DSE Biology Practice Paper: Genetics And Molecular Genetics (12 Questions, 50 Marks, 75 Minutes)

DSE Biology Practice Paper: Genetics and Molecular Genetics (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Biology: Genetics and Molecular Genetics

DSE Biology: 12 questions, 50 marks, 75 minutes on monohybrid and dihybrid crosses, sex linkage, DNA base composition, translation, mutation and Hardy-Weinberg.

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Paper details　Level: S6 | Subject: Biology | Time: 75 minutes | Marks: 50 | Questions: 12

Section A (2 to 3 marks each, 14 marks)

Answer all questions; for genetics questions show genotypes and genetic diagrams.

#	Question	Answer and marking notes
1	Two heterozygous mice (Aa x Aa) are crossed, where A is dominant and a is recessive. (a) State the genotypes and their ratio in the offspring. (b) Find the probability that an offspring shows the recessive phenotype.	(a) The offspring genotypes are AA : Aa : aa = 1 : 2 : 1. (b) Only aa shows the recessive phenotype, so the probability is 1/4 = 0.25 (25%). Marking: genotypes 1 mark; probability 1 mark. Common error: treating 1:2:1 as a phenotypic ratio.
2	A test cross is carried out between a heterozygote (Bb) and a homozygous recessive (bb). (a) Write the genetic diagram with genotypes. (b) Find the probability that an offspring is Bb.	(a) The cross Bb x bb gives Bb : bb = 1 : 1, one half of each. (b) The probability of Bb is 1/2 = 0.5. Marking: genetic diagram 1 mark; probability 1 mark. Common error: confusing a test cross with selfing.
3	A dihybrid cross AaBb x AaBb is made, with the two gene pairs on different chromosomes. (a) State the phenotypic ratio of the offspring. (b) If 320 offspring are produced, find the expected number showing aabb.	(a) The genes assort independently, so the phenotypic ratio is 9 : 3 : 3 : 1. (b) aabb accounts for 1/16, so 320 x 1/16 = 20 offspring. Marking: ratio 1 mark; number 1 mark. Common error: using 1/4 for the double recessive.
4	In the human ABO blood group, I^A and I^B are codominant and i is recessive. The father is I^A i (group A) and the mother is I^B i (group B). (a) List the possible blood groups of the children. (b) Find the probability of a group O child.	(a) The cross I^A i x I^B i gives I^A I^B, I^A i, I^B i and ii, that is blood groups AB, A, B and O. (b) Group O (ii) has probability 1/4 = 0.25. Marking: all four groups 1 mark; probability 1 mark. Common error: omitting group AB.
5	In a double-stranded DNA sample, adenine (A) makes up 30% of all bases. (a) Find the percentage of thymine (T) and state the reason. (b) Find the percentage of guanine (G) and of cytosine (C).	(a) A pairs with T and their numbers are equal on the two strands, so T is also 30%. (b) G + C = 100% – 30% – 30% = 40%, and G = C, so G and C are each 20%. Marking: T with reason 1 mark; G and C 1 mark each. Common error: forgetting G = C.
6	The coding region of a gene (excluding introns) is 900 nucleotides long. (a) Find the number of codons in this region. (b) Excluding the stop codon, find the number of amino acids translated.	(a) Number of codons = 900 / 3 = 300. (b) One codon is a stop codon and codes for no amino acid, so the number of amino acids is 300 – 1 = 299. Marking: codons 1 mark; subtracting the stop codon 1 mark; answer 1 mark. Common error: forgetting to subtract the stop codon.

Section B (3 to 5 marks each, 11 marks)

Show full working; give percentages to three significant figures.

