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.

Question 1

Multiple choiceEasy1 mark

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

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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 2

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 3

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?

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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 4

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 5

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 6

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?

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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 7

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
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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 8

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?

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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 9

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 10

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?

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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 11

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.

Question 12

Multiple choiceEasy1 mark

A reef exposed to a period of higher-than-usual seawater temperatures shows loss of colour in its corals. Which climate-change effect does this indicate?

Show answer and explanation

Answer: A. Increased water temperature is a cause of coral bleaching.

B: Confuses the effect of warming with ocean acidification, which is caused by increased carbon dioxide. C: Selects suppressed calcification, which is associated with increased carbon dioxide rather than warming. D: Gives an ecosystem-level consequence of coral loss, not the effect indicated here. Remember: warming causes bleaching; increased carbon dioxide causes ocean acidification and suppresses calcification.

Question 13

Paper 2Hard16 marks

Scientists investigated the effects of carbon dioxide concentration and water temperature on coral fragments. Fragments were kept for 10 weeks under four conditions. Other conditions were kept constant, and each condition was tested in six replicate tanks. The table gives mean results.

CO₂ conditionSeawater pHTemperature / °CMean calcification rate / mg CaCO₃ g⁻¹ day⁻¹Fragments showing bleaching / %
Ambient8.1274.05
Elevated7.8272.66
Ambient8.1303.948
Elevated7.8302.550
(a)

Calculate the percentage decrease in mean calcification rate when CO₂ is elevated at 27 °C.

[2]
(b)

Compare the effects of elevated CO₂ and increased temperature on calcification and bleaching.

[4]
(c)

A separate survey was carried out at one reef in Year 0 and again in Year 12. No comparison reef was surveyed. Fish species were grouped as those recorded using living coral for shelter and those not recorded doing so; the two groups together make up total fish species richness.

MeasureYear 0Year 12
Mean seawater pH8.17.8
Maximum summer water temperature / °C2730
Live coral cover / %5219
Algal cover / %1037
Total reef fish species2818
Fish species using living coral for shelter167
Fish species not recorded using living coral for shelter1211

Compare the changes in the two fish groups from Year 0 to Year 12, using data.

[2]
(d)

Evaluate the claim that rising CO₂ and water temperature could cause coral loss and lead to collapse of the reef ecosystem. Use evidence from both studies and give one limitation of the evidence.

[8]

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