HKDSE Biology Practice Paper: Genetics, DNA And Genetic Engineering (12 Questions, 100 Marks, 75 Minutes, With Answers)

HKDSE Biology Practice Paper: Genetics, DNA and Genetic Engineering (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-06

HKDSE Biology: Genetics, DNA and Genetic Engineering

S5 biology practice paper, 12 questions, 100 marks, 75 minutes: monohybrid and dihybrid crosses, test crosses, incomplete dominance and codominance, X-linked inheritance, DNA base compositi

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Section A: Mendelian Inheritance and Crosses (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	Mendel crossed pea plants for seed shape: round (R) is completely dominant over wrinkled (r). A pure-breeding round parent was crossed with a pure-breeding wrinkled parent to give the F1, and the F1 was self-pollinated to give 400 F2 seeds. (a) State the genotype of the F1. (b) Calculate the expected numbers of round and wrinkled seeds in the F2, showing your working. (c) Explain why observed numbers usually differ slightly from the expected numbers.	(a) The parents are RR x rr, producing only R and only r gametes, so all F1 are Rr. (2 marks) (b) Rr x Rr gives 1 RR : 2 Rr : 1 rr, i.e. round : wrinkled = 3 : 1. 400 x 3/4 = 300 round; 400 x 1/4 = 100 wrinkled. (3 marks) (c) Expected values are probabilities, not certainties: fertilisation is random and the sample is finite, so sampling error occurs; the environment may also affect seed survival. (1 mark)
2	A test cross is a standard tool in genetics. (a) Define a test cross. (b) A pea plant shows the dominant round phenotype. Explain how a test cross can decide whether it is RR or Rr, giving the expected phenotypic ratio of the offspring in each case. (c) A test cross gives 48 round and 52 wrinkled seeds. Identify the genotype of the plant and explain your reasoning.	(a) A test cross is a cross between an individual showing the dominant phenotype and a homozygous recessive individual (rr here), used to infer the unknown genotype from the offspring phenotypes. (2 marks) (b) RR x rr gives all Rr, i.e. 100% round; Rr x rr gives 1 Rr : 1 rr, i.e. 1 : 1 round : wrinkled. (2 marks) (c) 48 : 52 is close to 1 : 1, so the plant is heterozygous Rr; it produces R and r gametes in similar numbers, consistent with the law of segregation, and the small deviation is random error. (2 marks)
3	Two gene pairs A/a and B/b are on different chromosome pairs and both show complete dominance. A plant of genotype AaBb is self-pollinated and produces 320 offspring. (a) Calculate the expected number of each of the four phenotypes, showing your working. (b) Explain the genetic origin of the 9:3:3:1 ratio. (c) If the two genes were tightly linked on the same chromosome, explain how the phenotypic ratio of the offspring would change.	(a) The two gene pairs assort independently and each shows 3 : 1, so the combined ratio is 9 : 3 : 3 : 1. 320 x 9/16 = 180; 320 x 3/16 = 60; 320 x 3/16 = 60; 320 x 1/16 = 20; total 180 + 60 + 60 + 20 = 320. (3 marks) (b) During meiosis I the non-homologous chromosomes line up independently, so A/a and B/b alleles are distributed into four gamete types each at 1/4; random fusion of gametes then gives 9:3:3:1. (2 marks) (c) With tight linkage the parental gametes (AB and ab) predominate and recombinants (Ab, aB) are rare, so the offspring are mostly the two parental phenotypes, giving an excess of parental types rather than 9:3:3:1. (1 mark)
4	Compare incomplete dominance with codominance. (a) Give one example of each. (b) A pure red-flowered plant (CRCR) is crossed with a pure white-flowered plant (CWCW); all the F1 are pink. State the phenotypic ratio and the corresponding genotypic ratio of the F2 obtained by selfing the F1. (c) Explain why this example cannot be described by a 3:1 ratio.	(a) Incomplete dominance: the heterozygote shows an intermediate phenotype, e.g. red x white gives pink. Codominance: both alleles are fully expressed in the heterozygote, e.g. blood group AB shows both A and B antigens. (2 marks) (b) F1 is CRCW; selfing gives 1 CRCR : 2 CRCW : 1 CWCW, i.e. red : pink : white = 1 : 2 : 1, identical to the genotypic ratio. (3 marks) (c) The heterozygote is neither of the parental phenotypes but an intermediate, so there is no dominance to merge heterozygotes with one parental class and a 3 : 1 ratio cannot arise. (1 mark)

