Help with a VERY HARD genetics question PLEASE?

  • Thread starter paradoxium
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  • #1
I'm a first-year college student and I'm havnig a lot of trouble with a particular question. Here goes:

Manx cats are tailess and when crossed with one another produceo n averaeg one long-tailed cat for every 2 Manx. The allele is lethal in homozygous condition due to problems arisiging in development of the lower vertebrae.

Suppose an island were populated with Manx cats.

(a) What would the ferquency of Manx allele be in first generation zygotes produced by the founders? Indciate as completely as possible your argument for this and the next question?

(b) What would the frequency of the Manx allele be in breeding age individuals of thsi generation?

(c) What process is under study in this change in allele frequencies (one sentence or phrase only)?
 

Answers and Replies

  • #2
What are your own answers so far?
 
  • #3
For (a), my friends and I came up with:

Aa X aa --> 2 tailess (Aa) + 2 long-tailed (aa)
Manx allele = A (dominant)

Frequency = 2A/(6a + 2A) = 2/8 = 1/4 = 0.25

For (b):

The number of individuals can be calculated by:

2Aa = 2 X 0.25 (1-0.25) = 0.375

For (c):

Genetic drift??

I'm not sure if any of the our answers are right. Any help will be much appreciated. THANKS!!
 
  • #4
Moonbear
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I'll offer some help with part (a) and see how things go from there. You're right that Manx cats are heterozygotes, and in this example appears to be a dominant trait. However, take careful note of the founders for part (a). The problem states that the island is populated with Manx cats, not a mix of Manx and cats with tails. So your founders are all heterozygotes, not a cross of heterozygotes and homozygotes. See how that affects your calculations.
 

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