Ap Biology Lab Eight: Population Genetics

AP Biology Lab 8: Population Genetics Introduction G.H Hardy and W. Weinberg developed a theory that evolution could be described as a change of the frequency of alleles in an entire population. In a diploid organism that has gene a gene loci that each contain one of two alleles for a single trait t the frequency of allele A is represented by the letter p. The letter q represents the frequency of the a allele. An example is, in a population of 100 organisms, if 45% of the alleles are A then the frequency is .45. The remaining alleles would be 55% or .55. This is the allele frequency. An equation called the Hardy Weinberg equation for the allele frequencies of a population is p2+ 2pq+ q2 = 1. P represents the A allele frequency. The letter q represents the a allele. Hardy and Weinberg also gave five conditions that would ensure the allele frequencies of a population would remain constant. A. The breeding population is large. The effect of a change in allele frequencies is reduced. B. Mating is random. Organisms show no mating preference for a particular genotype. C. There is no net mutation of the alleles. D. There is no migration or emigration of organisms. E. There is no natural selection. Every organism has an equal chance for passing on their genotypes.

If these conditions are met then no change in the frequency of alleles or genotypes will take place. A simple class experiment will take place to serve as model of the evolutionary process in a stimulated population. This experiment is great in order to test a few of the basic parts of population genetics. In the experiment the class will place a piece of paper in their mouth to see if they can taste the chemical PTC which is phenythiocarbamide. People with the alleles AA, which is homozygous, and Aa, which is heterozygous, will be able to taste the PTC. People that can’t taste PTC are aa. Hypothesis By allowing a class to see if they can taste PTC and recording the results the Hardy Weinberg equation can be used to determine the allele frequencies of the class. Materials The materials used in this experiment are as follows: strips of PTC test paper, paper and a pencil. Methods Begin by placing a piece of the PTC test paper in your mouth. Tasters will have a bitter taste in their mouth. The frequency of tasters (p2 +2pq) is a found as a decimal by dividing the total number of tasters by the total number of students in the class. The

frequency of nontasters (q2 ) is found by dividing the number of tasters by the number of people in the class. Using the Hardy Weinberg equation the frequency of p and q can be found. q is found by taking the square root of q2. p is found by using the equation 1-q=p. Also calculate the frequencies of the North American population. Finally find 2pq that represents the percentage of the heterozygous tasters in the class. Record the results in table 8.1 Results Table 8.1 Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles Allele Frequencies

Phenotypes

Class Population

North American Population 1.

Tasters

Nontasters

P2 + 2pq

Q2

#

%

#

%

7

77.78

2

22.22

55

45

p

Q

.53

.47

.33

.67

What is the % of heterozygous tasters 2pq in your class?

49.82%

2. What % of the North American population is heterozygous for the taster trait? 44.15%

Case I Ideal Hardy Weinberg Populations Introduction In this experiment the entire class will represent an entire breeding population. In order to ensure random mating, choose another student at random. The class will simulate a population of randomly mating heterozygous individuals with an initial gene frequency of .5 for the dominant allele A and the recessive allele a and genotype frequencies of .25 AA, .50 Aa and .25 aa. Your initial genotype is Aa. Record this on the data page. Each member of the class will receive four cards. Two cards have a and two cards have A. The four cards represent the products of meiosis. Each “parent” contributes a haploid set of chromosomes to the next generation.

Hypothesis By conducting the experiment under ideal conditions we will be able to show an ideal Hardy Weinberg population. Materials The materials used in this experiment are as follows: cards labeled A and a, a pencil and a piece of paper. Methods Begin the experiment by turning over the four cards so the letters are not showing, shuffle them, and take the card on top to contribute to the production of the first offspring. Your partner should do the same. Put the two cards together. The two cards represent the alleles of the first offspring. One of you should record the genotype of this offspring in the Case I section on page 98. Each student pair must produce two offspring, so all four cards must be reshuffled and the process repeated to produce a second offspring. Then, the other partner should record the genotype. The very short reproductive career of this generation is over. Now you and your partner need to assume the genotypes of the two new offspring. Next, the students should obtain the cards requires to assume their new genotype. Each person should then randomly pick out another person to mate with on order to produce the offspring of the next generation. Follow the same mating methods used to produce offspring of the first generation. Record your data. Remember to assume your new genotype after each generation. The teacher will collect class data after each generation. Results Case I AA

Aa

aa

F1

1

5

2

F2

2

4

2

F3

1

6

1

F4

1

5

2

F5

1

5

2

Number of offspring with genotype AA =6x2= 12 A alleles

Number of offspring with genotype Aa = 25x1 = 25 A alleles Total = 37 A alleles P= .46 Number of offspring with genotype aa = 9x2 = 18 alleles Number of offspring with genotype Aa = 25 Total = 43 Q = .54 2pq=.49 .22+.49+..29 1. What does the Hardy Weinberg equation predict for the new p and q. It predicts that the new p and q will be determined by chance. 2. Do the results you obtained in this simulation agree? If not, why not? No the results do not agree because the population is not perfect. The population size is too small creating disequilibria. 3. What major assumptions were not strictly followed in this simulation? The assumption the population is large was not followed because in fact the breeding population used was very small.

