Answer to Question #343623 in Genetics for Andy

Question

Tay-Sachs disease is an inherited disease that is caused by a recessive allele. If a man and his wife, who are both carriers, have three children, what is the probability of each of the following? Show your calculations and give your answers as fractions.

a) All their three children do not have the disease.

d) At least one child is phenotypically normal without the disease.

b) One or more of their three children have Tay-Sachs disease.

c) All three children have the Tay-Sachs disease.

Answer 1

Let the gene responsible for the Tay-Sachs disease be a. Parents who are carriers of this gene are heterozygous and have the Aa genotype.

Let's perform the following cross:

Aa × Aa

↓

AA + Aa + Aa + aa

In this case, the probability that the child will be absolutely healthy, that is, will not be a carrier and will have the AA genotype, is 1/4. The probability that the child will be sick and will have the aa genotype is also 1/4. At the same time, the probability that the child will be a carrier and will have the Aa genotype is 1/2.

a) If all three children do not have the disease, then this implies that among them there can be both absolutely healthy and having the AA genotype, and carriers who have the Aa genotype.

Consider the different combinations of genotypes in these three children and calculate their probability:

• AA + AA + AA

1/4*1/4*1/4 = 1/64;

• AA + AA + Aa

1/4*1/4*1/2 = 1/32;

• AA + Aa + Aa

1/4*1/2*1/2 = 1/16;

• Aa + Aa + Aa

1/2*1/2*1/2 = 1/8.

d) If one child is phenotypically normal without the disease, then he can be either absolutely healthy, having the AA genotype, or a carrier, having the Aa genotype.

Consider the following combinations of genotypes:

• AA + aa + aa

1/4*1/4*1/4 = 1/64;

• Aa + aa + aa

1/2*1/4*1/4 = 1/32.

b) There may be several options here:

• one child has the Tay-Sachs disease, that is, he has the aa genotype;
• two children have the Tay-Sachs disease.

In the first case, there will be such combinations of genotypes:

• AA + AA + aa

1/4*1/4*1/4 = 1/64;

• AA + Aa + aa

1/4*1/2*1/4 = 1/32;

• Aa + Aa + aa

1/2*1/2*1/4 = 1/16.

In the second case, there will be such combinations of genotypes:

• aa + aa + AA

1/4*1/4*1/4 = 1/64;

• aa + aa + Aa

1/4*1/4*1/2 = 1/32.

c) If three children have the Tay-Sachs disease, then the combination of genotypes will be as follows:

• aa + aa + aa

1/4*1/4*1/4 = 1/64.