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
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 2
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?
Show answer and explanation
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 3
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?
Show answer and explanation
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 4
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 5
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?
Show answer and explanation
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 6
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 7
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 8
Multiple choiceHard1 markIn an insect population, wing length varies and is inherited. Across generations, individuals with shorter wings leave more offspring, and short wings become more common. Darwin’s contemporaries already accepted that species change; Lamarckism attributed change to traits acquired during an individual’s life and passed to offspring. Which account identifies both Darwin’s mechanism and why his theory was a paradigm shift?
Show answer and explanation
Answer: D. Natural selection acts on existing inherited variation: individuals that leave more offspring contribute more to later generations, so their trait becomes more common in the population. Darwin’s paradigm shift was not the acceptance that species change, which was already widespread; it was providing natural selection as a convincing mechanism in place of Lamarckism. A: uses inheritance of acquired traits and treats acceptance of species change as Darwin’s new contribution. B: gives the selection mechanism correctly but mistakes the already accepted idea of species change for the paradigm shift. C: identifies the historical shift but explains the population change through inheritance of acquired traits rather than differential reproduction. Remember: natural selection changes populations over generations; it does not make individuals acquire adaptive traits.
Question 9
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.
The best questions are still waiting
Unlock every question in this topic, then type, photograph or draw your answer and have it marked against the markscheme. Find out where you are dropping marks.