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)

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).

1. (1.2kw)x(15/60hours) = .3kwh --> a

2. (5kw)x(90/60hours)x(8cents/kwh) = $0.60 --> d

3. microwave --> 1200 watts = (1.2kw)x(6/60hours) = .12kwh

coffee-maker --> 900 watts = (.9kw)x(10/60hours) = .15kwh

therefore, it requires more energy for the coffee-maker to do the same amount of work.

so the answer is a.

the answer to your question is b