Blood type is characterized by the combination of alleles that an individual has at two different loci. Three alleles (I A, IB, and i) encode ABO blood group antigens. There are two co-dominant alleles (LM and LN) encode another set of antigens, the MN blood group (homozygous LM individuals have type M, homozygous LN individuals have type N, and heterozygous LMLN individuals have type MN). For the crosses below, what is the probability that a child will have a unique blood genotype (at BOTH loci) that is not the same as the blood group genotypes of either parent?

If the cross is LMLM x LNLN, the probability is 100%; If the cross is LMLN x LMLN, the probability is 25% for LMLM and 25% for LNLN.

Explanation: In genetics, one gene is considered dominant when the gene manifests itself over its other counterpart. However, there are cases in which an individual with two different genes express them simultaneously, To it, it's called Codominance.

As cited above, the MN blood group is a example of codominance. Due to it, when crossing two species, there can be individual with 1, 2 or 3 differents genotypes. If crossing a homozygous LM individual with a homozygous LN individual, their offspring will have a 100% chance of having a blood group different from their parents.

If crossing two heterozygous individual, the probability of having a child with blood genotype different from his or hers parents is 25% for type M and 25% for type N, because the ratio in codominance is 1:2:1.

the correct option is option a

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