X-linked recessive inheritance with affected father and carrier mother

Steps:

• Probability the baby is a girl: 1/2.
• A girl inherits the affected X chromosome from her affected father with probability 1.
• She inherits the affected X from her carrier mother with probability 1/2.
• Thus, probability she is homozygous affected (develops condition): (1/2) × 1 × (1/2) = 1/4.

Why B is correct:

• In X-linked recessive disorders, daughters of an affected father and carrier mother are obligate carriers but have a 1/2 chance of being homozygous affected, and with 1/2 chance of being female, the joint probability is 1/4.

Why the others are wrong:

• A: Incorrect, as inheritance allows a nonzero chance of the girl being affected.
• C: Incorrect, as being female alone gives 1/2 carrier status, but full expression requires the second allele (total 1/4).
• D: Incorrect, as the mother contributes a normal allele half the time, preventing certainty.

Final answer: B