HKDSE Biology Practice Paper: Genetics And Molecular Biology

HKDSE Biology Practice Paper: Genetics and Molecular Biology

2026-11-03

HKDSE Biology: Genetics and Molecular Biology

HKDSE Biology practice paper: genetics and molecular biology, 12 questions, 100 marks, 75 minutes, with full answers and marking points for A4 printing.

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Paper details　Level: S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12

Section A Mendelian inheritance and blood groups (Q1-Q4, 28 marks)

Section A has four compulsory questions.

#	Question	Answer and marking notes
1	Pure-breeding tall pea plants (TT) are crossed with pure-breeding dwarf plants (tt); all F1 plants are tall. The F1 plants are self-crossed to give 480 F2 plants. Calculate the expected number of plants of each genotype in F2 and state the F2 phenotypic ratio. (6 marks)	Answer: F1 genotype is Tt (1 mark). Selfing gives F2 genotypes in the ratio 1 TT : 2 Tt : 1 tt (2 marks), i.e. 120 TT, 240 Tt and 120 tt plants (2 marks). The F2 phenotypic ratio is tall : dwarf = 3 : 1 (1 mark).
2	A tall pea plant of genotype Tt is test-crossed with a dwarf plant, giving 400 offspring. Calculate the expected genotypes, phenotypes and numbers, and explain the use of a test cross in genetics. (7 marks)	Answer: Tt x tt gives 1 Tt : 1 tt (2 marks), i.e. 200 tall and 200 dwarf plants (2 marks); phenotypic ratio 1 : 1 (1 mark). A test cross uses a homozygous recessive partner, so the phenotype ratio of the offspring reveals whether an unknown dominant individual is homozygous (TT, all tall) or heterozygous (Tt, ratio 1 : 1) (2 marks).
3	Two pea plants heterozygous for seed shape and cotyledon colour (RrYy) are crossed, giving 480 offspring. Assuming independent assortment, calculate the expected number of each phenotype and state the phenotypic ratio. (7 marks)	Answer: Independent assortment gives 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (3 marks). Expected numbers: 480 x 9/16 = 270 (1 mark); 480 x 3/16 = 90 (2 marks); 480 x 1/16 = 30 (1 mark).
4	A father has blood group A (genotype IAi) and the mother blood group B (genotype IBi). State the possible blood groups of their children with the probability of each, and explain the relationship between alleles IA and IB. (8 marks)	Answer: Offspring genotypes are IAIB, IAi, IBi and ii, each with probability 1/4 (2 marks); the blood groups are AB, A, B and O respectively, each 25% (3 marks). IA and IB are codominant, so both are expressed when present together, giving group AB (2 marks); both are completely dominant over i (1 mark).

Section B Sex linkage, DNA structure and gene expression (Q5-Q8, 32 marks)

Section B has four compulsory questions.

#	Question	Answer and marking notes
5	Colour blindness is an X-linked recessive condition. A normal carrier woman (XBXb) marries a normal man (XBY). Calculate the probability that a son and a daughter are colour-blind, and explain why the incidence differs between the sexes. (8 marks)	Answer: Among sons, half are XBY (normal) and half are XbY (colour-blind) (3 marks); among daughters, half are XBXB and half are XBXb carriers, none affected (3 marks). Males have only one X chromosome, so one recessive allele is expressed; females have two X chromosomes and need two recessive alleles to be affected (2 marks).
6	A double-stranded DNA sample contains 2000 bases, of which adenine (A) is 30%. Calculate the number of each of the four bases and the total number of hydrogen bonds holding the two strands together. (8 marks)	Answer: A = 2000 x 30% = 600 bases (1 mark); by complementary base pairing A = T, so T = 600 (1 mark); G + C = 2000 – 1200 = 800, so G = C = 400 (2 marks). There are 600 A-T pairs with 2 hydrogen bonds each = 1200 (2 marks) and 400 G-C pairs with 3 hydrogen bonds each = 1200 (1 mark); total 2400 hydrogen bonds (1 mark).
7	One complete turn of the DNA double helix contains 10 base pairs and is 3.4 nm long. A DNA molecule is 34 micrometres long (1 micrometre = 1000 nm). Calculate the number of base pairs and the number of helical turns in the molecule. (8 marks)	Answer: Rise per base pair = 3.4 / 10 = 0.34 nm (2 marks); 34 micrometres = 34000 nm (1 mark); number of base pairs = 34000 / 0.34 = 100000 (3 marks); number of turns = 100000 / 10 = 10000 (2 marks).
8	A messenger RNA molecule has 600 nucleotides, one of which is part of a stop codon. Calculate the number of codons, the number of amino acids coded for and the number of peptide bonds formed. (8 marks)	Answer: Number of codons = 600 / 3 = 200 (2 marks); one is a stop codon that codes for no amino acid, so 199 amino acids are coded for (3 marks); number of peptide bonds = 199 – 1 = 198 (3 marks).

Section C Mutation, population genetics and linkage (Q9-Q12, 40 marks)

Section C has four compulsory questions.

#	Question	Answer and marking notes
9	The transcribed template strand of a gene contains 900 bases. Calculate the number of bases in the messenger RNA, the total number of bases in the double-stranded gene, and the maximum number of amino acids coded for (assume one stop codon). (10 marks)	Answer: The messenger RNA has 900 bases (2 marks); the double-stranded gene has 900 x 2 = 1800 bases (2 marks); there are 900 / 3 = 300 codons (2 marks); allowing one stop codon, at most 299 amino acids are coded for (2 marks), joined by 298 peptide bonds (2 marks).
10	Compare the effects of a base-pair substitution and a base-pair deletion on protein structure, and explain why a substitution may leave the protein unchanged. (10 marks)	Answer: A substitution affects only one codon and may change one amino acid (2 marks); the genetic code is degenerate, so several codons may code for the same amino acid and the substitution can be silent, leaving the protein unchanged (3 marks). A deletion shifts the reading frame (2 marks), so every codon after the deletion is regrouped (2 marks) and the amino acid sequence is extensively altered, usually abolishing protein function (1 mark).
11	A recessive disorder affects 9% of a population that is in Hardy-Weinberg equilibrium. Calculate the frequency of the recessive allele q, the dominant allele p, and the proportions of carriers and of the dominant phenotype. (10 marks)	Answer: q squared = 0.09, so q = 0.3 (2 marks); p = 1 – 0.3 = 0.7 (2 marks); carrier frequency 2pq = 2 x 0.7 x 0.3 = 0.42, i.e. 42% (3 marks); dominant phenotype = p squared + 2pq = 0.49 + 0.42 = 0.91, i.e. 91% (3 marks).
12	Two genes in Drosophila are linked. A heterozygote is test-crossed with a double recessive individual, and 120 of 1000 offspring are recombinant. Calculate the recombination frequency and the map distance, and explain the significance of a frequency below 50%. (10 marks)	Answer: Recombination frequency = 120 / 1000 x 100% = 12% (3 marks); map distance = 12 cM (2 marks). A frequency below 50% shows that the two genes are on the same chromosome, i.e. linked (3 marks); crossing over between homologous chromosomes produces recombinants only when it occurs between the genes, so linked genes give fewer recombinants than the 50% expected under independent assortment (2 marks).

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