Hardy weinberg equilibrium
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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 : 20080318
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 : 20080227
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 : 20080318
%
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 : 20080318
Re: Hardy weinberg equilibrium
q^2 = 0.00000001 or 0.000001%
Alla Instructor
 Posts : 103
Join date : 20080227
Online Science Education :: Subjects :: Biology :: Genetics :: 1 on 1 Tutoring
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