A Level H1 Biology Genetics Inheritance Quiz

A-Level Biology H1 Quiz - Genetics Inheritance

Answer Key and Marking Guide

Section A: Multiple Choice

1. A – 1/4 (homozygous recessive = tt, 1 box in Punnett square out of 4)
2. B – lagging strand only
3. D – missense mutation
4. D – 100% (father X^c Y, mother X^C X^C → all daughters are X^C X^c, carriers)
5. C – Crossing over between non-sister chromatids

(5 × 1 = 5 marks)

Section B: Short Answer

6. Answer: DNA ligase catalyses the formation of phosphodiester bonds between adjacent Okazaki fragments / between the newly synthesised DNA fragments, sealing the sugar-phosphate backbone. [2]
   Award [1] for “joins Okazaki fragments” or “seals DNA backbone”; [2] for mentioning phosphodiester bonds.

7. Answer: Genotype is the genetic makeup (combination of alleles) of an organism; phenotype is the observable physical or biochemical characteristics produced by the genotype and environment. [2]
   Accept any appropriate distinction; award [1] for each defined clearly.

8. Answer: Codominance is a condition in which both alleles in a heterozygote are expressed fully in the phenotype. Example: ABO blood group system in humans, where allele Iᴬ and Iᴮ produce both A and B antigens on red blood cells (blood type AB). [2]
   Award [1] for definition, [1] for correct example.

9. Answer: A frameshift mutation (insertion or deletion of nucleotides not in multiples of three) shifts the reading frame, altering all subsequent codons and often introducing a premature stop codon. This usually leads to a completely non-functional protein. A substitution typically changes only one amino acid, which may have a milder effect. [2]
   Award [1] for explaining reading frame shift; [1] for comparison with substitution.

10. Answer: DNA helicase. [2]
    Award [1] if name is misspelt but recognisable.

11. Answer: A test cross is a genetic cross between an individual showing the dominant phenotype but of unknown genotype and an individual that is homozygous recessive. The offspring ratios reveal whether the unknown parent is homozygous dominant or heterozygous. [2]
    Accept any clear explanation; [1] for crossing with homozygous recessive, [1] for determining genotype from offspring.

(6 × 2 = 12 marks)

Section C: Data-Based and Diagram Interpretation

12. Answer: (a) II-2 is heterozygous (Aa). [1]
    (b) II-2 is unaffected, so she must have at least one dominant allele (A). Her daughter (III-1) is unaffected, but her father (II-1) is homozygous recessive (aa) because III-1 has an affected sibling (unseen) – or deduced from pedigree: the presence of affected offspring in her generation indicates she carries the recessive allele. Since her father (I-1) must be Aa (he has an affected child), II-2 could inherit a; thus she is Aa. [2] (All logical deduction accepted.)
    (c) Probability III-1 is a carrier = 2/3. (She is unaffected, her parents are Aa × aa. Offspring: Aa, aa. Unaffected: Aa, Aa. Carrier = Aa → 2/3). [1]

13. Answer: On the lagging strand, DNA polymerase can only synthesise DNA in the 5’→3’ direction, but the strand is opened in the opposite direction. Therefore, synthesis is discontinuous: short fragments (Okazaki fragments) are made and later joined. Okazaki fragments are necessary because the replication fork opens in the 3’→5’ direction on this strand, obliging the polymerase to periodically re-start as more template is exposed. [4]
    Award up to [2] for explaining direction and discontinuous synthesis, [2] for necessity related to fork movement.

14. Answer: (a) Expected ratio for a dihybrid heterozygote self-cross (RrYy × RrYy) is 9:3:3:1. [1]
    (b) Total = 556. Expected: 9/16 × 556 = 312.75 round yellow; 3/16 × 556 = 104.25 round green; 3/16 × 556 = 104.25 wrinkled yellow; 1/16 × 556 = 34.75 wrinkled green.
    χ² = Σ (O–E)²/E
    = (315–312.75)²/312.75 + (108–104.25)²/104.25 + (101–104.25)²/104.25 + (32–34.75)²/34.75
    = (2.25)²/312.75 + (3.75)²/104.25 + (–3.25)²/104.25 + (–2.75)²/34.75
    = 5.0625/312.75 + 14.0625/104.25 + 10.5625/104.25 + 7.5625/34.75
    ≈ 0.0162 + 0.1349 + 0.1013 + 0.2177 ≈ 0.4701.
    Critical value = 7.82. Since 0.47 < 7.82, the difference is not significant; the observed results fit the expected ratio. [3]
    Award [1] for correct expected numbers, [1] for calculation of χ², [1] for correct conclusion with reference to critical value.

15. Answer: RNA polymerase binds to the promoter region of the gene. It unwinds the DNA double helix by breaking hydrogen bonds between complementary bases, exposing the template strand. RNA polymerase moves along the template strand in the 3’→5’ direction, adding free RNA nucleotides complementary to the template (A–U, C–G). It catalyses the formation of phosphodiester bonds between adjacent ribonucleotides, building a single-stranded mRNA molecule in the 5’→3’ direction. The growing mRNA peels away from the template and the DNA rewinds behind. [4]
    Award [1] for binding/promoter, [1] for base pairing, [1] for direction (template 3’→5’, mRNA 5’→3’), [1] for phosphodiester bond formation / rewinding.

16. Answer: The mutation changes the mRNA codon from GAG (for glutamic acid) to GUG (for valine). Glutamic acid is hydrophilic, whereas valine is hydrophobic. This substitution alters the primary structure of the β‑globin polypeptide. In low oxygen, the hydrophobic valine causes haemoglobin molecules to aggregate into long fibres. This changes the quaternary structure, distorting red blood cells into a sickle shape. This reduces oxygen-carrying efficiency and can block capillaries. [4]
    Award [1] for identifying the amino acid change; [1] for contrasting properties (hydrophilic → hydrophobic); [1] for aggregation/fibre formation; [1] for effect on quaternary structure and cell shape.

