## I need help w/ two Hardy Weinberg Problems plz

Genetics as it applies to evolution, molecular biology, and medical aspects.

xcuteJinax3
Garter
Posts: 8
Joined: Sun Feb 26, 2006 6:31 am

### I need help w/ two Hardy Weinberg Problems plz

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?

---------------------------------------------

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!

Enzyme
Coral
Posts: 336
Joined: Mon Dec 26, 2005 9:05 pm

### My answer to the first problem

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!

MrMistery
Inland Taipan
Posts: 6832
Joined: Thu Mar 03, 2005 10:18 pm
Location: Romania(small and unimportant country)
Contact:
The second one is pure textbook HW.
q^2=25.1
p+q=1

The rest is math
"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

xcuteJinax3
Garter
Posts: 8
Joined: Sun Feb 26, 2006 6:31 am

### Re: My answer to the first problem

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?

Khaiy
Coral
Posts: 158
Joined: Tue Feb 28, 2006 2:37 am
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.

Enzyme
Coral
Posts: 336
Joined: Mon Dec 26, 2005 9:05 pm
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 .

### Who is online

Users browsing this forum: No registered users and 4 guests