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Assuming Hardy-Weinberg conditions are met: a rare disorder is caused by an autosomal recessive mutation. Calculate...

Assuming Hardy-Weinberg conditions are met: a rare disorder is caused by an autosomal recessive mutation. Calculate the percentage of heterozygous carriers in a population if the disease affects 1 in 15,000 individuals (show your work).

Expert Solution

The equation that can be used for calculating percentage of heterozygous carriers (or the frequency of heterozygous carriers) for an autosomal recessive trait is

p²+ 2pq+ q² = 1

Assuming "A" is the wild type allele and "a" is the mutant allele. Now is the disease is autosomal recessive, It will appear in the genotypes "aa" and the heterozygous condition "Aa" will be the disease carriers in the population.

If this poputaion is in H-W equilibrium, allelic frequencies can be expressed as

A2 + 2Aa+ a2 =1

(Which is similar to the p²+ 2pq+ q² = 1)

For the disease, the frequency of "aa" in the popylation is = 1 in 15000

q2 = 1/15000

q = 1/122.48 =0.00816

Since q + p = 1 ,

p = 1 - q

p= 0.99184

Carrier frequency= 2 pq = 2 x(0.99184 x 0.00816)

2 pq =0.01619

This means 0.016 in 1 individual will be carriers. So in a population of 15,000 number of carrier individuals will be = 0.016 x 15000 = 240

Hence Percentage carriers (Heterozygous carriers) = (240 / 15000) x 100 =1.62 %

Therefore, the percentage of heterozygous carriers in a population if the disease affects 1 in 15,000 individuals will be 1.62 %

gladiator answered 2 years ago

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