Equilibrium Hardy-Weinberg


Hardy-Weinberg equilibrium law states that allele and genotype frequencies in a population will remain constant from one generation to next generation in the absence of disturbing factors. In this calculator, Hardy-Weinberg equilibrium can be used to calculate the expected common homozygotes, expected heterozygotes, expected rare homozygotes and the frequency range of the 2 (p and q) alleles from the observed genotypes.

Calculator of Hardy-Weinberg equilibrium

Common Homozygotes
Heterozygotes
Rare Homozygotes
Chi-squared (chi2)
Common Homozygotes (Genotype)
Expected (CH)
Observed (CH)
Heterozygotes (Genotype)
Expected (H)
Observed (H)
Rare Homozygotes (Genotype)
Expected (RH)
Observed (RH)
Frequency Range
p Allele Frequency
q Allele Frequency

Formula of Hardy-Weinberg equilibrium

pfreq = (CH + (0.5 * H)) / (CH + H + RH)

qfreq = (RH + (0.5 * H)) / (CH + H + RH)

ECH = pfreq * pfreq * (CH + H + RH)

EH = 2 * pfreq * qfreq * (CH + H + RH)

ERH = qfreq * qfreq * (CH + H + RH)

chi1 = ((CH – ECH) * (CH – ECH)) / ECH

chi2 = ((H – EH) * (H – EH)) / EH

chi3 = ((RH – ERH) * (RH – ERH)) / ERH

chi2 = chi1 + chi2 + chi3

  • CH = Common Homozygotes
  • H = Heterozygotes
  • RH = Rare Homozygotes
  • ECH = Expected Common Homozygotes
  • EH = Expected Heterozygotes
  • ERH = Expected Rare Homozygotes
  • pfreq = p Allele Frequency
  • pfreq = q Allele Frequency

It is also known as Hardy-Weinberg principle, model, theorem.

Example of Hardy-Weinberg equilibrium

If the observed common homozygotes is 21, heterozygotes is 10 and rare homozygotes is 2, then

Step 1 : p Allele Frequency

pfreq =

= (21 + (0.5 * 10)) / (21 +10 + 2)

= (21 + 5) / 33

= 0.7879

Step 2 : q Allele Frequency

qfreq =

= (2 + (0.5 * 10)) / (21 + 10 + 2)

= 7 / 33

= 0.2121

Step 3 : Expected CH

Expected CH =

= 0.78 * 0.2121 * (21 + 10 + 2)

= 20.48

Step 4 : Expected H

Expected H =

= 2 * 0.78 * 0.2121 * (33)

= 11.03

Step 5 : Expected RH

Expected RH = 1.4848

Step 6 : Chi Square

chi1 = ((CH – ECH) * (CH – ECH)) / ECH

chi1 = ((21 – 20.4848) * (21- 20.4848)) / 20.4848

chi1=0.0129

chi2 = ((H – EH) * (H – EH)) / EH
chi2 = ((10 – 11.0303) * (10 – 11.0303)) / 11.0303
chi2 = 0.0962

chi3 = ((RH – ERH) * (RH – ERH)) / ERH
chi3 = ((2 – 1.4848) * (2 – 1.4848)) / 1.4848
chi3 = 0.1787

chi2 = chi1 + chi2 + chi3
chi2 = 0.0129+0.0962+0.1787
chi2 = 0.2879

Leave a Reply 8

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Yurany Granada

Yurany Granada

Buenas tardes, quiero saber como debo citarlos. Gracias

Adriana Mena

Adriana Mena

How are the df calculated?

Oladejo Aliu Abiola

Oladejo Aliu Abiola

How do you calculate the expected value for the alleles

Daniel

Daniel

Si una enfermedad es recesiva y la frecuencia del alelo que la genera es de
0,08, ¿cómo serán las frecuencias genotípicas?

Nada

Nada

I have one question please, the equation was as follows;
ECH = pfreq * pfreq * (CH + H + RH)
yet you have used the frequency of the rare allele in the following example ,
Step 3 : Expected CH

Expected CH =

= 0.78 * 0.2121 * (21 + 10 + 2)

can you explain why, please?
Plus,
what do you mean exactly by the following;
chi2 = chi1 + chi2 + chi3

Aaron Castle

Aaron Castle

I love you, thank you for saving my grade in bio and letting me keep an A. Thank you!

hamzah H kzar

hamzah H kzar

thank you so much to helping me by your this website

kamna srivastava srivastava

kamna srivastava srivastava

very useful. with you examples anybody can learn easily and population genetics.