In an X-linked recessive condition, if the mother is a carrier and the father has no disease, what is the probability that a DAUGHTER will become a CARRIER?

Correct answer: A. 50%

• A. CORRECT. For X-linked recessive inheritance: Mother is X^c X (carrier, one affected X allele, one normal). Father is X Y (normal, unaffected). Daughters receive one X from each parent. 50% of daughters will inherit the mother's X^c (carrier allele) → become carriers. 50% will inherit the mother's normal X → be completely normal. Therefore, there is a 50% chance a daughter will be a carrier.
• B. 0% is incorrect. Since the mother carries an affected X allele, there is always a chance daughters will inherit it.
• C. 25% is the probability that a CHILD (regardless of sex) will be AFFECTED with an X-linked recessive disease (specifically, that a SON will be affected). For daughters and carrier status, 50% is the correct answer.
• D. 100% carrier probability would occur only if the mother were affected (homozygous X^c X^c), which the question specifies she is not — she is a carrier.