Sixtyfour percent of a population is able to taste a chemical called PTC. Tasters are either homozygous dominant or heterozygous for the trait.
1. Calculate the percentage of the population that carries the recessive allele
Biology Hardy Weinberg calculation HELP!!
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The question is harder than it looks. I got 0.6 as the answer using this method
If the trait is caused by a dominant allele and 64% of the population have it then using
p^{2}+2pq+q^{2}=1
p ^{2}+2pq=0.64
therefore 10.64=0.36=q^{2}
the frequency of the recessive allele is q so square root 0.36 to get q
q=
not 100% sure though 
(Original post by The Illuminati)
The question is harder than it looks. I got 0.6 as the answer using this method
If the trait is caused by a dominant allele and 64% of the population have it then using
p^{2}+2pq+q^{2}=1
p ^{2}+2pq=0.64
therefore 10.64=0.36=q^{2}
the frequency of the recessive allele is q so square root 0.36 to get q
q=
not 100% sure though 
(Original post by I'mBadAtMaths)
This is correct (the answer is 60%)
lol at you username and the nature of the question...irony 
Ok, thats what I thought I had to do ! I just wasnt sure if I had to add the 2pq value, and the q^2 value because they both have the recessive allele.
P=0.4
p+q= 1.00
q= 1.00p
q=1.000.4
q=0.6
I get so confused with these questions when to do what steps and when to stop. but I think the answer being 60 would be logical. 
(Original post by The Illuminati)
The question is harder than it looks. I got 0.6 as the answer using this method
If the trait is caused by a dominant allele and 64% of the population have it then using
p^{2}+2pq+q^{2}=1
p ^{2}+2pq=0.64
therefore 10.64=0.36=q^{2}
the frequency of the recessive allele is q so square root 0.36 to get q
q=
not 100% sure though
Didn't you work out percentage of recessive alleles, not percentage of the population who have recessive alleles? The allele would be carried by those in the q^2 and 2pq groups(?)
If q = 0.6, p = 0.4, therefore 2pq = 0.64  0.4^2 = 0.48(?)
q^2 = 0.36
So therefore wouldn't the answer be 0.36 + 0.48 = 84%?
EDIT: I think that makes sense anyway... if homozygous dominant is p^2 and p = 0.4, then 0.16 of the population are homozygous dominant so 0.84 of the population have to have at least one recessive allele... Sorry if I waffled a bit :P 
(Original post by uninspirational)
"1. Calculate the percentage of the population that carries the recessive allele"
Didn't you work out allele frequency, not percentage of the population? The allele would be carried by those in the q^2 and 2pq groups
If q = 0.6, p = 0.4, therefore 2pq = 0.64  0.4^2 = 0.48(?)
q^2 = 0.36
So therefore wouldn't the answer be 0.36 + 0.48 = 84%?
what you have worked out is the frequency of the genotypes that each have a recessive allele which sounds similar but isnt the same thing 
(Original post by The Illuminati)
q is number of people with the recessive allele aka the frequency of the recessive allele (just a not aa) in the population.
what you have worked out is the frequency of the genotypes that each have a recessive allele which sounds similar but isnt the same thing
Afaik q isn't the % of people with the allele, its the % of recessive alleles in that gene pool? One person can have either one or 2 recessive alleles so 60% recessive alleles doesn't mean 60% of people have them.
What I mean is: if he population is made up of: homozygous dominant individuals (p^2), homozygous recessive individuals (q^2) and heterozygotes (2pq), then the % of people with recessive alleles is 2pq+ q^2, is it not? 
(Original post by uninspirational)
Sorry I edited my previous post btw :')
Afaik q isn't the % of people with the allele, its the % of recessive alleles in that gene pool? One person can have either one or 2 recessive alleles so 60% recessive alleles doesn't mean 60% of people have them.
What I mean is: if he population is made up of: homozygous dominant individuals (p^2), homozygous recessive individuals (q^2) and heterozygotes (2pq), then the % of people with recessive alleles is 2pq+ q^2, is it not?
