Mendelian Genetics

Question 1

(a) (4 marks)

• Law of Independent Assortment: Alleles of different genes segregate independently of one another during gamete formation (so the inheritance of one gene does not affect another). (2)
• Cellular basis: during meiosis I, homologous chromosome pairs align at the metaphase plate in random orientations; genes on different (non-homologous) chromosomes assort independently, making the transmission of alleles at each locus statistically independent — hence the product rule applies. (2)

(b) (8 marks) Each single-gene $Aa \times Aa$ cross gives:

• $P(\text{dominant phenotype}) = \tfrac{3}{4}$, $P(\text{recessive}) = \tfrac{1}{4}$. (1 for stating both)

By the product rule (independence):

• (i) purple & tall $= \tfrac34 \times \tfrac34 = \tfrac{9}{16}$ (1)
• (ii) purple & short $= \tfrac34 \times \tfrac14 = \tfrac{3}{16}$ (1)
• (iii) white & tall $= \tfrac14 \times \tfrac34 = \tfrac{3}{16}$ (1)
• (iv) white & short $= \tfrac14 \times \tfrac14 = \tfrac{1}{16}$ (1)

Ratio $= \tfrac{9}{16}:\tfrac{3}{16}:\tfrac{3}{16}:\tfrac{1}{16} = 9:3:3:1$. (2 for correct proof) Sum check: $9+3+3+1=16 \Rightarrow$ probabilities sum to $1$. (1)

(c) (4 marks) For one locus $Aa\times Aa$: offspring genotypes $\tfrac14 AA, \tfrac12 Aa, \tfrac14 aa$. $P(\text{homozygous at that locus}) = \tfrac14 + \tfrac14 = \tfrac12$. (2) By independence, homozygous at both loci $= \tfrac12 \times \tfrac12 = \tfrac14$. (2)

(d) (6 marks) "Heterozygous at at least one locus" is the complement of "homozygous at both loci."

• $P(\text{homozygous at both}) = \tfrac14$ (from part c). (2)
• $P(\text{het at ≥1 locus}) = 1 - \tfrac14 = \tfrac34$. (3)
• The claim of $\tfrac12$ is FALSE; correct value is $\tfrac34$. (1)

Question 2

(a) (6 marks)

from fractions import Fraction
 
def punnett_ratio(p1, p2):
    counts = {}
    for a in p1:
        for b in p2:
            geno = ''.join(sorted([a, b], key=str.lower))
            counts[geno] = counts.get(geno, 0) + 1
    total = sum(counts.values())        # = 4
    return {g: Fraction(c, total) for g, c in counts.items()}

• Iterate over allele combinations (each parent contributes 1 of 2 gametes → 4 equally likely). (2)
• Canonicalise genotype ordering (e.g. "aA"→"Aa"). (2)
• Return fractions summing to 1. (2) Example: punnett_ratio("Aa","Aa") → {'AA':1/4, 'Aa':1/2, 'aa':1/4}.

(b) (5 marks)

from math import gcd
def phenotype_ratio(p1, p2):
    genos = punnett_ratio(p1, p2)        # fractions
    dom, rec = 0, 0
    for g, prob in genos.items():
        if any(ch.isupper() for ch in g):   # at least one dominant allele
            dom += prob
        else:
            rec += prob
    # convert to integer ratio over denom 4
    d, r = int(dom*4), int(rec*4)
    g = gcd(d, r)
    return (d//g, r//g)

• Classify each genotype: dominant phenotype if any uppercase allele present. (2)
• Sum probabilities and reduce. (2)
• Output for $Aa\times Aa$: dominant $=\tfrac34$, recessive$=\tfrac14$ → ratio $(3,1)$. (1)

(c) (6 marks) Expected under $1:1$ for $N=200$: $E=100$ purple, $100$ white. (1) $\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(102-100)^2}{100} + \frac{(98-100)^2}{100} = \frac{4}{100}+\frac{4}{100}=0.08$ (3) Since $0.08 < 3.84$ (df$=1$), we fail to reject the null. (1) The data are consistent with the unknown being $Aa$ (test cross ratio $1:1$). (1)

(d) (5 marks)

• The recessive parent ($aa$) can only donate a recessive allele, so every offspring's phenotype is determined solely by the allele from the unknown parent. (2)
• Thus offspring phenotypes are a direct 1:1 map to the unknown's gametes: all-dominant offspring ⇒ unknown is $AA$; a $1:1$ ratio ⇒ unknown is $Aa$. (2)
• A heterozygous test partner ($Aa$) contributes dominant alleles too, masking the recessive contribution — dominant offspring could arise from either parent's dominant allele, so the ratio no longer uniquely reveals the unknown genotype. (1)

Question 3

(a) (5 marks)

• II-2 and II-3 are both unaffected yet produce affected III-1. (1)
• If the trait were dominant, at least one parent would have to carry (and thus express) the dominant allele — contradiction, since both are unaffected. (2)
• Therefore the trait is recessive; the affected child is $aa$, receiving one $a$ from each parent, so both II-2 and II-3 must carry $a$ while being unaffected ⇒ genotype $Aa$ (carriers). (2)

(b) (4 marks) III-2 is unaffected, parents both $Aa$. Offspring of $Aa\times Aa$: $\tfrac14 AA:\tfrac12 Aa:\tfrac14 aa$. (1) Conditioning on "unaffected" (not $aa$): eligible outcomes $AA(\tfrac14)$ and $Aa(\tfrac12)$, total $\tfrac34$. (1) $P(Aa\mid \text{unaffected}) = \frac{1/2}{3/4} = \frac{2}{3}$ (2)

(c) (7 marks) III-2 partner is $Aa$. Child affected ($aa$) requires an $a$ allele from each parent.

• $P(\text{III-2 is }Aa)=\tfrac23$; if $Aa$, she passes $a$ with prob $\tfrac12$. If $AA$ (prob $\tfrac13$), she passes $a$ with prob $0$. (2)
• Partner ($Aa$) passes $a$ with prob $\tfrac12$. (1)
• $P(\text{affected}) = P(\text{III-2 passes }a)\times P(\text{partner passes }a)$ $= \left(\tfrac23\cdot\tfrac12 + \tfrac13\cdot 0\right)\times \tfrac12 = \tfrac13 \times \tfrac12 = \tfrac{1}{6}$ (3)
• Answer: $\dfrac{1}{6}$. (1)

[
  {"claim":"9:3:3:1 dihybrid probabilities sum to 1", "code":"p=[Rational(9,16),Rational(3,16),Rational(3,16),Rational(1,16)]; result = sum(p)==1"},
  {"claim":"P(het at >=1 locus)=3/4 in AaBb x AaBb", "code":"p_hom_both=Rational(1,2)*Rational(1,2); result = 1-p_hom_both==Rational(3,4)"},
  {"claim":"Chi-square for 102:98 vs 100:100 equals 0.08", "code":"chi=Rational((102-100)**2,100)+Rational((98-100)**2,100); result = chi==Rational(8,100)"},
  {"claim":"P(first child affected)=1/6", "code":"p=(Rational(2,3)*Rational(1,2)+Rational(1,3)*0)*Rational(1,2); result = p==Rational(1,6)"},
  {"claim":"P(III-2 carrier | unaffected)=2/3", "code":"result = (Rational(1,2)/Rational(3,4))==Rational(2,3)"}
]