and their amino acids relevant to these sequences are also shown.

SequenceCoding-strand DNA, 5′→3′
ReferenceGAA CCT TGG
PGAG CCT TGG
QGAA CAT TGG
RGAA CCT TGC
mRNA codonAmino acid
GAAglutamic acid
GAGglutamic acid
CCUproline
CAUhistidine
UGGtryptophan
UGCcysteine
(a)

For P, Q and R, state the codon and nucleotide substitution. Explain why these differences are SNPs.

[4]
(b)

For each variant, state the amino acid encoded at the changed codon and whether this amino acid differs from that in the reference sequence.

[6]
(c)

Explain why the substitution in P does not change the amino acid.

[3]
(d)

A student concludes that variants Q and R must produce a different phenotype from the reference individuals. Evaluate this conclusion using the information given.

[3]

Question 7

Multiple choiceEasy1 mark

A child with Down syndrome has three copies of chromosome 21. Which meiotic event could produce a gamete with two copies of chromosome 21, which then fuses with a gamete containing one copy?

Show answer and explanation

Answer: D. Nondisjunction is the failure of chromosomes or chromatids to separate during meiosis. This can produce a gamete with two copies of chromosome 21; fusion with a gamete containing one copy gives a zygote with three copies, associated with Down syndrome. A: crossing over exchanges chromosome segments but does not produce an extra copy of chromosome 21. B: a mutation changes DNA sequence, not the number of chromosome 21 copies. C: normal separation gives each gamete one copy of chromosome 21. Remember: nondisjunction can change the number of chromosomes in a gamete.

Question 8

Paper 2Medium16 marks

Figure 1 shows possible chromosome outcomes in a eukaryotic species with a diploid chromosome number of 2n = 4. The parent nucleus has completed DNA replication before either pathway begins. Chromosomes of the same length are homologous; solid and dashed lines distinguish the two homologues in each pair. The central dot on each chromosome represents a centromere. Crossing-over is not shown.

replicated parentnucleusMitosisMeiosisafter meiosis Iafter meiosis II
(a)

State the number of chromosomes and the number of chromatids in the replicated parent nucleus in Figure 1.

[2]
(b)

Using Figure 1, compare the number of nuclei formed and the chromosome number per nucleus after mitosis and meiosis.

[4]
(c)

Explain the role of DNA replication before mitosis and meiosis, referring to chromatids and their separation during these processes.

[4]
(d)

Explain two processes by which meiosis can generate genetically different nuclei. Explain why nuclear division should be completed before cytokinesis.

[6]

Question 9

Multiple choiceEasy1 mark

In a cell, cytosines in the promoter of gene P are methylated, while amino acids in histones in nucleosomes associated with gene H are methylated. Which interpretation of transcription is consistent with these tags?

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Answer: C. Cytosine methylation in a promoter represses transcription of the downstream gene. In contrast, methylation of histone amino acids in nucleosomes can either repress or activate transcription. A: Extends the two possible effects of histone methylation to promoter cytosine methylation. B: Reverses the effect of promoter cytosine methylation, although its histone statement is correct. D: Treats histone methylation as invariably repressive, excluding its possible activating effect. Remember: the effect of methylation depends on whether the tag is on promoter DNA or histones.

Question 10

Multiple choiceEasy1 mark

Four equal samples of cells lacking cell walls were placed in different media. Their volumes after 5 min are shown as percentages of their initial volumes. Which medium would give the cells the greatest risk of bursting?

MediumCell volume after 5 min (% of initial volume)
P130
Q112
R100
S88
Show answer and explanation

Answer: A. Water entering cells causes them to swell, and cells without cell walls can burst in a hypotonic medium. The greatest increase in volume is in P, at 130% of the initial volume, so this medium gives the greatest risk of bursting.

B: recognizes swelling in Q but overlooks that its volume increase is smaller than in P.
C: treats unchanged volume as swelling; it indicates no net change in cell volume.
D: reverses the direction of water movement: the decrease in volume indicates shrinkage, not swelling.

Remember: water entry swells cells without walls; water loss shrinks them.

Question 11

Multiple choiceEasy1 mark

A population of plants is already well adapted to a stable environment. Which reproductive strategy would most directly maintain that adaptation as the population increases?

Show answer and explanation

Answer: C. Asexual reproduction produces genetically identical offspring, so an adaptation suited to the existing environment can be retained as the population increases.

A: confuses fertilization with cloning; fusion of gametes combines alleles from two gametes.
B: assigns meiosis to asexual reproduction; meiosis occurs in sexual life cycles and breaks up parental allele combinations.
D: reverses the role of meiosis, which breaks up rather than preserves parental allele combinations.
Remember: asexual reproduction retains successful combinations, while sexual reproduction generates new combinations that can contribute to variation.

