32. In a plant, male sterility is caused by the presence of a dominant allele for a nuclear gene (Ms). There are also lines which carry a gene whose dominant allele (F) restores male fertility. These are called restorer lines. A cross is made between a male sterile line and a homozygous restorer line which does not contain the Ms allele. Of the Fl progeny, only those that carry the Ms allele are allowed to self-pollinate. What is the probability of obtainjng a male sterile line in the resulting progeny? (The Ms and F genes assort independently),
(1) 1/4         (2) 3/16
(3) 3/4         (4) 1/16

Step-by-step solution

Step 1: Define parental genotypes

Male sterility gene: Ms (dominant = sterile), ms (recessive = normal) [web:69].

Restorer gene: F (dominant = restores fertility in Ms plants), f (recessive, no restoration).

Male-sterile line: Ms ms ff (sterile).

Restorer line: ms ms FF.

Cross: Ms ms ff × ms ms FF

Gametes: Sterile → Ms f or ms f (½ each); Restorer → ms F only.

F₁: Half carry Ms ms Ff (fertile); all F₁ fertile due to F [web:73].

Step 2: Self selected F₁ (Ms ms Ff × Ms ms Ff)

Treat Ms and F loci independently:

Ms locus:

• 1/4 MsMs
• 1/2 Msms
• 1/4 msms

P(Ms–) = 3/4 (MsMs or Msms → male-sterile if not restored).

F locus:

• 1/4 FF
• 1/2 Ff
• 1/4 ff

Only ff fails to restore (P(ff) = 1/4).

Male sterility probability

Male-sterile plants must be: Ms– and ff

de>P(male sterile) = P(Ms–) × P(ff) = (3/4) × (1/4) = 3/16

Thus 3/16 of F₂ progeny will be male-sterile lines [web:69].

Option-wise check

• 1/4: would require Ms– and ff each at 1/2
• 3/16: correct – product of 3/4 (Ms–) and 1/4 (ff)
• 3/4: fraction carrying Ms but ignoring F (most fertile)
• 1/16: underestimates (only MsMs ff, ignores Msms ff)