The Huntington Library, Art Collections, and Botanical Gardens
Bean There, Done That: A Hardy-Weinberg Simulation
This lesson plan was adapted from San Francisco State University’s Bio 240:
http://userwww.sfsu.edu/~biol240/labs/lab_02hardyweinberg/pages/lab_01expt.html
Overview
Students will look at a population of beans to examine conditions under which the
Hardy-Weinberg equilibrium applies.
Introduction
In the first part of this exercise we will demonstrate the contention of the HardyWeinberg Theorem that genetic recombination as a result of sexual reproduction will not,
by itself, cause any change in allele frequencies or genotype frequencies in a population
from one generation to the next. Obviously, it is difficult to demonstrate this principal
using real organisms: depending on the organism, it could take days, weeks, months or
even years to show that allele and genotype frequencies remained constant for several
generations - or even for a single generation. Furthermore, the Hardy-Weinberg
equilibrium depends on the absence of any forces operating which might change allele
frequencies in a population. These forces—mutation, migration, selection, and chance
effects—are almost never absent from natural populations and it would be very difficult
even in the laboratory to erect a set of conditions that might reliably exclude them.
The Hardy-Weinberg Law can be mathematically demonstrated in the following
table. If p equals the frequency of allele A and q is the frequency of allele a, union of
gametes would occur as follows:
p
q
p p2 pq
q pq q2
One of the predictions of the Hardy-Weinberg Law refers to the genotypic
frequencies after one generation of random mating. In the above table, the genotypic
frequencies for AA is p2, the genotypic frequency for Aa is 2pq and the genotypic
frequency for aa will be q2. These are the values that are predicted by the law. The
second prediction is that the frequencies of the two alleles will remain the same from
generation to generation. To prove it, finish the Hardy-Weinberg derivation worksheet.
Assuming Hardy-Weinberg equilibrium is in effect we can calculate the allelic
frequencies from the genotypic frequencies. To calculate the allelic frequency from the
genotypic frequency, first figure out from the genotypes where all the A alleles are found,
(AA and ½ of the alleles of Aa individuals will be A). So (where f stands for frequency), p
= f(AA) + ½ f (Aa). Since p2 represents the frequency of AA and 2pq represents the
frequency of Aa then when we substitute those frequencies into the equation we get, p =
p2 + ½(2pq). Using analogous calculations we can calculate the frequency of the a (q)
allele. In absence of any factors that change the allelic frequencies, the genotypic and
allelic frequencies will remain the same from generation to generation.
The Hardy-Weinberg equilibrium also assumes random mating in the population,
and once again this is a situation that is rarely found in nature and difficult to achieve in
the lab. Therefore, for the purposes of our demonstration, we are going to work not with
real organisms but with a model—in this case dried beans.
Motivation
What causes organisms to change from one generation to the next? We know that
environmental pressures are a key force in natural selection. If we could eliminate natural
selection, for instance, would mating organisms change over time? What about mutation?
Imagine a large population of breeding fruit flies, completely isolated. Over a long period
of time, do you think the population would change based solely on the recombination of
genes that happens with sex? It’s almost impossible to test such a thing with living
organisms, but the Hardy-Weinberg Equilibrium says that if you remove mutation,
migration, selection, and chance effects from a mating population, the allele frequencies
will stay the same: you won’t see a change in the genotypes in the population. This lab
uses beans rather than living organisms to simulate such a set up. How right is HardyWeinberg?
Objectives
Upon completion of the lab, students should be able to
1. Derive the Hardy-Weinberg Equation.
2. Calculate genotypic and allelic frequencies.
3. Discuss the conditions under which a population is in Hardy-Weinberg Equilibrium.
4. Determine if a population is in Hardy-Weinberg Equilibrium.
5. Design and perform a test of a disruption to Hardy-Weinberg Equilibrium.
6. Identify why the Hardy-Weinberg Equilibrium is difficult or impossible to
demonstrate with living organisms.
