Answer and Explanation:
It is a dihybrid cross, where:
B= black (dominant)
b= brown
S= solid (dominant)
s= spotted
a) The genotype of the parents are:
homozygous black spotted= BBss ⇒ Gametes: Bs, Bs, bS, bS
homozygous brown solid= bbSS ⇒ Gametes: bs, bs, bs, bs
The F1 mice is the resultant progeny from the cross BBss x bbSS. We combine the games of the parents and all combinations are : BbSs. All F1 genotype is BbSs (heterozygous black and solid mice).
b) The testcross is carried out between F1 mice (BbSs) and brown spotted mice (bbss)
BsSs ⇒ Gametes: BS, Bs, bS, bs
bbss ⇒ Gametes: bs, bs, bs, bs
From the gametes, we obtain all the possible combinations:
BSbs = BbSs (black, solid) ⇒ ratio= 1/2 x 1/2= 1/4
Bsbs= Bbss (black, spotted) ⇒ ratio= 1/2 x 1/2= 1/4
bSbs= bbSs (brown solid) ⇒ ratio= 1/2 x 1/2= 1/4
bsbs= bbss (brown spotted) ⇒ ratio= 1/2 x 1/2= 1/4
So, we obtain 1/4 of black solid mice (BbSs), 1/4 of black spotted mice (Bbss), 1/4 of brown solid mice (bbSs) and 1/4 of brown spotted mice (bbss).
