Some forms of cataracts and deaf-dumbness in humans are autosomal recessive signs. What is the probability of having healthy children in a family if the parents are diheterozygous?
Let’s designate the deaf-mute gene as a, the cataract gene as b, healthy alleles – A and B. The genotypes of both parents will be AaBb.
R: AaBb x AaBb – healthy parents, carriers of disease genes;
G: AB, Ab, ab, ab x AB, Ab, ab, ab – possible gametes;
F1: Let’s use a Pennett lattice for a dihybrid crossing. Analyzing the genotypes presented in it, we find the probability of having a healthy child: 9/16 or 56.25%.
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