Materials
• Opaque Tupperware containers or coffee cans with lids
• Three different colors of beans that are approximately the same size (e.g., brown, white
and brown/white speckled). For each group you will need 2,000 beans (180 white
beans, 840 speckled beans, and 980 red beans)
Associated California Biology Standards
7b. Students know why alleles that are lethal in a homozygous individual may be carried
in a heterozygote and thus maintained in a gene pool.
7e. Students know the conditions for the Hardy-Weinberg equilibrium in a population
and why these conditions are not likely to appear in nature.
7f. Students know how to solve the Hardy-Weinberg equation to predict the frequency of
genotypes in a population, given the frequency of phenotypes.
Procedure
1. Count 100 beans of one color by hand and weigh them.
2. Place a cup on a scale and zero the scale. Fill it with beans until it reaches the weight
of the beans in step 1. Estimate this as 100 beans.
3. Use this procedure to count out 180 beans of type #1, 840 of type #2, and 980 of type
#3 (according to the student handout). Do one color bean for all containers before
moving to a second bean. You can count out the 40 beans for type #2 separately, and
for the last counts of types #1 and #3, weigh 100 beans and pull out 20 from that
container.
4. Distribute and explain copies of Hardy Weinberg simulation (see below).
5. Divide the students into groups of 5 or 6, depending on class size, and distribute
materials.
6. Hand out bean containers.
7. Use the following summary and test preparations sections to review the main ideas of
the lesson.
Evaluation
The following questions are listed under the Analysis section of the student handout and
maybe used as part of a report, class discussion, or assessment
1. What conditions must be present for Hardy-Weinberg Equilibrium to be in effect?
2. Many deleterious genes are recessive and therefore are expressed only in the
homozygote. Many of them are also lethal and many of them have also been present
in the human gene pool for hundreds of thousands and possibly even millions of
years. How do you explain the persistence of these genes over such long periods of
time in the face of such intense selective pressure against them?
3. Explain why evolution can not happen if the Hardy-Weinberg equilibrium is true in a
population.
Extension Activity
1. Have your students derive the Hardy-Weinberg Equation. Use the attached handout to
help them along the way.
Test Preparation
1. According to the Hardy-Weinberg equation, the dominant trait is represented by
(A) p
(B) q
(C) q2
(D) p2
(E) 2pq
2. In a population of 1,000 people, 90 have blue eyes. What percent of the population
has hybrid brown eyes?
(A) 3%
(B) 9%
(C) 21%
(D) 42%
(E) 49%
3. In terms of the algebraic symbols used in the Hardy-Weinberg equation (p and q), the
most likely effect of assortative mating on the frequencies of alleles and genotypes
for a locus would be
(A) a decrease in p2 compared to q2
(B) a trend toward zero for q2
(C) convergence of p2 and q2 toward equal values
(D) a change in p and q, the relative frequencies of the two alleles in the gene pool
(E) a decrease in 2 pq below the value expected by the Hardy-Weinberg
theorem
Name:__________________
Student Handout—Bean There, Done That: A Hardy-Weinberg Simulation
Each of these containers contains approximately 2,000 dried beans, representing a
population of 2,000 diploid organisms. The beans are of three colors -white, brown and
speckled - and each color represents one of the three possible genotypes for a certain trait.
Nine percent of the beans are white, and these represent individuals that are homozygous
recessive for the trait (genotype aa). Forty-two percent of the beans are speckled and
these represent individuals that are heterozygous for the trait (genotype Aa). Finally,
forty-nine percent of the beans are brown, and these represent individuals that are
homozygous dominant for the trait (genotype AA). (This starting information is
summarized in Table 1.)
Table 1. Bean colors, genotypes, and
genotype frequencies.
Color
Genotype
Relative
frequency %
White
aa
9
Speckled
Aa
42
Brown
AA
49
Procedure
(A) Calculating allele frequencies p and q arithmetically.
The first thing we have to do is to calculate the frequency p of the dominant allele
A and the frequency q of the recessive allele a. This can be done either by using simple
arithmetic or by using the Hardy-Weinberg equation. To calculate p by simple arithmetic,
first of all determine the total number of alleles for a given gene in a population of 2000
diploid individuals.
