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S5 Biology Practice Paper: Genetics, DNA and Gene Expression (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Biology: Genetics, DNA and Gene Expression
A 12-question, 100-mark, 75-minute paper on DNA replication, transcription, translation, mutation, gene expression and inheritance, with fully worked solutions.
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Paper details Level: S5 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: DNA Structure and Replication (32 marks)
Questions 1 to 4
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A double-stranded DNA sample contains 1200 base pairs, and adenine (A) makes up 20 percent of all bases. Find (a) the number of A bases, (b) the number of guanine (G) bases, (c) the total number of hydrogen bonds, given that A pairs with T by 2 hydrogen bonds and G pairs with C by 3. | (a) Total bases = 1200 x 2 = 2400, so A = 2400 x 20 percent = 480 bases (2 marks). (b) T is also 480 bases, so G and C together make 2400 – 960 = 1440 bases and G = 720 bases (3 marks). (c) Hydrogen bonds = 480 x 2 + 720 x 3 = 960 + 2160 = 3120 (1 mark). |
| 2 | A bacterial DNA molecule of 600 base pairs is replicated for 3 consecutive rounds in a nitrogen-containing medium. Find (a) the number of DNA molecules produced, (b) the number of molecules that contain an original parental strand, (c) the total number of nucleotides newly synthesised, and explain how this supports semi-conservative replication. | (a) The number of molecules doubles each round, so after 3 rounds there are 2 to the power 3 = 8 molecules (2 marks). (b) Under semi-conservative replication the two parental strands each remain in one molecule, so 2 molecules contain a parental strand (2 marks). (c) New base pairs = 600 x (8 – 1) = 4200 pairs, that is 8400 nucleotides (3 marks). Every daughter molecule has one old and one new strand, which is direct evidence for semi-conservative replication (1 mark). |
| 3 | Part of the template strand of a gene has the sequence 3′-TAC GGA TTC ACG ACT-5′. Find (a) the mRNA sequence transcribed from this template, showing the 5′ and 3′ ends, (b) the number of nucleotides in the mRNA, (c) the number of codons, (d) the number of amino acids obtained if a ribosome translates this stretch completely, naming the amino acid for each codon. | (a) The mRNA is complementary and antiparallel to the template, so it is 5′-AUG CCU AAG UGC UGA-3′ (3 marks). (b) 15 nucleotides (1 mark). (c) 5 codons (1 mark). (d) The stop codon UGA codes for no amino acid, so 4 amino acids are obtained (1 mark): AUG is methionine, CCU proline, AAG lysine and UGC cysteine (3 marks). |
| 4 | A polypeptide consists of 120 amino acids and its mature mRNA includes a start codon and a stop codon. Find (a) the minimum number of nucleotides in the mRNA, (b) the number of tRNA molecules involved in translation, (c) the number of peptide bonds in the polypeptide, (d) the number of water molecules released during its formation. | (a) Three bases code for each amino acid, plus 1 stop codon, so 120 x 3 + 3 = 363 nucleotides (3 marks). (b) One tRNA carries each amino acid, so 120 tRNAs (2 marks). (c) 120 amino acids are joined by 119 peptide bonds (2 marks). (d) One water molecule is released per peptide bond, so 119 water molecules (2 marks). |
Section B: Gene Expression and Protein Synthesis (36 marks)
Questions 5 to 8
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | The coding region of a eukaryotic gene has 5 exons with a total length of 903 base pairs and 3 introns with a total length of 597 base pairs. Find (a) the length of the pre-mRNA, (b) the length of the mature mRNA, (c) the percentage of base pairs removed during splicing, (d) the number of amino acids and peptide bonds produced, assuming the sequence contains 1 stop codon. | (a) Pre-mRNA = 903 + 597 = 1500 bases (2 marks). (b) After the introns are spliced out the mature mRNA has 903 bases (2 marks). (c) Percentage removed = 597 / 1500 x 100 = 39.8 percent (2 marks). (d) 903 / 3 = 301 codons, one of which is a stop codon, so 300 amino acids are produced with 299 peptide bonds (3 marks). |
| 6 | The anticodon of a tRNA is 3′-UAC-5′. Find (a) the codon on mRNA that pairs with it, showing the orientation, (b) the amino acid carried by this tRNA, (c) the number of stop codons and of sense codons out of the 64 codons in the genetic code, (d) explain why cells need only about 45 kinds of tRNA to serve all sense codons. | (a) The anticodon 3′-UAC-5′ pairs antiparallel with the codon 5′-AUG-3′ (2 marks). (b) AUG codes for methionine (2 marks). (c) There are 3 stop codons, so there are 64 – 3 = 61 sense codons (2 marks). (d) The third base of a codon shows wobble pairing, so one tRNA can recognise several synonymous codons and 61 kinds of tRNA are not needed (2 marks). |
