SEL - MATH: Hardy-Weinberg [LESSON]

MATH: Hardy-Weinberg

Hardy-Weinberg Equilibrium

The Hardy-Weinberg principle states that both allele and genotype frequencies in a population’s gene pool remain constant from generation to generation if the population remains in genetic equilibrium. A gene pool is made up of all the alleles present within a population. This principle provides a mathematical model for studying evolutionary changes in allelic frequency within a genetically stable population. This equilibrium is maintained by specific conditions: random mating, no mutations, no natural selection, large population size, and no genetic drift or gene flow. Use the chart below to help you remember the requirements for genetic equilibrium.

Force of evolution and corresponding requirements for equilibrium.

It is important to note that these conditions are unrealistic and that Hardy-Weinberg equilibrium is impossible in nature. However, genetic equilibrium is an ideal state that provides a baseline against which to measure change.

Do you remember what an allele is?

There are usually multiple variations of a given gene, called alleles. Offspring inherit one allele from each parent, and the combination of those two alleles determines the trait that is expressed in that individual. Alleles are often dominant or recessive. Dominant alleles are expressed with a capital letter, for example P. Recessive alleles are expressed with a lowercase letter, for example p.

Watch the Hardy-Weinberg Theory video below and take notes on the Hardy-Weinberg principle and the corresponding equations.

Hardy-Weinberg Equations

Hover over each piece of the equation to learn about what it represents in the Hardy-Weinberg Equation activity below.

Previously you learned that in Mendelian genetics, there are two variations, or alleles, for a given gene. One variation is dominant (A), while the other variation is recessive (a). If we mate two individuals that are heterozygous (Aa) for this trait, we find that

  1. 25% of their offspring are homozygous for the dominant allele (AA),
  2. 50% are heterozygous (Aa), and,
  3. 25% are homozygous for the recessive allele (aa).
Mendelian Genetics

 

A

a

A

AA

Aa

a

Aa

aa

 

AA = 1 out of 4 or 1/4 = 25%

Aa = 2 out of 4 or 1/2 = 50%

aa = 1 out of 4 or 1/4 = 25%

 

AA + Aa + Aa + aa = 100%

A2 + 2(Aa) + a2 = 1

 

 

 

 

 

 

 

Now let's take what we found and see if we can apply it to the Hardy-Weinberg equation.

Applied to the Hardy-Weinberg equation

 

p

q

p

pp

pq

q

pq

qq

 

pp = 1 out of 4 or 1/4 = 25%

pq = 2 out of 4 or 1/2 = 50%

qq = 1 out of 4 or 1/4 = 25%

 

pp + pq + pq + qq = 100%

p2 + 2pq + q2 = 1

 

 

 

 

 

 

 

So now that we understand how the equation was developed, let's look closer at what it means, and what we are supposed to do with it! p2 = percentage of homozygous dominant individuals, q2 = percentage of homozygous recessive individuals, and 2pq = percentage of heterozygous individuals in the population. Furthermore, p = frequency of the dominant allele in the gene pool and q = frequency of the recessive allele in the gene pool. Since we are assuming these are the only alleles in the gene pool for the trait, we can say that the frequency of the dominant allele plus the frequency of the recessive allele equals the entire gene pool. In other words, p + q = 1.

Ok, now what? In the example of our heterozygote individuals, Aa x Aa, we found the percentage of homozygous dominant individuals to be 25%, the percentage of homozygous recessive individuals to be 25%, and the percentage of heterozygotes to be 50%. That means that p2 = 25%, q2 = 25%, and 2pq = 50%. But what is the frequency of the alleles? To answer that question, we must take the square root of p2 and q2. If p2 = 25%, or 0.25, then p = 0.5. If q2 = 25%, or 0.25, then q = 0.5. Let's plug in our values into the equations and see if they make sense...

 

p + q = 1

(0.5) + (0.5) = 1

True

 

p2 + 2pq + q2 = 1

(0.5)2 + 2(0.5)(0.5) + (0.5)2 = 1

0.25 + 0.5 + 0.25 = 1

True

 

 

 

 

 

 

 

 

 

 

But the frequency of two alleles in an entire population of organisms is unlikely to be exactly the same. Say that the frequency A was 0.8 instead of 0.5? Since p + q = 1, that would mean that the frequency of a must be 0.2.

0.8 +0.2 = 1

If these are the new frequencies, what is the percentage of individuals that are homozygous dominant, heterozygous, and homozygous recessive?

p = 0.8

q = 0.2

 

p2 = .64 = 64%

2pq = 2(0.8)(0.2) = 0.32 = 32%

q2 = 0.04 = 4%

64% + 32% + 4% = 100%

 

 

 

 

 

 

 

 

Watch the Hardy-Weinberg Practice Problems video below.

Let's look at a few sample problems and see if we can apply the Hardy-Weinberg equation to determine allelic frequencies.

Try this! Answer the two practice questions below, select the correct answer then click the arrow.

Let’s try some more practice!

Click through the Hardy-Weinberg Presentation below for more information.

[CC BY 4.0] UNLESS OTHERWISE NOTED | IMAGES: LICENSED AND USED ACCORDING TO TERMS OF SUBSCRIPTION