With the help of the given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has…

Question

With the help of the given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in the F3 generation. (The pedigree shows a sex-linked recessive disorder: an unaffected carrier female mates with an unaffected male.)

1. A. 1/4 (Correct answer)
2. B. 1/2
3. C. 1/8
4. D. Zero

Correct Answer

Option A — 1/4

Detailed Solution & Explanation

The correct answer is 1/4.

Key Points

• Since a carrier female and a normal (non-carrier) male produced an affected male child, the disorder is X-linked recessive.
• Cross of carrier female (X$^c$X) with normal male (XY): offspring are X$^c$X$^c$? No — the Punnett square gives X$^c$X (carrier daughter), XX (normal daughter), X$^c$Y (affected son), XY (normal son).
• Out of the 4 offspring, only one is a disease-free carrier (X$^c$X), so the probability is $\dfrac{1}{4}$.