In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a+). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t+) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that exactly two will have apricot eyes and tan coats? Use three decimal places for the answer.

Homozygous means having only a single type of allele for a trait. The mouse is homozygous for brown eyes (a+a+) and black coat (t+t+) color which means its alleles are a+a+t+t+.

The mouse with apricot eyes (aa) and tan coat (tt) has an allele set of : aatt

When a+a+t+t+ mouse is crossed with aatt mouse, all of the resulting F1 mice will be having an allele set of aa+tt+.

Then, two mice having allele set of aa+tt+ are crossed to produce F2. The parent set of alleles is a, a+, t, t+ and a, a+, t, t+

The possible allele sets for the eyes can be: a+a+, aa+, a+a, aa

The possible allele sets for the coat color can be: t+t+, tt+, t+t, tt

So, the probability of a mouse with apricot eyes and tan coat is:

P(aa) . P(tt)

From the above listed possible sets of alleles, P(aa) = 1/4 and P(tt) = 1/4

So, the probability of a mouse having apricot eyes and tan coat is:

1/4 x 1/4 = 1/16

To calculate the probability that exactly 2 out of 8 mice will have apricot eyes and tan coat can be calculated using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where p = probability of a mouse having apricot eyes and tan coat

q= 1-p

n = total number of mice

x = number of mice having apricot eyes and tan coat

We have p = 1/16, q= 15/16, n = 8, x = 2. So,

P(X=2) = ⁸C₂ (1/16)² (15/16)⁶

= (28) (1/256) (0.6789)

= 0.074

The probability that exactly two out of eight mice will have apricot eyes and tan coats is 0.074