Hardy-Weinberg Problem I

[bionewbie]

Given two equal size populations that are both in Hardy-Weinberg equillibrium.

Population A p=0.4 q=0.6
Population B p=0.8 q=0.2

What would the expected frequency of heterozygous if these populations were combined into a single population?

[MrMistery]

Well, from what i can imagine all you have to do is add them and then divide by 2:
p=(0.4+0./2= 0.6
q=(0.6+0.2)/2= 0.4

From there, it's math

Hope i am right..

[2810712]

If the pops are mixed before breeding season-
Answer=[pA/2*qA/2+ pB/2*qB/2 + pA/2*qB/2 + pB/2*qA/2]

If pops are mixed after breeding season-
Answer=[pA/2*qA/2+ pB/2*qB/2 ]


pA= allele freq. of p in A similarly qA, pB, qB.






hrushikesh

[Dr.Stein]

Duh, I hate numbers... Can't help, sorry

[sdekivit]

Given two equal size populations that are both in Hardy-Weinberg equillibrium.

Population A p=0.4 q=0.6
Population B p=0.8 q=0.2

What would the expected frequency of heterozygous if these populations were combined into a single population?

if we assume population A eand B to count 100 organisms, then te new allel frequency for p will be:

120/200 = 0,6 and q: 80/200 = 0,4

--> thus heterozygous: 2 * 0,6 * 0,4 = 0,48 = 48%

[2810712]

Please have a review of my edited message and tell ur opinions...



hrushikesh