In cocker spaniels, black coat color (b) is dominant over red (b), and solid color (s) is dominant over spotted (s).
(a) if a red solid male was crossed with a black solid female to produce a red spotted puppy, the genotypes of the parent (with male genotype first) would be?
(b) if red spotted male was crossed with a black solid female and all the offspring from several crosses, expressed only the dominant traits, the genotype of the female would be?

a) male : bbSs , female : BbSs

b) BBSS

Explanation:

a) Male is red and solid so it can either have bbSS genotype or bbSs genotype. The puppy is spotted though (ss) which means it got one recessive s allele from male parent. Hence, male's genotype will be bbSs.

Female is black and solid. Both of these are dominant traits so she can either be homozygous for both, heterozygous for one or heterozygous for both. Since the puppy is red and spotted (bbss) it got recessive alleles b and s for both the genes from female parent too. Hence, female has genotype BbSs.

b) Black solid female can either be homozygous for both traits, heterozygous for one or heterozygous for both. She mated with a red spotted male (bbss). All the offspring will express only dominant traits when the female has dominant homozygous genotype for both traits.

BBSS X bbss :

       BS

bs  BbSs

As evident, resultant progeny has BbSs genotype which means all of them will be only black and solid. Hence, genotype of female would be BBSS.

answer: a) a compound microscope has a greater magnification ability than a simple microscope.

explanation: