In mice an allele for apricot eyes a is recessive to an alle

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 (r) 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 two will have apricot eyes and tan coats?

Solution

Recessive Apricot eyes :a

Dominant Brown eyes:a+

Recessive Tan coat color :t

Dominant Black coat color :t+

1.Homozygous brown eyes and black coat color will have genotype :a+a+t+t+

2,Apricot eyes and tan coat will have genotype :aatt

Cross between 1 and 2 will give F1 : a+a t+ t.......... all offsprings will be brown eyes and black coat color.

F2: a+a t+t * a+a t+t Paternal

Let the probability of apricot eyes and tan coat being formed be denoted as a = 1/16

Let the probability of other phenotype be denoted as b= 15/16

Binomial expression will give the probability that two will have apricot eyes and tan coats as follows:

(a + b)⁸ = a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸

From the above binomial expression,the term 28a²b⁶ term gives the probability of having two apricot eyed and tan coated mice.The probability of a is 1/16 and the probability of b is 15/16. Therefore, the probability is 28(1/16)²(15/16)⁶ = 0.074. (ans )

maternal	a+t+	a+t	a t+	a t
a+t+	a+a+t+t+	a+a+t+t	a+a t+t+	a+a t+t
a+t	a+a+t+t	a+a+t t	a+a t+t	a+a t t
a t+	a+a t+t+	a+a t+t	aa t+t+	aa t+t
a t	a+a t+t	a+a t t	aat+t	aa tt apricot eyes and tan coats