Question

Hardy Weinberg Question:

You sampled 550 students and found that 352 have a widow’s peak (a dominant trait).
a. What are the frequencies of alleles W and w?
b. What are the genotype frequencies (WW, Ww, and ww)?
c. If the expected allele frequencies are equal, what are the expected genotype frequencies?
d. What does the difference between the observed and expected genotype frequencies is tell you about the population?

So far this is what I have...
2a. 352/550 = 64% Ww or WW
36% = ww
q^2 = 0.36 so q =√(0.36) = 0.6
so p = 0.4

The frequencies of alleles W and w would be:
0.60^2 = 36% for w
0.40^2 = 16% for W

2b. 0.40^2 + 2(0.4)(0.6)+0.60^2 = 1
16% = WW
48% = Ww
36% = ww

2c.

2d.

Answer 1

First, that is a lie about widow's peak.

\[q^2+2pq+q^2=1\] q= frequency of allele 1 (let's use the dom allele) and p= frequency of recessive allele.

352 have dom, so \[550-352=x\] x, is the number of individuals with rec. allele. That means there are 198*2 individuals homozygous dominant, 296 total rec alleles in that portion of the pop.

Can you go from here?

Answer 2

crap, I see I should have read the full post.

Answer 3

You have everything you need to answer that question.

Answer 4

@mrdoldrum can you explain c and d to me? i don't really understand.

Answer 5

@eatacat Well, what are the expected frequencies in a simple dom/rec trait? You have probably done a Punnett square to find the expected genotype frequencies. You can then use that to find the expected allele frequencies.

For D, recall that HW makes certain assumptions about the group in question, like random mating, infinite population size, no selection acting on alleles in question, etc

Answer 6

@mrdoldum
i don't really know how to do c, but i'll figure it out later but for d

d. the differences between observed and expected genotypes may differ because of chance or may indicate that the sample population is not in Hardy-Weinberg equilibrium. under the HW model it has 5 basic principles, and a population's genotype and allele frequencies will remain unchanged over multiple generations - thus not in HW equilibrium [not evolving]. If there are any differences it means that there is a violation of one of principles.

most populations would be under the influence of natural selection and populations will undergo changes in their genetic makeup due to migration/drift/gene flow etc. Mutattions can alter the gene pool too, but most have little effect. Lastly, individuals often mate selecteivly rather than randomly; sexual selection.

Answer 7

If the allele frequencies are equal the genotype frequencies is the normal punnett square 1:2:1 dom/rec Now, yes you could have only WW and ww and fulfill the conditions of C, but that would be a special case and not what we would expect in most systems as it would only be stable for one generation if mating choice was not 100% enforced and had any randomness. So, we fall back on the 1:2:1 genotype ratio that gives a 1:1 allele ratio. Notice in the square that is 4 W alleles and 4 w alleles.

D) If we don't have the expected frequencies than we have violated the strict and unrealistic assumptions of the HW equation, eg, 100% random mating, infinite population size, no mutations, no gene flow, no selection.