Question 12

Paper 2Medium3 marks

Two individual plants, A and B, are of the same species. An insect visits flowers on both plants. The table shows when pollen is available and when stigmas are receptive.

PlantPollen availableStigmas receptive
ADays 1–2Days 4–5
BDays 4–5Days 1–2
(a)

State which plant’s pollen could be transferred to a receptive stigma on day 1.

[1]
(b)

Explain how the different maturation times and one feature of an insect-pollinated flower promote cross-pollination between the plants.

[2]

Question 13

Multiple choiceEasy1 mark

In pea plants, the P generation is YY × yy. F1 plants are self-pollinated to produce F2 plants; assume the gamete combinations are equally likely. The figure shows the F2 Punnett grid. What fraction of F2 zygotes have genotype Yy?

Allele in pollen gameteYyAllele in ovule gameteYyYYYyYyyy
Show answer and explanation

Answer: B. Each zygote combines one allele from a pollen gamete with one allele from an ovule gamete. Two of the four equally likely grid outcomes are Yy, so the fraction is 2/4.

A: counts only one Yy cell, missing the second combination of these alleles.
C: counts the three cells containing at least one Y, rather than only the heterozygous cells.
D: treats every fertilization as producing a heterozygote, overlooking the YY and yy cells.
Remember: each gamete contributes one allele to the zygote.

Question 14

Paper 2Medium16 marks

In a beetle species, allele A gives a dark body and allele a gives a pale body. Allele B gives long antennae and allele b gives short antennae. A and B are completely dominant to a and b, respectively.

A homozygous dark-bodied, long-antennaed beetle (AABB) is crossed with a homozygous pale-bodied, short-antennaed beetle (aabb). An F₁ beetle is testcrossed with an aabb beetle. The following offspring are recorded.

Offspring phenotypeNumber
Dark body, long antennae416
Pale body, short antennae404
Dark body, short antennae92
Pale body, long antennae88
(a)

State the genotype of the F₁ beetle and the arrangement of its alleles on its homologous chromosomes.

[2]
(b)

Assuming the genes assort independently, determine the gamete types produced by the F₁ beetle and the expected offspring genotypes and phenotypes from the testcross. State the expected phenotypic ratio.

[6]
(c)

Using the observed offspring, identify the recombinant gamete types from the F₁ beetle and their corresponding offspring genotypes and phenotypes.

[4]
(d)

Identify the parental offspring phenotype and genotype classes. Compare the observed results with the expected outcome if the genes assort independently, and infer whether the genes are linked.

[4]

Question 15

Multiple choiceEasy1 mark

After a person’s body temperature rises in a warm room, which response provides water at the skin surface for evaporative cooling?

Show answer and explanation

Answer: A. Sweat released onto the skin can evaporate, removing heat from the body and helping lower its temperature.

B: Vasodilation increases heat transfer to the skin but does not provide the water for evaporation.
C: Shivering contractions generate heat rather than directly increasing evaporative heat loss.
D: Hair erection helps retain an insulating layer of air, reducing heat loss.

Remember: sweating promotes evaporative cooling, whereas vasodilation promotes heat transfer to the skin.

Question 16

Paper 2Hard16 marks

Researchers investigated how active sodium-ion transport in the ascending limb of the loop of Henle relates to water reabsorption in the collecting ducts. They compared three groups of mammals: normal transport, reduced transport, and transport restored after being reduced. Each group contained eight animals. Water intake, the volume of fluid entering the collecting ducts (2.0 mL h⁻¹ per animal), and collecting-duct permeability to water were the same in all groups. The treatment used to reduce transport acted selectively on active sodium-ion transport in the ascending limb. Values are group means.

ConditionRelative rate of active Na⁺ transport in ascending limb (normal = 100)Osmotic concentration at cortex (arbitrary units)Osmotic concentration at outer medulla (arbitrary units)Osmotic concentration at inner medulla (arbitrary units)Water reabsorbed from collecting ducts (%)Urine flow (mL h⁻¹)
Normal transport1003006001000700.60
Reduced transport40300420560421.16
Transport restored90300580960680.64
(a)

Calculate the percentage decrease in osmotic concentration at the inner medulla when transport is reduced compared with normal transport.

[2]
(b)

Compare the osmotic concentration profiles and collecting-duct outcomes in the three conditions.

[3]
(c)

Explain how active sodium-ion transport in the ascending limb can affect water reabsorption from the collecting ducts and urine flow.

