Genetic question

[299678]

Advertisement - Members don't see this ad

Q. Assume a population in Hardy Weinberg equilibrium for a character trait with these genotypic frequencies AA = 0.25, Aa = 0.50, and aa = 0.25. If you remove all the homozygous dominants and allow the remaining population to reproduce, what will be the frequency of homozygous dominants in the next generation?

 

[Maygyver]

Q. Assume a population in Hardy Weinberg equilibrium for a character trait with these genotypic frequencies AA = 0.25, Aa = 0.50, and aa = 0.25. If you remove all the homozygous dominants and allow the remaining population to reproduce, what will be the frequency of homozygous dominants in the next generation?

Ok so, the new population you will have 50 A and 75 a alleles. The new total is 125 alleles so your p=50/125=0.4 and your q=75/125=0.6 From there, your new homozygous dominants will be p^2=0.4^2=0.16

 

[sfoksn]

Q. Assume a population in Hardy Weinberg equilibrium for a character trait with these genotypic frequencies AA = 0.25, Aa = 0.50, and aa = 0.25. If you remove all the homozygous dominants and allow the remaining population to reproduce, what will be the frequency of homozygous dominants in the next generation?

Hm... I think its (1/3)^2...

Just because each aa or AA contribute 2 to the counting of the alleles... What is the answer kpark?

 

[Maygyver]

Hm... I think its (1/3)^2...

Just because each aa or AA contribute 2 to the counting of the alleles... What is the answer kpark?

But if you remove all the AA, you are only left with .50 Aa to contribute A's.

 

[sfoksn]

I am really rusty on these stuff, so correct me if i do it wrong.

0.25 = AA, 0.5 = Aa , 0.25 = aa.

well i'm just gonna say there are 25 AA individuals, 50 Aa individuals, and 25 aa individuals.

So lets count the alleles: there are 2(25)+50 A, and 2(25)+50 a. so we have 100 A, and 100 a. Now we are removing all the Homozygous AA, right? Homozygous dominants contributed 50 allele counts towards the A.

So now we would have 50 A, and 100 a.

So frequency of A would be 50/150 = (1/3).. and homozygous dominant genotypic frequency would be (1/3)^2...

 

[Maygyver]

I am really rusty on these stuff, so correct me if i do it wrong.

0.25 = AA, 0.5 = Aa , 0.25 = aa.

well i'm just gonna say there are 25 AA individuals, 50 Aa individuals, and 25 aa individuals.

So lets count the alleles: there are 2(25)+50 A, and 2(25)+50 a. so we have 100 A, and 100 a. Now we are removing all the Homozygous AA, right? Homozygous dominants contributed 50 allele counts towards the A.

So now we would have 50 A, and 100 a.

So frequency of A would be 50/150 = (1/3).. and homozygous dominant genotypic frequency would be (1/3)^2...

Hah yeah, that's right. I think I misadded the recessives. I guess I should probably use scrap paper with more room. Hopefully I don't do that on the real thing.

 

[299678]

solution manual said
1/2 x 1/4 = 1/8 , so about 0.12%

 

[Maygyver]

You really should type out more of what the solution manual says when you ask for help and expect people to understand the solution when you don't give what the manual says. If not, there's really no point in asking on a forum if you can just understand the manual's solution.

 

[299678]

lol Maygyver
I wrote exactly what solution manual said. It isn't sufficient at all. That's why I posted it here for asking further explanation.

 

[Strikeout13GB]

its pretty simple actually (i think) since the AA will be gone theres only one way to get another AA and thats if the two Aa come together, so if that happens theres a 1/4 chance it will be AA (punnet square). As for the 1/2 theres a 50/50 chance that the Aa will pair with the other Aa, it could also pair with aa so 1/2 * 1/4= 1/8

 

[299678]

Thanks

 

[sfoksn]

I wonder why my method did not work?

Can anyone explain the mistake I made in my calculation?

 

[UCB05]

Perhaps it's because you can't just say there's a .33 A frequency and .66 a frequency and find homozygous dominants by (1/3)^2. Being able to use that setup requires hardy weinberg allele distributions, and having no AA in a generation implies some sort of selection going on. Since the actual allele frequency is lopsided-ly distributed into Aa and aa with no AA, it doesnt make sense to use the HW equation p^2 to find the next generation's distributions.

Don't ask me how they got the correct answer because the way I would have solved it was:
(probability of Aa meeting Aa)(probability of AA as a result of cross) =
(2/3)(2/3)(1/4) = 1/9, which coincidentally leads to the same answer as your setup, but apparently is wrong.