4.13: Hardy-Weinberg Equilibrium
- Page ID
- 62241
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.