Looking at the definitions of p and q is the only way to work this out without freaking out
p=frequency of A q=frequency of a
The question asks for the frequency of the recessive allele
sinse q=the frequency of the recessive allele that is what we try and find.....
That's the best way I can think of explaining it 
(Original post by The Illuminati)
The question asks for the frequency of the recessive allele
"1. Calculate the percentage of the population that carries the recessive allele"
So it's asking what percentage of people have at least one recessive allele aka. what percentage of people are not homozygous dominant.
If 50% of alleles in a gene pool are dominant and 50% are recessive, it does NOT mean that 50% of the population have recessive genes as 100% of people could have one dominant and one recessive gene.
If you're right then I'm sorry but I think this question may be worded in a ridiculous manner haha :') 
(Original post by uninspirational)
But it doesn't....
"1. Calculate the percentage of the population that carries the recessive allele"
So it's asking what percentage of people have at least one recessive allele aka. what percentage of people are not homozygous dominant.
If 50% of alleles in a gene pool are dominant and 50% are recessive, it does NOT mean that 50% of the population have recessive genes as 100% of people could have one dominant and one recessive gene.
If you're right then I'm sorry but I think this question may be worded in a ridiculous manner haha :')
The only thing I can say is that you are working out genotype frequency. Genotype means combination of alleles therefore you are working out who has a combination of alleles that include a recessive allele.
I on the other hand am working out the allele frequency for the recessive allele 
Predicting Allele Frequency (what the Q asks):
p (T) + q (t) = 1
Predicting Genotype Frequency
p^2 (TT) + 2pq (Tt) + q^2 (tt) = 1
The question gives you info on genotype frequency, so look to the 2nd formula first. The info tells you that p^2 + 2pq = 64%, so q^2 (tt) must be 1  64 = 36%.
The recessive allele frequency is simply q (using 1st formula), so square root q^2, which is 0.6, in % it's 60%.
Hope this helps 
(Original post by The Illuminati)
This is one of those horrid times where I completely understand what you're saying and why and I'm strugglining to find reasons as to why you're wrong, even though I know you are.
The only thing I can say is that you are working out genotype frequency. Genotype means combination of alleles therefore you are working out who has a combination of alleles that include a recessive allele.
I on the other hand am working out the allele frequency for the recessive allele
Like you mention in your first post that:
"If the trait is caused by a dominant allele and 64% of the population have it then using
p2+2pq+q2=1
p^{2}+2pq=0.64"
Wouldn't the same apply if we are to find the % of the population that carries the recessive allele. Just like 64% was the % of the population carrying the dominant allele.
So instead of p^{2}+2pq=0.64
We are finding 2pq+q^{2}= x
Where we know p and q.
A mark scheme for this question would also help to confirm. 
(Original post by Evanesyne)
Predicting Allele Frequency (what the Q asks):
p (T) + q (t) = 1
Predicting Genotype Frequency
p^2 (TT) + 2pq (Tt) + q^2 (tt) = 1
The question gives you info on genotype frequency, so look to the 2nd formula first. The info tells you that p^2 + 2pq = 64%, so q^2 (tt) must be 1  64 = 36%.
The recessive allele frequency is simply q (using 1st formula), so square root q^2, which is 0.6, in % it's 60%.
Hope this helps
1. Calculate the percentage of the population that carries the recessive allele
Which uninspirational worked out to be 84% which can be confirmed just as well with 1 p^{2} which is 10.16 = 0.84 
I copied and pasted the question onto google, answers given from varius others range from 0.48 to 0.6 to 0.84
i think my brain just exploded 
(Original post by TheGrinningSkull)
But the number of people who carry the recessive allele (As the question asks) would be Tt and tt.
Which uninspirational worked out to be 84% which can be confirmed just as well with 1 p^{2} which is 10.16 = 0.84
Another thought, this could also be easily misread as the % of carriers, where the 0.48 answer came from I suppose.
I haven't seen a question on HW that is worded as awkwardly as this.
I be getting back to unit 5 now
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