S5 Biology Practice Paper: Genetics, DNA And Gene Expression (12 Questions, 100 Marks, 75 Minutes) With Answers

S5 Biology Practice Paper: Genetics, DNA and Gene Expression (12 Questions, 100 Marks, 75 Minutes) with Answers

2026-11-09

HKDSE Biology: Genetics, DNA and Gene Expression

S5 Biology paper: 12 questions, 100 marks, 75 minutes. Mendelian crosses, DNA replication, transcription, mutation, sex linkage, PCR and genetic engineering. Answers included.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Give the genotype and the phenotype, and show each cross with a Punnett square or a branch diagram.

Give numerical answers to three significant figures unless stated otherwise.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A — Basic Principles of Inheritance

Give the genotype and the phenotype, and show each cross with a Punnett square or a branch diagram.

#	Question	Answer and marking notes
1	In garden peas the tall allele T is dominant to the dwarf allele t. Two heterozygous tall plants (Tt) are crossed and produce 400 offspring. (a) State the gametes produced by each parent and their ratio; (b) use a Punnett square to give the genotypic and phenotypic ratios of the offspring; (c) calculate the expected number of tall offspring; (d) find the proportion of homozygous tall plants among the tall offspring.	Answer: (a) Each parent produces gametes T and t in the ratio 1:1. (b) The genotypes are TT:Tt:tt = 1:2:1, so the phenotypes are tall:dwarf = 3:1. (c) Number of tall plants = 400 × 3/4 = 300. (d) Homozygous tall TT is 1/4 of all offspring, that is 100 plants, so among tall plants the proportion is 100 ÷ 300 = 1/3. Marking: (a) 1 mark for both gametes; (b) 1 mark for the genotypic ratio and 1 mark for the phenotypic ratio; (c) 1 mark for the working and value; (d) 2 marks for recognising a conditional probability and reaching 1/3.
2	A test cross is carried out between a heterozygous tall pea plant (Tt) and a dwarf plant (tt), giving 200 offspring. (a) State the purpose of a test cross; (b) predict the phenotypic ratio of the offspring; (c) calculate how many tall and how many dwarf plants are obtained; (d) state the genotype of the tall parent if all the offspring are tall.	Answer: (a) A test cross determines whether an individual showing the dominant phenotype is homozygous or heterozygous. (b) Tt × tt gives Tt and tt in the ratio 1:1, so tall:dwarf = 1:1. (c) 200 × 1/2 = 100 of each. (d) If every offspring is tall, the parent passed T to all of them, so its genotype must be homozygous TT. Marking: (a) 1.5 marks; (b) 1.5 marks; (c) 1.5 marks for the working and values; (d) 1.5 marks.
3	A double-stranded DNA sample contains 2,000 base pairs, and adenine (A) makes up 30% of all the bases. (a) Find the percentages of thymine (T), guanine (G) and cytosine (C); (b) find the numbers of A and G in the sample; (c) find the total number of hydrogen bonds in the molecule.	Answer: (a) By complementary base pairing A equals T, so T = 30%. The remaining 100% − 60% = 40% is shared equally by G and C, giving 20% each. (b) Total bases = 2,000 × 2 = 4,000, so A = 4,000 × 30% = 1,200 and G = 4,000 × 20% = 800. (c) There are 1,200 A-T pairs with 2 hydrogen bonds each and 800 G-C pairs with 3 each, so the total is 2 × 1,200 + 3 × 800 = 4,800 hydrogen bonds. Marking: (a) 1 mark for T, 1 mark each for G and C; (b) 1 mark for the total number of bases, 0.5 mark each for A and G; (c) 1 mark for the bonding rule and 1 mark for the total.
4	In peas the round allele R is dominant to wrinkled r, and the yellow cotyledon allele Y is dominant to green y. The two genes are on different chromosomes. Two round yellow plants (RrYy) are crossed and give 320 offspring. (a) State the gametes produced by the parents; (b) give the phenotypic ratio of the offspring; (c) calculate the numbers of round yellow and wrinkled green plants; (d) find the number of wrinkled yellow plants.	Answer: (a) The genes assort independently, so the gametes are RY, Ry, rY and ry, 1/4 of each. (b) The phenotypic ratio is round yellow:round green:wrinkled yellow:wrinkled green = 9:3:3:1. (c) Round yellow = 320 × 9/16 = 180 and wrinkled green = 320 × 1/16 = 20. (d) Wrinkled yellow = 320 × 3/16 = 60. Marking: (a) 1.5 marks for the four gametes; (b) 1.5 marks for the ratio; (c) 1 mark for each working and value; (d) 1 mark.

