Tay-Sachs disease is caused by an allele for a recessive trait. A DNA test on blood can determine if a healthy individual is homozygous or heterozygous. One member of a particular couple is homozygous dominant, and the other member is heterozygous. Refer to the information above. What is the probability that this couple will have a child with Tay-Sachs disease

Zero (0)

Explanation:

Let the allele for Tay-Sachs disease be t.

Homozygous dominant = TT

Heterozygous dominant = Tt

Since the disease is caused by recessive allele, both TT and Tt are normal and free from the disease, although Tt is a carrier.

Crossing the two:

TT x Tt = TT, Tt, TT and Tt

50% of the offspring have TT genotype while the remaining 50% have Tt genotype. Both TT and Tt are normal.

Hence, the probability that this couple will have a child with Tay-Sachs disease is 0.

the probability is Zero

Explanation:

Tay-sachs is an autosomal disorder. That is the gene variants or allele for the disorder is located on other chromosomes in the body excluding the sex chromosmes. Therefore it inheritance is in autosomal recessive manner.

Let T - rep. the dominant allele for Tay-Sachs,

while t-rep. the recessive allele for this diseases.

Assuming the Husband is homozygous dominant represented as TT.

and the wife is heterozygous as Tt

All the children produced assuming Mendelian monohybrid cross.

TT X Tt

are;

TT - free TT,-free Tt-carrier Tt.-carrier.

Evidently none will have this diseases, thus Zero probability. Since recessive allele of a disorder must pair with another recessive allele to exhibits its effects. However the husband is dominant therefore lack the recessive allele.