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all of the above are true

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).