contestada

In mice, black coat color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F1 mice are black and solid. A testcross is then carried out by mating the F1 mice with brown, spotted mice. a. Give the genotypes of the parents and the F1 mice. b. Give the genotypes and phenotypes, along with their expected ratios, of the progeny expected from the testcross. Testcross: Black, solid (BbSs) X Brown, spotted (bbss)

Respuesta :

jbferrado

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).

Otras preguntas