Hardy weinberg equilibrium

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Hardy weinberg equilibrium

Post by georgina_009 on Mon May 12, 2008 10:30 am

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

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Re: Hardy weinberg equilibrium

Post by Alla on Tue May 13, 2008 4:14 am

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.

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Re: Hardy weinberg equilibrium

Post by georgina_009 on Tue May 13, 2008 5:32 am

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

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Re: Hardy weinberg equilibrium

Post by Alla on Tue May 13, 2008 5:54 am

You are welcome!

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%

Post by georgina_009 on Tue May 13, 2008 11:12 am

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

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Re: Hardy weinberg equilibrium

Post by Alla on Wed May 14, 2008 1:23 am

q^2 = 0.00000001 or 0.000001%

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Re: Hardy weinberg equilibrium

Post by georgina_009 on Wed May 14, 2008 6:20 am

thanks

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Re: Hardy weinberg equilibrium

Post by Alla on Tue May 20, 2008 6:45 am

You are welcome!

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