Tay-Sachs disease is an autosomal recessive neurological disorder that is fatal

Question

Tay-Sachs disease is an autosomal recessive neurological disorder that is fatal in infancy. Despite its invariably lethal effect, Tay-Sachs disease occurs at very high frequency in some Central and Eastern European (Ashkenazi) Jewish populations. In certain Ashkenazi populations, 1 in 650 infants has Tay-Sachs disease. Population biologists believe the high frequency is a consequence of genetic bottlenecks caused by pogroms (genocide) that have reduced the population multiple times in the last several hundred years.

Part A

In the population described, what is the frequency of the recessive allele that produces Tay-Sachs disease?

Part B

Assuming mating occurs at random in this population, what is the probability a couple are both carriers of Tay-Sachs disease?

Explanation / Answer

A. p = the frequency of the dominant allele q = the frequency of the recessive allele p^2 = the frequency of the homozygous dominant individuals 2pq = the frequency of the heterozygous individuals q^2 = the frequency of the homozygous recessive individuals p+q=1 p^2 + 2pq + q^2 = 1 Since 1 in 650 are born with Tay-Sachs, so that is the frequency of aa (homozygous recessive). However, you want to find the frequency of allele in the entire population, so you need to also know how many are heterozygous. So, you know that q^2 = 1/650, and you want to find q. You can do this by simply taking the square root of 1/650. So, q = 0.039 B. To find this, you first must ask what the probability is that one person is heterozygous. Now that you know q, you can find p using p+q=1. p=1-q p=1-0.039 p=0.961 2pq = 2(0.039)(0.961) = 0.075 This is the proportion of the population that is heterozygous (i.e. the probability that one person would be heterozygous). The probability that two people have the mutation, then, is 0.075 x 0.075 = 0.057 This is the probability that two people in a random couple are both carriers of Tay-Sachs disease.

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