Materials: 3 colors of beans, totaling 4000 for each group.  Each group has 49% one color, 42% the second color and 9% the 3rd color.


Purpose: To simulate the microevolution model with populations of colored beans, illustrating random mating and the effects of selection and genetic drift.


Background: Populations, not individuals, evolve by gradual changes over time in the frequency of alleles that are found at genetic loci.  These changes result from mutation, selection, migration, or genetic drift.  Collectively, these processes comprise microevolution.  Mechanisms of microevolution are well understood.  In fact, one of these mechanisms, selection, has been used for centuries to increase the productivity of crops and livestock.

According to the microevolution model, a population of organisms can be considered to be a gene pool, which is composed of all the copies of every allele in the population at a given moment.  In diploid organisms, the genes in the gene pool occur in pairs in each individual, and individuals may be homozygous or heterozygous for a particular trait.

In 1908, G.H. Hardy and G. Weinberg showed that no matter how many generations elapsed, sexual reproduction in itself couldn’t change the frequencies of alleles in a gene pool.  Changes if they occur, must be due to the action of other factors---the real agents of evolution.

Example….a population of plants with red and white flowers.  Red flowers have the dominant allele and white flowers are found in individuals with only the recessive alleles.     A study  of 10,000 plants indicated that 84% of the plants had red flowers and 16% had white flowers.  Of the red flowered plants, 36% of the total plants were homozygous for the red allele and 48% were heterozygous.

The genotypic frequencies were:

36% AA          Where A = red allele

48% Aa                       a = white allele

16% aa



The frequencies of alleles can be obtained by the following:

Every individual with the genotype AA contributes two A alleles to the gene pool, and every individual with the genotype Aa contributes one A allele so given 10000 individuals, the frequency of the A allele in the gene pool equals:

2 x 36% x 10,000     +     48% x 10,000      =     12,000 A alleles


In the same population, the frequency  a in the gene pool can be calculated by similar reasoning (aa individual contributes two a, and Aa contributes 1 a.  Therefore,


2 x 16% x 10,000   + 48% x 10,000   = 8,000 a alleles


On a percentage basis, 60%  (12,000/12,000 + 8,000) of the alleles of the gene pool are A and 40% (8,000/ 12,000 + 8,000) are a.


This information describes the gene pool at the time of the study.   But, what happens to the allele frequencies when this population reproduces sexually?


The frequency of the dominant allele is p and that of the recessive allele is q, in any population where there are only two alleles at one gene locus:

P + q = 1   and     p = 1 – q    or     q = 1 – p


Sooo, a population in Hardy-Weinberg equilibrium, the homozygote dominant will be found p2% of the time, the heterozygote 2 pq% of the time and the homozyous recessive q2% of the time.

(p2%     36% AA )   (q2% …    16% aa)  (2 pq%     48% Aa)


Lab: Your lab population is 4000 beans in a coffee can.  There are 3 colors: 49% are one color, 42% another color and 9% a third color.  These percentages correspond to genotypes in hypothetical populations in which there are two alleles at a given locus.  Indicate what colors correspond to the genotypes  below:

Genotype                                        %                               Color                                          Number (% x 4000)

AA                                            49                                                                                            .

Aa                                            42                                                                                            .

aa                                             9                                                                                              .


Determining starting allele frequency-Accounting method

  # of AA individuals x 2 =3920

+# of Aa individuals x 1 =1680

# of A alleles in gene pool =5600


Do the same for the a allele


Determining starting allele frequency-Hardy-Weinberg Calculation  method

To use this  there is the major assumption that the genotypes are in proportion to what is expected by the Hardy-Weinberg equilibrium.

(which is that AA=p2 and aa=q2 )

Calculate the p and q values for the population in the can and enter it below.

P=________    q=________   Remember p+q=1   Are the relative frequencies of A and a the same by the accounting and calculation methods?_____

1. To simulate random mating in your population, mix up the can and withdraw two beans.

These are your breeding pair.

2. Repeat this 100 times...recording the colors drawn each time.


3. Record the info on the table below and list the Genotypes of the new breeding group.




Total 200

Are these frequencies the same as the total population of the can?

Compare them using p and q


Mating Pairs (color x color)                   Corresponding pair     Tallys                  Total of tallys