Noesis
Level 4 — ApplicationMendelian Genetics

Mendelian Genetics

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time Limit: 60 minutes Total Marks: 60

Answer all questions. Show all working, Punnett squares, and reasoning. Assume standard Mendelian inheritance unless stated otherwise.

Question 1 — Monohybrid & Test Cross (12 marks)

In rabbits, coat colour is controlled by a single gene. Black (B) is dominant to brown (b). A breeder has a black rabbit of unknown genotype.

(a) Explain what a test cross is and which genotype you would cross the unknown rabbit with. (3)

(b) The breeder crosses the black rabbit with a brown rabbit and obtains 8 offspring: 4 black and 4 brown. State the genotype of the unknown parent and justify your answer using a Punnett square. (5)

(c) A second, separate cross of the same black rabbit with a brown rabbit produced 6 black offspring and 0 brown. Does this prove the black rabbit is homozygous? Explain your reasoning in terms of probability. (4)

Question 2 — Dihybrid Cross (15 marks)

In pea plants, seed shape (Round R dominant to wrinkled r) and seed colour (Yellow Y dominant to green y) assort independently.

(a) State Mendel's Law of Independent Assortment. (2)

(b) A plant heterozygous for both genes (RrYy) is self-crossed. Using a Punnett square or the product rule, determine the phenotypic ratio of the offspring. (6)

(c) From this cross, out of 320 offspring, how many are expected to be round and green? (3)

(d) What proportion of the round-yellow offspring are expected to be fully homozygous (RRYY)? (4)

Question 3 — Probability, Product & Sum Rules (13 marks)

Two parents are both heterozygous for three unlinked genes: AaBbCc × AaBbCc. Each gene shows complete dominance.

(a) Calculate the probability that a single offspring shows the dominant phenotype for all three traits. (3)

(b) Calculate the probability that an offspring is homozygous recessive at all three loci (aabbcc). (3)

(c) Calculate the probability that an offspring shows the dominant phenotype for exactly two of the three traits. (4)

(d) Using the sum rule, calculate the probability that an offspring is either aaBBCC or AABBcc. (3)

Question 4 — Pedigree Analysis (12 marks)

A pedigree tracks a genetic condition (shaded = affected):

  • Generation I: Male I-1 (unaffected) × Female I-2 (unaffected).
  • Generation II: Their children are II-1 (affected female), II-2 (unaffected male). II-2 marries II-3 (unaffected female, unrelated).
  • Generation III: II-2 × II-3 have III-1 (affected male) and III-2 (unaffected female).

(a) Deduce whether the condition is dominant or recessive, and state one piece of evidence. (3)

(b) Assuming autosomal inheritance, state the genotypes of I-1, I-2, II-2, and II-3 (use D/d). (4)

(c) If II-2 and II-3 have a fourth child, what is the probability it is an unaffected carrier? (3)

(d) Explain why the condition being X-linked recessive is unlikely given this pedigree. (2)

Question 5 — Integrated Reasoning (8 marks)

In a plant species, flower colour is controlled by one gene: Purple (P) dominant to white (p). A gardener crosses two purple plants and obtains 148 purple and 52 white offspring.

(a) Determine the genotypes of the two parent plants, showing your reasoning with the observed ratio. (4)

(b) State Mendel's Law of Segregation and explain how it accounts for the appearance of white offspring from two purple parents. (4)

End of Paper

Answer keyMark scheme & solutions

Question 1 (12 marks)

(a) (3)

  • A test cross is crossing an individual of unknown genotype with a homozygous recessive individual (bb, brown). (2)
  • Rationale: recessive parent contributes only recessive alleles, so offspring phenotypes directly reveal the unknown's alleles. (1)

(b) (5)

  • Cross is Bb × bb (unknown is heterozygous). (1)
b b
B Bb Bb
b bb bb
  • Ratio 1 black : 1 brown = 50% black, 50% brown. (2)
  • The observed 4 black : 4 brown matches the 1:1 ratio, so the unknown must be Bb (heterozygous). If it were BB, all offspring would be black. (2)

(c) (4)

  • No, it does not prove homozygosity. (1)
  • If the rabbit were Bb, each offspring has probability ½ of being black. Probability all 6 are black by chance = (1/2)6=1/64≈0.0156(1/2)^6 = 1/64 \approx 0.0156. (2)
  • Since ~1.6% chance exists that a heterozygote produces 6 black offspring, absence of brown offspring is strong but not conclusive evidence. (1)

Question 2 (15 marks)

(a) (2) Law of Independent Assortment: alleles of different genes segregate/assort into gametes independently of one another (during meiosis), so the inheritance of one trait does not affect another. (2) (Valid for genes on different chromosomes.)