Case II Selection Hypothesis Using this experiment we will be able to simulate natural selection and use the Hardy Weinberg equation to determine the frequencies of the alleles. Introduction In this case you will modify the simulation to make it more realistic. In the natural environment, not all genotypes have the same rate of survival; that is, the environment might favor some genotypes while selecting against others. An example is the human condition, sickle cell anemia. It is a disease caused by a mutation on one allele, homozygous recessives often die early. For this simulation, you will assume that the homozygous recessive individuals never survive, and that heterozygous and homozygous dominant individuals survive ever time. Materials The materials used in this experiment are cards labeled A and a, a pencil and a piece of paper. Methods Once again start with your initial genotype and produce fertile offspring as in Case I. There is an important change in this experiment. Every time an offspring with the genotype aa is produced it dies. The parents must continue to reproduce until two fertile

offspring are produced. As in Case I proceed through five generations, but select against the aa every time. Results

Case II AA

Aa

aa

F1

2

6

0

F2

6

2

0

F3

6

3

0

F4

6

3

0

F5

4

4

0

Number of offspring with AA alleles 22x2= 44 Number of offspring with Aa alleles x1 = 18 Total = 62 P = .775 Number of offspring with genotype aa x2= 0 Number of offspring with genotype Aa – 18 Total a alleles = 18 Q = .225 2pq=.35 .60+.35+.05=1 1. How do the new frequencies of p and q compare to the initial frequencies in Case I? Through natural selection individuals with the genotype aa are eliminated causing a decline in the number of a alleles in this case. So the p frequencies in this case are higher than those in Case I and the q frequencies are lower than the q frequency in Case I. 2. How has the allele frequency of the population changed? In the first trial p was .46 and q was .54. The frequency of dominant A alleles is higher and the frequency of Aa is smaller because the total number of a alleles in the gene pool was reduced when the individuals with the genotype aa were selected against and died before they could reproduce.

3. Predict what would happen to the frequencies of p and q if you simulated another five generations. The frequency of q will continue to decrease, but it will not reach zero because the heterozygous Aa remain. 4. In a large population would it be possible to completely eliminate a deleterious recessive allele? Explain. No, it is impossible to completely eliminate a deleterious, harmful, recessive allele. Even though some people that express the trait because they are heterozygous recessive may die before they can pass the trait on to offspring, the gene pool will always have this allele because carriers are able to live normal lives and pass these alleles on to there offspring. An example is hemophilia.

Case III Heterozygote Advantage Hypothesis Using this experiment we will be able to show the advantage of heterozygotes in a population undergoing natural selection. Introduction From Case II, it is easy to see that the lethal recessive allele rapidly decreases in the population. However in a real population there is an unexpectedly high frequency of the sickle cell allele in some populations. Case II did not accurately depict a real situation. In the real world heterozygotes have an advantage over homozygous dominant organisms. This is accounted for in Case III. In this Case everything is like Case II except if your offspring is AA, flip a coin. If it is heads the individual dies, id it is tails it lives. Materials The materials used in this Case are cards marked A and a, and a coin. Methods Once again simulate five generations, staring again with the initial genotype form Case I. Again the genotype aa never survives. However the genotype AA will have a fifty-fifty chance of living. Determine if it survives by flipping a coin. Tails it lives and heads it dies. Finally total the class genotypes and calculate the p and q frequencies.

Results

Case III AA

Aa

aa

F1

3

5

0

F2

1

7

0

F3

2

6

0

F4

2

6

0

F5

4

4

0

Number of individuals with the genotype AA =12x2=24 Number of individuals with genotype Aa= 28x1=28 Total = 52 P=.65 Number of individuals with genotype aa=0 Number of individuals with genotype Aa = 28 Q= .35 2pq = .455 1. Explain how the changes in p and q frequencies in Case II compare with Case I and Case III. The frequency of p in Case II was higher than p in case I because in Case II and q was lower in Case II than Case I because in Case II aa were selected against. In Case III the frequency of p is lower than the p is case II because unlike in Case II in Case III the individuals with AA did not always survive. The q increased because in Case III more heterozygotes survived than in Case II. This displays the heterozygote advantage. 2. Do you think the recessive allele will be completely eliminated in either Case II or Case III? No, the recessive allele will not be eliminated because there will always be heterozygotes. 3. What is the importance of heterozygotes, the heterozygote advantage, in maintaining genetic variation in populations? The heterozygotes have both alleles, which is needed for genetic variation. Heterozygotes are essential for there to be genetic variation in a population.

Case IV Genetic Drift Hypothesis By using this experiment we will be able to simulate genetic drift in an isolated population. Materials The materials used in this experiment are cards labeled either A or a. Methods The simulation used in these experiments can be used to look at genetic drift. Then go through five generations like Case I but do not switch mates. Record the genotypic frequencies of p and q for the class after the fifth generation. Results

Case IV AA

Aa

Aa

F13

1

1

0

F2

1

1

0

F3

1

1

0

F4

2

0

0

F5

2

0

0

Number of individuals with the phenotype AA= 7x2=14 Number of individuals with Aa =18 Total = 32 P=0.4 Number of individuals with the phenotype aa= 15x2=30 Number of individuals with the phenotype Aa= 18 Total =48 Q=0.6 2pq= .48 .16+0.48+.36=1 1. Explain how the initial genotypic frequencies of the populations compare. The original % of the hybrid was 100. After the % of the hybrid was 50.

2. What do your results indicate about the importance of population size as an evolutionary force? The small determined that this is not natural selection. Error Analysis In each of the cases rounding decimals could have contributed to inaccurate results. Also inaccurately converting the data into terms of p and q could have cased errors in the results. Finally, the small size of the breeding population used increased the likelihood of errors taking place. Conclusion After completing the four Cases a few conclusions have been developed. When heterozygotes for a certain allele die it contributes to the decrease of the frequency of that allele in the gene pool. When a large number of individuals survive and are able to produce viable offspring then frequencies for the alleles they contained will increase. Also for several traits that can cause death the heterozygote has the advantage in surviving. Finally, when individuals of a population breed with the same mate for several generations genetic drift can take place and there will be more of a certain allele than that of another allele.