Question

an autosomal recessive disease is carried in the heterozygous form by 1 in 30 people. what is the frequency of the disease in the population?

Answer 1

If the incidence of an autosomal recessive disorder is known, then it is possible to calculate the carrier frequency.
From the given data, carrier incidence equals 1 in 30 then 2pq = 1/30 = 0.0333.
pq = 0.0333 / 2 = 0.01666.
p^2 + 2pq + q^2 should be = 1
p^2 + q^2 + 0.0333 = 1
Then p^2 + q^2 = 1 - 0.033 = 0.9667
If pq = 0.01666 and p^2 + q^2 = 0.9667, and
p + q should be = 1
The value of p = 0.0166 / q
The value of q = 0.0166 / p
Then p^2 + q^2 = (0.0166/q)^2 + (0.0166 / p)^2 = 0.9667
The frequency of disease occurrence is 1 - 0.0166/p

Answer 2

If the disease is autosomal recessive, then it will only manifest in individuals who have two copies of the mutant allele.

Let p be the frequency of the mutant allele in the population, and q be the frequency of the wild-type allele. Since the disease is recessive, individuals who are heterozygous (Aa) will not show the disease, but can pass on the mutant allele to their offspring.

According to the problem, the frequency of heterozygous carriers is 1 in 30, or 1/30. Thus, we can set up an equation relating the frequency of carriers to the allele frequencies:

2pq = 1/30

We can also use the fact that the frequencies of all alleles in the population must add up to 1:

p + q = 1

We can solve for q by rearranging the second equation:

q = 1 - p

Substituting this expression for q into the first equation, we get:

2p(1-p) = 1/30

Simplifying and rearranging, we get a quadratic equation in p:

2p^2 - 2p + 1/30 = 0

Using the quadratic formula, we can solve for p:

p = [2 ± sqrt(4 - 42(1/30))]/4

p = [2 ± sqrt(4/15)]/4

p ≈ 0.031 or p ≈ 0.969

We can discard the second solution, since it implies that the frequency of carriers is greater than 1, which is not possible. Therefore, the frequency of the mutant allele (and hence the frequency of the disease) is approximately:

p ≈ 0.031

or about 3.1% of the population.