A family has an X‑linked recessive disorder. The mother is a carrier (XcX) and the father is normal (XY). They have three children: one son who is color‑blind, and two daughters who are normal. What is the probability that their next child (fourth) will be a son with color‑blindness?
Answer
1/4
The correct answer is 1/4 — Thus the chance that the next child will be a color‑blind son is 1/4.
Step-by-step solution
- Write the parental genotypes: mother XcX, father XY.
- List possible gametes: mother Xc, X; father X, Y.
- Form all offspring possibilities and their probabilities: - Sons: XcY (color‑blind) 1/4; XY (normal) 1/4. - Daughters: XcXc (carrier) 1/4; XcX (carrier) 1/4.
- Observe that the probability of a son who is color‑blind is 1/4, independent of past children.
- Thus the chance that the next child will be a color‑blind son is 1/4.
Common mistake: Thinking previous children change the probability for the next child
Concept tested: X‑linked recessive inheritance
Published · How this solution was prepared and checked
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