solving hardy weinberg equilibrium problems

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Hardy Weinberg Problemset

Hardy-Weinberg equilibrium (HWE) is a fundamental concept in population genetics that describes the theoretical relationship between allele frequencies and genotype frequencies in an ideal, non-evolving population.

Students can practice using the Hardy Weinberg equilibrium equation to determine the allele frequencies in a population. This set of 10 questions gives students just enough information to solve for p (dominant allele frequency) and q (recessive allele frequency). They will also calculate the percentage of heterozygous individuals (2pq).

I designed this worksheet for an AP Biology class and was revised April 2019. There is an older version that has many of the answers posted online, so this one has been changed slightly so that students cannot easily find the answers and take a shortcut.

The ten problems may be too many for some students, I usually ask my students to consider how much practice they need. If they are confident after doing 3-4 of the problems, then the rest can be optional.

Time Required: 20-30 minutes Grade Level: 11-12

HS-LS3-3 Apply concepts of statistics and probability to explain the variation and distribution of expressed traits in a population.

Shannan Muskopf

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The Hardy-Weinberg Equations And How To Use Them

What is the hardy-weinberg principle.

The Hardy-Weinberg principle, also referred to as the Hardy-Weinberg equilibrium, is a set of 5 assumptions which when satisfied can enable the determination of allele and genotype frequencies of a population. These frequencies will also remain constant for future generations. The principle was discovered by Godrey Hardy and Wilhelm Weinberg in 1908, based on Gregor Mendel’s Law of Segregation. To estimate the frequency of alleles and genotypes of a certain population, there is two simple formula that can be used.

The assumptions of the Hardy-Weinberg principle

There are 5 assumptions that are made when using the Hardy-Weinberg equations. These are:

• No natural selection : There are no evolutionary pressures which may favour a particular allele.
• Random mating: Each individual in a population mates randomly so that mating with an individual carrying a particular allele is not favoured.
• No mutations: There are no DNA mutations occurring for the alleles which may affect their function.
• A closed population: Individuals within the population do not leave and new individuals are not introduced to the population.
• Large population size: The population is considered large enough, at best infinite, so that major changes in allele frequencies do not cause a genetic drift.

If any of these assumptions are not satisfied, then the principle cannot be applied.

Determining the allele frequency

The first Hardy-Weinberg equation ( p + q = 1 ) concerns estimating the frequency of alleles in a population. Each gene usually has two alleles (diploid organism), one from each parent. These alleles are denoted as the dominant ( A ) and recessive ( a ) forms. These are represented as ‘ p ‘ and ‘ q ‘ is the equation below.

In a population, the combined frequency of both the alleles must equal 1 (100%).

In a population, there are two alleles for ear shape: having detached lobes (dominant, A) or having attached lobes (recessive, a). Determine the allele frequency of the recessive allele ‘a’ (attached lobes) given the frequency of the dominant allele ‘A’ (attached lobes) is 73%.

The recessive allele frequency is 27%.

Determining the genotype frequencies

The Hardy-Weinberg equation used to determine genotype frequencies is:  p 2 + 2pq + q 2 = 1 .

Again, if one genotype frequency is known, it is possible to use the Hardy-Weinberg equations to work out the others.

Let’s use the same example above regarding earlobes (detached lobes and attached lobes). Determine the genotype frequency of the homozygous dominant (AA) allele for having detached earlobes given the frequency of the attached earlobe (aa) phenotype is 6%.

The homozygous dominant genotype frequency is 57%.

4. If you wanted to go one step further and work out the frequency of the heterozygous genotype ( Aa ), so ‘ 2pq ‘ in the Hardy-Weinberg equation, you can do. Since we know the value of ‘ p ‘ (0.755) and ‘ q ‘ (0.245). All that is required is to multiply 2 by 0.755 and by 0.245. Doing this will give 0.37. So, 37% of the population will have the heterozygous genotype ( Aa ).

Notice, if you add all of the genotype frequencies together, this equals 1 (0.57 + 0.37 + 0.06 = 1).

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AP®︎/College Biology

Course: ap®︎/college biology   >   unit 7.

• Allele frequency
• Allele frequency & the gene pool
• Hardy-Weinberg equation

Applying the Hardy-Weinberg equation

• Discussions of conditions for Hardy-Weinberg
• Mechanisms of evolution
• Hardy-Weinberg equilibrium

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Hardy Weinberg equilibrium Problems and Solutions

Population genetics: hardy-weinberg equilibrium .

