Answered By Tyleigh M

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.

The .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. 

1. Calculate the frequency of each allele. Show all your work and express your answer as a value between 0 and 1 rounded to two decimal places

P will equal the square root of p²: √p²= √ (.541) = .74

So the frequency of the A allele is .74

Q will equal 1-p: 1-(.74)= .26

So the frequency of the a allele is .26

2. What percentage of the mouse population is expected to be carriers of the lethal allele?

This is asking for the % of the population that have the yellow coat ie: Aa

This is represented by 2pq in the HW equation.

So 2pq = 2(.74 x .26) = .38 rounded to .4

.4 x 100% = 40% of the population.

3. How many mice will die during fetal development?

This is asking for the proportion of aa in the population. This is represented by q² in the HW equation.

So q^{2 =} (.26)² = .0676

To determine how many mice this accounts for, simply times 0.0676 by the mouse population (2000)

2000x0.0676= 135 mice will die during fetal development.

** Remember: you can verify your work by using the HW equation. If you add your calculations together, you should always receive 1. 

EX: p²+2pq+q²=1

(.541)+(.4)+(.0676)= 1

*Note: If using rounded numbers, you will receive a value that is close, but not equal to 1.