I need help w/ two Hardy Weinberg Problems plz

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xcuteJinax3
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I need help w/ two Hardy Weinberg Problems plz

Post by xcuteJinax3 » Sun Feb 26, 2006 6:37 am

1) A metabolic disease of man called phenylketonuria is the result of a recessive gene. If the frequency of phenylketonuria is 1/10,000, what is the probability that a person is a carrier for this trait?

2) When the blood samples of 999 students were tested with anti-Rh serum, 74.9% were positive and 25.1% negative. Assuming a single pair of alleles R and r, what proportion of the students would be expected to be RR, Rr and rr?


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For the first problem I have no idea what to do at all and the second problem I think I would use punnet squares or a formula but im not sure at all. Please help me~ thanks!

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My answer to the first problem

Post by Enzyme » Mon Feb 27, 2006 2:54 pm

MY ANSWER TO THE FIRST PROBLEM:

If PKU (phenylketonuria) is an autosomic recessive desease and we suppose that the population is in balance for this gen (A,a) we can know:

a) The frequency of the individuals who suffer PKU (recessive homozygous):

As PKU is due to the effect of the recessive allele of an autosomic gen, all the individuals who suffer this disease have this genotype: aa (recessive homozygous). Therefore, if we found in this population one individual affected for every (each) ten thousand individuals, the frequency of recessive homozygous is:

Frequency (PKU) = q^2 = 1/10 000 = 0.0001

b) The frequency (q) of the recessive allele which produces PKU.

If we suppose that the population is in Hardy-Weinberg balance and the frequency of affected individuals is 0.0001, we can calculate the frequency of the recessive allele (q) which produces PKU as the squared root of the frequency of ill individuals: q = 0.01

c) The frequency of healthy individuals and carriers of this trait.

In a population in which an autosomic locus with two alleles (one dominant and another recessive) is segregated, the individuals with dominant phenotype (healthy individuals, in this case) are divided in:

- Dominant homozygous (AA).
- Heterozygous (Aa): The carriers of the recessive allele which produces the desease.

In a balanced population, the frequency of heterozygous is 2pq and, in the case in which one of the alleles have a very little frequency, the frequency of the other allele will be aproximately 1 and, as a consequence, the frequency of heterozygous is aproximately two times the frequency of the rare allele. I want to mean:

Frequency (Heterozygous) = 2pq = (9 999/10 000) · (1/10 000). So the result is 0.00019998, aproximately 0.0002 = 2q.


Do you understand? If you already have any doubt, ask me ;). See ya!
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Post by MrMistery » Mon Feb 27, 2006 6:28 pm

The second one is pure textbook HW.
q^2=25.1
p+q=1

The rest is math
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xcuteJinax3
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Re: My answer to the first problem

Post by xcuteJinax3 » Tue Feb 28, 2006 12:27 am

Enzyme wrote:MY ANSWER TO THE FIRST PROBLEM:

If PKU (phenylketonuria) is an autosomic recessive desease and we suppose that the population is in balance for this gen (A,a) we can know:

a) The frequency of the individuals who suffer PKU (recessive homozygous):

As PKU is due to the effect of the recessive allele of an autosomic gen, all the individuals who suffer this disease have this genotype: aa (recessive homozygous). Therefore, if we found in this population one individual affected for every (each) ten thousand individuals, the frequency of recessive homozygous is:

Frequency (PKU) = q^2 = 1/10 000 = 0.0001

b) The frequency (q) of the recessive allele which produces PKU.

If we suppose that the population is in Hardy-Weinberg balance and the frequency of affected individuals is 0.0001, we can calculate the frequency of the recessive allele (q) which produces PKU as the squared root of the frequency of ill individuals: q = 0.01

c) The frequency of healthy individuals and carriers of this trait.

In a population in which an autosomic locus with two alleles (one dominant and another recessive) is segregated, the individuals with dominant phenotype (healthy individuals, in this case) are divided in:

- Dominant homozygous (AA).
- Heterozygous (Aa): The carriers of the recessive allele which produces the desease.

In a balanced population, the frequency of heterozygous is 2pq and, in the case in which one of the alleles have a very little frequency, the frequency of the other allele will be aproximately 1 and, as a consequence, the frequency of heterozygous is aproximately two times the frequency of the rare allele. I want to mean:

Frequency (Heterozygous) = 2pq = (9 999/10 000) · (1/10 000). So the result is 0.00019998, aproximately 0.0002 = 2q.


Do you understand? If you already have any doubt, ask me ;). See ya!




Okay I understand but ... so all we need in this problem would be to q^2 and 2pq, then you use p+q=1 to find out p, to put into the 2pq formula.


just want to know basically if what I said is right?

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Post by Khaiy » Tue Feb 28, 2006 6:03 pm

Yeah, that sounds about right. You need to find q^2, which is the number of people expressing the trait divided by the total number of people. Then, you take the square root of that number to get q. 1-q=p, and then you should have enough information for the rest of the equations.

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Post by Enzyme » Sat Mar 04, 2006 9:22 am

Yes, firstly you need to find q^2. Then you have to do the squared root of the result and after these operations, you'll find the rest ;).
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