Hardy Weinberg equation help me!

[murt]

-If he dark color allele is recessive and if the population is in equilibrium, how many light colored alleles are there in the collecting site?

Total=80 Light= 37 Dark=43

 

I got 54 but it was wrong.

[imatfaal]

hmm this is the first time I have ever looked at this - so perhaps my meanderings can help or maybe not

 

you have the equation

 

p^2 + q^2 + 2pq = 1

 

or using d and l for dark and light

 

d^2 + l^2 + 2dl = 1

 

we know that 43/80 are dark and are d^2 cos it is recessive

 

(43/80) + l^2 +2dl = 1

 

l^2 +2dl = 37/80

 

we know d = sqrt(43/80) so you have a quadratic in one unknown that you can solve to get l

 

- feel free to tell me I am barking up wrong tree

[chadn737]

Its rather simple. If you have 43 dark colored individuals, then q² = 43/80 (the frequency of homozygous dark alleles). So if you want the value of q you take the _____ of q² (fill in the blank with the appropriate function).

 

Now recall that p + q = 1. So to get p we use the equation p = 1 - q.

 

 

EDIT: earlier I forgot to put in the fact that q² equals the frequency of homozygous dark alleles, not the number of individuals. However, once the value of q (frequency of dark colored alleles) is known, then you automatically know the value p (frequency of light colored alleles) as both should total 1.

 

Then simply plug these frequencies in to figure out the number of alleles.

Edited January 29, 2014 by chadn737

[overtone]

To get 54, it looks as if you set aside the 86 dark alleles in the dark morph from the 160 total; then subtracted 37 alleles in the light morph to account for the light, multiplied the remaining 37 alleles by the ratio of dark morphs, rounded up to 20 and added that to the 86 for your total dark alleles. There would be 54 alleles left to be light - is that how that number came up?

 

That procedure errs in taking the probability of the hidden alleles in the light morph being dark to be the same as the probability of a morph being dark, and would also err if the somewhat more alert attempt of using the overall probablity of a dark allele (root 43/80) were made. The source of the trouble is in the trickiness of calculating the probability of a hidden allele being dark, given the fact that it can be either or none of the pair - if one sets aside half the alleles as being light, the probability calculation acquires complications.

 

Imtfaal has a better approach - note the significance of the 2dl term in the probability calculation.