Number of unique alleles = ___________
Total number of alleles = ______________
Next, how many of these are dominant (A) alleles? These alleles are found in
homozygous dominants and in heterozygotes, so we have to know how many of these
individuals we have.
Homozygous dominant individuals = (0.49) x 2,000 =
Heterozygous individuals = (0.42) x 2,000 =
Homozygous recessive individuals = (0.09) x 2,000 =
How many dominant (A) alleles are there in this population? Remember, each
homozygous dominant individual has 2 such alleles, but each heterozygote has only 1.
Total number of A alleles = _____________
Total number of a alleles = _____________
Finally, p, the frequency of A alleles in the population, can be determined by dividing the
total number of A alleles by the total number of alleles.
p=
Total number of A alleles = _____________
Total number of alleles
Now repeat this procedure to get q, the frequency of the recessive allele a in the
population, by simple arithmetic.
q=
Total number of a alleles = _____________
Total number of alleles
(B) Calculating allele frequencies p and q from the Hardy-Weinberg equation
We can calculate p using the frequencies of the genotypes from Table 1.
p = the frequency of AA genotypes + 1/2 the frequency of Aa genotypes
In our populations, p calculated in this way would be ______________.
Recall that the Hardy-Weinberg equations tells us that 1=p2+2pq+q2. This is equivalent to
(p+q)=1 (If you need to prove that to yourself, square both sides of the p+q=1 equation
and see what you get).
Knowing that p+q =1. Calculate q using the value for p you calculate above.
q =____________
You can check you answer by calculating q the same way we calculated p above.
q = the frequency of aa genotypes + 1/2 the frequency of Aa genotypes
In our populations, q calculated in this way would be ______________.
Do we get the same answer for p and q by using simple arithmetic (Section A) as we get
by using the Hardy-Weinberg equation (Section B)? Is p + q = 1 in both cases?
For genes that show dominance, it is often impossible to distinguish between the
phenotypes of homozygous dominants and heterozygotes, because they look the same. In
these cases, we might take the square root of the frequency of homozygous recessives to
determine q, and the estimate the value of p by the equation p=1 – q.
(C) The simulation.
We are now ready to begin our random mating simulation. Shake your can
vigorously to make sure your beans are thoroughly mixed. Then, without looking, reach
into the can and draw out the first two beans your fingers encounter. Use a tally mark to
record the colors of the beans you have withdrawn, and the genotypes these colors
represent, on Table 2. Put the beans back in the can and draw out another pair. Tally the
colors and genotypes of this pair, and put them back into the can. Repeat this procedure
until you have tallied a total of twenty-five matings on Table 2.
Table 2. Random mating tallies.
Mating Pairs
Genotypes Mating
(color x color)
Tallies
brown x brown
AA x AA
brown x speckled
AA x Aa
brown x white
AA x aa
speckled x
Aa x Aa
speckled
speckled x white
Aa x aa
white x white
aa x aa
(D) Using the results to calculate genotype frequencies.
Next, use Punnett squares and your knowledge of Mendelian genetics to
determine the genotypic ratios that we would expect to find in the offspring that would be
produced by each of the six possible mating combinations in our population. Express
these genotypic ratios as fractions (1, 1/2, 1/4, or 0) and record them in Table 3. The
second one is filled in for you. Then fill out Table 4 with the fractions of each genotype
and the number of times you drew each of the six possible mating types. Determine the
representation of each genotype among the offspring produced by these matings. To do
this, assume that each mating produces four offspring, and multiply the number of times
each mating type was drawn by four to get the total number of offspring produced by
matings of that particular type. Finally, multiply that number by the fractional expected
genotypic ratio for each genotype. This will give you the number of offspring of a
particular genotype that we can expect to be produced from matings of each of the six
possible mating types. Total the number of offspring of each of the three possible
genotypes, and enter these totals in Table 4.
