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.

Question

Mathematics

Tobias Mejia

27 January, 21:24

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.

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Kaelyn Allison · Answer 1

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

Step-by-step explanation:

Apricot Eyes = a (Recessive Allele)

Brown Eyes = a + (Dominant Allele)

Tan Coat = t (Recessive Allele)

Black Coat = t + (dominant Allele)

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