Genes calculating frequencies (Hardy weinburg equilibrium)

[motivationG]

Ok so I have the question and the answer but I dont understand how the answer was achieved..anyone who can help me I will appreciate it so muchh!!


1) (a) 1. A1 and A2 are two alleles at an autosomal locus. The three genotypes in a population are distinguishable by their phenotypes.

(a) Calculate the frequency of the A1 allele in populations I and II

A1A1 A1A2 A2A2

I 47 26 7

II 19 42 39



And the answer given to be is:
Population I, p = ((2 × 47) + 26)/160 = 0.75 = freq. A1
Population II, p = ((2 × 19) + 42)/200 = 0.40 = freq. A1

I'm so confused where do 160 and 200 (in bold) come from??? Also why are we multiplying by 2??

I'm soo stressed this is not making sense

[darkido]

hello there!!

THis is what i make out of it:

160 and 200 are the total number of alleles in the populations I and II respectively. This can be obtained by multiplying 2 for all genotypes in each population.(ie each genotype (eg. A1A1) is made up of two alleles) So total genotypes for population I is 80 and 80x2=160. Population II, 100x2=200.

We multiply 47 and 19 by 2 because they are the frequency for genotype A1A1 and we are trying to find the allele frequency for A1. (ie. 2 A1 alleles in a A1A1 genotype). However in an A1A2 genotype, we get only one A1 allele, so we simply use the frequency of genotype A1A2. (ie. 1 A1 allele in a A1A2 genotype).

Add the A1 frequencies up and then divide it by the total number of alleles you have in the population then you get your A1 allele frequency in each population.

hmmm that should be how it goes...hope it helps!

[kotoreru]

You've just got the follow the numbers on this one, really. I still hate these questions.

But as previous poster says:

160 comes from (47x2)+(26x2)+(7x2).
Same for the 200.

You're just trying to find the frequency of the allele in that number.

You count the 47 twice because both parts are A1, and the 26 once because it's half A2. Dividing by the total gives you the frequency.