Table 3: Predicted Outcomes From Potential Crosses
AA x AA
AA xAa
AA x aa
Aa x Aa
Aa x aa
aa x aa
AA Aa
AA Aa
Genotypic ratios expressed as fractions:
AA __
AA ½
AA___
Aa __
Aa ½
Aa___
aa __
aa 0
aa___
AA___
Aa___
aa___
AA___
Aa___
aa___
AA___
Aa___
aa___
Table 4. Representation of various genotypes among offspring.
Mating
Type
Times
Drawn
Offspring
per mating
AA X AA
4
AA X Aa
4
AA X aa
4
Aa X Aa
4
Aa X aa
4
aa X aa
4
Total
offspring
Ratio
Ratio
Ratio
#
#
#
AA
Aa
aa
AA
Aa
aa
Ratio = genotypic ratios of offspring expressed as fractions.
# = numbers of offspring of various genotypes.
Since you drew 25 mating pairs and each mating produced four offspring, the
grand total of all offspring produced is equal to 100. Therefore, the genotype frequency
expressed as a percentage for each of the three possible genotypes will be numerically
equal to the number of offspring for each of these genotypes. Enter these genotype
frequencies in Table 5. Do they correspond to the genotype frequencies with which we
began, as recorded in Table 1?
Table 5: Genotypic frequencies
Genotype
Number of Offspring
Relative Frequency (%)
AA
Aa
aa
Value p______________ Value q ________________
Now calculate the frequency p of the dominant allele A among the offspring
produced by our simulated random matings, and the frequency q of the recessive allele a
in the same group. You may do this either arithmetically or by using the Hardy-Weinberg
equation. Has either p or q changed in the new generation? Have we demonstrated
successfully that genetic recombination as a result of sexual reproduction can not by itself
result in any change in allele frequencies or genotype frequencies from one generation to
the next?
Remember that our demonstration was designed to exclude the effects of various
factors that can alter allele frequencies. How successfully has this been done? How would
you be sure that statistically you had been successful? Have we, for example, been
entirely successful in excluding the influence of chance effects on allele and genotype
frequencies? If you don't think that we have, suggest some ways in which we might
change our procedure so as to eliminate or at least reduce the impact of these effects on
our results.
(E) Demonstrating factors that skew Hardy-Weinberg predictions.
Suggest an experiment or exercise using our beans and coffee cans that would
demonstrate the effect of migration on allele and genotype frequencies. How could we
demonstrate the impact of selective pressures? Of non-random mating? (This matter of
non-random mating is particularly intriguing. What would happen if, for example, you
prohibited matings between beans representing different genotypes and permitted only
beans representing the same genotype to mate? Would this procedure alter genotype
frequencies in the next generation, and if so how? What if you did the opposite and
permitted only beans representing different genotypes to mate? Would this procedure
alter genotype frequencies in the next generation, and if so how? Would either procedure
have any effect on allele frequencies?).
Design and carry out at least one of these exercises and report your results to the
class. (Whatever you do, make sure that you leave the coffee cans as you found them. Put
any beans that you remove from a can back in the same coffee can they came from. Do
not add beans to a coffee can that were not in that can to begin with. THIS IS VERY
IMPORTANT!)
Analysis
Answer the following questions on a separate sheet of paper.
1. What conditions must be present for Hardy-Weinberg Equilibrium to be in effect?
2. Many deleterious genes are recessive and therefore are expressed only in the
homozygote. Many of them are also lethal and many of them have also been present
in the human gene pool for hundreds of thousands and possibly even millions of
years. How do you explain the persistence of these genes over such long periods of
time in the face of such intense selective pressure against them?
3. Explain why evolution can not happen if the Hardy-Weinberg equilibrium is true in a
population.
Hardy-Weinberg Law Derivation
The Hardy-Weinberg Law: 1=p2+2pq+q2
Recall that alleles are alternative forms the same gene, which is located at a
specific position on a specific chromosome. The genetic composition of each individual
is that individual’s genotype. Each individual will carry two alleles of every gene, one
from their mother and one from their father. This composition of alleles for all
individuals in a population is called the gene pool.
