A woman types as Rh-positive with an anti-c titer of 32. What is the father's most likely Rh phenotype if their baby shows no hemolytic disease?

The correct answer is that the father's most likely Rh phenotype is R1r. In this scenario, the woman is Rh-positive and has an anti-c titer of 32, which indicates the presence of antibodies against the c antigen, part of the Rh blood group system. Given that she is Rh-positive, she can possess one of the following genotypes: R1R1, R1r, R2R2, R2r, or rr.

The fact that the baby shows no signs of hemolytic disease suggests that the baby likely inherited a Rh phenotype that does not provoke an immune response from the mother’s anti-c antibodies. For this to happen, the baby should either not express the c antigen or have a compatible Rh genotype.

If the father were homozygous (rr), it would mean the baby must inherit the r allele from the father, resulting in a phenotype that would express the c antigen (c positive), leading to a greater likelihood of hemolytic disease due to the mother’s anti-c antibodies. Therefore, it is less likely for the father to have the rr phenotype.

If the father's phenotype were R2r (having the R2 allele, which leads to the c antigen), there could again