Question

- Vi tal:ros. b The agouti gene determines coat colour in mice. Heterozygous mice have yellow coats, while homozygous dominant mice have black coats. However, having two copies of the recessive alleles is lethal. In a population of 2000 mice, 1082 mice have black coats. Calculate the frequency of each altele. Show all your work and express your answer as a value between 0 and 1 rounded to two decimal places. (3 marks) B 15 E U ch X2 x 를 들 E 2 <> XQ

Answer 1

The Agouti gene determines coat colour in mice. We can call this gene "A" for Agouti, where A is dominant and a is recessive.

Heterozygous mice have yellow coats: So mice with "Aa" are Yellow.

Homozygous dominant mice have black coats: So mice with "AA" are black.

Homozygous recessive genotypes are lethal for mice: SO mice with "aa" will not survive.

In a population of 2000 mice, 1082 have Black coats (or AA genotypes).

(1082)/(2000)= proportion of .541

*Remember the Hardy Weinberg Equation (p²+2pq+q²=1) and p+q=1 always.

P will equal the square root of p²: √p²= √ (.541) = 0.74

So the frequency of the A allele is 0.74

Q will equal 1-p: 1-(.74)= 0.26

So the frequency of the a allele is 0.26

The 0.541 proportion is representative of the mouse population with Black coats, or AA population. The Homozygous dominant genotype is represented in the HW equation as p². So, we can find p by square rooting p², and q by manipulating the p+q=1 equation.