Hardy weinberg equilibrium
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Online Science Education :: Subjects :: Biology :: Genetics :: 1 on 1 Tutoring
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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
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.
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!
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
%
Just one more thing -
For the second problem - does that mean that the final answer is 0.00000001% or 0.0001%?
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
Online Science Education :: Subjects :: Biology :: Genetics :: 1 on 1 Tutoring
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