Mendel Genetics?

Question

In Labrador dogs, the dominant gene B determines black coat color and bb produces brown. A seperate gene E shows dominant epistasis over the B and b alleles, resulting in a golden coat. The recessive e allows the expression of B and b. Determine gene types
a. Golden female(dog 1) X golden male (Dog 2)
offspring - 7 gold, 1 black, 1 brown
b. black F(dog 3) X golden male (dog 2)
offspring 8 gold, 5 black, 2 brown

Answer 1

a. Golden female(dog 1) X golden male (Dog 2)
offspring - 7 gold, 1 black, 1 brown
b. black F(dog 3) X golden male (dog 2)
offspring 8 gold, 5 black, 2 brown

I'd look at this problem by drawing a pedigree. Generation I on the pedigree has dogs 1, 2, and 3 with 2 in the middle because dog 2 is involved in both crosses. Connect the parents and write the offspring in generation II like you always do on a pedigree. Now for the logic:

1. Dogs 1 and 2 are golden, so we know they each have one E allele. _ _ E _ x _ _ E _.
2. Dogs 1 and 2 have two offspring that are not golden (ee), so we know the parents each have one e allele. _ _ Ee x _ _ Ee.
3. Dogs 1 and 2 have one brown offspring (bb), so we know they each have one b allele. _ b Ee x _ b Ee.
4. Dogs 1 and 2 have one black offspring (B_), so we know that at least one of the parents has B, but we don't know which one. So we have to leave that blank for now.
5. Dog 3 is black, so it must be B_ee.
6. Now we are looking at the parents 2 and 3: _bEe x B_ee. Half of their offspring are golden, and that fits a cross of Ee x ee, so the golden alleles are correct.
7. Lets look at their non-golden offspring which are 5 black and 2 brown. That's close to a 3:1 ratio. Remember that we get 3:1 when we cross two heterozygous parents: Bb x Bb. Fit those alleles into the blanks for dogs 2 and 3: BbEe x Bbee. Their offspring are all logical for these parental genotypes.
8. Go back to the cross involving dogs 1 and 2, assuming that dog 2 is BbEe -- which we just reasoned out in step 7.
Dog 1 x 2: _ bEe x BbEe.
9. Among the non-golden offspring from dogs 1 and 2 there is one black and one brown. This isn't enough to be certain, but we work with what we have. To get a 1:1 ratio, one parent must be heterozygous and the other must be homozygous recessive. We've already decided that dog 2 is heterozygous Bb. So dog 1 must be homozygous recessive bb. Dog 1 x 2: bbEe x BbEe.

There are your answers:
Dog 1 bbEe
Dog 2 BbEe
Dog 3 Bbee

These problems are like a little treasure hunt. Just start with the most obvious, little, easy bit; fill it in; and go from there. It's usually easiest to start with the recessives because they are homozygous for sure, but each problem presents a little different scenario.