Tay-Sachs disease is inherited as an autosomal recessive. In a certain large eas

Question

Tay-Sachs disease is inherited as an autosomal recessive. In a certain large eastern European population, the frequency of Tay-Sachs disease is 1 percent.

a) If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs, what is the frequency of the allele that causes Tay-Sachs?

b) What would be the frequency of heterozygotes?

c) What is the probability of two heterozygotes marrying?

d) In children of such marriages, what would be the frequency of Tay-Sachs disease?

e) What proportion of all Tay-Sachs births are produced by such marriages?

Explanation / Answer

A. The frequency of the q allele can be calculated as:

Taking 100 as population, 1% gives 0.01. This is q2. To find out q we take square root of 0.01 which gives 0.1. This gives 10% as the frequency of the q allele. The p allele would be 90%.

B. The frequency of heterozygotes would be 2pq.

= 2 x 0.1 x 0.9

This gives 0.18 or 18%.

C. The probability that two heterozygotes to marry would be: 18% x 18% = 3.24%

D. When two heterozygotes are crossed, the punnet square would be:Aa x Aa

The genotypes of possible offpsring would be AA, Aa,aa in the ratio of 1:2:1. Hence the probability would be 0.25%.

E. All the people who suffer from this disease occur by marriage between heterozygotes. If a heterozygote marries a homozygous dominant, the offspring will have atleast one dominant allele. Two homozygous recessive cannot marry.Also a heterozygote and a homozygous recessive cannot marry. The reason is that homozygous recessive individuals usually die before the age of 10. Hence, they cannot marry and pass the allele. Only heterozygotes can pass the allele.

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