# Hardy weinberg equilibrium ## Hardy weinberg equilibrium

Thanks for your help with restriction mapping i eventually managed to figure out what i did wrong in that last one and used how you did the easier one to map it.

I just wanted know if you could tell me how to do/ tell me if i'm right for these qns:

There are only two alleles for a particular gene in a pop. The frequency of heterozygotes is 20% Assuming hardy weinberg equilibrium calculate the allelic frequencies in this pop.

Haemophilia A is a rare reccesive sex linked trait. If in a given population in equilibrium, 0.01% of males have this phenotype what % of females would show it?

PKU occurs in approx 1 in 10,000 people. It is due to an autosomal recessive allele. What proportion of normal individuals are carriers for this allele?
My answer - 0.0198

You don't have to answer them if you don't want to, just give me a point in the right direction.
Thanks

georgina_009
Student Posts : 25
Join date : 2008-03-18 ## Re: Hardy weinberg equilibrium

Let's start with the easy:

3rd problem you solved correctly.

2nd problem:

Males have only 1 allele either X or x, so 0.01% males that have Haemophilia A have xY genotype and 99.99% have XY genotype. Thus, 0.01% = f(q) = 0.0001 (because there is no second X chromosome and f(y) in males = 1) and females will have f(q^2)= 0.0001^2

1st problem:

Since heterozygous are observed alleles are probably codominant.

Heterozygous is 2pq=0.2 (or 20%). p^2 + 2pq + q^2 = 1. Therefore,

p^2 + q^2 = 1  2pq = 0.8;
pq = 0.1; p= 0.1/q; p^2 =0.01/(q^2 );
0.01/(q^2 ) + q^2 = 0.8 or q^4  0.8q^2 + 0.01=0

Solve quadratic equation:

q^4  0.8q^2 + 0.01=0 or ax^2 + bx^2 + c=0

q^2 = [0.8 ± √(0.8^2 -4(0.01))] /2 or x = [-b ± √(b^2  4ac)]/2a

q^2 = either 0.013 or 0.787. Note that the other number is p^2.

Check the solution: p^2 + 2pq + q^2 = 1; 0.013 + 0.2 + 0.787 = 1.

Alla
Instructor Posts : 103
Join date : 2008-02-27 ## Re: Hardy weinberg equilibrium

Thankyou Alla,
I have been working through these with your answers and it makes much more sense now!

georgina_009
Student Posts : 25
Join date : 2008-03-18 ## Re: Hardy weinberg equilibrium

You are welcome!

Alla
Instructor Posts : 103
Join date : 2008-02-27 ## %

Just one more thing -
For the second problem - does that mean that the final answer is 0.00000001% or 0.0001%?

georgina_009
Student Posts : 25
Join date : 2008-03-18 ## Re: Hardy weinberg equilibrium

q^2 = 0.00000001 or 0.000001%

Alla
Instructor Posts : 103
Join date : 2008-02-27 ## Re: Hardy weinberg equilibrium

thanks

georgina_009
Student Posts : 25
Join date : 2008-03-18 ## Re: Hardy weinberg equilibrium

You are welcome!

Alla
Instructor Posts : 103
Join date : 2008-02-27 ## Re: Hardy weinberg equilibrium

Sponsored content Permissions in this forum:
You cannot reply to topics in this forum