Download Hardy-Weinberg Theorem - Effingham County Schools

Excerpts from Chapter 23
Population Genetics
• Population – a localized group of individuals
belonging to the same species
• Species – a group of populations whose
individuals can interbreed and produce
fertile offspring
• Gene pool – total aggregate of genes (all
alleles at all gene loci)
Hardy-Weinberg Theorem
•
•
The gene pool of a non-evolving population
will remain constant over generations
For the theorem to work, the following
conditions must be met:
1.
2.
3.
4.
5.
Large population size
No gene flow (no immigration or emigration)
No mutations
Random mating
No natural selection
Hardy-Weinberg Theorem
Two formulas for Hardy-Weinberg must be learned:
p2 + 2pq + q2 = 1
and
p+q=1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2 pq = percentage of heterozygous individuals
Hardy-Weinberg Theorem
Problem:
You have sampled a population in which you know that
the percentage of the homozygous recessive
genotype (aa) is 36%. Using that 36%, calculate
the following:
1. The frequency of the “aa” genotype
2. The frequency of the “a” allele
3. The frequency of the “A” allele
4. The frequencies of the genotypes “AA” and “Aa”
5. The frequencies of the two possible phenotypes if
“A” is completely dominant over “a”
Hardy-Weinberg Theorem
“the percentage of the homozygous recessive
genotype (aa) is 36%”
1. The frequency of the “aa” genotype
Given: 36% or 0.36
Hardy-Weinberg Theorem
“the percentage of the homozygous recessive
genotype (aa) is 36%”
2. The frequency of the “a” allele
The frequency of aa (q2) is 0.36; therefore
the frequency of a (q) is √0.36 = 0.60
Hardy-Weinberg Theorem
“the percentage of the homozygous recessive
genotype (aa) is 36%”
3. The frequency of the “A” allele
The frequency of aa (q2) is 0.36; therefore
the frequency of a (q) is √0.36 = 0.60
If q = 0.60 and p + q = 1; then p = 1 – 0.60 =
0.40
Hardy-Weinberg Theorem
“the percentage of the homozygous recessive
genotype (aa) is 36%”
4. The frequencies of the genotypes “AA” and
“Aa”
If p = 0.40 and q = 0.60
AA = p2 = 0.16
Aa = 2pq = (2)(0.40)(0.60) = 0.48
Hardy-Weinberg Theorem
“the percentage of the homozygous recessive
genotype (aa) is 36%”
5. The frequencies of the two possible
phenotypes if “A” is completely dominant
over “a”
Dominant phenotype – AA and Aa
0.16 + 0.48 = 0.64
Recessive phenotype – aa = 0.36