A Level H2 Biology Genetics Inheritance Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=tencent/hy3:free; model_label=Tencent HY3 Free; generated=2026-07-09; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

A-Level Biology H2 Quiz - Genetics Inheritance

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 60

Duration: 75 minutes
Total Marks: 60

Instructions:

• Answer all 20 questions.
• Section A: Short structured questions (1–8). Section B: Data and cross analysis (9–14). Section C: Extended reasoning (15–20).
• Show all working for calculation questions.
• Use clear biological terminology.

---

Section A: Short Structured Questions (1–8)

1. Define the term codominance and give one named example in human genetics. [2]

---

---

2. State the genotype of a human male with normal colour vision, given that the allele for colour blindness ($X^c$) is recessive to the normal allele ($X^C$). [1]

---

3. In a test cross, an organism showing a dominant phenotype is crossed with a homozygous recessive individual. Explain why this is done. [2]

---

---

4. A gene undergoes a substitution mutation. State one possible effect on the encoded protein. [1]

---

5. Distinguish between discontinuous and continuous variation, giving one example of each. [2]

---

---

6. What is the purpose of the chi-squared ($\chi^2$) test in genetics? [1]

---

7. In meiosis, explain the importance of crossing-over between homologous chromosomes. [2]

---

---

8. Define epistasis and state the typical modified dihybrid ratio produced by recessive epistasis. [2]

---

---

---

Section B: Data and Cross Analysis (9–14)

9. In Drosophila, red eye ($R$) is dominant to white eye ($r$) and the gene is X-linked. A red-eyed female heterozygous for the allele is crossed with a red-eyed male.
(a) State the genotypes of the parents. [1]
(b) Give the expected phenotypic ratio of offspring. [2]

(a) _______________________________________________________
(b) _______________________________________________________

10. In mice, coat colour depends on two genes, $A$ and $B$. Functional pigment requires at least one dominant allele at both loci (complementary epistasis). A cross $AaBb \times AaBb$ yields 9 coloured : 7 albino.
(a) State the genotypes that produce coloured coat. [1]
(b) Calculate the expected fraction of albino offspring. [2]

(a) _______________________________________________________
(b) _______________________________________________________

11. A population of 1000 plants has 160 showing a recessive dwarf phenotype ($tt$). Assume Hardy–Weinberg equilibrium.
(a) Calculate the frequency of the recessive allele $t$. [2]
(b) Calculate the frequency of heterozygous tall plants ($Tt$). [2]

(a) _______________________________________________________
(b) _______________________________________________________

12. The table below shows observed and expected numbers for a genetic cross expecting a 3:1 ratio from 200 offspring.

Phenotype	Observed	Expected
Dominant	155	150
Recessive	45	50

Using $\chi^2 = \sum \frac{(O-E)^2}{E}$, calculate $\chi^2$ to 2 decimal places. [3]

---

---

13. A couple are both carriers of the sickle cell allele ($Hb^A Hb^S$).
(a) Draw a genetic diagram to show the possible genotypes of their children. [3]
(b) State the probability of a child having sickle cell anaemia. [1]

(a) _______________________________________________________
(b) _______________________________________________________

14.

Image pending generation: graph for Q14.

With reference to Fig. Q14-fig1, identify which individual is heterozygous for sickle cell trait and explain your reasoning. [3]

---

---

---

Section C: Extended Reasoning (15–20)

15. Explain how non-disjunction during meiosis can lead to Down syndrome (trisomy 21). Include the stage at which it may occur. [4]

---

---

---

16. Describe the process of semi-conservative DNA replication and state the role of DNA polymerase. [4]

---

---

---

17. A plant species shows incomplete dominance for flower colour: $RR$ red, $Rr$ pink, $rr$ white. A pink-flowered plant is selfed.
(a) State the expected genotype and phenotype ratios. [3]
(b) Explain why this differs from a typical Mendelian monohybrid ratio. [2]

(a) _______________________________________________________
(b) _______________________________________________________

18. Discuss how environmental factors can affect phenotype using one genetic example. [4]

---

---

---

19. Compare and contrast autosomal linkage and sex linkage, with respect to inheritance patterns. [4]

---

---

---

20. The gene for fur colour in a mammal shows dominant epistasis: $Y\_$ yields yellow regardless of $B/b$; $yyB\_$ black; $yybb$ brown. A $YyBb \times YyBb$ cross produces 12 yellow : 3 black : 1 brown.
(a) Explain the genetic basis of this ratio. [4]
(b) State the expected number of brown offspring from 64 total. [1]

(a) _______________________________________________________
(b) _______________________________________________________

<!-- TuitionGoWhere generation metadata: stage=5-1; model=tencent/hy3:free; model_label=Tencent HY3 Free; generated=2026-07-09; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

A-Level Biology H2 Quiz - Genetics Inheritance (Answer Key)

Total Marks: 60
Topic: Genetics & Inheritance (Syllabus 9477 Core Idea 2)

---

Section A Answers (1–8)

1. [2 marks]

• Codominance: both alleles in a heterozygote are fully expressed, neither is dominant or recessive. [1]
• Example: ABO blood group $I^A I^B$ gives blood group AB where both A and B antigens present. [1]
  Teaching note: Contrast with incomplete dominance (blended). Mark both parts separately.

