07 1.7 Mendel's second law (law of independent assortment)
- When two purebred individuals that differ in two or more pairs of contrasting traits are crossed, each pair of traits is inherited independently of the others.
- Analysed separately, each trait appears in F₂ in the ratio 3 : 1.
The dihybrid cross: seed shape and seed colour
R = round (dominant), r = wrinkled; Y = yellow (dominant), y = green. P: pure round yellow (RRYY) × wrinkled green (rryy) Gametes: RY and ry F₁: all RrYy, round yellow (100%) — both dominant traits appear.
F₁ × F₁: RrYy × RrYy. Each parent produces four types of gametes with equal probability: RY, Ry, rY, ry. Fusion gives 16 combinations:
- RRYY
- RRYy
- RrYY
- RrYy
- RRYy
- RRyy
- RrYy
- Rryy
- RrYY
- RrYy
- rrYY
- rrYy
- RrYy
- Rryy
- rrYy
- rryy
Count of phenotypes in F₂ (out of 16):
- round yellow (both dominant): 9
- round green (first dominant, second recessive): 3
- wrinkled yellow (first recessive, second dominant): 3
- wrinkled green (both recessive): 1
Phenotype ratio 9 : 3 : 3 : 1.
Check each trait alone: round = 9 + 3 = 12, wrinkled = 3 + 1 = 4 → 12 : 4 = 3 : 1. Yellow = 9 + 3 = 12, green = 3 + 1 = 4 → 3 : 1. So shape is inherited independently of colour.
Worked example 4 (check your understanding)
T = tall, t = short; G = green pod, g = yellow pod. Two plants heterozygous for both traits (TtGg × TtGg) are crossed. What fraction are tall with green pods? From the 9 : 3 : 3 : 1 pattern, tall green = 9/16 (56.25%). You can also multiply independent probabilities: P(tall) = 3/4 and P(green) = 3/4, so 3/4 × 3/4 = 9/16.
Worked example 5: a test cross in humans
- Curly hair (H) is dominant over straight (h); brown eyes (B) dominant over blue (b).
- A man with curly hair and brown eyes (HhBb) marries a woman with straight hair and blue eyes (hhbb).
- Man's gametes: HB, Hb, hB, hb.
- Woman's gametes: only hb.
- HhBb
- Hhbb
- hhBb
- hhbb
Phenotypes: curly brown : curly blue : straight brown : straight blue = 1 : 1 : 1 : 1 (25% each).