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18.6: The Hardy-Weinberg Equilibrium

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• John W. Kimball
• Tufts University & Harvard

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If two individuals mate that are heterozygous (e.g., Bb ) for a trait, we find that

• 25% of their offspring are homozygous for the dominant allele ( BB )
• 50% are heterozygous like their parents ( Bb )
• 25% are homozygous for the recessive allele ( bb ) and thus, unlike their parents, express the recessive phenotype.

This is what Mendel found when he crossed monohybrids. It occurs because meiosis separates the two alleles of each heterozygous parent so that 50% of the gametes will carry one allele and 50% the other and when the gametes are brought together at random, each B (or b )-carrying egg will have a 1 in 2 probability of being fertilized by a sperm carrying B (or b ). (Left table)

However, the frequency of two alleles in an entire population of organisms is unlikely to be exactly the same. Let us take as a hypothetical case, a population of hamsters in which 80% of all the gametes in the population carry a dominant allele for black coat ( B ) and 20% carry the recessive allele for gray coat ( b ).

Random union of these gametes (right table) will produce a generation:

• 64% homozygous for BB (0.8 x 0.8 = 0.64)
• 32% Bb heterozygotes (0.8 x 0.2 x 2 = 0.32)
• 4% homozygous ( bb ) for gray coat (0.2 x 0.2 = 0.04)

So 96% of this generation will have black coats; only 4% gray coats.

Will gray coated hamsters eventually disappear? No. Let's see why not.

• All the gametes formed by BB hamsters will contain allele B as will one-half the gametes formed by heterozygous ( Bb ) hamsters.
• So, 80% (0.64 + .5*0.32) of the pool of gametes formed by this generation with contain B .
• All the gametes of the gray ( bb ) hamsters (4%) will contain b but one-half of the gametes of the heterozygous hamsters will as well.
• So 20% (0.04 + .5*0.32) of the gametes will contain b .

So we have duplicated the initial situation exactly. The proportion of allele b in the population has remained the same. The heterozygous hamsters ensure that each generation will contain 4% gray hamsters. Now let us look at an algebraic analysis of the same problem using the expansion of the binomial ( p + q ) 2 .

\[(p+q)^2 = p^2 + 2pq + q^2\]

The total number of genes in a population is its gene pool .

• Let \(p\) represent the frequency of one gene in the pool and \(q\) the frequency of its single allele.
• \(p^2\) = the fraction of the population homozygous for \(p\)
• \(q^2\) = the fraction homozygous for \(q\)
• \(2pq\) = the fraction of heterozygotes
• In our example, p = 0.8, q = 0.2, and thus \[(0.8 + 0.2)^2 = (0.8)^2 + 2(0.8)(0.2) + (0.2)^2 = 064 + 0.32 + 0.04\]

The algebraic method enables us to work backward as well as forward. In fact, because we chose to make B fully dominant, the only way that the frequency of B and b in the gene pool could be known is by determining the frequency of the recessive phenotype (gray) and computing from it the value of q .

q 2 = 0.04, so q = 0.2, the frequency of the b allele in the gene pool. Since p + q = 1, p = 0.8 and allele B makes up 80% of the gene pool. Because B is completely dominant over b , we cannot distinguish the Bb hamsters from the BB ones by their phenotype. But substituting in the middle term ( 2pq ) of the expansion gives the percentage of heterozygous hamsters. 2pq = (2)(0.8)(0.2) = 0.32

So, recessive genes do not tend to be lost from a population no matter how small their representation.

Hardy-Weinberg law

So long as certain conditions are met (discussed below), gene frequencies and genotype ratios in a randomly-breeding population remain constant from generation to generation. This is known as the Hardy-Weinberg law.

The Hardy-Weinberg law is named in honor of the two men who first realized the significance of the binomial expansion to population genetics and hence to evolution. Evolution involves changes in the gene pool. A population in Hardy-Weinberg equilibrium shows no change. What the law tells us is that populations are able to maintain a reservoir of variability so that if future conditions require it, the gene pool can change. If recessive alleles were continually tending to disappear, the population would soon become homozygous. Under Hardy-Weinberg conditions, genes that have no present selective value will nonetheless be retained.

