Question

Two populations of frogs are in Hardy-Weinberg equilibrium for a gene with two alleles, M and m. The
frequency of m alleles in population 1 is 0.2 and 0.4 in population 2. If there are 100 frogs in each
population, what is the difference in the number of heterozygous frogs between the two populations?

Answer 1

The answer given in 16, but I got 8..
Anyone show me how the ans is 16?

Answer 2

Please show us your math. Were you explicitly given the frequency of the recesive allele in the problem or is that your calculation or interpretation of some other data?

Answer 3

I'm juz self learning this principle, so I guess I did it wrong somehow....

so m is 0.2
mm is 0.04
M is 0.8
MM is 0.64
Mm is 0.16 ( 0.8-0.64)

Answer 4

the frequency is given , but i am not sure whether it is for mm or just m itself

the 2nd population

m is 0.4
mm is 0.16
M is 0.6
MM is 0.36
Mm is 0.24

Answer 5

You have to use the castle hardy weinburg (CHW) equation:

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

where p^2 is the frequency of homozygous dominant (MM) individuals, 2pq is the frequency of heterozygotes (Mm) and q^2 is the frequency of homozygous recessive (mm) individuals.

Here the interpretation gets sticky. If the question explicitly told you that that "the frequency of the recessive allele is 0.2 and 0.4" then q = 0.2 and 0.4. But if the question told you that "there frequency of homozygous recessive individuals is 0.2 and 0.4 respectively" then q^2 = 0.2 and 0.4.

So we are all stay on the same page, could you confirm that for me...

Answer 6

yea i know . but I watched the khanacademy video, he basically juz subtract MM from M to get Mm

..and yea..give me some time to understand what u are trying to say :D

Answer 7

well, then i think it is the same as what i did
q is 0.2

but the equation is q^2, so u need to square that, which is 0.04
since q is 0.2, p is 0.8 , square that u get 0.64

Answer 8

So it told you that "the frequency of m is 0.2 in population 1 and 0.4 in population 2." Cool. Thanks.

In that case you sub the values for q into the CHW equation and solve it for p. You do that both populations.

Population 1.

p^2 + 2pq + q^2 = 1
p^2 + 2(0.2)p + 0.2^2 = 1
p^2 + 0.4p + 0.04 = 1

p1 = 0.8

Population 2.

p^2 + 2pq + q^2 = 1
p^2 + 2(.4)p + 0.4^2 = 1
p^2 + 0.8p + 0.16 = 1

p2 = 0.6

Now you need to calculate the frequency of heterozygotes. That is the 2pq part of the equation.

For population 1 you have q = 0.2, p = 0.8 so:

2pq = 2*0.2*0.8 = 0.32

For population 2 you have q = 0.4 and p = 0.6 so:

2pq = 2*0.4*0.6 = 0.48.

Now you interpret the frequencies: if you have a hundred organisms in population 1 and 2, then in population 1 there are 32 heterozygotes and in population 2 there are 48 heterozygotes.

The problem wants you to compute the difference, so

48 - 32 = 16.

Clear?

Answer 9

ohhh yea i get it !! thanks!
I will try to work on my hardy weinburg eqm questions :D

Answer 10

Cool, any time.