Solve Trihybrid Cross Problem: Phenotypic Ratio 27:9:9:9:3:3:3:1

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The discussion revolves around solving a trihybrid cross problem involving rabbit coat color, determined by three independently assorting loci. The expected phenotypic ratio from the cross AaTtBb X AaTtBb is confirmed to be 27:9:9:9:3:3:3:1, with the main phenotypes being colorless, tan, and black. Participants suggest using a Punnett square to visualize the combinations, while also discussing alternative methods to streamline the process. The importance of identifying dominant alleles to quickly determine phenotypes is emphasized, particularly for efficient problem-solving in exam conditions. Ultimately, the participant successfully applies the method and confirms their understanding of the phenotypic ratios.
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[SOLVED] Trihybrid Cross Problem

Hi, I need help trying to figure out how to solve this question:

Assume that in rabbits there are three different independently assorting autosomal loci that affect coat color. A colorless-pigment precursor V is converted to a colorless precursor W by action of the A allele. W is converted to a tan pigment by action of the T allele, and the tan pigment is converted to black pigment by action of the B allele:

A- T- B- (reaction proceeds)
V-----> W-----> tan-----> black
aa tt bb (reaction blocked)

The homozygous recessive condition at each locus results in loss of enzyme activity for the reaction controlled by that gene. A cross of trihybrid bunnies (AaTtBb X AaTtBb) would be expected to give rise to what phenotypic ratio in the offspring?


I know the phenotypic ratio is 27:9:9:9:3:3:3:1. I can't figure out how to work this out. Can someone help me?
 
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Have you tried setting up a Punnett Square yet?

Also, the phenotypic ratio will not be the same as if each loci was encoding a completely different trait. Instead, the alleles at each loci are dependent upon the other two loci for the phenotypic expression pattern. You only have three possible phenotypes: colorless, tan, black.
 
Yes I did set up a punnett square (with 64 squares). I got these set up on the left and top side of it: ATB, Atb, AtB, ATb, aTB, atb, aTb, atB. Did I do this entirely wrong? Also, would the ratio be something like 27:9:28...since there are only 3 possible phenotypes? I don't understand how to get the final answer.
 
You have the right set up for the Punnett square.

Yes, you're much closer now on the ratio. Basically, you have to look at each of the genotypes for the offspring and evaluate in order:
Does it have A_ genotype? Yes, then continue. No, then colorless.
Does it have T_ genotype? Yes, then continue. No, then colorless (Keep in mind that either A or a are still colorless if T and B are not present.)
Does it have B_ genotype? Yes, then black. No, then tan.

Now see if you can figure out the ratio for each color.
 
Also, is there any other way to solve this without using the Punnett square? It took a lot of time to do that.
 
predentalgirl1 said:
Also, is there any other way to solve this without using the Punnett square? It took a lot of time to do that.

The mathematicians around here would likely tell you yes, there's another way, but it's not likely to be any faster than just sketching out a quick Punnett square. The trick on one like this is not to spend a lot of time writing out every genotype for the offspring. Instead, you can run down a row very quickly by looking at the alleles for a single parent and determining phenotype as you go. For example, in the row and column where each parent has ATB as their alleles, you can mark the entire row as black (the contribution from the second parent is irrelevant if the other is dominant for all three traits). Then, if your next column has ATb for one parent's alleles, you only need to look at whether the other parent has B or b to determine if the offspring are Black or Tan. But, while still learning, I encourage you to keep using Punnett squares so you really see how the combinations are arising...it's easier to double check for mistakes that way too.
 
Moonbear said:
You have the right set up for the Punnett square.

Yes, you're much closer now on the ratio. Basically, you have to look at each of the genotypes for the offspring and evaluate in order:
Does it have A_ genotype? Yes, then continue. No, then colorless.
Does it have T_ genotype? Yes, then continue. No, then colorless (Keep in mind that either A or a are still colorless if T and B are not present.)
Does it have B_ genotype? Yes, then black. No, then tan.

Now see if you can figure out the ratio for each color.


Ok, I use your method with the punnett square and I got 27/64 black, 28/64 colorless, and 9/64 tan. Is this right?
 
Moonbear said:
The mathematicians around here would likely tell you yes, there's another way, but it's not likely to be any faster than just sketching out a quick Punnett square. The trick on one like this is not to spend a lot of time writing out every genotype for the offspring. Instead, you can run down a row very quickly by looking at the alleles for a single parent and determining phenotype as you go. For example, in the row and column where each parent has ATB as their alleles, you can mark the entire row as black (the contribution from the second parent is irrelevant if the other is dominant for all three traits). Then, if your next column has ATb for one parent's alleles, you only need to look at whether the other parent has B or b to determine if the offspring are Black or Tan. But, while still learning, I encourage you to keep using Punnett squares so you really see how the combinations are arising...it's easier to double check for mistakes that way too.

Ok...thanks. I will do that the next time I have a trihybrid problem because writing everything down took up a lot of time and I won't have that much time when I take my test. Thanks very much. :)
 
predentalgirl1 said:
Ok, I use your method with the punnett square and I got 27/64 black, 28/64 colorless, and 9/64 tan. Is this right?

Yep, you got it! :smile:
 
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Moonbear said:
Yep, you got it! :smile:

Yes! Thanks very much for your help. I was stumped on this question for over an hour. Thanks again. :)
 
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