When the Hardy-Weinberg Law Fails

To see what forces lead to evolutionary change, we must examine the circumstances in which the Hardy-Weinberg law may fail to apply. There are five:

• genetic drift
• nonrandom mating
• natural selection

The frequency of gene B and its allele b will not remain in Hardy-Weinberg equilibrium if the rate of mutation of B -> b (or vice versa) changes. By itself, this type of mutation probably plays only a minor role in evolution; the rates are simply too low. However, gene (and whole genome) duplication - a form of mutation - probably has played a major role in evolution. In any case, evolution absolutely depends on mutations because this is the only way that new alleles are created. After being shuffled in various combinations with the rest of the gene pool, these provide the raw material on which natural selection can act.

Many species are made up of local populations whose members tend to breed within the group. Each local population can develop a gene pool distinct from that of other local populations. However, members of one population may breed with occasional immigrants from an adjacent population of the same species. This can introduce new genes or alter existing gene frequencies in the residents.

In many plants and some animals, gene flow can occur not only between subpopulations of the same species but also between different (but still related) species. This is called hybridization . If the hybrids later breed with one of the parental types, new genes are passed into the gene pool of that parent population. This process is called introgression . It is simply gene flow between species rather than within them.

Comparison of the genomes of contemporary humans with the genome recovered from Neanderthal remains shows that from 1–3% of our genes were acquired by introgression following mating between members of the two populations tens of thousands of years ago.

Whether within a species or between species, gene flow increases the variability of the gene pool.

Genetic Drift

As we have seen, interbreeding often is limited to the members of local populations. If the population is small, Hardy-Weinberg may be violated. Chance alone may eliminate certain members out of proportion to their numbers in the population. In such cases, the frequency of an allele may begin to drift toward higher or lower values. Ultimately, the allele may represent 100% of the gene pool or, just as likely, disappear from it.

Drift produces evolutionary change, but there is no guarantee that the new population will be more fit than the original one. Evolution by drift is aimless, not adaptive.

Nonrandom Mating

One of the cornerstones of the Hardy-Weinberg equilibrium is that mating in the population must be random. If individuals (usually females) are choosy in their selection of mates, the gene frequencies may become altered. Darwin called this sexual selection .

Nonrandom mating seems to be quite common. Breeding territories, courtship displays, "pecking orders" can all lead to it. In each case certain individuals do not get to make their proportionate contribution to the next generation.

Assortative mating

Humans seldom mate at random preferring phenotypes like themselves (e.g., size, age, ethnicity). This is called assortative mating . Marriage between close relatives is a special case of assortative mating. The closer the kinship, the more alleles shared and the greater the degree of inbreeding . Inbreeding can alter the gene pool. This is because it predisposes to homozygosity . Potentially harmful recessive alleles — invisible in the parents — become exposed to the forces of natural selection in the children.

It turns out that many species - plants as well as animals - have mechanisms be which they avoid inbreeding. Examples:

• Link to discussion of self-incompatibility in plants.
• Male mice use olfactory cues to discriminate against close relatives when selecting mates. The preference is learned in infancy - an example of imprinting. The distinguishing odors are controlled by the MHC alleles of the mice and are detected by the vomeronasal organ (VNO).

Natural Selection

If individuals having certain genes are better able to produce mature offspring than those without them, the frequency of those genes will increase. This is simply expressing Darwin's natural selection in terms of alterations in the gene pool. (Darwin knew nothing of genes.) Natural selection results from differential mortality and/or differential fecundity .

Mortality Selection

Certain genotypes are less successful than others in surviving through to the end of their reproductive period. The evolutionary impact of mortality selection can be felt anytime from the formation of a new zygote to the end (if there is one) of the organism's period of fertility. Mortality selection is simply another way of describing Darwin's criteria of fitness: survival .

Fecundity Selection

Certain phenotypes (thus genotypes) may make a disproportionate contribution to the gene pool of the next generation by producing a disproportionate number of young. Such fecundity selection is another way of describing another criterion of fitness described by Darwin: family size . In each of these examples of natural selection, certain phenotypes are better able than others to contribute their genes to the next generation. Thus, by Darwin's standards, they are more fit . The outcome is a gradual change in the gene frequencies in that population.

