help with genetics problem

Question

  A horse’s coat color is controlled by many different gene pairs. A dominant B will give the horse a black color, and a b will give the horse a chestnut color. However, when a dominant W is present in the second gene pair, the horse will be white no matter what. If the second gene pair is ww, then the horse will be the color of whatever the first gene pair is made of.

   If a heterozygous white (BbWw) mare is crossed with a heterozygous white (BbWw) stallion, what could the offspring's phenotype be?

Answer 1

Hi Ign. Genetics is fun!

First you need to look at your cross:

a heterozygous white (BbWw) mare is crossed with a heterozygous white (BbWw) stallion

The Punnett square for the cross would look like this:

CROSS	BW	bW	Bw	bw
BW	BBWW	BbWW	BBWw	BbWw
bW	BbWW	bbWW	BbWw	bbWw
Bw	BBWw	BbWw	BBww	Bbww
bw	BbWw	bbWw	Bbww	bbww

When you count up:

1. 12 / 16 or 75% would have at least 1 dominant W, so they would be white
2. 3 /16 or 18.75% would have the recessive ww pair, so the color would be determined by the Bb, so they would be black
3. 1 / 16 or 6.25% would be double recessive bbww, so they would be chestnut.

Answer 2

So when both parents are heterozygotes, the crossing will always end with a 9:3:3:1 ratio of offspring. This means that nine of the offspring will have the two dominants alleles, three will have one dominant allele and one recessive allele, three will have the reverse of the other three, and one will be recessive with both genes. So, the genotypes for this example would be:

9: BBWW, BbWW, BbWW, BBWw, BBWw, BbWw, BbWw, BbWw, BbWw

3: Bbww, Bbww, BBww

3: bbWw, bbWw, bbWw

1: bbww

We know that when a dominant W is present, the horse will be white. So, 12 of the 16 above will be white. 3 will be black, and one will be chestnut. Here's a link that shows what the Punnett square would look like for a cross of two heterozygotes regarding two genes:

https://www.nature.com/scitable/topicpage/some-genes-are-transmitted-to-offspring-in-6524945/