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Two genes assort independently. A prospective mother is homozygous recessive for gene "A" and homozygous dominant for gene "B." The father is heterozygous for gene "A" and homozygous recessive for gene "B." What is the probability that their child will be heterozygous for both the "A" and "B" genes?

Respuesta :

Answer:

50%

Explanation:

According to the given information, the genotype of the mother who is homozygous recessive for gene "A" and homozygous dominant for gene "B"= "aaBB". Likewise, the genotype of the father who is heterozygous for gene "A" and homozygous recessive for gene "B"= "Aabb".

A cross between aaBB and Aabb would produce a progeny in following genotype ratio= 1/2 AaBb : 1/2  aaBb

Therefore, there are 50% chances for this couple to have a child heterozygous for both A and B genes (AaBb).

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