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ESAT Challenge Biology Esat-bio-challenge-2

10 questions10 marks20Updated August 2026

The ESAT Challenge Biology Esat-bio-challenge-2 paper in full: all 10 questions, each with its answer. ESAT is the Engineering and Science Admissions Test. Sit it cold under exam timing, mark it, then work back through anything you missed using the solutions below.

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Question 1

1 mark
A genetic condition caused by a recessive allele affects 11 person in 1000010\,000. Assuming the population is in Hardy-Weinberg equilibrium, approximately what proportion of the population are carriers?
  • A.11 in 2525
  • B.11 in 5050
  • C.11 in 100100
  • D.11 in 200200
  • E.11 in 50005000

Answer: B

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Question 2

1 mark
Two unaffected parents have a child with an autosomal recessive condition, and later a second child who is unaffected. That second child has children with an unrelated person who is known to be a carrier. What is the probability that their first child is affected?
  • A.12\tfrac12
  • B.13\tfrac13
  • C.14\tfrac14
  • D.16\tfrac16
  • E.18\tfrac18

Answer: D

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Question 3

1 mark
A fly heterozygous at two linked loci is test crossed. The 10001000 offspring fall into four phenotypic classes, of sizes 415415, 385385, 112112 and 8888. What is the map distance between the two loci?
  • A.1010 cM
  • B.1212 cM
  • C.2020 cM
  • D.4040 cM
  • E.5050 cM

Answer: C

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Question 4

1 mark
Complete oxidation of one mole of glucose releases about 2870kJ2870\,\text{kJ}. Suppose 3232 moles of ATP are synthesised per mole of glucose, and that ATP hydrolysis releases about 30.5kJ mol130.5\,\text{kJ mol}^{-1}. What is the approximate efficiency of energy capture?
  • A.60%60\%
  • B.12%12\%
  • C.24%24\%
  • D.34%34\%
  • E.48%48\%

Answer: D

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Question 5

1 mark
A plant cell has a solute potential of 800kPa-800\,\text{kPa} and a pressure potential of +300kPa+300\,\text{kPa}. It is placed in a solution of water potential 700kPa-700\,\text{kPa}. What happens?
  • A.Water enters the cell and its pressure potential rises.
  • B.There is no net movement of water.
  • C.Solute leaves the cell down its concentration gradient.
  • D.Water leaves the cell and it becomes fully plasmolysed.
  • E.Water leaves the cell and its pressure potential falls.

Answer: E

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Question 6

1 mark
An enzyme obeys rate=Vmax[S]Km+[S],\text{rate}=\frac{V_{\max}[S]}{K_m+[S]}, with Vmax=100μmol min1V_{\max}=100\,\mu\text{mol min}^{-1} and Km=2.0mmol dm3K_m=2.0\,\text{mmol dm}^{-3}. What is the rate, in μmol min1\mu\text{mol min}^{-1}, at a substrate concentration of 6.0mmol dm36.0\,\text{mmol dm}^{-3}?
  • A.2525
  • B.5050
  • C.6060
  • D.7575
  • E.100100

Answer: D

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Question 7

1 mark
A person has a tidal volume of 500cm3500\,\text{cm}^3, a breathing rate of 12min112\,\text{min}^{-1}, and an anatomical dead space of 150cm3150\,\text{cm}^3. Calculate the alveolar ventilation rate in dm3min1\text{dm}^3\,\text{min}^{-1}.
  • A.1.81.8
  • B.4.24.2
  • C.6.06.0
  • D.7.87.8
  • E.4242

Answer: B

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Question 8

1 mark
An ecologist marks and releases 100100 beetles. In a later sample of 120120 beetles, 2020 are marked. It is then discovered that marking makes a beetle slower and therefore easier to catch. Which statement is correct?
  • A.The estimate is 600600, and the true population is probably larger.
  • B.The estimate is 600600, and the true population is probably smaller.
  • C.The estimate is 600600, and it is unbiased.
  • D.The estimate is 240240, and the true population is probably larger.
  • E.The estimate is 24002400, and it is unbiased.

Answer: A

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Question 9

1 mark
A bacterium has a single circular chromosome of 6.0×1066.0\times10^{6} base pairs, replicated from one origin, bidirectionally, with each replication fork advancing at 10001000 base pairs per second. Grown in a rich medium, the same bacterium is observed to divide every 2020 minutes. Which statement is correct?
  • A.One round of replication takes 100100 minutes, so the observed division time must be an error.
  • B.One round of replication takes 5050 minutes, so the forks must in fact move faster than stated.
  • C.One round of replication takes 5050 minutes, and a new round begins before the previous one has finished.
  • D.One round of replication takes 2020 minutes, because replication starts at many origins at once.
  • E.One round of replication takes 5050 minutes, so each daughter cell inherits an incomplete chromosome.

Answer: C

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Question 10

1 mark
Isolated membrane vesicles containing ATP synthase are equilibrated in a buffer at pH 88, then transferred abruptly into a medium at pH 44. A burst of ATP synthesis is observed, with no electron transport taking place at any stage. What does this demonstrate?
  • A.Electron transport is required for ATP synthesis.
  • B.A proton gradient alone is sufficient to drive ATP synthesis.
  • C.ATP synthase catalyses electron transfer directly.
  • D.The ATP was made by substrate-level phosphorylation.
  • E.Low pH denatures ATP synthase, releasing bound ATP.

Answer: B

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ESAT Challenge Biology Esat-bio-challenge-2: Questions & Worked Solutions | esat.fyi