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.
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Draw a Punnett square for the F\textsubscript{2} parents (Aa x aa) to visualize the probabilities of the F\textsubscript{3} genotypes.
The pedigree shows an autosomal recessive trait. In the F\textsubscript{2} generation, the affected male (shaded square) must have the genotype "aa," and the carrier female (half-filled circle) must have the genotype "Aa." Their children in the F\textsubscript{3} generation can have the following genotypes: Aa (carrier, no disease) or aa (affected). The probability of an F\textsubscript{3} child being a carrier (Aa) is 1/2. Since the question asks for the probability of any of their children being a carrier, and there's only one child being considered in F\textsubscript{3} from this particular couple, the answer is 1/2 1/2 = 1/4 (Note: If the questions asked what is the probability of having a daughter who is a carrier that will become 1/2 1/2 1/2 = 1/8 because girls are also 1/2 probability.)
