What would be the predicted carrier frequency if the incidence of a disease is 1/6400?

What would be the predicted carrier frequency if the incidence of a disease is 1/6400?

• A. 1/80
• B. 1/40
• C. 1/20
• D. 1/10

Answer: (B)

To determine the predicted carrier frequency when given the incidence of a recessive disease, the Hardy-Weinberg principle is typically applied. The incidence of the disease represents the frequency of individuals who are homozygous recessive (aa) for the trait. In this case, with a disease prevalence of 1 in 6400, this implies that q² = 1/6400, where q is the frequency of the recessive allele. To find q, the square root of the incidence is calculated: \[ q = \sqrt{1/6400} = 1/80 \] The frequency of the dominant allele (p) is then calculated using the Hardy-Weinberg equation, which states that p + q = 1. Therefore, p is: \[ p = 1 - q = 1 - (1/80) = 79/80 \] The carrier frequency (heterozygotes, Aa) is represented by 2pq: \[ 2pq = 2 \times (79/80) \times (1/80) = (79/80) \times (1/40) = 79/3200 \] In terms of carrier frequency, this

To determine the predicted carrier frequency when given the incidence of a recessive disease, the Hardy-Weinberg principle is typically applied. The incidence of the disease represents the frequency of individuals who are homozygous recessive (aa) for the trait. In this case, with a disease prevalence of 1 in 6400, this implies that q² = 1/6400, where q is the frequency of the recessive allele.

To find q, the square root of the incidence is calculated:

[

q = \sqrt{1/6400} = 1/80

]

The frequency of the dominant allele (p) is then calculated using the Hardy-Weinberg equation, which states that p + q = 1. Therefore, p is:

[

p = 1 - q = 1 - (1/80) = 79/80

]

The carrier frequency (heterozygotes, Aa) is represented by 2pq:

[

2pq = 2 \times (79/80) \times (1/80) = (79/80) \times (1/40) = 79/3200

]

In terms of carrier frequency, this