4.13: Hardy-Weinberg Equilibrium
- Page ID
- 62241
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The alleles in the equation are defined as the following:
- In a two allele system with dominant/recessive, we designate the frequency of one as \(p\) and the other as \(q\) and standardize to:
\(p\) = dominant allele
\(q\) = recessive allele
\(p + q = 1.00\) (100% or total population)
- Therefore the total frequency of all alleles in this system equal 100% (or 1)
- Likewise, the total frequency of all genotypes is expressed by the following quadratic where it also equals 1:
\(p^2\) = number of homozygous dominant individuals in a population (frequency of homozygous dominant) ex. AA
\(q^2\) = number of homozygous recessive individuals in a population (frequency of homozygous recessive) ex. Aa
\(2pq\) = number of heterozygous individuals in a population (frequency of heterozygous) ex. Aa
\(p^2 + 2pq + q^2 = 1.00\) (100% or total population)
- This equation is the Hardy-Weinberg theorem that states that there are no evolutionary forces at play that are altering the gene frequencies.
**Please see below for my walk-through of how to do a problem.
Here is the equation:
\(p^2 + 2pq + q^2 = 1.00\) (100% of population)
\(p^2\) = all individuals who are homozygous dominant
\(q^2\) = all individuals who are homozygous recessive
\(2pq\) = all individuals who are heterozygous
Also important: \(p + q = 1.00\)
\(p\) = the dominant allele
\(q\) = the recessive allele
Problem: a recessive trait (let’s say R) is seen in 16% of a population. First, we must determine the frequencies of the other organisms.
Step 1: identify recessive individuals (\(q^2\))
\(q^2\) here equals 16% or 0.16
This means that 16% of the population is homozygous recessive for the trait
Step 2: use \(q^2\) to find \(q\) (we want to look at just the allele, not a whole organism) To do this, take the square root of 0.16
The square root of 16 = 4, so the square root of 0.16 = 0.4
This means that the recessive allele is found in 40% of the population
You can use calculators!!
Step 3: Use \(q\) to find \(p\) (here we use the recessive allele \(q\) to find the dominant allele \(p\))
\(p + q = 1.00\)
\(p + 0.4 = 1.00\)
\(p\) must equal 0.6
This means that the dominant allele is found in 60% of the population
Step 4: Use \(p\) to find \(p^2\) (just square 0.6)
6 x 6 = 36 so 0.6 x 0.6 = 0.36
This means that 36% of the population is homozygous dominant for the trait
Step 5: plug into the equation:
\(p^2 + 2pq + q^2 = 1.00\)
\(0.36 + 2pq + 0.16 = 1.00\)
\(2pq\) must equal 0.48
This means that 48% of the population is heterozygous
If we came back in another generation, these numbers would all be different
Therefore, the frequencies are:
RR = (\(p^2\)) = 36% R = (\(p\)) = 60%
rr = (\(q^2\)) = 16% r = (\(q\)) = 40%
Rr = (\(2pq\)) = 48%
If we did this again in generation 2, as long as any of these numbers changed, we have evidence of evolution.