in a certain set of lab mice, black hair is dominant to white hair, and course hair is dominant to fine hair.?

Question

Determine the % of black, fine-haired offspring that will result if 2 mice heterozygous for both of these traists are mated. Draw a pummet Square to support your answer.

Answer 1

The percentage is 18.75%, or 3/16. I don't feel bad giving you that answer, because it's the easy part. The hard part is drawing the Punnett square (not the "pummet Square"). I'll give you a little help with that, too.

Heterozygous means that each parent has one chromosome with the dominant allele and one with the recessive allele. For color, we'll call the dominant allele "B" for black and the recessive allele "b" for white. For texture, we'll call the dominant allele "C" for coarse, and the recessive will be "c" for fine.

(Why not "w" for white and "f" for fine? Because everyone usually does it my way, most likely including your teacher, so you'll have to do it this way, too.)

So if each parent is heterozygous for both traits, then each parent's genotype is BbCc.

In meiosis, each parent will produce gametes with one of four possible genotypes: BC, bC, Bc, or bc. Each will occur with the same frequency.

So the question is, how many possible combinations of the two parents' gametes are there? That's where the Punnett square comes in, and that's where I'll stop.

You take it from here.

Answer 2

This sounds like a question from a school biology book. Read back over it and figure it out on you're own.

Answer 3

BH bh There is a zero percent chance.

BH BBHH BbHh

bh BbHh bbhh

B and b represent hair color
H and h represent hair type