Section B: DNA, Gene Expression and Genetic Engineering (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	Red-green colour blindness is caused by a recessive allele (b) on the X chromosome. A colour-blind father (XbY) marries a woman who has normal vision but is a carrier (XBXb). (a) Use a Punnett square to list the four possible genotypes of the children and their proportions. (b) State the probability that a son is colour blind and the probability that a daughter is colour blind. (c) Explain why colour blindness is much more common in males than in females in the general population.	(a) The mother XBXb gives XB and Xb eggs; the father XbY gives Xb and Y sperm. Daughters: XBXb (carrier) 1/4 and XbXb (colour blind) 1/4; sons: XBY (normal) 1/4 and XbY (colour blind) 1/4. (4 marks) (b) Half of the sons carry Xb, so the probability is 1/2; half of the daughters are XbXb, also 1/2, while the other half are carriers. (3 marks) (c) Males have only one X chromosome (XY), so a recessive allele is expressed directly with no normal allele to mask it; females need two copies of b to be affected, so the frequency in males is far higher. (2 marks)
6	Base analysis of a double-stranded DNA sample shows that adenine (A) makes up 30% of all bases. (a) Calculate the percentage of thymine (T), guanine (G) and cytosine (C), showing your working. (b) State the base-pairing rule you used. (c) If the fragment is 1,200 base pairs long, calculate the actual number of G bases.	(a) A pairs with T in equal numbers, so T = 30%. (1 mark) The remaining 100% – 30% – 30% = 40% is shared equally by G and C, so G = 20% and C = 20%. (2 marks) (b) Complementary base pairing: A pairs with T by two hydrogen bonds and G pairs with C by three, so in a double helix the amount of A equals T and the amount of G equals C (Chargaff rule). (3 marks) (c) 1,200 base pairs = 2,400 bases; G = 2,400 x 20% = 480. (3 marks)
7	The DNA template strand reads 3-TACGGATTC-5 from left to right. (a) Write the mRNA sequence produced by transcription, labelling the 5 and 3 ends. (b) List the three codons on the mRNA and the amino acid each specifies (AUG = methionine, CCU = proline, AAG = lysine). (c) Compare the effect of a base substitution with that of a base insertion on protein structure, and explain why an insertion is usually more serious.	(a) The template 3-TACGGATTC-5 is complemented by mRNA (A to U, T to A, G to C, C to G) giving 5-AUGCCUAAG-3. (3 marks) (b) Codons: AUG = methionine (start), CCU = proline, AAG = lysine. (3 marks) (c) A substitution changes one codon only and, because the code is degenerate, may still specify the same amino acid (silent mutation), so the effect can be nil; an insertion shifts the reading frame so every later codon is regrouped, the amino acid sequence changes over a long stretch and a stop codon often appears early, so the protein usually loses function. (3 marks)
8	Recombinant DNA technology is used to produce human insulin on a large scale. (a) List the four main steps in the correct order. (b) Explain why the same restriction enzyme must be used to cut both the target gene and the plasmid vector. (c) State two uses of gel electrophoresis in this procedure.	(a) (1) Cut the DNA containing the target gene and the plasmid vector with a restriction enzyme; (2) join the gene to the vector with DNA ligase to form a recombinant plasmid; (3) introduce the plasmid into a host cell such as E. coli (transformation); (4) screen the transformed cells, culture them on a large scale and purify the product. (4 marks) (b) The same enzyme produces identical sticky ends on both DNA molecules, so the cut bases can pair by complementary base pairing and ligase can seal the join; different enzymes give incompatible ends. (3 marks) (c) First, it separates the digested fragments by size to confirm that the target fragment was cut out; second, it checks that the recombinant plasmid carries an insert of the expected size, and the band can be excised to purify the fragment. (2 marks)

Section C: Population Genetics, Ecology and Evaluation (40 marks)

Each question carries 10 marks; give full explanations or evaluations.

