How would you do a cross between...?

Question

an heterozygous individual for both Huntingtons (autosomal dominant trait) and PKU (autosomal recessive trati) that marries an individual without Huntingtons, but is heterozygous for PKU.

Lets say that
H= Huntington allele
h=non Huntington allele
P=no PKU
p=PKU allele

Answer 1

The heterozygous individual for both Huntingtons and PKU has a genotype of HhPp. As for the individual without Huntingtons, but is heterozygous for PKU her genotype is hhPp.

So nce again, the first thing is to make out the different types gametes. The first produces HP Hp hP hp spermatozoids and the second produces only hP and hp eggs.

Your Punnett square will have two columns and four rows.
____|__hP__|__hp_
HP_|_HhPP_|_HhPp
Hp_|_HhPp_|_Hhpp
hP_|_hhPP_|_hhPp
hp_|_hhPp_|_hhpp

There you go. All that is left to do is the interpretation of the 8 different possibilities.

Good Luck!

Answer 2

Setting up this cross is pretty standard. Good job setting up the traits, you are right on there.

Our first individual, heterozygous (otherwise known as a het) for both would take one of each form of the allele. That is, for Huntington they would be able to generate a H and an h gamete, so they are Hh. We also know they are a het for PKU, so they would be Pp. (this dominant and recessive pattern is standard for a het.) This makes your first individual the following: HhPp.

Your second individual does not have Huntingtons, so they would have to be recessive in both alleles, which would be hh. (A single dominant allele in that case would cause the disease, such as Hh or HH). For the PKU, we know they are heterozygous, so that makes them Pp (note that we always put the dominant allele first in a heterozygous pair). This individual would be described as hhPp then.

This makes your final cross to be: HhPp x hhPp

I'll let you figure out how to do the Punnett square.

Answer 3

Hhpp + hhPp
Hhpp ------> 1/2 Hp & 1/2 hp
hhPp ------> 1/2 hP & 1/2 hp

Kids: 1/4HhPp, 1/4Hhpp, 1/2hhpp