which solution is really true?

Question

A population has 3 genotypes distributed as follows:
FF200, Ff0, ff800. What would be the Hardy-Weinberg equilibrium genotype frequencies in this population? Show values for p and q?
someone show me how to do this but each was different, which one is really right?
F2+Ff+f2
400/2000 =.2 F
1600/2000=.8F
.2^2= .04FF
2 x .2 x .8= .32 Ff
.8^2= .64 ff
------------------------------... or this one?
p2+ 2pq+q2=1
p+q=1
200+800 = 1000
200/1000= .20 (p2)
800/1000= .80 (q2)
2pq= 0
p= .44 (root of .2)
q=.89 ( root of .8)

Answer 1

The first one is correct.

The error in the second one is that it assumes that the population is already in equilibrium: notice that there are no heterozygotes in the final answer? It assumes that 2pq = 0 in the H-W equilibrium equation, which is incorrect. All populations are not necessarily at equilibrium; they rarely are, in fact, thanks to environmental pressures, disease and human interference.

The first one is correct because it determines allele frequency first and then finds the genotypes at equilibrium. Why use 400/2000 and not 200/1000: 400 represents the number of F alleles. In a population with 1000 genotypes, there are 2*1000 = 2000 alleles.

I think there's a type; it should be 1600/2000 = 0.8f, not 0.8F.

Anyway, this explanation uses the frequencies in the H-W equation, which then gives you some heterozygotes. Ta-da! Remember, always find frequencies first, then find genotypes.

Answer 2

The two answers hinge on whether or not the population is currently at Hardy-Weinberg equilibrium or not. You have to decide if the question is asking if this population is in equilibrium now or if it is *going* to be in equilibrium and you want to know the frequencies when it gets there.

If the population is not currently at Hardy-Weinberg equilibrium and is going to be, then the first answer is correct. You are looking for allele frequencies and calculating what they will be at equilibrium. Lamictalfan explained this well.

However, if you assume the population is at Hardy-Weinberg equilibrium (for instance, if the heterozygote is lethal), then, the second answer is almost correct. You got the correct answer for q, but not for p. However, there is a problem when you get through the calculations, which we'll see shortly. Here they are:

FF 200
Ff 0
ff 800
Total: 1000

let p be the frequency of the F allele and q be the frequency of the f allele, so:

q^2 = 800/1000 = 0.8
q = 0.8944

since p + q = 1
p = 1 - q = 1 - 0.8944 = 0.1056

Therefore, in this population, the frequency of the F allele is 0.1056 and the frequency of the f allele is 0.8944

However, as ecolink (and Lam) mentioned, the frequency of heterozygotes should be 2pq. The problem says the frequency of Ff is 0. So, 2pq = 0, but from the calculations, 2pq = 2 * 0.1056 * 0.8944 = 0.1889, which clearly does not equal zero. So, the population must not yet be at equilibrium and so, the first answer is the one to use and is the correct one.

Answer 3

Here's my take on this question.

If the population has 200 FF and 800 ff, but NO Ff ... then it looks like the Ff individuals just did not survive. There's no Punnett square you can make that has only FF and ff in the boxes. So I don't think this population can be represented by Hardy-Weinberg unless you figure in all the heterozygous Ff that didn't survive.

I think that the first solution you have printed above has a problem. The question says that FF is 200/1000, not F=200/1000. So p^2 =0.2 and p = sqrt 0.2. Same for ff.

The second solution you have printed above also has a problem. I agree that p^2 = 0.20 and q^2 = 0.80. But mathematically 2pq cannot be 0 because neither p nor q is equal to zero. Also, if p= 0.44 and q=0.89, p + q does not add up to 1.

So ask your teacher if this was a question without a solution since the numbers do not fit Hardy-Weinberg at all.

Answer 4

The second one im pretty sure,
:D ^_^