Question

bo blood type is controlled by three alleles of a single gene. blood type o is associated with the recessive allele. blood types a, b, and ab are associated with the codominant ( i^a ) and ( i^b ) alleles. the pedigree shows the inheritance of abo blood types in a family. which two genotypes are possible for the individual indicated with the question mark? a ( i^a ) b ( i^b i^b ) c ( i^a ) d ( i ) e ( i^b )

Explanation:

• Blood type is determined by alleles \(I^A\), \(I^B\), and \(i\). \(I^A\) and \(I^B\) are codominant, and \(i\) is recessive.
• The parents of the individual (II - 4) have blood types B (\(I^B\) - ) and A (\(I^A\) - ). Their children (III) have blood types A (\(I^A I^A\) or \(I^A i\)) and B (\(I^B I^B\) or \(I^B i\)).
• For the children to have \(I^A I^A\) and \(I^B\) - genotypes, the parents (II - 3 and II - 4) must contribute \(I^A\) and \(I^B\) or \(i\) alleles.
• If we consider the possible genotypes of the parents:
• Let's assume the parent with blood type B (II - 3) has genotype \(I^B i\) (since it can pass on \(i\) to children with blood type A if the other parent contributes \(I^A\)).
• The parent with the question mark (II - 4) must be able to pass on \(I^A\) (for children with blood type A) and \(i\) (if we consider the recessive allele inheritance). So possible genotypes for II - 4 are \(I^A i\) (option C) and \(I^A I^A\) (but not in the options). Wait, re - evaluating:
• Another approach: The children have blood types A and B. The parent with blood type B (II - 3) can have \(I^B i\) or \(I^B I^B\). If we assume simple Mendelian inheritance for codominant alleles.
• The parent with the question mark (II - 4) must have alleles that can combine with \(I^B\) (from II - 3) to produce \(I^A I^-\) (A blood type) and \(I^B I^-\) (B blood type). The only way is if II - 4 has \(I^A i\) (so that when combined with \(I^B i\) (assumed for II - 3) can give \(I^A I^B\) (AB, not in children but \(I^A i\) (A) and \(I^B i\) (B)) or if II - 3 has \(I^B I^B\) and II - 4 has \(I^A i\) (gives \(I^A I^B\) (not in children) but also \(I^B i\) (B) and \(I^A I^B\) (not in children, but maybe a miscalculation). Wait, no:
• The key is that the child with blood type A in III must get \(I^A\) from II - 4. The child with blood type B in III must get \(I^B\) from II - 3. So II - 4 must have \(I^A\) (to give to A - type children) and can have \(i\) (to combine with \(I^B\) from II - 3 to make \(I^B i\) (B - type child)). So the genotype of II - 4 is \(I^A i\) (option C). But wait, another check:
• Blood type A can be \(I^A I^A\) or \(I^A i\), blood type B can be \(I^B I^B\) or \(I^B i\). Since the children have A and B, and assuming II - 3 is \(I^B i\) (because if it was \(I^B I^B\), all children would get \(I^B\) from it). Then II - 4 must be \(I^A i\) (so that when crossed with \(I^B i\):

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But we see \(I^A i\) (A) and \(I^B i\) (B) in children.

Answer:

C. \(I^A i\)