# 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

Frequency of homozygous recessive genotype,

q2 = (10,000 - 9,898)/10,000 &#8776; 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

#### enfuego

10+ Year Member
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

OP
A

#### arginine1

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