| 7 | Part of the coding strand of a gene is 5′-ATG GAA TGC TTT-3′, and its transcript is translated into 4 amino acids. If the second base of the third codon changes from G to A, find (a) the mutant mRNA sequence, (b) the number of amino acids translated from it, (c) the type of mutation (silent, missense or nonsense) with an explanation, (d) the percentage reduction in the length of the protein. | (a) The wild-type sequence is 5′-ATG GAA TGC TTT-3′; changing the second base of the third codon TGC from G to A gives TAC, so the mutant mRNA is 5′-AUG GAA UAC UUU-3′ (3 marks). (b) UAC still codes for an amino acid (tyrosine), so 4 amino acids are still produced (2 marks). (c) This is a missense mutation, because one base substitution makes the codon code for a different amino acid while the length of the protein is unchanged (2 marks). (d) The percentage reduction in length is 0 percent (1 mark). |
| 8 | The expression of a gene is measured by fluorescence: 200 units before induction and 1400 units two hours after adding the inducer. Find (a) the fold increase in expression, (b) the percentage increase, (c) explain how transcription factors regulate gene expression in eukaryotic cells and why this matters for cell differentiation. | (a) Fold increase = 1400 / 200 = 7 (2 marks). (b) Increase = (1400 – 200) / 200 x 100 = 600 percent (2 marks). (c) Transcription factors bind to promoter or enhancer sequences upstream of the gene and help or block RNA polymerase, switching transcription on or off (3 marks). Different cells contain different combinations of transcription factors, so one genome can give different proteins and cell differentiation results (2 marks). |
Section C: Inheritance and Mutation (32 marks)
Questions 9 to 12
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Two heterozygous black mice (Bb) are crossed and produce 32 offspring in a year. Assuming black (B) is completely dominant over white (b), find (a) the gametes produced by each parent and their ratio, (b) the expected ratio of black to white offspring, (c) the expected numbers of black and white offspring, (d) the expected number of homozygous black offspring. | (a) Both parents are Bb, so each produces B and b gametes in a 1:1 ratio (2 marks). (b) The genotypes are 1 BB : 2 Bb : 1 bb, so black to white is 3:1 (2 marks). (c) Black = 32 x 3 / 4 = 24, white = 8 (2 marks). (d) Homozygous black = 32 x 1 / 4 = 8 (2 marks). |
| 10 | A father has blood group A (genotype IAi) and the mother has blood group B (genotype IBi). Find (a) the gametes each parent can produce, (b) the possible blood groups and genotypes of the children, (c) the theoretical ratio of the four blood groups, (d) the probability that a child has blood group O, as a percentage. | (a) The father IAi produces IA and i gametes; the mother IBi produces IB and i gametes (2 marks). (b) The children can be IAIB (AB), IAi (A), IBi (B) or ii (O) (3 marks). (c) The ratio of the four blood groups is 1:1:1:1 (2 marks). (d) The probability of group O is 1 / 4 = 25 percent (1 mark). |
| 11 | Red-green colour blindness is an X-linked recessive trait. A woman with normal vision, whose father was colour blind, marries a man with normal vision. Find (a) the genotype of the woman, (b) the probability that a son is colour blind, (c) the probability that a daughter is colour blind, (d) the expected number of colour-blind children if they have 2 sons and 2 daughters. | (a) The woman has normal vision but her father was colour blind and her X chromosomes come from both parents, so she is a carrier (2 marks). (b) A son receives one of her X chromosomes, so there is a 50 percent chance that he is colour blind (2 marks). (c) A daughter receives the normal X chromosome of her father, so the chance that she is colour blind is 0 percent (2 marks). (d) Of 2 sons 1 is expected to be colour blind and neither daughter is affected, so 1 child is expected to be colour blind (2 marks). |
| 12 | In a population of 1000 people, 160 are affected by a disease caused by an autosomal recessive allele. Assuming Hardy-Weinberg equilibrium, find (a) the frequency q of the recessive allele, (b) the frequency p of the dominant allele, (c) the number of carriers (heterozygotes), (d) the number of homozygous dominant individuals. | (a) q squared = 160 / 1000 = 0.16, so q = 0.4 (2 marks). (b) p = 1 – q = 0.6 (2 marks). (c) Carrier frequency = 2pq = 2 x 0.6 x 0.4 = 0.48, so 0.48 x 1000 = 480 carriers (2 marks). (d) Homozygous dominant frequency = p squared = 0.36, so 360 individuals (2 marks). |
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