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A **carrier woman** (X^c X) marries a **normal man** (X^c Y) for colour blindness. What proportion of their offspring are expected to be colour-blind?
AAll daughters; no sons
BAll sons; no daughters
CHalf of the sons; none of the daughters
DAll children equally
Answer & Solution
Correct answer: C. Half of the sons; none of the daughters
Mother's gametes: 1/2 X^C, 1/2 X^c. Father's gametes: 1/2 X^C, 1/2 Y. Offspring: 1/4 X^C X^C (normal F), 1/4 X^C X^c (carrier F), 1/4 X^C Y (normal M), 1/4 X^c Y (colour-blind M). Daughters: 50% normal, 50% carriers — none colour-blind (need two X^c). Sons: 50% normal, 50% colour-blind. Hence half of the sons, none of the daughters.
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