Mendelian Genetics

Question 1 (8 marks)

(a) Definitions (4 × 1):

• Gene — a unit of heredity; a DNA segment coding for a trait (e.g. the gene for flower colour). (1)
• Allele — an alternative version of a gene (e.g. $P$ = purple, $p$ = white). (1)
• Genotype — the allelic makeup of an organism (e.g. $Pp$). (1)
• Phenotype — the observable trait resulting from the genotype (e.g. purple flowers). (1)

Why: the four terms nest — a gene has multiple alleles; the allele combination is the genotype; the genotype expressed is the phenotype.

(b) (2): Homozygous = two identical alleles ($PP$ or $pp$); heterozygous = two different alleles ($Pp$). (1) Because the dominant allele masks the recessive, $Pp$ looks identical to $PP$ (both purple) but never to $pp$ (white) — dominance hides the recessive allele. (1)

(c) (2): Law of Segregation: the two alleles of a gene separate during gamete formation so each gamete carries only one allele. (1) Corresponds to the separation of homologous chromosomes during anaphase I of meiosis. (1)

Question 2 (9 marks)

(a) (3): Parents: $PP$ (true-breeding purple) $\times$ $pp$ (white). (1) F1 genotype: all $Pp$. (1) F1 phenotype: all purple. (1)

(b) (4): $Pp \times Pp$:

(1 for square) Genotypic ratio $1\ PP : 2\ Pp : 1\ pp$. (1.5) Phenotypic ratio $3$ purple $: 1$ white. (1.5)

(c) (2): Among purple F2, genotypes are $1\ PP : 2\ Pp$, total 3. (1) Probability heterozygous $= 2/3$. (1)

Question 3 (10 marks)

(a) (2): Law of Independent Assortment: alleles of different genes assort independently into gametes. (1) Because each gene pair segregates independently, the combined ratio is the product of two independent $3:1$ monohybrid ratios, giving $9:3:3:1$. (1)

(b) (5): Each gene individually: $Rr\times Rr \to 3$ round : 1 wrinkled; $Yy\times Yy \to 3$ yellow : 1 green. (1) Apply product rule to combine:

• Round Yellow: $3/4 \times 3/4 = 9/16$
• Round Green: $3/4 \times 1/4 = 3/16$
• Wrinkled Yellow: $1/4 \times 3/4 = 3/16$
• Wrinkled Green: $1/4 \times 1/4 = 1/16$ (3 for correct products)

Phenotypic ratio 9 : 3 : 3 : 1. (1)

(c) (3): Round and green $= P(\text{round}) \times P(\text{green}) = 3/4 \times 1/4 = 3/16$. (2 method, 1 answer)

Question 4 (7 marks)

(a) (3): Cross the unknown tall plant with a homozygous recessive short plant ($tt$). (1) The recessive parent contributes only $t$ gametes, so offspring phenotypes directly reveal the tall parent's second allele. (1) If any short offspring appear, the parent carried a $t$ (was $Tt$). (1)

(b) (2):

• If $TT \times tt$: all $Tt$, all tall (100%). (1)
• If $Tt \times tt$: $1\ Tt : 1\ tt$, 1 tall : 1 short (50:50). (1)

(c) (2): 21 tall : 19 short ≈ 1:1. (1) The appearance of short offspring proves a $t$ allele; the ~1:1 ratio matches the $Tt \times tt$ expectation, so the parent is heterozygous $Tt$. (1)

Question 5 (8 marks)

(a) (2): Product rule: probability of independent events both occurring = product of their individual probabilities. (1) Sum rule: probability of either of two mutually exclusive events = sum of their probabilities. (1)

(b) (2): $Aa\times Aa$ gives $1/4$ recessive ($aa$). (0.5) All three recessive: $(1/4)^3 = 1/64$. (1.5)

(c) (4): $P(\text{recessive})=1/4$, $P(\text{not})=3/4$. (1) Exactly one of three: choose which child $\times$ probabilities: $\binom{3}{1}\left(\tfrac14\right)^1\left(\tfrac34\right)^2 = 3 \times \tfrac14 \times \tfrac{9}{16} = \tfrac{27}{64}.$ (2 for binomial setup, 1 for answer $27/64 \approx 0.422$).

Question 6 (8 marks)

(a) (3): Key: □ = unaffected male, ○ = unaffected female, filled = affected; horizontal line = mating, vertical = offspring.

• Gen I: □ × ○ (both unaffected). (1)
• Gen II: ●(affected daughter), □, □ (two unaffected sons); affected daughter married to unaffected □. (1)
• Gen III: ●(affected son), ○(unaffected daughter). (1)

(b) (2): The trait is recessive. (1) Justification: two unaffected Gen I parents produced an affected child — the trait skipped/hidden in parents, only possible if it is recessive and both parents were carriers. (1)

(c) (3): Gen I parents: both $Bb$ (carriers, needed to produce affected $bb$ child). (1×2) Affected Gen II daughter: $bb$. (1)

[
  {"claim":"F2 heterozygous fraction among purple = 2/3","code":"result = Rational(2,3) == Rational(2,4)/Rational(3,4)"},
  {"claim":"Round-green dihybrid probability = 3/16","code":"result = Rational(3,4)*Rational(1,4) == Rational(3,16)"},
  {"claim":"All three children recessive = 1/64","code":"result = Rational(1,4)**3 == Rational(1,64)"},
  {"claim":"Exactly one of three recessive = 27/64","code":"from sympy import binomial\nresult = binomial(3,1)*Rational(1,4)**1*Rational(3,4)**2 == Rational(27,64)"}
]