[4]
(d)

In a separate test, collecting-duct segments with identical water permeability were used. The osmotic concentration of the fluid inside each segment was 300 units. The surrounding solution had the concentration shown below; its solute could not cross the duct wall. The percentage of the initial water volume inside the segment that crossed out after 10 minutes was measured.

Osmotic concentration of surrounding solution (arbitrary units)Water crossing out of segment (%)
3000
60024
100048

Deduce what this test adds to the interpretation of the first experiment.

[2]
(e)

A researcher concludes: “These results establish that active sodium-ion transport in the ascending limb is the only factor controlling water reabsorption from the collecting ducts.” Evaluate this conclusion using both experiments.

[5]

Question 17

Multiple choiceEasy1 mark

The offspring of a cross between two plants carry combinations of alleles at two genes that were not together in either parent, although every allele was present in at least one parent. What process explains this variation?

Show answer and explanation

Answer: B. Sexual reproduction combines alleles from the parents, creating new combinations of alleles already present. The alleles themselves need not be new for offspring to show a new combination.

A: confuses a new combination of alleles with a new allele, which mutation can produce.
C: treats a change in allele frequency through natural selection as the source of the offspring’s combination.
D: selects transmission from one parent, rather than the combination of alleles from two parents.

Remember: mutation generates new alleles; sexual reproduction generates new combinations of alleles.

Question 18

Paper 2Medium16 marks

In a moth population, wing pattern is controlled by two alleles, B and b. The B allele gives dark wings and is dominant; moths with genotype bb have pale wings. Treat the counts below as the complete populations, so ignore sampling error.

GenerationPopulation information
11000 moths; 160 have pale wings
21000 moths; 400 BB, 400 Bb and 200 bb

Assume generation 1 is in genetic equilibrium. Allele counting in generation 2 gives p = 0.60 and q = 0.40. Records of the mating pairs that produced generation 2 show that moths preferentially mated with partners having the same wing pattern.

(a)

State what p and q represent for this gene and give their relationship.

[2]
(b)

Calculate p and q for generation 1, and the expected frequencies of BB, Bb and bb.

[5]
(c)

Calculate the observed genotype frequencies in generation 2. Determine whether they fit the Hardy–Weinberg equation for the allele frequencies given.

[4]
(d)

Explain how the mating pattern could account for the genotype frequencies in generation 2.

[2]
(e)

State three conditions, other than random mating, required for a population to remain in genetic equilibrium.

[3]

Question 19

Paper 2Medium16 marks

A kelp-forest survey recorded the following data. Sea otters feed on sea urchins, which graze on kelp. Kelp forests provide food and habitat for many species.

YearSea otters counted along a 10 km coastlineSea urchins (individuals m⁻²)Kelp cover (%)
201840882
2021291563
2024162443

Managers also estimated annual replacement and harvest for two renewable resources. Annual replacement is the number of individuals entering the harvestable stock during the year through growth or recruitment.

ResourceAnnual replacementAnnual harvest
Scots pine, a terrestrial plant150 trees year⁻¹180 trees year⁻¹
Atlantic cod, a marine fish30 000 fish year⁻¹24 000 fish year⁻¹
(a)

Describe the patterns in the sea otter, sea urchin and kelp data between 2018 and 2024.

[3]
(b)

Using the data and feeding relationships, explain how a decline in sea otters could affect ecosystem stability and support identifying sea otters as a keystone species.

[4]
(c)

Calculate the percentage of annual replacement harvested for each resource. Show your working.

[4]
(d)

Using your results from (c), assess whether harvesting meets the sustainability criterion for each resource in the year shown.

[3]
(e)

Explain why the results for one year cannot by themselves establish whether each harvest is sustainable over the long term.

[2]

Question 20

Multiple choiceHard1 mark

A record for one region covers four consecutive intervals, each 1 million years long. In each interval, the record identifies the ecosystem type and 10 species groups. The final column shows how many groups are also recorded in the immediately preceding interval. Which conclusion is supported by the complete record?

Time interval (million years ago)Ecosystem typeSpecies groups shared with preceding interval
4–3Forest—
3–2Forest8
2–1Forest3
1–0Forest6
Show answer and explanation

Answer: A. All four consecutive intervals are identified as forest, spanning four million years. Fewer than 10 groups are shared at each transition, so species-group composition changed while the ecosystem type remained forest.
B: Confuses the same number of groups in each interval with the same groups; the overlap values are below 10.
C: Uses species-group turnover to infer ecosystem-type change, although every interval is identified as forest.
D: Stops at the most recent two intervals and overlooks the forest classifications in the earlier intervals.
Remember: continuity of ecosystem type does not require identical species composition.

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