# Hardy-Weinberg Equilibrium

Edited by Jessie, Jen Moreau, Sharingknowledge, SarMal

**4 Parts:**

### Evolution

Evolution is defined as a change in the frequency of alleles in a **population** over time.

Let's break this sentence down: A allele is an alternative form of a gene. For example, fur color in mice could be black or white, where black is dominant (B) and white is recessive (b). A population is a group of individuals of the same species in a particular geographic area that can interbreed. In this example, we could use a population of mice along the California coast. There are many mice, which can have the phenotype black (BB or Bb genotype) or white fur (bb genotype). The percentage of mice that have BB, Bb, or bb genotypes are called the genotype frequencies of the population. The percentage of mice with the allele B or b, are called the **Allele Frequencies**. Finally, evolution is happening over time. This means that any changes that occur happen over the course of generations of mice. Evolution is when the frequency of B or b changes in the whole population of mice over the course of generations.

### A Population Undergoing Evolution

To predict whether evolution is occurring over generations of a population a mathematical model developed by G. H. Hardy and Wilhelm Weinberg called the Hardy-Weinberg Equilibrium can be used. This model states that allele and genotype frequencies in a population will remain the same from generation to generation without other evolutionary influences (i.e. bottlenecks, mutation, selection etc.) So when a population is in Hardy-Weinberg Equilibrium evolution is **not** happening.

## Conditions for Hardy-Weinberg Equilibrium

For a population to be in Hardy-Weinberg Equilibrium 5 assumptions must be met.

- 1This means that all of the individuals in the population have an equal ability to survive and produce offspring.No natural selection.

Advertisement - 2No new alleles are formed; the alleles that are in the population remain the same.No mutation.

- 3This means there is no gene flow. The number of a total number of alleles in the population remains the same; no new individuals enter or leave the population.No migration.

- 4This reduces the effects of genetic drift so that the allele frequencies are not subject to changes due to random chance.Population is infinitely large.

- 5Each individual in the population is equally likely to mate no matter the phenotype or genotype. When non-random mating occurs there is sexual selection. For example, when peacocks mate, females pick males with the largest, brightest tail. Those will smaller tails will not get to mate as often.Random Mating.

## The Mathematical Model

When a population is in Hardy-Weinberg Equilibrium

**Frequency of Alleles:** p and q are the frequency of alleles. Back to the example with mice, p will be the frequency of the B allele and q the frequency of the b allele. The frequency of all alleles in the population is 100%, so the frequency of p and q equals 1. This can be written as:

p = f(B) = frequency of B allele

q = f(b) = frequency of b allele

p + q = 1 = frequency of all alleles

**Frequency of Genotypes:** In the mice population there are 3 genotypes BB (black mouse, homozygous), Bb (black mouse, heterozygous), and bb (white mouse, homozygous). The frequency of BB, Bb, or b can be written:

p^{2} = f(BB) = frequency of black homozygous BB

2pq = f(Bb) = frequency of black heterozygous Bb

q^{2} = f(bb) = frequency of white homozygous bb

### Examples

**Example 1**

Sickle cell anemia is a rare disorder seen in recessive homozygotes (aa). Individuals with the disease have red blood that forms a characteristic sickle shape. In North America, this occurs in 1 in 400 African-American births. What is the frequency of carriers of this disease (Aa genotype)?

We know that 1 in 400 births are the genotype aa. This means:

q^{2} = f(aa) = 1/400 = 0.0025

From this, we can find q.

sqrt (q^{2}) = q = sqrt(0.0025) = 0.05

If q = 0.05, then we can find p.

p + q = 1

p + 0.05 = 1

p = 0.95

Now that we know p=0.95 and q=0.05, we can determine the frequency of carriers (Aa).

2pq = f(Aa) = 2*0.95*0.05 = .095 so 9.5% of the individuals are heterozygotes (carriers).

**Example 2**

Within a population of mice, the color black (B) is dominant to white (b) and 60% of the mice are black. What percentage are heterozygotes? What is the frequency of homozygous recessive mice?

We know 60% of the mice are either BB or Bb, we don't know which. So:

p^{2} + 2pq + q^{2} = 1 and p^{2} = f(BB), 2pq = f(Bb), and q^{2} = f(bb). We can replace part of the equation to get:

f(BB) + f(Bb) + q2 = 1, and we know f(BB) + f(Bb) = 0.6.

0.6 + q2 = 1
q^{2} = 0.4, so the frequency of homozygous recessive mice is 0.4, or 40%.

To find the percentage of heterozygotes we now find p and q.

q^{2} = 0.4

q = 0.63, now find p.

p + q = 1

p + 0.63 = 1

p = 0.37, now find the frequency of heterozygotes.

f(Bb) = 2pq = 2*0.37*0.63 = 0.47. So, the frequency of heterozygotes is 0.47 or 47%.

**Example 3**

There is a population of 800 black mice and 200 white mice. White is recessive to black. What are the allele and genotype frequencies? What is the number of heterozygous you would predict in this population? If the next generation has 1500 new mice, how many would you predict are black and white?

There are 1000 total mice. 800 black. 200 white. And white is recessive (bb). From that, we get:

f(aa) = q^{2} = 200/1000 = 0.2. Now find q.

sqrt(q^{2})= q = sqrt(0.2) = 0.45. Now find p.

p + q =1

p + 0.45 = 1

p = 0.55, now you have all the information to find genotype and allele frequencies.

allele frequencies:

f(B) = p = 0.55, or 55%.

f(b) = q = 0.45, or 45%

genotype frequencies:

f(BB) = p^{2} = 0.552 = 0.3, or 30%.

f(Bb) = 2pq = 2*0.55*0.45 = 0.5, or 50%.

f(bb) = q^{2} = 0.452 = 0.2, or 20%.

What is the number of heterozygotes you would predict in this population?*what is the frequency of heterozygotes? 0.5 what is the size of the population? 1000.*

frequency of heterozygotes x size of population = predicted # of heterozygotes.

0.5*1000 = 500 heterozygotes.

If the next generation has 1500 new mice, how many would you predict are homozygous recessive?

Similar to the last question:*the population size? 1500. what do you want to know? frequency of black (BB and Bb) and white (bb)*

frequency black = 0.3*1500 + 0.5*1500 = 450 + 750 = 1200 black mice

frequency white = 0.2*1500 = 300 white mice.

## Referencing this Article

If you need to reference this article in your work, you can copy-paste the following depending on your required format:

__APA (American Psychological Association)__

Hardy-Weinberg Equilibrium. (2017). In *ScienceAid*. Retrieved Jan 20, 2019, from https://scienceaid.net/HardyWeinberg_Equilibrium

** MLA (Modern Language Association)**
"Hardy-Weinberg Equilibrium."

*ScienceAid*, scienceaid.net/HardyWeinberg_Equilibrium Accessed 20 Jan 2019.

** Chicago / Turabian**
ScienceAid.net. "Hardy-Weinberg Equilibrium." Accessed Jan 20, 2019. https://scienceaid.net/HardyWeinberg_Equilibrium.

If you have problems with any of the steps in this article, please ask a question for more help, or post in the comments section below.

## Comments

## Article Info

Categories : Biology

Recent edits by: Sharingknowledge, Jen Moreau, Jessie