A man has the genotype AaBbCc, in other words he is heterozygous in three uncoupled loci on three different chromosomes. A, B and C are the dominant alleles while a, b and c are the recessive alleles.

1. How many gametes can he form?

The answer is 8 and I can see how you derive that by just manually writing each and every one down but what is the mathematical method to derive this? I tried permutations but none I came up with worked.

2. He has a child with a woman of the same genotype (AaBbCc) how many different genotypes can the offspring have?

Here the answer is 27. I'm not sure how they came to this conclusion either?

3. What is the probability that the offspring has the genotype AaBbCc?

This is about the time I start flipping tables because it seems to me that it's 1 out of 27. There are 27 different genotypes and this is ONE of them?!

But the answer is 1/8 as there's a 50% chance of Aa, Bb and Cc when mixing apparently so it becomes 1/2*1/2*1/2. Why isn't it just 1 of the possible genotypes?


1 Answer 1

  1. Since there are 2 alleles of each gene and 3 different genes, 2EE3 = 8? Let's test this by making a table, for 1 gene there are two different kinds of gametes, so 2EE1 = 2, for 2 genes there are 4 different kinds of gametes (AB, Ab, aB, ab) so 2EE2 = 4, and you already did the calculation yourself for 3 genes, and 2EE3 = 8.

Why don't you try working it out yourself for 4 genes and see if you come up with 16 different kinds of gametes?

  1. Two words: Punnett Square. In this case for a trihybrid cross:

triple-het x triple-het cross found on Google Image

  1. That is a Bernoulli coin-flipping probability. If you are flipping a true coin three times in a row, how often will you get three heads in a row? 1/8? All the events are independent, right? what about the Punnett square? Does it give the same result somehow?

Here is the answer worked out for you.


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