Question

A man with blood group AB marries a woman with blood group O. What is the probability that their child will have blood group A?

• A. \( 0\% \)
• B. \( 25\% \)
• C. \( 50\% \)
• D. \( 75\% \)

Hint:

When solving blood group inheritance problems, use a Punnett square to determine the possible genotypes. Remember that the \( i \) allele is recessive, so blood group O only appears when the genotype is \( ii \).

Answer 1

The Correct Option is C

Human blood groups are determined by the ABO gene, which has three alleles: \( I^A \), \( I^B \), and \( i \). The \( I^A \) and \( I^B \) alleles are codominant, while \( i \) is recessive. The genotypes for the blood groups are: - Blood group AB: \( I^A I^B \) - Blood group O: \( ii \) The man with blood group AB has the genotype \( I^A I^B \), and the woman with blood group O has the genotype \( ii \). To find the probability of their child having blood group A, we use a Punnett square to determine the possible genotypes of their offspring. The man (\( I^A I^B \)) can contribute either an \( I^A \) or an \( I^B \) allele. The woman (\( ii \)) can only contribute an \( i \) allele. The possible genotypes of the offspring are: \[ \begin{array}{c|cc} & I^A & I^B
\hline i & I^A i & I^B i
\end{array} \] - \( I^A i \): Blood group A (since \( I^A \) is dominant over \( i \)). - \( I^B i \): Blood group B (since \( I^B \) is dominant over \( i \)). The possible blood groups of the offspring are: - Blood group A (\( I^A i \)): 1 out of 2 possibilities. - Blood group B (\( I^B i \)): 1 out of 2 possibilities. Thus, the probability that their child will have blood group A is: \[ \frac{1}{2} = 50\% \]