Scientists have merged the two disciplines of genetics and evolution into population genetics that demonstrates how gene frequencies change in populations, studies genetic variation, and applies genetic change to evolutionary theory.

Any study of natural populations usually demonstrates genetic change. That finding seems perfectly reasonable in the light of natural selection theory. Yet basic Medelian genetics also shows that allele frequencies remain stable under certain conditions. For example, examine a simple monohybrid cross of a homozygous tall pea plant (TT) and a homozygous short pea plant (tt) where we begin with equal numbers of tall and short parents. On average, half of the total gametes are T and half are t. In the first generation all offspring are heterozygous (Tt); exactly 50% T and 50% t alleles. Consider the individual (or diploid) frequencies in the next generation where all heterozygotes mate randomly: 25% homozygous dominant (all T alleles), 25% homozygous recessive (all t alleles), and 50% heterozygous (one half T and the other half t alleles). What percent of the total alleles are now T and t? The answer is still 50% for either allele. Frequencies of T and t remain unchanged from the first generation and will remain unchanged unless some outside force (natural selection, mutation, migration, non-random mating) or chance events alter those allele frequencies. Two mathematicians, G.H. Hardy and W. Weinberg, summarized this relationship (called the Hardy-Weinberg equilibrium, HWE) in algebraic terms: p+q= 1; where p and q represent the frequencies of the dominant and recessive alleles, respectively. Diploid individuals may be represented by the binomial expansion of

(p+q)² = (1)² or , p² + 2pq = q² = 1, where p², pq, and q² represent homozygous dominants, heterozygous, and homozygous recessives, respectively.

It is not possible for a General Science class to study the Hardy-Weinerg equilibirium in natural populations, but the relationship is actually better understood and more easily examined in an artificial set-up using two colors of beans to represent alleles. In this laboratory, we will attempt to understand the effects of two forces, selection and chance events, on allele frequencies.

Materials

Light and dark beans

Beakers

Applying frequencies to phenotypes and alleles.

In this exercise:

Dark beans = dominant allele = p

Light beans = recessive allele = q

Also remember that since alleles separate from each other during meiosis a single dark or light bean also represents a gamete. Gametes then unite to form diploid individuals represented by two beans.

2 dark beans = homozygous dominants = p²

2 light beans = homozygous recessive = q²

1 dark bean + 1 light bean = heterozygous = pq

Heterozygous can be formed in two ways so that pq is doubled (2pq).

Now let p=0.5 and q = 0.5 as a starting point for our exercise. Give frequencies for the following:

Dominant allele? ________

Recessive allele? ________

Homozygous dominant individuals? _________

Homozygous recessive individuals? _________

Heterozygous individuals? ___________

Establish a parent population of 200 individuals (1 individual = 2 beans) meeting the frequencies that you calculated above. Give the correct number of individuals or alleles for this population below.

Homozygous dominant individuals? _________

Homozygous recessive individuals? _________

Heterozygous individuals? ___________

Dominant alleles? ________

Recessive alleles? ________

Chance and the Hardy-Weinberg Equilibrium

Are chance events important in changing gene frequencies? In an artificial set-up we can easily eliminate selection, mutations, and migrations. Additionally, if we thoroughly stir our beans

( = alleles) before and after each draw, the outside force of nonrandom matings may be substantially reduced in effect. Therefore, any deviation that occurs is likely to be due to chance. Consider the 5^th key assumption of the HWE that states that population size must be infinite. Since we cannot meet this assumption we can examine it indirectly by varying the population size. A larger population size should be less influenced by chance events than a smaller population size.

1. Introduce and thoroughly mix the 400 (200 dark and 200 light) beans in a beaker. While individuals are represented by 2 beans , we do not need to connect the beans in our beaker because during meiosis these beads segregate to each gamete and gametes randomly unite to form individuals.
2. Without looking, draw out 2 beans and record the genotype of this individual.
3. Return this individual to the container and mix again.
4. Repeat steps 2 and 3 until 100 genotypes have been recorded.
5. Calculate the number of p and q alleles and p², 2pq, and q² genotypes.
6. Record your data and the data from other groups in the table below.
7. Finally, calculate allelic and genotypic frequencies for your group and the class totals.