Section B — Molecular Genetics and Gene Expression

Show the working and the unit for each calculation; probabilities may be given as fractions or percentages.

#	Question	Answer and marking notes
5	An mRNA molecule is made of 900 nucleotides and contains one start codon and one stop codon. (a) Find the number of codons; (b) find the number of amino acids in the polypeptide produced; (c) explain the function of the stop codon; (d) state the number of amino acids if 3 codons are skipped during translation.	Answer: (a) Number of codons = 900 ÷ 3 = 300. (b) The stop codon does not code for an amino acid, so the number of amino acids is 300 − 1 = 299. (c) A stop codon is not recognised by any tRNA, so no tRNA enters the ribosome; the ribosome releases the finished polypeptide and translation ends. (d) Three fewer amino acids gives 299 − 3 = 296. Marking: (a) 2 marks; (b) 2 marks with the subtraction of the stop codon explained; (c) 2 marks; (d) 2 marks.
6	A DNA molecule with 4,000 base pairs replicates semi-conservatively in a cell. After 3 rounds of replication: (a) find the number of DNA molecules produced; (b) find the total number of strands; (c) find the number of molecules that still contain an original strand; (d) describe the centrifugation evidence for semi-conservative replication.	Answer: (a) The number doubles each round, so after 3 rounds there are 2³ = 8 molecules. (b) Each DNA has 2 strands, so 8 × 2 = 16 strands. (c) The two original strands are each kept in one daughter molecule, so 2 molecules contain an original strand. (d) DNA labelled with 15N gives one heavy band on centrifugation; after one round in 14N medium all the DNA forms a single intermediate band, showing that every daughter molecule has one old and one new strand; the heavy and light bands appear only after two rounds. Marking: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) 2 marks for the banding sequence and 2 marks for the old-plus-new explanation.
7	The template strand of a gene has the sequence 3 prime TAC CGT AAA 5 prime. (a) Write the mRNA sequence produced and state its direction; (b) give the amino acid sequence (AUG is methionine, GCA is alanine and UUU is phenylalanine); (c) if the second codon of the template strand changes from CGT to CAT, write the new second mRNA codon and name the type of mutation; (d) explain the effect on the polypeptide if one base is inserted before the second codon.	Answer: (a) By complementary base pairing and antiparallel orientation the mRNA is 5 prime AUG GCA UUU 3 prime. (b) The amino acids are methionine, alanine and phenylalanine. (c) The template CAT is transcribed as GUA, a missense mutation, so the second amino acid changes from alanine to valine. (d) Inserting one base shifts the reading frame, so every later codon is re-divided, the amino acid sequence changes greatly and the protein is usually non-functional; this is a frameshift mutation. Marking: (a) 2 marks, 1 for direction and 1 for sequence; (b) 2 marks; (c) 2 marks for the codon and 2 marks for the mutation type; (d) 1 mark for the reading frame and 1 mark for the consequence.
8	Red-green colour blindness in humans is caused by a recessive allele on the X chromosome. A carrier woman (XCXc) marries a man with normal colour vision (XCY) and they have 400 children. (a) Use a Punnett square to give the genotypes and their ratio; (b) find the proportion of sons who are colour blind; (c) estimate the number of colour blind sons among the 400 children; (d) explain whether a daughter can be colour blind.	Answer: (a) The genotypes are XCXC, XCXc, XCY and XcY, 1/4 of each. (b) Half of the sons carry Xc, so 1/2 of the sons are colour blind. (c) There are about 200 sons and half are affected, that is 100. (d) With this cross no daughter can be colour blind, because the father passes his normal XC to every daughter, so each daughter has at least one dominant allele and is either normal or a carrier. Marking: (a) 3 marks; (b) 2 marks; (c) 2 marks for the working and value; (d) 1 mark for the conclusion and reason.