17. Answer: (a) Restriction endonuclease (restriction enzyme). [1]
    (b) When the same restriction enzyme cuts both the plasmid and the insulin gene, it creates complementary single-stranded overhangs (sticky ends). These sticky ends allow base pairing between the gene and the plasmid, facilitating the insertion and the action of DNA ligase to seal the phosphodiester backbone. [2]
    (c) The antibiotic resistance gene serves as a selectable marker; only bacteria that have taken up the recombinant plasmid will survive and grow on a medium containing the antibiotic, allowing identification of successfully transformed cells. [1]

(6 × 4 = 24 marks)

Section D: Extended Response

18. Answer:
    Mitosis: DNA replicates once (S phase) producing identical sister chromatids; during mitosis, chromatids separate, and each daughter cell receives an identical set of chromosomes. This maintains genetic stability – daughter cells are genetically identical to the parent cell. Important for growth (increasing cell number without genetic change), repair (replacing damaged cells with identical copies), and asexual reproduction (producing clones).
    Meiosis: DNA replicates once, then two divisions: homologous chromosomes separate in anaphase I; sister chromatids separate in anaphase II. This halves chromosome number, generating haploid gametes. Genetic variation arises from: (i) crossing over in prophase I (exchange of alleles between non-sister chromatids), (ii) independent assortment of homologous pairs at metaphase I (random alignment leading to different combinations of maternal and paternal chromosomes in gametes), (iii) random fusion of gametes during fertilisation. Thus meiosis generates genetically unique individuals, which is vital for evolution.
    [6 marks]
    Award: [1] for explanation of mitosis producing identical cells; [1] for roles in growth/repair/asexual repro; [2] for how meiosis generates variation (crossing over and independent assortment) with clear description; [1] for linking variation to evolution; [1] for comparison and concluding statement.

19. Answer:
    (a) Parental genotypes: Female Xᴼ Xᵒ (tortoiseshell), Male Xᴼ Y (orange). Gametes: Female produces Xᴼ, Xᵒ; Male produces Xᴼ, Y.
    Punnett square: | | Xᴼ (male) | Y (male) | |--|---------|----------| | Xᴼ (♀) | Xᴼ Xᴼ (orange ♀) | Xᴼ Y (orange ♂) | | Xᵒ (♀) | Xᴼ Xᵒ (tortoiseshell ♀) | Xᵒ Y (non‑orange ♂) | Phenotypic ratios: 50% orange females, 50% tortoiseshell females; 50% orange males, 50% non‑orange males. [4]
    Award [1] for correct parental genotypes and gametes; [1] for correct Punnett square; [2] for correct phenotypes and ratios.
    (b) Tortoiseshell coat colour requires two X chromosomes with different alleles (Xᴼ and Xᵒ). Males have only one X chromosome (XY), so they can only express the single allele they carry. Thus, a male tortoiseshell would need an extra X chromosome (XXY genotype), which is a rare chromosomal abnormality. [2]
    Award [1] for linking trait to two X chromosomes; [1] for explaining rarity (requires abnormal chromosome complement).

20. Answer:
    Using sickle cell anaemia: A point mutation (substitution) changes the DNA triplet CTC to CAC in the β‑globin gene. This leads to the substitution of glutamic acid (hydrophilic) by valine (hydrophobic) at position 6 of the β‑globin polypeptide (primary structure change). In low oxygen conditions, the hydrophobic valine promotes aggregation of haemoglobin molecules into stiff fibres, which distort red blood cells into a sickle shape. This damages the quaternary structure and reduces the protein’s ability to carry oxygen effectively. Consequences for the organism include anaemia, blockage of capillaries causing pain and organ damage, and increased susceptibility to infections. (Alternatively, using cystic fibrosis: a deletion of three nucleotides (ΔF508) causes loss of phenylalanine, resulting in a misfolded CFTR protein that does not reach the cell membrane, disrupting ion transport. Consequences: thick mucus, lung infections, digestive problems.) [6]
    Award: [1] for naming the mutation type; [1] for describing the change at the DNA and amino acid level; [1] for explaining effect on primary structure; [1] for explaining effect on higher-order structure (quaternary/tertiary) and function; [1] for linking to protein non-function; [1] for at least two consequences.

(3 × 6 = 18 marks)

Total = 5 + 12 + 24 + 18 = 59 marks (note: Section C question 12 answer allocated 4 marks, summing to 24 correct; overall 60 as designed: MCQ 5, Short Answer 12, Data-based 24, Extended 18 = 59; the missing 1 mark can be added by making question 12 worth 5 marks – but the quiz header states /60. Adjust marking scheme: question 12 allocated 4 marks; the total is indeed 59. To make exactly 60, award an extra mark for well-structured explanation in any extended answer or increase question 12 to 5 marks. For simplicity, the quiz states 60 marks and these 59 can be accepted as a design; but ensure total marks = 60 by adding 1 mark to, say, question 18 making it 7 marks. However, to strictly match the quiz header, I’ll allocate 5 marks to question 12: (a) 1, (b) 2, (c) 2 = 5, making Section C total 25 marks. Then total = 5+12+25+18 = 60. I’ll adjust the answer key for question 12 to 5 marks.)

Revised marking for 12: (a) 1 mark; (b) 2 marks; (c) 2 marks (1 for correct probability, 1 for explanation). That yields 5 marks, making total 60.