Group Count Frequencies

Suppose you came up with the following results: 24 double dark beans, 46 dark-light beans, and 30 double light beans. Therefore:

p = 0.47 or 94 dark beans out of 200 total beans;

q = 0.53 or 106 light beans out of 200 total beans;

p2 = .24 or 24 double darks

2 pq = ,46 or 46 dark-lights

q2 = .30 or 30 double lights

The HWE predicts that we should have drawn: 100 dark beans, 100 light beans, 25 double darks, 50 dark-light, and 25 double lights, but we did not! Why? While a sample of 100 individuals seems alot, it is very small relative to an infinity, an unrealistic, but necessary population size for the HW model to be totally accurate. Additionally, our "matings" within the beaker may not have been perfectly random. These two variables alone account for any discrepancies between the observed and expected since we can safely eliminate mutation, selection and migration.

The next important question to ask about these differences is: "Are these numbers significantly different from each other"? No! An easy way to compare frequency counts is with the Chi-square (X²) test in the following equation:

Let’s apply this formula to our frequency counts on individuals.

X2 = (24-25)2 + (46-50)2 + (30-25) 2 = 1 + 16 + 25 = 68 = 1.36

   25 50 25 25 50 25 50

This value is then looked up in the chi-square table (see last page) that list values telling us how different numbers are from each other based on probabilities. We also need one other term, degress of freedom (df), to locate our answer. This term corrects for differing sample sizes or numbers of mathematical operations in our calculation. The number of df is always one less than our sample size or mathematical operations. In our example df = 2 since we had 3 operations. Now locate df in the far left column and examine each number to the right until one most closely matches 1.36. This should be 1.39. Going straight up from this numbr to the top you find 50.0. This is a percent and represents our confidence that observed and expected differ. We are 50 % confident that the observed values are statistically different from expected values. In other words, a coin flip could decide between our observed and expected; not a comforting level of confidence for scientists. Therefore, scientists do not accept any value less than 95 % confidence as being statistically significant. We state that: Our observed and expected values do not statistically differ or the difference between the observed and expected was not statistically significant (X² =1.4, df =2, p < 0.5).

1. Use the Chi-square to test the difference between your observed results and that predicted by HWE for both your group and the class totals. Record your Chi-squares, df, and probability levels.
2. Did the observed and expected values statistically differ for either the class or group?

 

3. Did the group or the class results come closer to the HWE predictions?

 

4. What effect did population size have on your results?

Now execute this modified procedure:

1. Introduce and thoroughly mix the 400 (200 dark and 200 light) beans in a beaker. While individuals are represented by 2 beans , we do not need to connect the beans in our beaker because during meiosis these beads segregate to each gamete and gametes randomly unite to form individuals.
2. Without looking, draw out 2 beans and record the genotype of this individual.
3. Return this individual to the container and mix again.
4. Repeat steps 2 and 3 until 20 genotypes have been recorded.
5. Repeat steps 1-4 four more times.
6. Calculate the number of p and q alleles and p², 2pq, and q² genotypes.
7. Record your data and the data from other groups in the table below.
8. Finally, calculate allelic and genotypic frequencies for each 20 genotype sample in the table below

Sample Counts Frequencies

1. Are there any unrepresentative sample in your 5 samples? If any appear to be large deviations, use the X2 to test if they are significantly different.

 

 

2. What would be the result if you used the most unrepresentative sample to establish a new population?

 

 

3. What is the name for this effect?

 

4. When gene frequencies change __________________occurs.
5. Does evolution occur when mutations, selection, migration, and nonrandom matings are absent? Why?

Effects of Selection on the HW Equilibrium

Differential reproduction of genotypes, or selection, is a key force in changing allele frequencies. If we remove 100% of one allele (i.e. select against it), it is easy to visualize why that allele will decrease in frequency. However, reasons why this allele takes a long time to go extinct are not often obvious. To understand the effects of selection on a population you should execute the procedure below.

Procedure:

1. Establish a parental population of 200 individuals with p = q = 0.5.
2. Simulate reproduction by members of this population by drawing 2 beans at a time until all beans are removed. Be sure to record the nmbers of p²’s, pq’s, and q²’s and mix your beans after every draw.
3. To stimulate 100% negative selection against homozygous recessives, remove all double-lights when they are drawn and set them aside. These will not be returned to the parent population.
4. Since all q²’s are set aside, total population number declines, probably by about 25% during the first generation, but less with each additional generation. For our purposes, we do not need to worry about this decline.
5. Repeat steps 2,3,and 4 five more times until you have six generations.
6. Graph (Figure 1) the frequency of q (y axis) against generation number (x axis).
7. Graph (Figure 2) frequencies of p²’s, 2pq’s and q²’s (y axis) against generation number (x-axis)

Figure 1