A male carrier of a recessive somatic mutation marries a female without this mutation. What is the probability that their child will be a carrier?
A. 0%
B. 25%
C. 50%
D. 100%
The probability that their child will be a carrier can be calculated using Punnett square analysis.
If the male carrier has the genotype Aa (A representing the normal allele and a representing the recessive allele), and the female does not have the mutation (genotype AA), the possible genotypes of their child will be:
- Aa: 50% chance (child inherits one recessive allele from the male carrier)
- AA: 50% chance (child inherits one normal allele from the male carrier)
Since the child has a 50% chance of inheriting the recessive allele from the male carrier, the probability that the child will be a carrier is 50%.
Therefore, the answer is C. 50%.