genetic equilbrium

[Eri3]

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I am having a trouble with Hardy-Weinberg equilibrium and on most of my practice exams I have seen this question come up.

Does anyone have a good grasp on this concept- I am having trouble translating the info given in the problem and using the proper formula ( P+Q=1 or P^2 + 2PQ + Q^2 = 1). I have read numerous sources but none of them make sense to me for some reason.

For example, how would you think through this problem?

In a large isolated community it is found that 84% of the people have an autosomal dominant disease. What is the frequency of the recessive allele?

 

[Koalafied]

See below

 

[yko0584]

Well instead of working with the dominant allele do the recessive. q^2 = 0.16 would be the recessive.
sqrt (0.16) = 0.4 ??? Not really sure its been a while, but I work it out like this.

 

[Koalafied]

See below

 

[SafetyDog]

yko0584 has it right...

It should be 0.84 = p^2 + 2pq, not 0.84 = p*2 because the heterozygotes still express the dominant trait. However, yko0584's method is what you should do first since it's easiest.

This leads to p = 0.6 and q = 0.4.

 

[Koalafied]

See below

 

[SafetyDog]

So (0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 1

2PQ = 0.48
P^2 = 0.36
Q^2 = 0.16

Adds up to 1.

 

[Eri3]

yko0584 has it right...

It should be 0.84 = p^2 + 2pq, not 0.84 = p*2 because the heterozygotes still express the dominant trait. However, yko0584's method is what you should do first since it's easiest.

This leads to p = 0.6 and q = 0.4.

i think that's correct bc autosomal dominant pertains to the homo and hetero phenotype but i'm having the most trouble with translating the words into the proper values.

 

[saintguitar]

There are two formulas related to the Hardy-Weinberg.

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

Once you know what each letter means, it makes solving the problem much easier.

First of all, the first formula is for ALLELE and its frequency. So whenever the problem asks you for a frequency of an "allele," you need to immediately think that the answer will be either p, or q.
P is the frequency of a dominant allele while q stands for the frequency of a recessive allele.

The second formula is for PHENOTYPES, and if the problem mentions "phenotypes" or "expressed," then you know its the frequency for a specific phenotype. It gets a little tricky here because you have to consider "heterozygous population," which is 2pq in the formula.
So p^2 + 2pq is the frequency of the dominant phenotypes, and q^2 is the freuency of the recessive phenotypes.

In the case of your problem, it says that, "it is found that 84% of the people have an autosomal dominant disease. What is the frequency of the recessive allele?"

So when you read the question, you have to see what it wants, this case the frequency of the recessive allele, which is "q".
We are given the frequency of an autosomal dominant disease, which is a "phenotype", meaning we have not just p^2, but p^2 + 2pq, as the heterozygous population will also show the dominant phenotype. So from the second formula, p^2 +2pq + q^2 = 1, we can change it like this

1 - (p^2 + 2pq) = q^2
1 - 0.84 (the dominant phenotype frequency given in this question; its 84%, meaning 0.84) = 0.16

We are not done yet. 0.16 is the frequency of the recessive "phenotype", and we want the "allele" frequency, q. So we just take a square root of 0.16, which is 0.4.

So the answer is 0.4 and we go get the point.

 

[Eri3]

There are two formulas related to the Hardy-Weinberg.

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

Once you know what each letter means, it makes solving the problem much easier.

First of all, the first formula is for ALLELE and its frequency. So whenever the problem asks you for a frequency of an "allele," you need to immediately think that the answer will be either p, or q.
P is the frequency of a dominant allele while q stands for the frequency of a recessive allele.

The second formula is for PHENOTYPES, and if the problem mentions "phenotypes" or "expressed," then you know its the frequency for a specific phenotype. It gets a little tricky here because you have to consider "heterozygous population," which is 2pq in the formula.
So p^2 + 2pq is the frequency of the dominant phenotypes, and q^2 is the freuency of the recessive phenotypes.

In the case of your problem, it says that, "it is found that 84% of the people have an autosomal dominant disease. What is the frequency of the recessive allele?"

So when you read the question, you have to see what it wants, this case the frequency of the recessive allele, which is "q".
We are given the frequency of an autosomal dominant disease, which is a "phenotype", meaning we have not just p^2, but p^2 + 2pq, as the heterozygous population will also show the dominant phenotype. So from the second formula, p^2 +2pq + q^2 = 1, we can change it like this

1 - (p^2 + 2pq) = q^2
1 - 0.84 (the dominant phenotype frequency given in this question; its 84%, meaning 0.84) = 0.16

We are not done yet. 0.16 is the frequency of the recessive "phenotype", and we want the "allele" frequency, q. So we just take a square root of 0.16, which is 0.4.

So the answer is 0.4 and we go get the point.

thank you- this is what i needed. Thinking of p^2, 2pq and q^2 in terms of phenotypes makes a lot more sense to me

 

[Koalafied]

There are two formulas related to the Hardy-Weinberg.

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

Once you know what each letter means, it makes solving the problem much easier.

First of all, the first formula is for ALLELE and its frequency. So whenever the problem asks you for a frequency of an "allele," you need to immediately think that the answer will be either p, or q.
P is the frequency of a dominant allele while q stands for the frequency of a recessive allele.

The second formula is for PHENOTYPES, and if the problem mentions "phenotypes" or "expressed," then you know its the frequency for a specific phenotype. It gets a little tricky here because you have to consider "heterozygous population," which is 2pq in the formula.
So p^2 + 2pq is the frequency of the dominant phenotypes, and q^2 is the freuency of the recessive phenotypes.

In the case of your problem, it says that, "it is found that 84% of the people have an autosomal dominant disease. What is the frequency of the recessive allele?"

So when you read the question, you have to see what it wants, this case the frequency of the recessive allele, which is "q".
We are given the frequency of an autosomal dominant disease, which is a "phenotype", meaning we have not just p^2, but p^2 + 2pq, as the heterozygous population will also show the dominant phenotype. So from the second formula, p^2 +2pq + q^2 = 1, we can change it like this

1 - (p^2 + 2pq) = q^2
1 - 0.84 (the dominant phenotype frequency given in this question; its 84%, meaning 0.84) = 0.16

We are not done yet. 0.16 is the frequency of the recessive "phenotype", and we want the "allele" frequency, q. So we just take a square root of 0.16, which is 0.4.

So the answer is 0.4 and we go get the point.

You're right, the statement: 84% of the people have an autosomal dominant disease is inclusive of both the homozygous population (P^2) and the heterozygous population (2PQ), and that's because we inherit an allele from both parents.

So that 84%, in this case, is P^2 + 2PQ. Then it's correct to subtract 0.84 from 1 and get 0.16. And taking the SQRT(0.16) would give you 0.4. This is common trap many testers fall into, including me just now. I've hope my slip serves to y'all as a reminder on how tricky these questions may be worded, but at the same time very simple to answer.

thank you- this is what i needed. Thinking of p^2, 2pq and q^2 in terms of phenotypes makes a lot more sense to me

The frequency of the recessive allele is indeed 0.4 => 40%.

So here it is:
P = dominant allele frequency
Q = recessive allele frequency

P^2 = dominant population
Q^2 = recessive population
2PQ = heterozygous population
P^2+2PQ = dominant phenotype population
P^2+2PQ+Q^2 = total population