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A particular recessive genetic disorder is polygenic. It is caused by 2 separate, unlinked genes (Genes A and B). To have the disorder, a person must be homozygous recessive for BOTH of these genes. A mother is heterozygous for Gene A and homozygous recessive for Gene B. The father is heterozygous for gene Gene A and homozygous dominant for Gene B. If they have 2 children, what is the probability that 1 of them will have this disorder

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luisvaschetto

Answer:

1/4

Explanation:

1-Mother:

The probability to pass the disorder to her progeny is: 1 x 1/2 = 1/2

Father:

2-The probability to pass the disorder to his progeny is: 1 x 1/2 = 1/2

Therefore, the probability to have the disorder is equal to the product of both independent events, i.e., 1/2 x 1/2 = 1/4 or 25%  

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