P 2pq Q 1.0 And P Q 1

AP BiologyNameAP Lab Two: Mathematical Modeling, Hardy-WeinbergIn 1908 G.H. Hardy and W. Weinberg independently suggested a scheme whereby evolutioncould be viewed as changes in the frequency of alleles in a population of organisms. In this scheme, if“A” and “a” are alleles for a particular gene locus and each diploid individual has two such loci, then “p”can be designated as the frequency of the “A” allele and “q” as the frequency of the “a” allele. Thus, in apopulation of 100 individuals (each with two loci) in which 40% of the alleles are “A”, “p” would be 0.40.These are referred to as allele frequencies. The frequency of the possible diploid combinations of thesealleles (AA, Aa, aa) is expressed asP2 2pq q2 1.0andp q 1.0Hardy and Weinberg argued that if five conditions are met, the populations’ allele and genotypefrequencies will remain constant from generation to generation. These conditions are:1. The breeding population is large. The effect of chance on changes in allele frequencies isthereby greatly reduced.2. Mating is random. Individuals show no mating preference for a particular phenotype.3. There is no mutation of the alleles. No alteration in the DNA sequence of alleles.4. No differential migration occurs, immigration or emigration.5. There is no selection. All genotypes have an equal chance of surviving and reproducing.The Hardy-Weinberg equation describes a situation in which if all five conditions are met then nochanges will occur in either allele or genotype frequencies in the population. If changes do occur, thenan evolution must be in the mix.Lab 2A: Estimating Allele Frequencies for a Specific Trait.In this exercise, you will use the entire class as a sample population. The ability to taste the bitterqualities of PTC is evidence of a dominant allele, which means the taster is either homozygous dominantor heterozygous. The inability to taste anything on the PTC paper is evidence of the recessive allele, andthe non-taster is homozygous recessive.1. Using PTC test papers, tear off a short strip and press it to your tongue tip. PTC tasters will sensea bitter taste. For the purposes of this exercise these individuals are considered tasters.2. A decimal number representing the frequency of tasters (p2 2pq) should be calculated bydividing the number of tasters in the class by the total number of students in the class. Adecimal number representing the frequency of nontasters (q2) can be obtained by dividing thenumber of nontasters by the total number of students. You should then record these numbersin Table 2.1.3. Use the Hardy-Weinberg equation to determine the frequencies of the two alleles. Thefrequency “q” can be calculated by taking the square root of (q2). Once “q” has been

determined, “p” can be determined because 1 – q p. Record these values in Table 2.1 for theclass and also calculate and record the values of p and q for the North American population.Lab 2B: Case Studies.In this exercise, the entire class will represent a breeding population. In order to ensure random mating,choose another student at random. If there are an odd number of students, Mr. Marshall will alsoparticipate. For the purpose of the study, assume “mating” is possible regardless of gender.Case 1: Ideal Hardy-Weinberg PopulationThe entire class will be given cards, two for each individual. The beginning gene frequency for thepopulation will be 0.5 for the dominant allele, A, and 0.5 for the recessive allele, a.1. Before beginning, write a hypothesis for what you think will occur in this case study.2. Every participant will be given a heterozygous genotype. You will have four cards, two labeled“A” and the others labeled “a”. These will represent your parent gametes.3. Turn the four cards over so that the letters do not show and shuffle them.4. Everyone randomly moves about the area until time is called. When this occurs, the person youare standing closest to is your new mate. You may not repeat mates in back-to-backgenerations. Remember, genders and genotypes are not a factor.5. Each person randomly selects one card from their stack. The two cards are then compared toreveal the genotype of the first offspring. One of you will record this genotype in the Fgeneration on Table 2.2.6. Reshuffle the cards and pick again with your partner. The second partner will record thisgenotype as their offspring and record in their Table 2.2.7. The reproductive season is over. You will now take on the genotype of the offspring youcreated. Go to the main counter and retrieve four cards that will be your new gametesa. If your new genotype is AA, you need four “A” cardsb. If your new genotype is Aa, you need two “A” cards and two “a” cardsc. If your new genotype is aa, you need four “a” cards8. Repeat steps 3-7 for a total of five generations.9. Calculate new frequencies in Table 2.3Case 2: Directional SelectionIn this exercise we will model what happens if the environment has a selection for a specific trait versesanother. For the purpose of this study, we will assume that the recessive trait does not enable theindividual to survive. Thus, anyone with a homozygous recessive genotype will die.1. Before beginning, write a hypothesis for what you think will occur in this case study.2. Start with the initial genotype as you did for case 1.3. Perform steps 3-7 as previously. This time, however, if the offspring is “aa,” it will notreproduce. Assume the partners would mate until a child is born, so reshuffle the cards andchoose again. Record your results in Table 2.4.