Section C — Genetic Analysis and Biotechnology

State the principle used and the evidence on which your conclusion is based.

#	Question	Answer and marking notes
9	A family pedigree shows that a healthy couple in the first generation had one affected daughter and one healthy son; the healthy son married a healthy woman and they had two affected sons and one healthy daughter. (a) Deduce the mode of inheritance and justify it; (b) give the genotypes of the first-generation couple; (c) find the probability that the son and his wife have another affected child; (d) state what genetic counselling should tell the family.	Answer: (a) Autosomal recessive. Normal parents have an affected daughter; if the allele were X-linked recessive the father of an affected daughter would have to be affected, which the pedigree contradicts, so the gene is autosomal and both parents are carriers. (b) The first-generation couple is Aa × Aa. (c) The son is a healthy carrier Aa, and if his wife is also Aa the chance of an aa child is 1/4. (d) Explain that every pregnancy still carries a 1 in 4 risk, that prenatal testing can be considered, that carriers look normal but can pass the allele on, and that close relatives should not marry. Marking: (a) 1 mark for the mode and 2 marks for the reason; (b) 3 marks; (c) 3 marks; (d) 1 mark.
10	A circular plasmid is cut with a restriction enzyme and analysed by gel electrophoresis, giving five fragments of 850, 600, 450, 300 and 150 base pairs. (a) Find the total length of the plasmid; (b) explain how the distance moved in the gel depends on fragment length; (c) find the ratio of the shortest to the longest fragment; (d) a second sample gives only two fragments of 1,150 and 1,200 base pairs with the same total length; explain the difference.	Answer: (a) Total length = 850 + 600 + 450 + 300 + 150 = 2,350 base pairs. (b) Shorter fragments move further through the gel because the gel is a mesh and small fragments pass through the pores more easily. (c) 150 : 850 = 3 : 17. (d) The same total length with fewer fragments means the second sample has three fewer cut sites, perhaps because point mutations removed existing sites or because those sequences were never present. Marking: (a) 2 marks; (b) 3 marks; (c) 2 marks; (d) 1.5 marks for the difference and 1.5 marks for the reason.
11	Each cycle of the polymerase chain reaction (PCR) has denaturation for 30 seconds, annealing for 30 seconds and extension for 60 seconds, and doubles the amount of target DNA. (a) Find the time for one cycle; (b) find how many times the target DNA is multiplied after 20 cycles; (c) calculate the total time for 30 cycles; (d) explain why PCR needs a heat-stable DNA polymerase.	Answer: (a) One cycle takes 30 + 30 + 60 = 120 seconds, that is 2 minutes. (b) After 20 cycles the amount is 2 to the power 20, that is 1,048,576 times. (c) Total time = 30 × 2 = 60 minutes. (d) Denaturation heats the sample to about 95 degrees to separate the strands, and an ordinary DNA polymerase would denature and lose activity; a heat-stable enzyme such as Taq polymerase stays active at that temperature, so it can be reused each cycle without adding fresh enzyme. Marking: (a) 2 marks; (b) 2 marks for the expression and 2 marks for the value; (c) 2 marks; (d) 2 marks.
12	A plasmid of 4,000 base pairs is cut with one restriction enzyme, a gene of 800 base pairs is inserted and the fragments are joined by DNA ligase. (a) Find the length of the recombinant plasmid; (b) explain why the same enzyme is used on the plasmid and on the gene; (c) explain the role of sticky ends in forming the recombinant DNA; (d) state what happens if the gene is inserted in the opposite direction.	Answer: (a) The recombinant plasmid is 4,000 + 800 = 4,800 base pairs long. (b) The same enzyme produces identical sticky ends on both molecules, so the plasmid and the gene can pair by complementary bases and DNA ligase then forms the phosphodiester bonds. (c) Sticky ends are single-stranded overhangs that hydrogen bond with complementary sequences, holding the two fragments together long enough for ligase to join them. (d) If the gene is inserted the other way round its transcription runs away from the plasmid promoter, so RNA polymerase cannot transcribe it and the gene is not expressed. Marking: (a) 2 marks; (b) 3 marks; (c) 3 marks; (d) 2 marks.

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