The Hardy-Weinberg Law (HWL) states that when a population is in
equilibrium, the genotypic frequencies will be in the proportion p2, 2pq and q2. In a
hypothetical population where the frequency of allele A is p and the frequency of allele a
is q, each genotype passes on both alleles that it possesses with equal frequency.
Therefore in a population with just 2 alleles of a gene, the possible combinations are
Female
Gametes
pA
qa
Male
pA
p2 AA
pq Aa
qa
pq Aa
q2 aa
This table shows the relationship between the allelic frequencies (p and q) and the
genotypic frequencies (p2, 2pq, and q2), which form the basis of the HWL. For example,
the frequency of the genotype AA is p2; the frequency of the genotype Aa is 2pq.
The HWL states that the genotypic frequencies remain constant generation after
generation if the population is large, mates randomly, and is free from evolutionary
forces. For the above example, that would mean that after many generations the
frequency of AA is still p2 and the frequency of Aa is still 2pq.
We can demonstrate this by considering a hypothetical randomly mating
population from table above. To do this, first consider all the possible matings from every
genotypic outcome above. These matings are listed in Column A below.
Next, since we are assuming that this population is subject to HWL, we can
assume that all matings are random and therefore will occur with equal frequency. For
instance, AA x AA matings don’t occur more often than aa x aa matings. This means that
the frequency of mating between any two genotypes is the product of how frequently
these genotypes are present in the population. Therefore, the mating frequency of AA x
AA, for instance, is p2 x p2 or p4, which is already filled in for Column B.
Now that we know how frequent each of these matings are, we can determine the
alleles that the offspring of that mating will contribute to the gene pool. Columns C-E are
the genotypic frequencies resulting in the next generation. In our example of AA x AA,
100% of the offspring will have the genotype AA, so the frequency of that genotype in
the next generation is p4.
Fill in the chart below. Afterwards, tally down to get the totals in each column. If
Hardy-Weinberg is correct, the total of Column B should equal the totals of D, C, and E
combined, which should come out the same as the frequencies of the original generation.
Put another way, the combined totals of C, D, and E, which makes up the entire
population of the next generation, should still result in the same Hardy-Weinberg
equation: 1=p2+2pq+q2.
A
Type of mating
Male x female
p2 AA x p2 AA
p2 AA x 2pq Aa
2pq Aa x p2 AA
p2 AA x q2 aa
q2 aa x p2 AA
B
Mating
frequency
p4
C
AA
p4
D
Offspring frequencies
Aa
-
E
aa
-
-
2pq Aa x 2pq Aa
2pq Aa x q2 aa
q2 aa x 2pq Aa
q2 aa x q2 aa
total
-
-
p2 AA +2pq Aa + q2 aa = 1
Implications of HWL
When the frequency of one allele is low, the homozygote for that allele is the
rarest of the genotypes. This rare allele will be maintained not only in the homozygote
but in the heterozygote population too. This point is also illustrated by the distribution of
genetic diseases in humans, which are usually rare and recessive.
An example: tyrosine-negative albinism is present in 1 in 40,000 individuals, or
0.000025. If q2 = 0.000025 then q= 0.005 and p= 1-q = 0.995. The heterozygote
frequency is 2pq = 2 x 0.995 x 0.005 = 0.00995 (almost 1% of the population). The
heterozygote carrier will be 1 out of every 100 people. This can illustrate how a rare
allele is maintained in a population.
To determine if HWL is in effect in a population, obtain gene frequencies and
calculate the expected genotypic frequencies and compare those frequencies with the
observed frequencies of the genotypes using a chi- square test. (Recall that the chi-square
test gives us the probability that the difference between what we observed and what we
expected under the HWL is due to chance.)
Hardy-Weinberg Law Derivation
The Hardy-Weinberg Law: 1=p2+2pq+q2
Recall that alleles are alternative forms the same gene, which is located at a
specific position on a specific chromosome. The genetic composition of each individual
is that individual’s genotype. Each individual will carry two alleles of every gene, one
from their mother and one from their father. This composition of alleles for all
individuals in a population is called the gene pool.