#	Question	Answer and marking notes
9	DNA profiling is widely used in forensic science and paternity testing. (a) Explain the principle, including why band patterns differ between individuals (except identical twins). (b) Give two practical applications. (c) A child has five bands; three match the mother and two match the alleged father. Explain how paternity is assessed from this result and state one risk that could lead to a wrong conclusion.	(a) Short tandem repeats (STR/VNTR) vary in repeat number between individuals; PCR amplifies them and gel electrophoresis separates the products by size, giving an individual-specific band pattern. Identical twins share the same pattern because their genomes are identical. (4 marks) (b) Any two: paternity testing, matching crime-scene samples, tissue matching before transplantation, reuniting separated relatives, identifying species or the origin of a sample. (2 marks) (c) Every band in a child must come from either the mother or the father: three maternal bands account for the mother, and if the remaining two both match the alleged father the result supports paternity; any band present in neither parent excludes him. (3 marks) Risk: sample contamination, subjective band scoring or a chance match in a database can mislead, so tests should be repeated and supported by other evidence. (1 mark)
10	A recessive autosomal disease affects 40 people in a community of 1,000. Assume the population is in Hardy-Weinberg equilibrium. (a) Calculate the recessive allele frequency q, the dominant allele frequency p and the number of carriers, showing your working. (b) State two conditions required for Hardy-Weinberg equilibrium. (c) If affected individuals die before reproductive age, explain how the allele frequency would change.	(a) q squared = 40/1000 = 0.04, so q = 0.2; p = 1 – 0.2 = 0.8; carrier frequency 2pq = 2 x 0.8 x 0.2 = 0.32, i.e. 1,000 x 0.32 = 320 carriers. (4 marks) (b) Any two: large population, random mating, no mutation, no migration, no natural selection acting on the trait. (2 marks) (c) Death of affected individuals removes recessive alleles from the gene pool, so q falls; but carriers are unaffected and still pass the allele on, so the decline is very slow and the allele is unlikely to disappear. (4 marks)
11	In a grassland ecosystem, producers fix 50,000 kJ per square metre per year, and the efficiency of energy transfer between trophic levels is 10%. (a) Calculate the energy available each year per square metre to the primary consumers and to the secondary consumers. (b) Explain why food chains rarely have more than four or five trophic levels. (c) Compare the shape of a pyramid of energy with that of a pyramid of biomass, and explain why a pyramid of biomass can be inverted in some aquatic systems.	(a) Primary consumers: 50,000 x 10% = 5,000 kJ; secondary consumers: 5,000 x 10% = 500 kJ per square metre per year. (4 marks) (b) About 90% of the energy at each level is lost through respiration, undigested material and parts not eaten, so only about 10% passes on; energy falls sharply at each step and by the fourth or fifth level it can no longer support a viable population. (3 marks) (c) A pyramid of energy is always upright because energy decreases along the chain, as required by thermodynamics; a pyramid of biomass can be inverted in aquatic systems where phytoplankton have a small standing crop but reproduce and are consumed so rapidly that they support a larger biomass of zooplankton. (3 marks)
12	An insect-resistant genetically modified maize (Bt maize) is planted over a large area. (a) State two potential benefits for farmers or the environment. (b) State two potential risks, each with a specific mechanism. (c) Suggest two field-management measures that reduce gene flow and delay the evolution of pest resistance, and explain the principle behind each.	(a) Any two: lower pesticide use, lower cost and higher yield for farmers, less fungal toxin contamination after insect damage, less pesticide exposure for farm workers. (3 marks) (b) Any two with a mechanism: pests evolve resistance under strong selection by the Bt toxin; Bt pollen carried by wind or insects to wild relatives causes gene flow; non-target insects such as butterfly larvae are harmed by feeding on the pollen; resistant weeds or secondary pests emerge. (4 marks) (c) Measure one: plant a refuge of about 20% non-Bt maize nearby so that susceptible pests mate with resistant ones, diluting resistance alleles. Measure two: keep isolation strips or stagger flowering dates so that pollen cannot reach related species, reducing gene flow; combine with resistance monitoring and rotation. (3 marks)

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