Black fur in mice (b) is dominant to brown fur (b). short tails (t) are dominant to long tails (t). what fraction of the progeny of crosses bbtt × bbtt will be expected to have black fur and long tails?
a) 1/16
b) 3/16
c) 3/8
d) 1/2
e) 9/16

8/16 ~ 1/2

Explanation:

This cross is a dihybrid cross involving two genes: one coding for fur colour (B) and the other for tail length (T).

The allele for black fur (B) is dominant over the allele for brown fur (b) while the allele for short tail (T) is dominant over the allele for long tail (t).

In a cross involving BbTt × BBtt, according to the law of independent assortment of genes, the alleles of the two genes get sorted into gametes independent of one another.

BbTt undergoes meiosis to produce BT, Bt, bT, and BT

BBtt undergoes meiosis to produce Bt, Bt, Bt, and Bt

The gametes are crossed using a punnet square (see attachment) to produce 16offsprings.

The progenies expected to have Black fur(B) and long tail (t) will have these genotypes: BBtt and BBtt

Since the long tail (t) is recessive, it will only be expressed when the two alleles are homologous.

Looking at the punnet square from the attachment, one will realize that 8 (circled) out of 16 progenies are black fured and long tailed i.e 1/2 .

hope this !

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