Transcript 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
•Proposed in 1908
•Use mathematics to describe allele
frequencies in gene pools
•Based on following assumptions
•
(places to start from)
HW Model assumptions pt 1
 1.
Organisms are diploid (2 genes/trait)
 2. Reproduction is sexual
 3. Generations do not overlap

 4.
Parents die before children reproduce
No mutations
HW model assumptions pt 2
 5.
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Gametes unite at random
Mating is random
One allele is not favored over another for
reproduction
 6.
NO migration in or out of population
 7. Populations size is very large

Remember: probability works best with large
sample sizes
HW First Prediction
p
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p= allele frequency of dominant allele
q= allele frequency of recessive allele
 p2
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+q=1
X 2pq X q2 = 1
p2 (p X p) =frequency of homozygous dominant
2pq= frequency of heterozygotes
q2 (q X q) = frequency of homozygous recessive
HW Model Second Prediction
 Allele
frequencies will remain stable over
time - The Population is said to be in;

“Hardy Weinberg Equilibrium”
 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 – to be in HW
equilibrium
 Genetic frequencies still tend to stay close
to stable between generations
Using HW to calculate frequences
 Frequencies

p+q =1
 Frequencies
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of alleles:
of genotypes:
p squared= homozygous dominant
2pq = heterozygous
q squared = homozygous recessive
Example: Brown vs. Blue eyes
 Often,
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B allele(dominant)
b allele(recessive)
 Our
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only can look at phenotypes
class has 18 students
17 brown eyes
1 non-brown eyes
Step 1
 Step
1: Calculate frequency of
homozygous recessive genotype (aa)
 Why?

q2 is homozygous recessive (bb)
• The only ones you can tell their genotype just by
looking at them

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
q2 = 1/18
q2 = 0.0555
(we’re saying that around 5 percent of class has nonbrown eyes and are bb)
Step 2
 Step
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2: calculate allele frequency of q
q2 = (1/18)=(0.0555)
q = 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: 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 p2 and 2 pq

p2 (homozygous dominant) =
• p2 =0.584
• (around 58 % is homozygous BB)

2pq (heterozygous)=
• 2 pq= 0.3554
• (around 35 % is Bb)
Step 5:
 Double
check your genotype frequencies

(p2 )
+ (2pq)
+ (q2 )
=1
(0.584) + (0.3554) + (0.0555) = 1
1=1

Yah, we’re correct!!!
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