Extensions of Mendelian Genetics

Question 1 (12 marks)

(a) (3 marks)

• Multiple alleles: three alleles ($I^A$, $I^B$, $i$) exist for the gene in the population, though any individual carries only two. (1)
• Codominance: $I^A$ and $I^B$ are both fully expressed in the heterozygote ($I^A I^B$ = type AB), neither masks the other. (1)
• $i$ is recessive to both — combining both concepts in one system. (1)

(b) (4 marks)

• Mother AB → genotype must be $I^A I^B$ (only one possibility). (1)
• Her gametes carry either $I^A$ or $I^B$ — never $i$. (1)
• A type O child is genotype $ii$, requiring an $i$ allele from each parent. (1)
• The AB mother cannot contribute $i$, so she cannot produce a type O child at all. (1)

(c) (5 marks)

• Since the type O child requires $i$ from both parents, and the AB mother cannot supply $i$, no candidate father can produce a type O child with this mother. (2)
• Candidate A ($I^A i$ or $I^A I^A$): even if $I^A i$, child would be A or O only if mother supplied i — she can't. Excluded. (1)
• Candidate B (similar reasoning): Excluded. (1)
• Candidate AB: carries no $i$ either. Excluded. (1)
• (Conclusion: The maternal claim itself is inconsistent with a type O baby — no father fits.)

Question 2 (14 marks)

(a) (3 marks)

• Gametes: CCRR → CR; ccrr → cr. (1)
• F₁ genotype = CcRr. (1)
• Phenotype: C present + Rr → pink. (1)

(b) (7 marks)

• CcRr × CcRr → 9 C_R_ : 3 C_rr : 3 ccR_ : 1 ccrr. (2)
• Assign phenotypes:
  • 9 C_R_ → but R_ splits by incomplete dominance: within C_ , RR : Rr : rr = 1:2:1. (1)
  • Coloured (C_) classes: $\frac{3}{16}$ CCRR/CcRR type = red (RR), $\frac{6}{16}$ = pink (Rr), $\frac{3}{16}$ C_rr = white. (2)
  • cc classes (ccR_ + ccrr = 4/16) → all white (epistasis). (1)
• Final ratio: Red : Pink : White = 3 : 6 : 7. (1)

(c) (4 marks)

• 9:3:3:1 assumes two genes with simple dominance and independent phenotypes. (1)
• Here epistasis (cc masks R) merges the 3 (C_rr) and 4 (cc__) into white. (1)
• Incomplete dominance further splits C_R_ into red/pink instead of one class. (1)
• Net effect: modified ratio 3:6:7 rather than 9:3:3:1. (1)

Question 3 (12 marks)

(a) (3 marks)

• Her father was haemophiliac → his X is $X^h$; he passes his only X to all daughters. (1)
• So she must inherit $X^h$ from him and (being unaffected) $X^H$ from her mother. (1)
• Genotype = $X^H X^h$ (carrier). (1)

(b) (6 marks) Cross: $X^H X^h$ (mother) × $X^H Y$ (unaffected father). (1)

(2)

• Daughters: $\frac{1}{2}$ $X^H X^H$ (normal), $\frac{1}{2}$ $X^H X^h$ (carrier) → all unaffected. (1.5)
• Sons: $\frac{1}{2}$ $X^H Y$ (normal), $\frac{1}{2}$ $X^h Y$ (affected). (1.5)

(c) (3 marks)

• An affected daughter is $X^h X^h$ — she needs $X^h$ from both parents. (1)
• Therefore the father must actually be haemophiliac (contradicting "unaffected" statement); his genotype = $X^h Y$. (2)

Question 4 (14 marks)

(a) (2 marks)

• Parental classes = the two most frequent: $abc$ (410) and $+++$ (400). (1)
• Total parental = 810 (81%), confirming they are non-recombinant. (1)

(b) (6 marks) Recombinants for each pair (a fly is recombinant for a pair if that pair's arrangement differs from parental $abc$/$+++$):

Double crossovers (rarest): $a+c$ (5) and $+b+$ (5) = 10. These recombine the middle gene.

a–b recombination: classes where a,b combos are recombinant = $ab+$(32) + $++c$(38) + $a+c$(5) + $+b+$(5) = 80 $RF_{a-b} = \frac{80}{1000} = 8.0\%$ (2)

b–c recombination: = $a++$(58) + $+bc$(52) + $a+c$(5) + $+b+$(5) = 120 $RF_{b-c} = \frac{120}{1000} = 12.0\%$ (2)

a–c recombination: = $ab+$(32) + $++c$(38) + $a++$(58) + $+bc$(52) = 180 $RF_{a-c} = \frac{180}{1000} = 18.0\%$ (2)

(c) (4 marks)

• Largest distance (a–c = 18) → a and c are outermost; b is in the middle. (1)
• Check: $RF_{a-b} + RF_{b-c} = 8 + 12 = 20$; a–c observed 18. Difference (2 cM) = double crossovers counted twice. (1)
• Map: a —8 cM— b —12 cM— c (a–c = 18 cM directly). (2)

(d) (2 marks)

• Expected DCO = $0.08 × 0.12 = 0.0096 → ×1000 = 9.6$. (0.5)
• Observed DCO = 10.
• Coefficient of coincidence $= 10/9.6 = 1.04$. (1)
• Interference $= 1 - 1.04 = -0.04$ (essentially none / slight negative interference). (0.5)

Question 5 (8 marks)

(a) (4 marks)

• Three gene pairs → number of "dark" alleles ranges 0–6 → 7 phenotypic classes. (2)
• Lightest = aabbcc (0 dark alleles) proportion $= \left(\frac{1}{4}\right)^3 = \frac{1}{64}$. (2)

(b) (4 marks)

• Identical twins share genotype, so genetic potential is identical. (1)
• Environment (UV/sunlight exposure) modifies expression — greater sun → more melanin. (1)
• Polygenic traits show continuous variation highly sensitive to environment (multifactorial). (1)
• Demonstrates phenotype = genotype + environment interaction; genes set a range (norm of reaction). (1)

[
  {"claim":"Q4 a-b recombination frequency = 8%","code":"rec=32+38+5+5; result=(rec/1000==0.08)"},
  {"claim":"Q4 b-c recombination frequency = 12%","code":"rec=58+52+5+5; result=(rec/1000==0.12)"},
  {"claim":"Q4 a-c recombination frequency = 18%","code":"rec=32+38+58+52; result=(rec/1000==0.18)"},
  {"claim":"Q4 coefficient of coincidence approx 1.04","code":"exp=0.08*0.12*1000; obs=10; cc=obs/exp; result=(abs(cc-1.0417)<0.01)"},
  {"claim":"Q5 lightest skin proportion = 1/64","code":"result=(Rational(1,4)**3==Rational(1,64))"},
  {"claim":"Q5 seven phenotypic classes for 3 additive gene pairs","code":"result=((2*3)+1==7)"}
]