Kathleen's brown cat has kittens. The mother cat is heterozygous for coat color and the father cat is also. White coats are recessive. What are the chances they will have a female, white kitten?

Question

I'm having trouble setting up this problem.

Answer 1

Hi Lexi!

First of all you'll need to set up a punnett square. Generally you do a capital letter for the dominant trait and a lowercase for the recessive trait. Both parents are brown, so they have at least one "brown gene," which we'll call "B" and since they're heterozygous, that means they have at least one "white gene" which we'll call "b" (gotta keep those letters consistent even if it looks kind of silly to call a white coat trait "b", haha). Ok, so let's set up the punnentt square:

| B | b |

B | | |

b | | | Alright, that's sort of a square––now we fill it out, like this:

| B | b |

B |BB| Bb|

b | bB| bb|

This means that 1/4 of the offspring should be homozygous dominant (BB), 2/4 or 1/2 of the offspring should be like their parents––heterozygous (Bb), and then 1/4 of them should be homozygous recessive (bb). If these dogs had 4 offspring, you could expect 3 of them to be brown (the two Bb's and the BB) and one of them to be white (bb). So there's a 1/4 probability or 25% chance they'll have a white puppy. We also know there's a 1/2 or 50% change the puppy will be female, so there will be a (0.5)(0.25) = 0.125 chance they'll have a female white puppy.

Answer 2

This is a classic Punnet Square problem with two ways to solve. Let's break it down:

Genotypes

Mother cat: Ww (heterozygote)

Father cat: Ww (heterozygote)

Now let's separate each allele from each parent and cross them to see all combinations.

(2 alleles x 2 alleles = 4 possible genotypes for hair color in the offspring):

W w

W WW Ww

w Ww ww

Because white coats are recessive, the only way white will be expressed is if the genotype is homozygous recessive for the trait (ww), which is 1/4 or 25% of the possible genotypes.

Multiply this by the 50% probability that the offspring is female: (.25) x (.5) = .125 or 12.5% chance.

Another way is to include the xx and xy chromosomes from the parents into the Punnet Square, making sure to include all possible combinations.

Wx Wx wx wx

Wx WWxx WWxx Wwxx Wwxx

Wy WWxy WWxy Wwxy Wwxy

wx Wwxx Wwxx wwxx wwxx

wy Wwxy Wwxy wwxy wwxy

Again, you're looking for homozygous recessive (ww) female (xx); there are 2 wwxx combinations out of 16 possible.

2/16 = 12.5%