Tetrahybrid Cross help!!!?

Question

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?
a.)aabbccdd
b.)AaBbCcDd
c.)AABBCCDD
d.)AaBbccDd
e.)AaBBCCdd

i know it will be a 16 x 16 punnett square. But I don't know what to cross. Can you please help me? thanks.

Answer 1

You can set up the 16x16 square with these gametes:
ABCD
ABCd
ABcD
ABcd
AbCD
AbCd
AbcD
Abcd
aBCD
aBCd
aBcD
aBcd
abCD
abCd
abcD
abcd

You could also figure these probabilities independently like this. Consider the first genotype aabbccdd. If you cross Aa x Aa, what's the chance you get aa? It's 1/4 if you look at or imagine a 2 x 2 square. What's the chance that Bb x Bb yields a bb? 1/4. Chance of cc? 1/4. Chance of dd? 1/4.
What's the chance of all four of these happening in the same individual? In probability "and" means to multiply.
Chance of aa AND bb AND cc AND dd in same individual?
= 1/4 * 1/4 * 1/4 * 1/4 = 1/256 That's the answer.

Let's do letter d.
Chance of Aa and Bb and cc and Dd in same individual?
= 1/2 * 1/2 * 1/4 * 1/2 = 1/32

That's way easier than the 16 x 16 square, and fewer chances of making a mistake.

Edit: the way bioguy did the problem (above my answer) is fine except he added 1/16 + 1/16 instead of multiplying them. You have to multiply if both parts have to be true. One parent donates ABCD AND other parent donates ABCD; 1/16 * 1/16 = 1/256.

Answer 2

If you really want to draw it out, you'd have to figure out the possible gametes:
ABCD, AbCD, ABcD, ABcD, etc.
You'd be better off calculating it mathematically.
There's a 1/2 chance of A being in a gamete, a 1/2 of a and so on. So ABCD's chances are 1/2 x 1/2 x/ 1/2 x 1/2 which = 1/16. That means the other parent's chances of creating a similar gamete is 1/16 as well. So to get choice C, AABBCCDD, is 1/32.

Answer 3

Aabbcc X Aabbcc Punnett Square