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.
Show Hint
Draw a Punnett square for the F2 parents (Aa x aa) to visualize the probabilities of the F3 offspring's genotypes.
The affected male is genotype aa as the disease is autosomal recessive, requiring two copies of the recessive allele for expression.
The female is a carrier, genotype Aa, possessing one normal allele (A) and one disease allele (a), rendering her asymptomatic but capable of transmitting the "a" allele.
Step 2: F3 Generation Genotypes (Offspring)
Possible genotypes of their offspring (F3 generation) can be determined using a Punnett square.
Punnett Square Analysis: The father (<strong>aa</strong>) can only contribute the a allele. The mother (<strong>Aa</strong>) can contribute either the A or the a allele. The resulting offspring genotypes are:
Aa (carrier, unaffected) — Inheritance of A from the mother and a from the father.
aa (affected) — Inheritance of a from both parents.
Consequently, the probability of an offspring being a carrier (Aa) is 1/2, and the probability of being affected (aa) is also 1/2.
Step 3: Probability of an F3 Carrier
The question requires the probability of a child from this couple (F3 generation) being a carrier (Aa).
Based on the preceding analysis, the probability of an offspring being a carrier (Aa) is 50%, or 1/2.
Step 4: Gender-Specific Probabilities
If the question specifies a daughter who is a carrier, the probability remains 1/2 for that individual child. Autosomal inheritance is not influenced by gender.
More complex scenarios involving multiple conditions or specific gender outcomes would alter the calculation but not the fundamental principles of inheritance.
Final Answer
The probability of any child being a carrier (Aa) is 1/2.
For a specific child, such as a daughter, the probability of being a carrier is also 1/2.
