Download Hardy-Weinberg Equilibrium

Hardy-Weinberg
Equilibrium
Life is a fine balance between
shared commonalities and
amazing diversity.
Hardy and Weinberg

Godfrey Hardy
 mathematician

Wilhelm Weinberg
 physician
Hardy-Weinberg Model
 Proposed
in 1908
 Uses mathematics to make assumptions
of allele frequencies in gene pools
 Based on several assumptions
HW Model assumptions pt 1
 1.
Organisms are diploid
 2. Reproduction is sexually
 3. Generations do not overlap

 4.
Parents die before children reproduce
No mutations
HW model assumptions pt 2
 Gametes


unite at random
Mating is random
One allele is not favored over another
 NO
migration in or out of population
 Populations size is very large

Remember: probability works best with large
sample sizes
HW First Prediction
p
+q=1
 p= allele frequency of dominant allele
 q= allele frequency of recessive allele
 p squared=frequency of homozygotes
dominant
 2pq= frequency of heterozygotes
 q squared= frequency of homozygote
recessive
HW Model Second Prediction
 Allele
frequencies will remain stable over
time
 Genetic variation does not disappear
 Polymorphic gene pools stay polymorphic
This is an ideal model..
 Extremely
rare for any population to meet
all requirements perfectly
 Genetic frequencies still tend to stay close
to stable between generations
Using HW to calculate frequences
 Frequencies

p+q =1
 Frequencies



of alleles:
of genotypes:
p squared= homozygous dominant
2pq = heterozygous
q squared = homozygous recessive
Example: Brown vs. Blue eyes
 Often,


p= B allele(dominant)
q = b allele(recessive)
 Our


only can look at phenotypes
class has 18 students
17 brown eyes
1 non-brown eyes
Step 1
 Step
1: Calculate genotype frequency of
homozygous recessive
 Why?

q squared is homozygous recessive
• The only ones you can tell their genotype just by
looking at them



q squared = 1/18
q squared = 0.0555
(we’re saying that around 5 percent of class has nonbrown eyes and are bb)
Step 2
 Step
2:Use genotype frequency of q
squared(homozygous recessive) to
calculate allele frequency of q




q squared = (1/18)=(0.0555)
q = the square root of (1/18)= the square root
of (0.0555)
q= 0.2356
(we’re saying that around 23 % of all the eye
color alleles in our class population are b)
Step 3
 Step
3: use allele frequency of q to
calculate allele frequency of p
 We know: p + q = 1

So
p=1–q
p = 1- 0.2356
p = 0.7644
Step 4:
 Step
4: You know the frequencies of the
alleles. You can now calculate the
frequency of p sqared and 2 pq

p squared= homozygous brown eyes=
• p squared =0.584
• (around 58 % is homozygous BB)

2pq = heterozygous brown eyes=
• 2 pq= 0.3554
• (around 35 % is Bb)
Step 5:
 Double
check your genotype frequencies

(p squared) + (2pq) + (q sqared) = 1
(0.584) + (0.3554) + (0.0555) = 1
1= 1

Yah, we’re correct!!!

