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)

2026-11-03

DSE Biology: Genetics and Molecular Genetics

DSE Biology: 12 original questions, 50 marks, 75 minutes on base pairing, crosses, sex linkage, transcription, mutation and PCR.

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

Section A: Short Questions (4 marks each)

Answer all questions. Show the working, substitution and units.

#	Question	Answer and marking notes
1	In a double-stranded DNA sample, adenine (A) makes up 30% of all bases. Find (a) the percentage of thymine (T); (b) the percentages of guanine (G) and cytosine (C); (c) the difference in the number of hydrogen bonds between A-T and G-C pairs, and its effect on the stability of DNA. (4 marks)	(a) A pairs with T in equal numbers, so T = 30%. (b) A + T = 60%, leaving 40% shared equally by G and C, so G = C = 20%. (c) An A-T pair has 2 hydrogen bonds while a G-C pair has 3; the higher the G-C content, the more energy is needed to separate the strands, so the DNA is more stable. Marking: 1 mark for each percentage, 1 mark for hydrogen bonds and stability.
2	A double-stranded DNA molecule contains 2000 base pairs, of which 600 are adenine. Find (a) the total number of bases; (b) the number of guanine bases; (c) the total number of hydrogen bonds in this DNA. (4 marks)	(a) Total bases = 2000 x 2 = 4000. (b) A = T = 600, so A + T = 1200 and G + C = 4000 – 1200 = 2800, hence G = C = 1400. (c) Hydrogen bonds = 600 x 2 + 1400 x 3 = 1200 + 4200 = 5400.
3	In peas, tall stem (T) is dominant to short stem (t). Two heterozygous tall plants are crossed. Find (a) the genotypic ratio of the offspring; (b) the phenotypic ratio; (c) the probability that a tall offspring is homozygous. (4 marks)	(a) Tt x Tt gives TT : Tt : tt = 1 : 2 : 1. (b) Both TT and Tt are tall, so tall : short = 3 : 1. (c) Tall plants are 3/4 of the offspring and homozygous TT is 1/4, so the required probability is (1/4) / (3/4) = 1/3.
4	A pea plant heterozygous for both round seeds and yellow cotyledons (RrYy) is self-pollinated, giving 320 seeds. Find the expected number of each of the four phenotypes, and state the law of inheritance involved and the condition for it to hold. (4 marks)	The two gene pairs assort independently, giving a 9 : 3 : 3 : 1 ratio. Round yellow = 320 x 9/16 = 180; round green = 320 x 3/16 = 60; wrinkled yellow = 60; wrinkled green = 320 x 1/16 = 20, totalling 320. This follows Mendel law of independent assortment, which holds when the two gene pairs are on different homologous chromosomes and separate independently in meiosis.
5	The coding region of a gene is 900 base pairs long. Find (a) the number of coding bases in the mRNA transcribed; (b) the number of codons; (c) the number of amino acids in the polypeptide, excluding the stop codon. (4 marks)	(a) The mRNA is transcribed from the template strand, so it has 900 coding bases. (b) Every three bases form a codon, so 900 / 3 = 300 codons. (c) The last codon is a stop codon and codes for no amino acid, so the polypeptide has 300 – 1 = 299 amino acids.
6	A human somatic cell has 46 chromosomes. Find (a) the chromosome number of a gamete; (b) the chromosome number, chromatid number and DNA content of each cell after meiosis I; (c) explain the importance of meiosis in keeping the chromosome number constant between generations. (4 marks)	(a) Meiosis halves the chromosome number, so a gamete has 23 chromosomes. (b) After meiosis I the homologous chromosomes have separated; each cell has 23 chromosomes, each with 2 chromatids, i.e. 46 chromatids, and a DNA content of 2C, the same as the somatic cell. (c) Meiosis halves the chromosome number of gametes and fertilisation restores 46, so the chromosome number stays constant between generations; crossing over and independent assortment also increase genetic variation.

