Question

An autosomal locus has alleles A and a.

The frequency of individuals with the autosomal recessive phenotype is given. Which statements are true? (pick all that are true)

HWE= Hardy-Weinberg equilibrium

1. We can calculate q=Freq(a) even if we don't assume Hardy-Weinberg Equilibrium

2. Even if we don't assume HWE, we can calculate the genotype frequencies that we weren't given

3. If we assume HWE, we can calculate the genotype frequencies that we weren't given

4. If we assume HWE, we can calculate both allele frequencies

5. We can calculate both allele frequencies even if we don't assume HWE.

Answer 1

Hardy Weinberg equilibrium (HWE) states that the frequency of alleles and genotype remain the same in successive generations, if no evolutionary force acts on the population.

The given statement is that there are two alleles, A and a, and the phenotypic frequency of recessive individuals is known.

Let us consider an example, out of 100 individuals, 36 express the recessive phenotype. This means, the frequency of aa individuals is 36/100. In other words, taking frequency of A = p, and frequency of allele a = q, we get,

q² = 36/100

q = √ (36/100)

q = 0.6

Now since there are only two alleles in a population for this trait (A, a) , thus the sum of their frequencies will be 1, even in the absence of HWE. The frequency of allele A, i.e.

p = (1-0.6)

p = 0.4

Now, we can calculate the genotypic frequencies of other classes using the above two results. Hence, the correct statements should be:

We can calculate q=Freq(a) even if we don't assume Hardy-Weinberg Equilibrium. Even if we don't assume HWE, we can calculate the genotype frequencies that we weren't given. We can calculate both allele frequencies even if we don't assume HWE..