4. Proceed through five generations, selecting against the homozygous recessive offspring 100% ofthe time. Then add up the genotype frequencies that exist in the population and calculate thenew p and q frequencies in Table 2.5.5. Submit your data for each generation on the survey. When all survey data has been collected,plot a graph showing the p/q values for each generation in Case 2.Case 3: Heterozygote AdvantageSometimes being a heterozygote is more advantageous to being homozygous. If one allele is healthy andthe other makes you susceptible for a specific disease, being a heterozygote can help you be healthywhile fighting off infections due to disease. For example, individuals who are heterozygotes for sicklecell anemia are slightly more resistant to malaria than homozygote individuals.1. Before beginning, write a hypothesis for what you think will occur in this case study.2. In this rounda. aa genotypes are still lethal due to sickle blood cells and the organism will not survive.Choose again.b. Aa genotypes are healthy with round alleles AND they are immune to malaria.Nothing additional needs to be done with these offspring.c. AA genotypes allow the individual to be born with normal blood types BUT leave themsusceptible to malaria. If this genotype is chosen, flip a coin.i. If heads up, the individual dies from malaria in childhood. Choose again.ii. If tails up, the individual avoids malaria and nothing else needs to be done.3. Simulate ten generations, starting with the original genotypes from Case I. Since we want tomaintain a constant population size, make sure you always retry if the genotype causes yourorganism to die. Record your genotypes for each generation in Table 2.6.4. Calculate genotypic frequencies at the end of the simulation in Table 2.7.Case 4: Disruptive SelectionSometimes being either version of homozygous for a trait is better than being heterozygous. If wingcolors in moths are codominant, then it would be better to be either all black-winged or all whitewinged; a spotted moth would have a difficult time blending in.1. Before beginning, write a hypothesis for what you think will occur in this case study.2. In this rounda. AA and aa genotypes are healthy. Nothing additional needs to be done.b. Aa genotypes leave moths with susceptible phenotypes. Flip a coini. If tails up, the individual avoids death and nothing else needs to be done.ii. If heads up, the individual is eaten by a predator. You will sit out theremainder of the activity *NOTE: You will still make two total children withyour current partner*

3. Simulate five generations, starting with the original genotypes from Case I. Since we want tomaintain a constant population size, make sure you always retry if the genotype causes yourorganism to die. Record your genotypes for each generation in Table 2.8.4. Calculate genotypic frequencies at the end of the simulation in Table 2.9.Post-Lab Questions1. How did our classroom frequencies for PTC compare to North America’s? Explain why you thinkthe numbers are what they are.2. For each of the scenarios, explain whether you believe “evolution” occurred. Use evidence fromthe scenario data to justify your claim.3. In a large population, is it possible to completely eliminate a deleterious recessive allele? Explain4. How can heterozygotes help a population maintain genetic variety but not phenotypic variety?5. Which Hardy-Weinberg rule did we break in scenario #4?6. Rh is an antigen for your blood (when someone says they are A or A-, the /- refers to thepresence of the Rh factor). In the United States about 16% of the population is Rh negative. Theallele for Rh negative is recessive to the allele for Rh positive. If the student population of a highschool is 2,000, how many students would you expect for each of the three possible genotypes?7. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. Whatis the frequency of the dominant allele?Lab Book Requirements1. Date of Lab2. Title of Lab3. Objective of Lab4. Manipulated and Responding variable for Lab 2B case study #25. At least three controls for Lab 2B case study #26. Hypothesis for 2B case study #27. Table 2.18. Table 2.29. Table 2.410. Table 2.611. Table 2.812. Graph of Table 2.413. Conclusion14. What is the next experiment you could perform?15. Answers to Post-Lab Questions

Table 2.1: Phenotypic Proportions of Tasters and Nontasters and Frequencies of the DeterminingAllelesAllele Frequency Based on theH-W EquationPhenotypesTasters(p2 2pq)#Nontasters(q2)%#p%Class PopulationNorth AmericanPopulation0.550.45Table 2.3: Allele Frequency for Case Study #1: Hardy-Weinberg EquilibriumNumber of A alleles present at the fifth generationNumber of offspring with genotype AA x 2 A allelesNumber of offspring with genotype Aa x 1 A allelesTotal A allelesp (Total number of A alleles) / (Total number of alleles in the population)Number of a alleles present at the fifth generationNumber of offspring with genotype aa x 2 a allelesNumber of offspring with genotype Aa x 1 a allelesTotal a allelesq (Total number of a alleles) / (Total number of alleles in the population)Table 2.5: Allele Frequency for Case Study #2: Directional SelectionNumber of A alleles present at the fifth generationNumber of offspring with genotype AA x 2 A allelesNumber of offspring with genotype Aa x 1 A allelesTotal A allelesp (Total number of A alleles) / (Total number of alleles in the population)Number of a alleles present at the fifth generationNumber of offspring with genotype aa x 2 a allelesNumber of offspring with genotype Aa x 1 a allelesTotal a allelesq (Total number of a alleles) / (Total number of alleles in the population)q