А- gene for normal body developmenta if the genotype of a woman АА:
a- brachydactyly
Aa- - A man with brachydactyly (heterozygothe) P. Aa x AA
AA, Aa -genotype of a healthy woman F1. AA AA Aa Aa
100% - healthy children
if the genotype of a woman Аa
P. Aa x Aa
F1. AA Aa Aa aa
1/4=75% - healthy children
If the father is heterozygote for the brachydactyly gene. and the mother is a dominant homozygote, the probability of having healthy children is 100%. If the mother is heterosegote, then the probability of having healthy children is 75%.
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