Honors Biology

Evolution can be described as the change in the allele frequencies of a gene pool over time. Natural selection can place pressure upon specific phenotypes and cause a change in the frequency of the alleles that produce the phenotypes. For example, predators often select as prey individuals that are different from the majority of a population. From a brood of mostly normal-colored black chickens, an eagle would probably be more likely to select a white chicken than a normal-colored one. The reason for this seems obvious – the white is different and stands out among the population. The eagle represents a natural-selection pressure that selects against the white chicken and the genes that produce white feathers.

Objectives    

Simulate natural selection by using beans of tow different colors. Calculate the frequencies of alleles subjected to selection pressure over five generations. Draw a graph that demonstrates the changes of allele frequencies over five generations. Describe how natural selection can affect allele frequencies over time.

Pre-Lab 1. Assuming random mating, calculate the expected frequency of Black and white feathered chickens. _______________________________________________________________________________ 2. How many of the Black feathered chickens will be homozygous dominant? __________________ How many will be heterozygous? ____________________________________________________

Materials Colored pencils (2) Cup Graph paper

Dark beads Light beads

Procedure Assume that the dark beads represent alleles that produce black chickens and the black feathers are dominant to white feathers. The light bead, then, represents a recessive allele that produces white chickens in the homozygous condition. Selection against black-colored chickens (BB or Bb) occurs 25 percent of the time. That is, predators remove 25 percent of all the normal phenotypes each generation. Selection against white chickens (bb) is 100 present. This means that predators eat all of the white chickens present in the population in each generation.

BB

Bb

bb

1. Place 50 dark beads and 50 light beads into the cup. 2. Shake up the beads. Without looking, reach into the cup and remove two beads. These represent an individual chicken’s genotype. Set these beads aside. Continue to remove beads, two at a time, until you have 50 “chickens.” Each with two alleles for feather color. 3. Arrange the beads on the table in two columns as shown in Figure 2. The two columns represent the two possible chicken phenotypes, black (BB or Bb) and white (bb). 4. In Table 1, record the number of chickens with each genotype. 5. Examine your columns of “chickens” and remove 100 percent of the white individuals and 25 percent of the normal colored chickens. If the number of chickens you calculate is a fraction, remove a whole “chicken.” Assume that it was injured, but escaped and died later. To calculate 8. Multiply the frequency by 100 to the number of “chickens” to be removed, multiply calculate the percent. 0.6 times 100, or the total number of “chickens” in the column by 60%. the percentage to be removed, and divide by 100. (15/25) X 100 = 60% For example 25% (.25) of 30 “chickens” equals 9. Begin the next generation by placing 7.5 “chickens.” You would then remove 8 100 beads into the cup. The proportions “chickens.” of dark and light beads should be the same as the percentages you calculated .25 X 30 = 7.5 in the previous step. In the given example, you would place 60 dark beads 6. Count the number of dark beads and light beads and 40 light beads into the cup. that remain on the table and record the number in 10. Repeat steps 2 through 7 until you have Table 2. generated data for five generations. 7. Calculate the frequencies of both types of bead 11. Graph the frequencies of each allele remaining and record your results in Table 1. To over five generations. Plot the calculate frequency, divide the number of beads frequency of the allele on the vertical of one type by the total number of beads. For axis and the number of the generations example, if there are 15 dark beads and 10 light on the horizontal axis. Use a different beads remaining, the frequency of dark beads is colored pencil for each allele. 15 divided by 25 (15 + 10) or 0.6.

Table 1 Number of Birds with each Genotype/Generation Generation BB Bb bb 1 50 2 3 4 5 Table 2 Allele Frequencies Allele B

Generation

Allele b

Start

Number

Frequency

Percentage

Number

Frequency

Percentage

1

50

0.50

50

50

0.50

50

2 3 4 5

ANALYSIS 1. Did allele frequencies change over time? _____________ Why or why not? __________________ _______________________________________________________________________________ 2. Did either of the alleles totally disappear? ______________ Why or why not? _______________ _______________________________________________________________________________ 3. What does shaking of the cup of beads simulate? _______________________________________ _______________________________________________________________________________ 4. Why do you remove two beads for each individual? ____________________________________ _______________________________________________________________________________ 5. What does your graph show about allele frequencies subjected to natural selection? ___________ _______________________________________________________________________________ _______________________________________________________________________________ 6. How does natural selection affect allele frequency? _____________________________________ _______________________________________________________________________________ _______________________________________________________________________________ 7. On the basis of the number of beads in the cup, what is the frequency of each gene (B and b) in the original population? ______________________________________________________________ _______________________________________________________________________________

8. Compare your data with class totals, how do they compare? Explain. ______________________ _______________________________________________________________________________

Evolutionary changes occur in populations rather than individuals. All the genes of a population are called the gene pool. Each gene in the gene pool is present in a certain proportion of the genes of the population. That proportion is called the gene frequency (allele frequency). Gene frequency =

the number of a specific type of gene__ the total number of genes in a gene pool

Changes in a species can be traced by observing changes in genotype frequency, the proportion of a specific genotype among the genotypes of a population. Genotype frequencies often can be calculated by observing and recording the phenotypes of a specific population and by using the Hardy-Weinberg Principle. Mathematically, the Hardy-Weinberg Principle states that p2 + 2pq + q2 = 1 and p + q = 1, where p and q represent the dominant and recessive alleles respectively for a trait with two possible alleles. The formulas can be used to determine the genotypic frequencies if the frequency of the genes or of a genotype are known. The Hardy-Weinberg Principle is used to measure evolutionary changes in populations. It predicts the changes that will occur in genotype frequencies (gene frequencies remaining the same) in succeeding generations of a population if certain conditions are met. Those conditions include the following: (1) the population must be large, (2) mating must be random and all offspring must have an equal chance of surviving, and (3) mutations, migration and other things that change gene frequencies must not occur. Because populations are sometimes small, they often do not mate randomly, all offspring often do not have an equal chance of surviving, and because factors such as mutation and migration often occur, the HardyWeinberg Principle is an oversimplification of what really happens in natural populations. However, it has some value in studying populations, especially those that are isolated. In real populations, gene frequencies do change. 9.

Apply the Hardy-Weinberg Principle using the class data. Find p + q = 1 You know q because it represents the receive allele

Use to Find p2 + 2pq + q2 = 1

10. Compare and contrast the Actual allele combinations from Table 1 to the calculated values of the Hardy-Weinberg Principle.