## Hardy-Weinberg Problem I

**Moderators:** honeev, Leonid, amiradm, BioTeam

### Hardy-Weinberg Problem I

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?

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
- Inland Taipan
**Posts:**6832**Joined:**Thu Mar 03, 2005 10:18 pm**Location:**Romania(small and unimportant country)-
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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..

p=(0.4+0./2= 0.6

q=(0.6+0.2)/2= 0.4

From there, it's math

Hope i am right..

"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter

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

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

Last edited by 2810712 on Mon Feb 06, 2006 2:03 am, edited 4 times in total.

### Re: Hardy-Weinberg Problem I

bionewbie wrote: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%

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