The Hardy-Weinberg Law (HWL) states that when a population is in
equilibrium, the genotypic frequencies will be in the proportion p2, 2pq and q2. In a
hypothetical population where the frequency of allele A is p and the frequency of allele a
is q, each genotype passes on both alleles that it possesses with equal frequency.
Therefore in a population with just 2 alleles of a gene, the possible combinations are
female
Gametes
pA
qa
male
pA
p2 AA
pq Aa
qa
pq Aa
q2 aa
This table shows the relationship between the allelic frequencies (p and q) and the
genotypic frequencies (p2, 2pq, and q2), which form the basis of the HWL. For example,
the frequency of the genotype AA is p2; the frequency of the genotype Aa is 2pq.
The HWL states that the genotypic frequencies remain constant generation after
generation if the population is large, mates randomly, and is free from evolutionary
forces. For the above example, that would mean that after many generations the
frequency of AA is still p2 and the frequency of Aa is still 2pq.
We can demonstrate this by considering a hypothetical randomly mating
population from table above. To do this, first consider all the possible matings from every
genotypic outcome above. These matings are listed in Column A below.
Next, since we are assuming that this population is subject to HWL, we can
assume that all matings are random and therefore will occur with equal frequency. For
instance, AA x AA matings don’t occur more often than aa x aa matings. This means that
the frequency of mating between any two genotypes is the product of how frequently
these genotypes are present in the population. Therefore, the mating frequency of AA x
AA, for instance, is p2 x p2 or p4, which is already filled in for Column B.
Now that we know how frequent each of these matings are, we can determine the
alleles that the offspring of that mating will contribute to the gene pool. Columns C-E are
the genotypic frequencies resulting in the next generation. In our example of AA x AA,
100% of the offspring will have the genotype AA, so the frequency of that genotype in
the next generation is p4.
Fill in the chart below. Afterwards, tally down to get the totals in each column. If
Hardy-Weinberg is correct, the total of Column B should equal the totals of D, C, and E
combined, which should come out the same as the frequencies of the original generation.
Put another way, the combined totals of C, D, and E, which makes up the entire
population of the next generation, should still result in the same Hardy-Weinberg
equation: 1=p2+2pq+q2.
A
Type of mating
Male x female
p2 AA x p2 AA
p2 AA x 2pq Aa
2pq Aa x p2 AA
p2 AA x q2 aa
q2 aa x p2 AA
2pq Aa x 2pq Aa
2pq Aa x q2 aa
q2 aa x 2pq Aa
q2 aa x q2 aa
total
B
Mating
frequency
p4
C
AA
p4
D
Offspring frequencies
Aa
-
4p3q
2 p3q
2 p3 q
-
2p2q2
-
2p2q2
-
4 p2q2
p2q2
2 p2q2
p2q2
4pq3
-
2 pq3
2 pq3
q4
(p2 +2pq + q2)2=1
2
2
E
aa
-
q4
2
p (p +2pq + q )
= p2
2
2
2pq (p +2pq + q )
= 2pq
2
2
q (p +2pq + q2)
= q2
p2 AA +2pq Aa + q2 aa = 1
Implications of HWL
When the frequency of one allele is low, the homozygote for that allele is the
rarest of the genotypes. This rare allele will be maintained not only in the homozygote
but in the heterozygote population too. This point is also illustrated by the distribution of
genetic diseases in humans, which are usually rare and recessive.
An example: tyrosine-negative albinism is present in 1 in 40,000 individuals, or
0.000025. If q2 = 0.000025 then q= 0.005 and p= 1-q = 0.995. The heterozygote
frequency is 2pq = 2 x 0.995 x 0.005 = 0.00995 (almost 1% of the population). The
heterozygote carrier will be 1 out of every 100 people. This can illustrate how a rare
allele is maintained in a population.
To determine if HWL is in effect in a population, obtain gene frequencies and
calculate the expected genotypic frequencies and compare those frequencies with the
observed frequencies of the genotypes using a chi- square test. (Recall that the chi-square
test gives us the probability that the difference between what we observed and what we
expected under the HWL is due to chance.)