EXAM TOMORROW! PLEASE HELP! 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? Show your work.
(1)aabbccdd
(2) AaBbCcDd
(3) AABBCCDD
(4) AaBBccDd

Question

EXAM TOMORROW! PLEASE HELP! 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? Show your work.
(1)aabbccdd
(2) AaBbCcDd
(3) AABBCCDD
(4) AaBBccDd

followthefez · Answer

Normally I would just draw a punnett square but with 4 genes?! That would take ages! And in an exam that's not good. 
Is there an easier way to do this?

anonymous · Answer

l.	Just remember that the probability of a heterozygote (Xx) = 2/4  or ½  and the probability of a homozygote XX or xx = ¼

Then just take each genotype and substitute the probability (either ½ for a heterozygote like Aa, and ¼ for the homozygote such as BB). After you have substituted the probabilities for each trait (ABCD) for each genotype, then multiply across to get the probability for the each one.
AaBbCcDd = ½ x ½ x ½ x ½ = 1/16 
AABBCCDD = ¼ x ¼ x ¼ x ¼ = 1/256
AaBBccDd= ½ x ¼ x ¼ x ½ = 1/64
AaBBCCdd = ½ x ¼ x ¼ x ¼ = 1/128

followthefez · Answer

You sure? Because I think the AaBbCcDd frequency is meant to be 0.4%?
Got that off a punnett square calculator. Could be wrong but want to make sure.

anonymous · Answer

Yes, I am sure - because I know that the odds of producing a tetrahybrid from a tetrahybrid cross are 0.0625 (or 1 in 16) and AaBbCcDd is a tetrahybrid. .4 would be EXTREMELY high - that means that 40% of the progeny would be tetrahybrids.

anonymous · Answer

First determine the frequency of inheriting each genotype individually:

Aa x Aa: AA (25%), aa (25%), Aa (50%)

Bb x Bb: BB (25%), bb (25%), Bb (50%)

Cc x Cc: CC (25%), cc (25%), Cc (50%)

Dd x Dd: DD (25%), dd (25%), Dd (50%)

The just multiply the frequencies together for each genotype to get the overall frequency:

1. aabbccdd = 25% x 25% x 25% x 25% = 0.4%  (1/256) 1 / 256 = .0039 
2. AaBbCcDd = 50% x 50% x 50% x 50% = 6.25% (1/16)  1 / 16 = .0625
3. AABBCCDD = 25% x 25% x 25% x 25% = 0.4%(1/256) 1 / 256 = .0039 
4. AaBBccDd = 50% x 25% x 25% x 50% = 1.6% (1/64) 1 / 64 = .015625
5. AaBBCCdd = 50% x 25% x 25% x 25% = 0.8% (1/128) 1/ 128 = .0078125

abb0t · Answer

Is this for your midterm or final?

followthefez · Answer

Okay, that makes sense. Cheers. Final. :(

followthefez · Answer

Everyone I've talked to so far has a final tomorrow! The world is full of miserable and worried students!