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Hardy-Weinberg Principle
1. 
Random mating
2. 
No mutations
3. 
No natural selection
4. 
Very large population size
5. 
No movement into or out of the population.
 
Are these common?
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 
One or more conditions is ALWAYS missing
 
Therefore it is impossible in nature!
 
It is an ideal baseline against which we can measure
genetic change.
 
Some populations, the rate of evolution is so low,
that they are close to equilibrium
 
We can use the Hardy-Weinberg Equilibrium to
estimate allele frequency
 
Allele frequency of A = p
 
Allele frequency of a = q
 
And p + q = 1
 
Frequency of AA = p2
 
Frequency of aa = q2
 
Frequency of Aa = 2pq
 
So, Hardy Weinberg Equation is
p2 + 2pq + q2 = 1
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 
 
You are studying the gene that regulates
interlocking fingers in a small isolated village with a
population of 2500 individuals. It is already known
that a dominant allele (F) causes one to interlock
their fingers in such a way that the left thumb is
nearly always on top, while a recessive allele (f) in the
homozygous conditions results in the right thumb
being on top
Data Collected:
 
 
 
Left thumb over right = 2275 individuals = FF, Ff
genotypes
Right thumb over left = 225 individuals = ff genotype
Data Collected:
 
 
Left thumb over right = 2275 individuals = FF, Ff genotypes
Right thumb over left = 225 individuals = ff genotype
 
Since there are 2500 individuals in the population, it is
understood that there are 5000 alleles in the population
 
Since we are assuming that the population is in H-W
equilibrium, then p2+ 2pq + q2 = 1, for FF, Ff & ff.
 
 
Recessive genotype frequency =
  q2 = (ff)/(total pop) = 225/2500 = 0.09 = 9% of pop.
Recessive allele frequency =
  q = (q2)1/2 = (0.09)1/2 = 0.3 = 30%
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 
Dominant allele frequency =
 
 
Homozygous dominant genotype frequency =
 
 
p2 = (0.7)2 = 0.49 = 49%
Heterozygous genotype frequency =
 
 
p = 1 – q = 1 – 0.3 = 0.7 = 70%
2pq = 2 (0.7)(0.3) = 0.42 = 42%
Number in the population:
 
 
 
ff: 225 individuals (9%)
FF: 49% of 2500 = 1225
Ff: 42% of 2500 = 1050
 
Do the population genotypes match the numbers
predicted by H-W, given the allele frequencies?
 
Example: If 4% of a population (4/100) has sicklecell anemia (aa), 60% (60/100) are heterozygous for
sickle-cells (Aa), and 36% (36/100) do not have the
sickle-cell allele (AA).
Find the allele frequencies:
1. 
• 
q2 = 0.04, so q = (.04)1/2 = 0.2
• 
p = 1 – q = 1 – 0.2 = 0.8
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Then, predict the genotype frequencies under H-W:
2. 
• 
• 
• 
aa: q2 = 0.04
Aa: 2pq = 0.32
AA: p2 = 0.64
Finally, comparing to the actual numbers, we have
40% aa, 60% Aa, and 36% AA in the actual
population.
3. 
• 
Therefore the population is NOT in H-W equilibrium.
 
What might cause this?
 
Where might this population be located?
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