Question

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 the cross BbTt × BBtt will have black fur and long tails?

a.1/16
d.1/2
b.3/16
e.9/16
c.3/8

Answer 1

Half of the progeny from the cross BbTt × BBtt will have both dominant traits of black fur and short tails.

Step-by-step explanation:

The question involves determining the phenotypic outcome of offspring from a genetic cross where black fur (B) is dominant to brown fur (b) and short tails (T) are dominant to long tails (t). Given the parents' genotypes, BbTt and BBtt, we want to find the fraction of progeny that will exhibit both dominant phenotypes: black fur and short tails.

First, we decipher the possible gametes these parents can produce. The BbTt mouse can produce BT, Bt, bT, and bt gametes, while the BBtt mouse can only produce Bt gametes. Setting up a Punnett square and combining these gametes show that all offspring will have at least one B allele (and thus black fur), due to one parent being homozygous dominant (BB) for the black fur trait.

For tail length, the cross between the Tt and tt alleles of the parents will result in 50% Tt (short tail) and 50% tt (long tail) offspring. Since all offspring will have black fur, we only consider the tail trait. Half of the offspring will have a short tail.

1/2 of the progeny will have black fur and short tails. This straightforward mendelian inheritance problem does not involve epistatic interactions, as none of the alleles presented in the question exert epistasis over the other.