2. [1 mark]

• $X^C Y$
  Teaching note: Male is XY; normal vision needs $X^C$ on his single X.

3. [2 marks]

• To determine the unknown genotype of the dominant-phenotype individual. [1]
• If any recessive offspring appear, the parent must be heterozygous; if all dominant, likely homozygous. [1]
  Teaching note: Test cross reveals hidden recessive allele.

4. [1 mark]

• May cause a single amino acid change (missense) or no change (silent) due to degenerate code.
  Accept: nonsense (stop codon) also valid. [1]

5. [2 marks]

• Discontinuous: few distinct categories (e.g., blood group). [1]
• Continuous: range of phenotypes from many genes + environment (e.g., height). [1]

6. [1 mark]

• To test whether observed ratios differ significantly from expected (null hypothesis). [1]

7. [2 marks]

• Crossing-over exchanges segments between non-sister chromatids. [1]
• Increases genetic variation in gametes. [1]

8. [2 marks]

• Epistasis: one gene masks expression of another gene. [1]
• Recessive epistasis typical ratio: 9:3:4. [1]

---

Section B Answers (9–14)

9. [3 marks]
(a) Female: $X^R X^r$; Male: $X^R Y$ [1]
(b) Offspring: females all red-eyed ($X^R X^R$, $X^R X^r$); males 1/2 red ($X^R Y$), 1/2 white ($X^r Y$). Overall: 2 red female : 1 red male : 1 white male, or phenotype ratio red : white = 3:1 with sex split. [2]
Mark: (a) 1, (b) 2

10. [3 marks]
(a) $A\_B\_$ (i.e., $AABB, AABb, AaBB, AaBb$) [1]
(b) Albino = $3/16 + 3/16 + 1/16 = 7/16$ [2: 1 for correct complement, 1 for sum]
Teaching: Complementary needs both dominants.

11. [4 marks]
(a) Recessive phenotype $tt = q^2 = 160/1000 = 0.16$; $q = \sqrt{0.16} = 0.4$. [2]
(b) $p = 1 - 0.4 = 0.6$; heterozygote $2pq = 2 \times 0.6 \times 0.4 = 0.48$ (48%). [2]
Show formulas clearly.

12. [3 marks]
$\chi^2 = \frac{(155-150)^2}{150} + \frac{(45-50)^2}{50} = \frac{25}{150} + \frac{25}{50} = 0.1667 + 0.5 = 0.67$ (2 dp). [3: 1 for each term, 1 for total]
Teaching: Use given formula; small $\chi^2$ means no significant difference.

13. [4 marks]
(a) Genetic diagram:
Parents: $Hb^A Hb^S \times Hb^A Hb^S$
Gametes: $Hb^A, Hb^S$ each
Offspring: $Hb^A Hb^A$ (normal), $Hb^A Hb^S$ (carrier), $Hb^S Hb^S$ (sickle cell anaemia) in ratio 1:2:1. [3]
(b) Probability = 1/4 or 25%. [1]

14. [3 marks]

• Individual III is heterozygous. [1]
• Shows two bands at 20 mm (HbA) and 35 mm (HbS), indicating both alleles present. [2]
  Image note: Bands at stated mm; heterozygote has both.

---

Section C Answers (15–20)

15. [4 marks]

• Non-disjunction = failure of homologous chromosomes (Anaphase I) or sister chromatids (Anaphase II) to separate. [2]
• Gamete receives two copy of chr21; fertilisation gives trisomy 21 (47, +21). [2]
  Marking: stage 2, consequence 2.

16. [4 marks]

• Strand separation by helicase; each acts as template. [1]
• DNA polymerase adds complementary nucleotides 5'→3'. [1]
• Semi-conservative: each new DNA has one old and one new strand. [1]
• DNA polymerase also proofreads. [1]

17. [5 marks]
(a) $Rr \times Rr$: genotypes 1 $RR$ : 2 $Rr$ : 1 $rr$; phenotypes 1 red : 2 pink : 1 white. [3]
(b) Incomplete dominance means heterozygote is intermediate, not dominant, so 1:2:1 not 3:1. [2]

18. [4 marks]

• Example: temperature affects fur colour in Himalayan rabbits (cool extremities dark). [2]
• Genotype fixed but phenotype varies with environment. [2]
  Accept other valid examples.

19. [4 marks]

• Autosomal linkage: genes on same non-sex chromosome, inherited together, fewer recombinants. [2]
• Sex linkage: gene on X/Y, shows sex-biased inheritance (e.g., more males affected). [2]

20. [5 marks]
(a) $Y$ dominant epistatic masks $B/b$. $Y\_$ = 3/4 yellow (12/16); $yyB\_$ = 3/16 black; $yybb$ = 1/16 brown → 12:3:1. [4]
(b) Brown = 1/16 of 64 = 4. [1]