Derivation Hardy–Weinberg principle

length of p, q corresponds allele frequencies (here p = 0.6, q = 0.4). area of rectangle represents genotype frequencies (thus aa : aa : aa = 0.36 : 0.48 : 0.16).

the different ways form genotypes next generation can shown in punnett square, proportion of each genotype equal product of row , column allele frequencies current generation.

the sum of entries p + 2pq + q = 1, genotype frequencies must sum one.

note again p + q = 1, binomial expansion of (p + q) = p + 2pq + q = 1 gives same relationships.

summing elements of punnett square or binomial expansion, obtain expected genotype proportions among offspring after single generation:

these frequencies define hardy–weinberg equilibrium. should mentioned genotype frequencies after first generation need not equal genotype frequencies initial generation, e.g. f1(aa) ≠ f0(aa). however, genotype frequencies future times equal hardy–weinberg frequencies, e.g. ft(aa) = f1(aa) t > 1. follows since genotype frequencies of next generation depend on allele frequencies of current generation which, calculated equations (1) , (2), preserved initial generation:

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1


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{\displaystyle {\begin{aligned}f_{1}({\text{a}})&=f_{1}({\text{aa}})+{\tfrac {1}{2}}f_{1}({\text{aa}})=p^{2}+pq=p(p+q)=p=f_{0}({\text{a}})\\f_{1}({\text{a}})&=f_{1}({\text{aa}})+{\tfrac {1}{2}}f_{1}({\text{aa}})=q^{2}+pq=q(p+q)=q=f_{0}({\text{a}})\end{aligned}}}

for more general case of dioecious diploids [organisms either male or female] reproduce random mating of individuals, necessary calculate genotype frequencies 9 possible matings between each parental genotype (aa, aa, , aa) in either sex, weighted expected genotype contributions of each such mating. equivalently, 1 considers 6 unique diploid-diploid combinations:

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{\displaystyle \left[({\text{aa}},{\text{aa}}),({\text{aa}},{\text{aa}}),({\text{aa}},{\text{aa}}),({\text{aa}},{\text{aa}}),({\text{aa}},{\text{aa}}),({\text{aa}},{\text{aa}})\right]}

and constructs punnett square each, calculate contribution next generation s genotypes. these contributions weighted according probability of each diploid-diploid combination, follows multinomial distribution k = 3. example, probability of mating combination (aa,aa) 2 ft(aa)ft(aa) , can result in aa genotype: [0,1,0]. overall, resulting genotype frequencies calculated as:

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{\displaystyle {\begin{aligned}&\left[f_{t+1}({\text{aa}}),f_{t+1}({\text{aa}}),f_{t+1}({\text{aa}})\right]=\\&\qquad =f_{t}({\text{aa}})f_{t}({\text{aa}})\left[1,0,0\right]+2f_{t}({\text{aa}})f_{t}({\text{aa}})\left[{\tfrac {1}{2}},{\tfrac {1}{2}},0\right]+2f_{t}({\text{aa}})f_{t}({\text{aa}})\left[0,1,0\right]\\&\qquad \qquad +f_{t}({\text{aa}})f_{t}({\text{aa}})\left[{\tfrac {1}{4}},{\tfrac {1}{2}},{\tfrac {1}{4}}\right]+2f_{t}({\text{aa}})f_{t}({\text{aa}})\left[0,{\tfrac {1}{2}},{\tfrac {1}{2}}\right]+f_{t}({\text{aa}})f_{t}({\text{aa}})\left[0,0,1\right]\\&\qquad =\left[\left(f_{t}({\text{aa}})+{\tfrac {1}{2}}f_{t}({\text{aa}})\right)^{2},2\left(f_{t}({\text{aa}})+{\tfrac {1}{2}}f_{t}({\text{aa}})\right)\left(f_{t}({\text{aa}})+{\tfrac {1}{2}}f_{t}({\text{aa}})\right),\left(f_{t}({\text{aa}})+{\tfrac {1}{2}}f_{t}({\text{aa}})\right)^{2}\right]\\&\qquad =\left[f_{t}({\text{a}})^{2},2f_{t}({\text{a}})f_{t}({\text{a}}),f_{t}({\text{a}})^{2}\right]\end{aligned}}}

as before, 1 can show allele frequencies @ time t+1 equal @ time t, , so, constant in time. similarly, genotype frequencies depend on allele frequencies, , so, after time t=1 constant in time.

if in either monoecious or dioecious organisms, either allele or genotype proportions unequal in either sex, can shown constant proportions obtained after 1 generation of random mating. if dioecious organisms heterogametic , gene locus located on x chromosome, can shown if allele frequencies unequal in 2 sexes [e.g., xx females , xy males, in humans], f′(a) in heterogametic sex ‘chases’ f(a) in homogametic sex of previous generation, until equilibrium reached @ weighted average of 2 initial frequencies.