Calculating the Effect of Natural Selection on Gene Frequencies

The effect of natural selection on gene frequencies can be quantified. Let us assume a population containing

• 36% homozygous dominants ( AA )
• 48% heterozygotes ( Aa ) and
• 16% homozygous recessives ( aa )

The gene frequencies in this population are \(p = 0.6\) and \(q = 0.4\). The heterozygotes are just as successful at reproducing themselves as the homozygous dominants, but the homozygous recessives are only 80% as successful. That is, for every 100 AA (or Aa ) individuals that reproduce successfully only 80 of the aa individuals succeed in doing so. The fitness (\(w\)) of the recessive phenotype is thus 80% or 0.8.

Their relative disadvantage can also be expressed as a selection coefficient , \(s\), where

\[s = 1 − w\]

In this case,

\[s = 1 − 0.8 = 0.2.\]

The change in frequency of the dominant allele (\(Δp\)) after one generation is expressed by the equation

\[Δp = \dfrac{s p_0 q_0^2}{1 - s q_0^2}\]

where \(p_0\) and \(q_0\) are the initial frequencies of the dominant and recessive alleles respectively. Substituting, we get

\[\begin{align} Δp & = \dfrac{(0.2)(0.6)(0.4)^2}{1 − (0.2)(0.4)^2} \\[5pt] &=\dfrac{0.019}{0.968} \\[5pt] &=0.02 \end{align}\]

So, in one generation, the frequency of allele A rises from its initial value of 0.6 to 0.62 and that of allele a declines from 0.4 to 0.38 (\(q = 1 − p\)).

The new equilibrium produces a population of

• 38.4% homozygous dominants (an increase of 2.4%) ( p 2 = 0.384)
• 47.1% heterozygotes (a decline of 0.9%)( 2pq = 0.471) and
• 14.4% homozygous recessives (a decline of 1.6%)( q 2 = 0.144)

If the fitness of the homozygous recessives continues unchanged, the calculations can be reiterated for any number of generations. If you do so, you will find that although the frequency of the recessive genotype declines, the rate at which a is removed from the gene pool declines; that is, the process becomes less efficient at purging allele a . This is because when present in the heterozygote, a is protected from the effects of selection.

Practice Hardy-Weinberg

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Hardy-weinberg equilibrium calculator.

The relationship between allele frequencies and genotype frequencies in populations at Hardy-Weinberg Equilibrium is usually described using a trait for which there are two alleles present at the locus of interest.

This calculator demonstrates the application of the Hardy-Weinberg equations to loci with more than two alleles. Visit the genetic drift and selection illustration for more on the Hardy-Weinberg Equilibrium.

Update the values by changing the allele frequency in the blue box below the graph. The calculator has a check that prevents the allele frequencies from summing to any value other than 1. To avoid having your values changed, make sure your values sum to one and enter them from top to bottoms (p then q then r ...)

• With 2 alleles, the number of genotypes is 1 + 2 = 3
• 3 alleles there are 1 + 2 + 3 = 6 genotypes
• 4 alleles there are 1 + 2 + 3 + 4 = 10 genotypes.

The general formula for finding the sum of a set of integers from 1 to n is:

Genotypes = n * n+1 / 2

The calculator does not go beyond 5 alleles and 15 possible genotypes. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles.

If a population has 10 alleles for a specific gene, the combined, total number of homozygous and heterozygous genotypes present in the population will be:

(10 * 11) / 2 = 55

This breaks down to 10 homozygous genotypes and 45 heterozygous genotypes. The sum of the allele frequencies would still need to equal 1 :

p + q + r + s + t + u + v + w + x + y = 1

As would the sum of the genotype frequencies:

p 2 + 2pq + 2pr + 2ps + 2pt + 2pu + 2pv + 2pw + 2px + 2py + q 2 + 2qr + 2qs + 2qt +2qu + 2qv + 2qw + 2qx + 2qy + r 2 + 2rs + 2rt + 2ru + 2rv + 2rw + 2rx + 2ry + s 2 + 2st+ 2su + 2sv + 2sw + 2sx + 2sy + t 2 + 2tu + 2tv + 2tw + 2tx + 2ty + u 2 + 2uv + 2uw +2ux + 2uy + v 2 + 2vw + 2vx + 2vy + w 2 + 2wx + 2wy + x 2 + 2xy + y 2 = 1

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