With the help of given pedigree, find out theprobability for the birth of a child having no diseaseand being a carrier (has the disease mutation in oneallele of the gene) in F3 generation.
Answer
Pedigree Analysis: X-linked Recessive Disorder
This pedigree shows an X-linked recessive disorder. In such disorders:
- Females with one mutant allele (XcX) are carriers.
- Males with the mutant allele (XcY) are affected.
Goal:
Determine the probability of a child in the F3 generation being a carrier (XcX) and unaffected.
Parental Genotypes in F2 Generation:
- Mother: Carrier Female (XcX)
- Father: Unaffected Male (XY)
Gametes:
- From Mother: Xc or X
- From Father: X or Y
Punnett Square:
| Xc (Mother) | X (Mother) | |
|---|---|---|
| X (Father) | XcX (Carrier Female) |
XX (Unaffected Female) |
| Y (Father) | XcY (Affected Male) |
XY (Unaffected Male) |
Offspring Possibilities:
- XcX – Carrier Female ✅
- XX – Unaffected Female
- XcY – Affected Male
- XY – Unaffected Male
Conclusion:
Out of the 4 possible outcomes, only 1 child is carrier and unaffected (XcX).
Therefore, the probability = 1/4
Correct answer: Option (1) – ¼