#	Question	Answer and marking notes
7	A recessive inherited disease affects 1 in 10,000 people in a population at Hardy-Weinberg equilibrium. (a) Find the frequency q of the recessive allele. (b) Find the frequency p of the dominant allele. (c) Find the percentage of carriers (heterozygotes) in the population, to three significant figures.	(a) q squared = 1/10000 = 0.0001, so q = 0.01. (b) p = 1 – 0.01 = 0.99. (c) 2pq = 2 x 0.99 x 0.01 = 0.0198, about 1.98%. Marking: q 1 mark; p 1 mark; 2pq and percentage 1 mark. Common error: taking the disease incidence as q instead of q squared.
8	A mother is a carrier of an X-linked recessive disease (X^H X^h) and the father is unaffected (X^H Y). (a) Find the probability that a son is affected and that a daughter is affected. (b) Find the overall probability of an affected child and explain why no daughter is affected.	(a) A son receives X^h from the mother with probability 1/2, so a son is affected with probability 0.5; every daughter receives X^H from the father, so the probability for a daughter is 0. (b) Sons and daughters are equally likely, so the overall probability is 1/2 x 1/2 = 1/4 = 0.25. Marking: son 1 mark; daughter zero with reason 1 mark; overall 1 mark. Common error: assuming half of the daughters are affected.
9	The coding region of an mRNA is 372 nucleotides long and ends with a stop codon. (a) Find the number of codons. (b) Find the number of amino acids in the polypeptide translated. (c) The first base of codon 40 changes from G to A but the amino acid sequence is unchanged. Name this type of mutation and explain why the sequence does not change.	(a) Number of codons = 372 / 3 = 124. (b) After subtracting one stop codon, the number of amino acids is 124 – 1 = 123. (c) This is a silent mutation. The genetic code is degenerate, so several codons code for the same amino acid and the sequence is unchanged. Marking: codons 1 mark; amino acids 1 mark; name and degeneracy 3 marks. Common error: calling a silent mutation a missense mutation.

Section C (Questions 10 to 12, 25 marks)

Long questions: answer in parts and explain the principles you use.

#	Question	Answer and marking notes
10	Two parents with normal phenotypes are both carriers of a recessive disease (Tt x Tt). (a) Find the probability that a child is phenotypically normal. (b) If they plan three children, find the probability that all three are normal, as a percentage to three significant figures. (c) Explain why the proportion of real families with three normal children may differ from the theoretical value.	(a) Each child is normal with probability 3/4 = 0.75. (b) Three normal children = (3/4)^3 = 0.421875, about 42.19%. (c) Each child is an independent event, so the theoretical value is a long-run average; real families are few, so random fluctuation makes the observed proportion differ. Marking: one child 1 mark; product for three 2 marks; percentage 1 mark; independence 3 marks. Common error: multiplying 3/4 by 3 instead of cubing it.
11	A closed population has 50000 people, of whom 500 have a recessive disease, and the population is at Hardy-Weinberg equilibrium. (a) Find q squared and q. (b) Find p and the carrier genotype frequency (2pq). (c) Find the number of carriers in the population. (d) State two conditions for Hardy-Weinberg equilibrium and explain how inbreeding changes the proportion of carriers.	(a) q squared = 500 / 50000 = 0.0100, so q = 0.1. (b) p = 1 – 0.1 = 0.9; 2pq = 0.1800. (c) Carriers = 50000 x 0.1800 = 9000. (d) Conditions: random mating, large population, no mutation, no migration, no natural selection (any two). Inbreeding increases homozygotes, so the proportion of carriers falls and the incidence of the recessive disease rises. Marking: q squared and q 2 marks; p and 2pq 2 marks; number 1 mark; conditions and inbreeding 3 marks. Common error: giving a frequency where a number is asked.
12	The coding region of a bacterial gene has 1200 base pairs and contains no introns. (a) Find the number of codons and the number of amino acids, excluding the stop codon. (b) Two bases are deleted at position 100 of the coding region. Name this type of mutation and find the first amino acid that is changed. (c) Explain why deleting two bases usually affects protein function more than deleting three. (d) Another strain has only a single substitution at position 100 that leaves the amino acid unchanged. Describe its effect on protein function and compare the two cases.	(a) Number of codons = 1200 / 3 = 400; excluding the stop codon there are 399 amino acids. (b) Deleting two bases is a frameshift mutation. Base 100 lies in codon 34, so every amino acid from number 34 onwards is changed. (c) Deleting two bases shifts the reading frame, so all later codons are read differently, the amino acid sequence is largely changed and a stop codon often appears early. Deleting three bases removes one amino acid but keeps the reading frame, so the effect is more local. (d) A substitution that leaves the amino acid unchanged is a silent mutation, so the protein structure and function are usually unaffected; the effect is far milder than a frameshift. Marking: codons and amino acids 2 marks; type of mutation 1 mark; first changed amino acid 2 marks; reading frame 3 marks; silent mutation comparison 2 marks. Common error: thinking a two-base deletion changes only one amino acid.

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