Consider a lethal recessive mutation (such as Tay-Sachs disease), in which indiv

Question

Consider a lethal recessive mutation (such as Tay-Sachs disease), in
which individuals homozygous for the allele do not survive to
reproduce. What proportion of the population is heterozygous?
1. Write down the genotypes and their fitnesses. We’ll call the allele for Tay-Sachs,
A2 and the other allele (which does not cause Tay-Sachs), A1.

2. What is s? Hint: what is the fitness of an individual that does not survive to
reproduce?

3. Use a ballpark of = 10-6

4. Use the mutation-selection balance formula to find equilibrium frequency of
recessive phenotype, q*

5. Assume that, at equilibrium, the population is at Hardy Weinberg proportions.
Now you can calculate p* and q*

6. What is f[A1A2] at equilibrium, the frequency of “carriers”?

Explanation / Answer

1. The number of heterozygous individuals = 2 * p * q

Where p is representing the frequency of dominant allele and q is representing the frequency of recessive allele.

S is denoting the “selection coefficient”. This coefficient represent the difference in phenotype fitness with reference to any standard fitness that have value of fitness= 1

Genotypes of homozygous = A1A1 and A2A2

Genotypes of heterozygous=A1A2

Fitness of dominant individual = W0

Fitness of homozygous individual= W1

Fitness of recessive individual= W2

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