3
$\begingroup$

I have a question from a past exam that asks the following:

"For the organism of genotype AABbCcDDEe (homozygous for 2 genes, heterozygous for 3 genes), how many different types of gametes will be made if the C gene and the E gene are linked on the same chromosome while the remaining genes are all on different chromosomes?"

However, I have no answer key, and these answers are the choices:

A. 2

B. 3

C. 4

D. 8

E. 9

Here's my approach to the question:

We have 5 traits, and each trait consists of 2 alleles. To make functional gametes, each gamete must receive one of each allele so that it can survive. Clearly, for AA and DD, there's only one choice: A, and D.

However, for Bb, Cc, and Ee, things are a little different.

C and E are linked on the same chromosome. c and e are on different chromosomes. Therefore, a gamete can have either CE, or c and e. Similarly, a gamete can have either B or b.

Therefore, shouldn't we be able to create 4 different types of gametes?:

ABCDE, ABcDe, AbCDE, AbcDe

So wouldn't the answer be C. 4?

Please show me any mistakes I might be making, or any gaps in my understanding. This would be very much appreciated.

Thank you!

$\endgroup$

1 Answer 1

2
$\begingroup$

You are correct

Assuming that by linked, the question assumes perfect linkage (no cross-over ever happen), then your answer is correct! Good job!

Comments

Here are a few comments though...

We have 5 traits, and each trait consists of 2 alleles.

There are 5 loci. A locus does not necessarily match to a single trait and a single trait does not necessarily match to a single locus. Actually, it is very very rarely (if ever) the case.

To make functional gametes, each gamete must receive one of each allele so that it can survive.

This is not necessarily true. A "normal meiosis" (not leading to variation in copy number) may create a non-viable genotype and a "abnormal meiosis" (leading to variation in copy number) may create a viable genotype. But, it is true that we will only assume "normal meiosis" for this answer.

$\endgroup$
1
  • $\begingroup$ Hi there, thank you very much for answering my question and giving additional advice. Sorry that I couldn't get around to accepting your answer and giving feedback straight away. Your answer helped put these concepts I learned into greater context since it seemed like I had a kind of naive understanding of meiosis. $\endgroup$ Nov 30, 2017 at 22:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .