33 out of 4300 people have autosomal recessive disease.What is allele frequency for each allele?

Question

Lister's Disease is an autosomal recessive disease and was studied in the Inuit population of Canada. It was found that 33 individuals have the disease of the 4300 individuals studied. What is the allele frequency in this population for each allele?
What is the frequency of carriers of the disease in this population?

Answer 1

Frequency of recessive allele r in pop 4300 (8600 alleles)

rr 33 people for 66 alleles
Rr & RR for 4267 people (8534 alleles) RR+Rr=8534

f(r)=(rr+1/2Rr)/(RR+Rr+rr)=(66+1/2Rr)/8600=q
f(R)=(RR+1/2Rr)/ (RR+Rr+rr)=(RR+1/2Rr)/8600=p

(66+1/2Rr)/8600+(RR+1/2Rr)/8600=1= p+q

f(Rr)=2pq

Answer 2

Sorry about this, this is gonna be a long answer, and I'm not sure if I can explain it easily without drawing a punnets square diagram. Right, If we first of all name the different alleles so you know what I'm talking about: XC = X chromosome, normal wings Xc = X chromosome, cut wings Y = Y chromosome (Because males have one X and one Y) F = Not Fuzzy f = Fuzzy Ok, first of all: Genotype of the male = Xc,Y,F,F it has cut wings, and a normal body, and is male. Genotype of the female = XC,XC,f,f it has normal wings and a fuzzy body Now, with these genotypes, the male can produce Xc,F and Y,F gametes in equal proportions, and the female produces all XC,f gametes. When you put these into a 4x4 punnets square you get an F1 generation all with the phenotype of normal wings and a normal body (the wild type), however they will have within their genotype the recessive genes that their parents carried, they will have the genotypes Xc,XC,F,f if they're female, and the genotype XC,Y,F,f if it's male, with equal proportions of male and female. When you cross the F1, your gonna have to cross male and female (obviously) so there will be male gametes XC,F XC,f Y,F and Y,f, and female gametes Xc,F Xc,f XC,F and XC,f. These can then also be put into a 4x4 punnets square and give offspring with the genotypes Xc XC F F, Xc XC F f, XC XC F F, Xc XC f f, XC XC F f, XC XC f f, Xc Y F F, Xc Y F f, XC Y F f, XC Y f f, Xc Y f f and XC Y F F in the ratios 1:3:1:1:1:1:1:2:2:1:1:1, giving the phenotypes of Normal, Cut wings-normal body, Normal wings-Fuzzy body and Cut wings-Fuzzy Body in the ratios 9:3:2:2. Hope that explains how to get to the answer well enough, This really isn't the easiest way to explain genetic crosses like that. I apologise now that there may be some error it my calculations, it is quite early here, and I was using a different notation when working it out, so many not have copied it onto here correctly. Edit: Looked at this when more awake, same as below, my working is right, but I agree with him, 9:3:3:1, but please, look through this for yourself, as he and I have both said, and done, this stuff is easy to trip up on

Answer 3

I am rusty on this, so I have to think it through:

let r be the proportion of the recessive gene, then rr is the proportion of the homozygous recessive in the population, rr we are given is 33/4300.

sqrt(33/4267)=0.0876

Now let R be the proportion of the dominate gene in the populations (assumed equal with men and women).

R = 1 - 0.0876 = 0.9124
RR in population is 0.9124^2 = 0.8324


the proportion of Rr is 2*R*r = 2*0.8324*0.0876 = 0.1458

Apparently the above is correct, my old genetics book is falling apart, so I resorted to Wiki and looked up the Hardy-Weinberg principle, so that is the topic which pertains to this problem. You may need more detail than given here to explain it.

Please check for errors, as I say, I am rusty on this.