Hardy weinberg help

[arginine1]

this is a question and answer from achiever test 2. Can someone explain this calculation to me?

A gene locus of two alleles belonging to a population of 10,000 members is in a state of Hardy-Weinberg equilibrium. What is the frequency of heterozygotes if 9,898 individuals collectively exhibit dominant trait?


A. 0.09
B. 0.10
C. 0.18
D. 0.90
E. 0.81


Correct Answer: C

Frequency of homozygous recessive genotype,

q2 = (10,000 - 9,898)/10,000 ≈ 0.01

Recessive allelic frequency,

q = (0.01)1/2 = 0.1

Dominant allelic frequency,

p = 1 - q = 1 - 0.1 = 0.9

From the Hardy-Weinberg equation, the frequency of heterozygotes can finally be computed as follows:

2pq = 2(0.9)(0.1) = 0.18

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[enfuego]

Use these equations:

allele frequency
p + q = 1
(p = dominant allele frequency, q = recessive allele frequency)

genotype frequency
p^2 + 2pq + q^2 = 1
(p^2 = homozygous dominant frequency, 2pq = heterozygous frequency, q^2 = homozygous recessive frequency)

So dominant trait = homozygous dominant + heterozygous
The remaining people must be homozygous recessive
So use that to calculate q^2
Take the square root to find q
p + q = 1
Subtract to find p (dominant allele)
Then plug p and q into 2pq to find the frequency of heteozygotes

 

[toothdoc0121]

this is the way i did it

dominant freq is p=0.9898


recessive freq is q=0.1212 because p+q=1 so 1-p=q

heterozygous freq is= 2*p*q

so 2*0.9898*0.1212= 0.18

hope this helps

 

[arginine1]

Yeah.. the equations you guys used I'm more familiar with as well. It makes sense the way you solved it. Thanks