Section B: Application Questions (4 marks each)

For genetics questions, show genotypes, gametes and a Punnett square; write explanations in complete sentences.

#	Question	Answer and marking notes
7	A purple-flowered pea plant is test-crossed with a white-flowered plant (pp); the offspring are 84 purple and 76 white. Find (a) the genotype of the purple parent; (b) explain the observed ratio; (c) predict the offspring if the parent had been PP. (4 marks)	(a) White offspring appear, so the purple parent must carry the recessive allele and is Pp. (b) Pp x pp gives 1 Pp : 1 pp, i.e. purple : white = 1 : 1; the observed 84 : 76 is close to 1 : 1 for a sample of 160, within random error. (c) PP x pp gives only Pp, so all offspring are purple.
8	Colour blindness is controlled by a recessive gene (b) on the X chromosome. A carrier mother (XᴮXᵇ) marries a normal father (XᴮY). Find (a) the probability of a colour-blind son among the children; (b) the probability that a son is colour blind; (c) the probability that a daughter is colour blind, with an explanation. (4 marks)	The four Punnett square combinations are XᴮXᴮ, XᴮXᵇ, XᴮY and XᵇY, each with probability 1/4. (a) Only XᵇY is colour blind, so the probability is 1/4. (b) A son is either XᴮY or XᵇY with equal chance, so half of the sons are colour blind. (c) A daughter always receives Xᴮ from her father and therefore has at least one dominant allele, so the probability is 0.
9	Part of the template strand of a gene reads TAC GCA TTG and is translated into Met-Arg-Asn. If one base pair is deleted in the DNA corresponding to the second codon, explain how the mutation affects the polypeptide and name this type of mutation. (4 marks)	Codons are read continuously in groups of three, so deleting one base pair shifts the reading frame by one base from that point on; every later codon is regrouped, the amino acid sequence changes greatly, and a stop codon may appear early, shortening the polypeptide. This is called a frameshift mutation.

Section C: Long Questions (14 marks in total)

Answer the long questions in points and use correct biological terms.

#	Question	Answer and marking notes
10	The polymerase chain reaction (PCR) starts from one DNA molecule and runs for 30 cycles. Find (a) the final number of DNA molecules, in standard form; (b) the three steps of each cycle and the main role of each step. (4 marks)	(a) Each cycle doubles the number of DNA molecules, so after 30 cycles there are 2^30 = 1,073,741,824, about 1.07 x 10^9 molecules. (b) Denaturation (about 95 C) breaks the hydrogen bonds so the two strands separate; annealing (about 55 C) lets the primers bind to complementary sequences at the ends of the templates; extension (about 72 C) uses a heat-stable DNA polymerase to synthesise new strands from the primers.
11	A 4500 bp plasmid is cut by a restriction enzyme at two sites, giving fragments of 1800 bp and 1200 bp. Find (a) the length of the third fragment; (b) which fragment moves furthest in gel electrophoresis, and explain why. (4 marks)	(a) The third fragment is 4500 – 1800 – 1200 = 1500 bp. (b) The 1200 bp fragment moves furthest. The gel forms a network of pores, and shorter fragments pass through more easily, so they travel further.
12	(a) Describe the complete process from transcription to translation of a gene in a eukaryotic cell (4 marks). (b) Explain how a nonsense mutation affects the protein produced (2 marks).	(a) Transcription: RNA polymerase binds the promoter and unwinds the double helix, then synthesises pre-mRNA from the template strand by complementary base pairing; the pre-mRNA is spliced to remove introns, giving mature mRNA that leaves the nucleus through a nuclear pore. Translation: the mRNA binds to a ribosome and the start codon AUG codes for methionine; tRNA molecules pair their anticodons with the codons and deliver amino acids one by one, the ribosome catalyses peptide bond formation and moves along the mRNA, and the polypeptide is released at a stop codon. (b) A nonsense mutation changes an amino acid codon into a stop codon, so translation stops early, and the shortened protein usually loses its function.

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