Extensions of Mendelian Genetics

Question 1

(a) Cross $I^AI^B\,Rr \times ii\,rr$. (5)

• ABO: $I^AI^B \times ii \Rightarrow \tfrac12\,I^Ai$ (type A), $\tfrac12\,I^Bi$ (type B). (1 for gametes, 1 for ABO ratio)
• Rh: $Rr \times rr \Rightarrow \tfrac12\,Rr$ (Rh+), $\tfrac12\,rr$ (Rh−). (1)
• Type B, Rh+ = $\tfrac12 \times \tfrac12 = \tfrac14 = 25\%$. (2)

(b) (4) ABO codominance: $I^A$ and $I^B$ each encode a functional transferase enzyme adding a different sugar (N-acetylgalactosamine vs galactose) to the H antigen; both products are made and expressed simultaneously → both A and B antigens on red cells (2). Rh: the $R$ allele produces a functional membrane antigen protein; $r$ produces no functional product, so a single $R$ produces enough antigen — the presence/absence of one product gives simple dominance (2).

(c) (7) Under HWE, genotype freqs from $(p+q+s)^2$:

• Type A = $I^AI^A + I^Ai = p^2 + 2ps = 0.09 + 2(0.3)(0.6)=0.09+0.36=0.45$ (2)
• Type B = $q^2 + 2qs = 0.01 + 2(0.1)(0.6)=0.01+0.12=0.13$ (2)
• Type AB = $2pq = 2(0.3)(0.1)=0.06$ (1)
• Type O = $s^2 = 0.36$ (1)
• Sum $=0.45+0.13+0.06+0.36=1.00$ ✓ (1)

(d) (4)

function abo_pheno_freqs(p, q, s):
    assert abs(p+q+s - 1) < 1e-9      # allele freqs sum to 1
    A  = p*p + 2*p*s
    B  = q*q + 2*q*s
    AB = 2*p*q
    O  = s*s
    return (A, B, AB, O)

Assertion: input allele frequencies must sum to 1 (2 for correct formulae, 2 for valid assertion).

Question 2

(a) (6) $BbEe \times BbEe$ → $9\,B\_E\_ : 3\,bb E\_ : 3\,B\_ee : 1\,bbee$.

• $ee$ masks colour → yellow = $B\_ee + bbee = 3+1 = 4$. (2)
• Black = $B\_E\_ = 9$; Brown = $bbE\_ = 3$. (2)
• Ratio 9 black : 3 brown : 4 yellow (recessive epistasis of $ee$ over locus B). (2)

(b)(i) $2\times3 = 6$ additive alleles possible → $k=0,1,2,3,4,5,6$ → 7 classes. (2)

(b)(ii) (6) Each parent is heterozygous at 3 loci → each contributes 3 alleles; per allele, probability of "additive" = $1/2$ independently. Total additive alleles in offspring = sum of 6 independent Bernoulli($1/2$) → Binomial$(n=6,\,p=1/2)$. Fraction with $k$ additive alleles $= \binom{6}{k}(1/2)^6 = \binom{6}{k}/64$. (3 derivation + 3 for binomial argument)

(b)(iii) (3) Darkest ($k=6$): $\binom{6}{6}/64 = 1/64$. Intermediate ($k=3$): $\binom{6}{3}/64 = 20/64 = 5/16$.

(c) (3) Polygenic: one trait controlled by many genes producing continuous variation, e.g. human skin colour/height (1.5). Pleiotropy: one gene affecting many traits, e.g. sickle-cell allele affecting RBC shape, anaemia, malaria resistance (1.5).

Question 3

(a) (6) Total = 1000.

• A–B recombinants = single(A–B) + doubles = $(44+46) + (4+6) = 90 + 10 = 100$. RF(A–B) = $100/1000 = 10\% = 10$ cM. (3)
• B–C recombinants = single(B–C) + doubles = $(48+52) + (4+6) = 100+10 = 110$. RF(B–C) = $110/1000 = 11\% = 11$ cM. (3)

(b) (7) Map (order A–B–C, since doubles are rarest confirming B central):

A ---10 cM--- B ---11 cM--- C

(2 for correct order + distances)

• Observed double recombinants = $10/1000 = 0.01$. (1)
• Expected doubles = RF(A–B)×RF(B–C) = $0.10 \times 0.11 = 0.011$. (1)
• Coefficient of coincidence $C = \dfrac{\text{observed}}{\text{expected}} = \dfrac{0.01}{0.011} = 0.909$. (2)
• Interference $I = 1 - C = 1 - 0.909 = 0.091$. (1)

(c) (4) Directly measuring A–C recombination counts only offspring showing recombination between the outer markers; double crossovers restore the parental arrangement of A and C (though B is swapped), so they are NOT scored as A–C recombinants. Summing the two intervals recovers these hidden double crossovers, giving a total slightly larger than direct A–C RF. Hence map distances are additive but direct long-range RF underestimates.

(d) (3) RF = 50% for genes on different chromosomes: independent assortment produces equal numbers of parental and recombinant gametes ($1:1:1:1$), so recombinants = 50% — the maximum observable RF.

[
  {"claim":"Q1a type B Rh+ = 1/4", "code":"result = (Rational(1,2)*Rational(1,2)) == Rational(1,4)"},
  {"claim":"Q1c ABO phenotype freqs sum to 1", "code":"p,q,s=Rational(3,10),Rational(1,10),Rational(6,10); A=p*p+2*p*s; B=q*q+2*q*s; AB=2*p*q; O=s*s; result = (A==Rational(45,100)) and (B==Rational(13,100)) and (AB==Rational(6,100)) and (O==Rational(36,100)) and (A+B+AB+O==1)"},
  {"claim":"Q2biii darkest=1/64 and intermediate=5/16", "code":"from sympy import binomial; d=binomial(6,6)/64; m=binomial(6,3)/64; result = (d==Rational(1,64)) and (m==Rational(5,16))"},
  {"claim":"Q3 coefficient of coincidence 0.909 and interference approx 0.091", "code":"C=Rational(10,1000)/(Rational(100,1000)*Rational(110,1000)); I=1-C; result = (abs(float(C)-0.909)<0.001) and (abs(float(I)-0.0909)<0.001)"}
]