Tay-Sachs disease (A) is a rare autosomal recessive disease. Individuals with it

Question

Tay-Sachs disease (A) is a rare autosomal recessive disease. Individuals with it lack the enzyme hexosaminidase A, and therefore accumulate gangliosides in their nervous system leading to paralysis, epilepsy, blindness, and eventual death. Heterozygous individuals show no symptoms. A second gene (B) independently sorted from the Tay-Sachs gene, is an autosomal dominant gene that causes brachydactylous (dramatically shortened fingers). A father is AaBb and the mother is Aabb.
a. What is the phenotype of the father? ____________________
b. What is the phenotype of the mother? ______________
c. What is the probability that their child will have short fingers? ___________
d. What is the probability that a child will have short fingers and be a carrier for Tay-Sachs disease? _____________
e. If the child has brachydactylous, what is the probability he is heterozygous for it? ____
f. If the couple has three kids, what is the probability that at least one of them will have the genes for both Tay-Sachs disease and brachydactylous? _________

Explanation / Answer

Recessive genes don't express in hetrozygous genotype. a) not having Tay-Sachs disease and having brachydactylous. b) not having Tay-Sachs disease and not having brachydactylous c) probability is 0.5 d) 0.125(0.25*0.5) e)1(only Bb and bb genotypes produced in progeny in almost equal numbers) e) possible genotypes in progeny causing no diseases = Aabb. so, probability that a child is born with no disease = 1/16 so, probability that all 3 kids are born with no disease = 1/16^3 so, probability that at least one kid is born with both disease = 1-1/16^3 = 255/256

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