(b) (6)

  • Each gene gives 3:1 ratio on self-cross; combined by product rule:
    • Round : wrinkled = 3:1; Yellow : green = 3:1. (2)
  • Phenotypic ratio = 9 Round Yellow : 3 Round green : 3 wrinkled Yellow : 1 wrinkled green. (4)

(c) (3)

  • Round green proportion = 3/163/16. (1)
  • 320×3/16=60320 \times 3/16 = 60. (2) → 60 offspring.

(d) (4)

  • Among round-yellow (R_Y_) offspring: fraction = 9/169/16 of total. RRYY = 1/161/16 of total. (2)
  • Proportion RRYY within round-yellow = 1/169/16=19\dfrac{1/16}{9/16} = \dfrac{1}{9}. (2)

Question 3 (13 marks)

(a) (3) Each gene: P(dominant phenotype) = 3/4. By product rule: (3/4)3=27/64(3/4)^3 = 27/64. (3)

(b) (3) Each gene: P(homozygous recessive) = 1/4. (1/4)3=1/64(1/4)^3 = 1/64. (3)

(c) (4)

  • P(dominant) = 3/4, P(recessive) = 1/4 per gene.
  • Exactly two dominant = choose which 2 of 3 are dominant: (32)(3/4)2(1/4)1=3×916×14=2764\binom{3}{2}(3/4)^2(1/4)^1 = 3 \times \frac{9}{16}\times\frac{1}{4} = \frac{27}{64} (4)

(d) (3)

  • P(aaBBCC) = (1/4)(1/4)(1/4)=1/64(1/4)(1/4)(1/4) = 1/64. (1)
  • P(AABBcc) = (1/4)(1/4)(1/4)=1/64(1/4)(1/4)(1/4) = 1/64. (1)
  • Mutually exclusive → sum rule: 1/64+1/64=2/64=1/321/64 + 1/64 = 2/64 = 1/32. (1)

Question 4 (12 marks)

(a) (3)

  • The condition is autosomal recessive. (1)
  • Evidence: two unaffected parents (I-1 × I-2) produced an affected child (II-1) — affected offspring from unaffected parents indicates recessive. (2)

(b) (4)

  • I-1 = Dd, I-2 = Dd (both carriers, since they have affected daughter dd). (2)
  • II-2 = unaffected but had affected child (III-1 = dd), so II-2 = Dd. (1)
  • II-3 = unaffected but had affected child → must carry d → Dd. (1)

(c) (3)

  • II-2 (Dd) × II-3 (Dd) → offspring: 1 DD : 2 Dd : 1 dd.
  • Unaffected carrier = Dd = 2/4 = 1/2. (3)

(d) (2)

  • If X-linked recessive, affected daughter II-1 (X^d X^d) would require her father I-1 to be affected (X^d Y). But I-1 is unaffected. (2) (Contradiction rules out X-linked recessive.)

Question 5 (8 marks)

(a) (4)

  • Observed ratio 148:52 ≈ 3:1 (148/52 ≈ 2.85 ≈ 3). (2)
  • A 3:1 ratio arises from Pp × Pp — both parents are heterozygous purple. (2)

(b) (4)

  • Law of Segregation: the two alleles of a gene separate during gamete formation, so each gamete carries only one allele; alleles pair up again at fertilisation. (2)
  • Each Pp parent produces P and p gametes. A p gamete from each parent can combine → pp (white) offspring, explaining white offspring from purple parents. (2)
[
  {"claim":"Q1c: probability 6 black offspring if heterozygous = 1/64","code":"result = (Rational(1,2)**6 == Rational(1,64))"},
  {"claim":"Q2c: round-green offspring out of 320 = 60","code":"result = (320*Rational(3,16) == 60)"},
  {"claim":"Q2d: proportion RRYY within round-yellow = 1/9","code":"result = (Rational(1,16)/Rational(9,16) == Rational(1,9))"},
  {"claim":"Q3a and Q3c both equal 27/64","code":"result = (Rational(3,4)**3 == Rational(27,64)) and (3*Rational(3,4)**2*Rational(1,4) == Rational(27,64))"},
  {"claim":"Q3d: sum of two homozygous classes = 1/32","code":"result = (Rational(1,64)+Rational(1,64) == Rational(1,32))"}
]