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NAME: ________________________________________ DATE: _________ BLOCK: _____ In order to study how gene frequencies can change over time (evolve), a benchmark is needed to compare evolving populations with non-evolving populations. Gene pools that are not evolving were described mathematically by the Hardy Weinberg Theorem in 1908. p2 + 2pq + q2 = 1 where p is the dominant allele frequency and q is the recessive allele frequency This theorem states that the allele frequencies of a population remain constant over time as long as certain criteria are met. The following conditions must be true for the Hardy Weinberg Theorem for non-evolving populations. Large population size No gene flow No mutations Random mating No natural selection In this lab you will use the Hardy Weinberg Theorem to calculate gene frequencies. Remember, a population has evolved if the gene frequencies of a population change over generations. Note that the Hardy Weinberg Theorem can only work for genes with two alleles. For example, this would not work for blood type because there are multiple alleles. There are several parts to this lab. Each part represents a real-life scenario so that you can see the possible effects of each on gene frequencies. Exercise A: Estimating allele frequencies for a specific trait 1. Use PTC test papers to determine if you are a taster (dominant trait) or a nontaster (recessive). 2. Use Hardy-Weinberg to determine the frequency of p and q. Complete the table below. Populations Class Population North American Population Allele Frequency Based on Hardy Weinberg Phenotypes Tasters (p2 + 2pq) # % Nontasters (q2) # % 0.55 0.45 1 p q PART ONE: Hardy Weinberg Equilibrium Assume that “A” represents a dominant allele for brown eyes and “a” represents the recessive allele for blue eyes. 1. Pick up two index cards: an “A” card and an “a” card. This means that everyone starts out heterozygous (Aa). 2. Find your first partner and mate to make your first offspring. 3. The offspring are made by randomly selecting one card from each person. The resulting baby will have a genotype based on the cards chosen. For example, the baby may be aa. Each person should take back the card that they contributed to the baby. 4. Next, repeat step 3 so that you are making a second baby for the other person. 5. Now, in nature this does not happen, but in order to simulate generations, you will become your offspring. In the example above, you would become aa. There will be times when you have to get different cards because your genotype changes. 6. Next, you find a new partner randomly and mate again. 7. Each person will do this for 5 generations (make five babies). 8. Record the genotypes of each generation (each baby that you make) in Table 1. 9. Return to your desk and report your individual data to Mrs. Flick as instructed. 10. Record the class data in the calculation chart. PART TWO: Natural Selection You will repeat mating steps as in part one with one difference. There has been a change in the environment. A new emergent virus from the rainforest has jumped from birds to humans. Scientists have found that people who are homozygous for blue eyes (aa) are more susceptible to the virus and more likely to die because the virus can only enter the brain through the blue-eyed pigment. In other words the environment is selecting for those people who are AA and Aa. To represent this in our simulation, babies who are aa die. 1. If you make an aa baby, it dies. You must keep trying to make a baby until you make one that survives. 2. Just as in Case One you must make 5 babies and become the baby you made for each round. 3. Record the genotypes of each generation (each baby that you make) in Table 1. 4. Return to your desk and report your individual data to Mrs. Flick as instructed. 5. Record the class data in the calculation chart. PART THREE: Homozygous Advantage You will repeat mating steps as in part one with the following difference. This time the “a” represents the recessive allele for sickle cell anemia and the “A” represents the dominant normal allele. It turns out that people who are heterozygous (Aa) for sickle cell are protected against the severe symptoms malaria. This means that people who are AA have a higher change of dying from malaria and people who are aa have higher chance of dying from sickle cell. 2 1. If you make an aa or AA baby, pick up a penny and flip it. If the penny lands heads up the baby dies. If the penny lands tails up, the baby survives. 2. If the baby survives you will become that baby (either aa or AA) in the next generation. If the baby dies, you must keep trying to make a baby that will survive. 3. Record the genotypes of each generation (each baby that you make) in Table 1. 4. Return to your desk and report your individual data to Mrs. Flick as instructed. 5. Record the class data in the calculation chart. PART FOUR: Genetic Drift You will repeat mating steps as in part one with the following difference. You must mate in small groups of 4 or 5 people. You may not mate with anybody outside of your group. Assume that each group represents a small founder population on separate islands. 1. Record the genotypes of each generation (each baby that you make) in Table 1. 2. Return to your desk and report your individual data to Mrs. Flick as instructed. 3. Record the class data in the calculation chart. Table 1: Baby Data Part One: H-W Equilibrium My initial genotype Aa Part Two: Natural Selection My initial genotype Aa F1 genotype F1 genotype F2 genotype F2 genotype F3 genotype F3 genotype F4 genotype F4 genotype F5 genotype F5 genotype Part Three: Heterozygote Advantage My initial genotype Aa Part Four: Genetic Drift My initial genotype F1 genotype F1 genotype F2 genotype F2 genotype F3 genotype F3 genotype F4 genotype F4 genotype F5 genotype F5 genotype 3 Aa QUESTIONS Show your calculations for all answers involving math! Exercise A: Tasters 1. What is the percentage of heterozygous tasters (2pq) in the class? 2. What percentage of the North American Population is heterozygous for the taster trait? PART ONE: HARDY-WEINBERG EQULIBRIUM 3. What does H-W predict for the new p and q (for the 5th generation)? 4. Do the results you obtained in this simulation agree with the above prediction? ____ Why or why not? 5. What major conditions were not strictly followed in this simulation? PART TWO: NATURAL SELECTION 6. How do the initial frequencies of p and q compare to the final frequencies? 7. Has evolution occurred in this simulation? Explain your answer. 4 8. Predict what would happen to the p and q frequencies if you simulated another 5 generations? 9. In a large population, would it be possible to completely eliminate a recessive allele – even one that is homozygous lethal? Explain your answer. PART THREE: HETEROZYGOUS ADVANTAGE 10. How do the initial frequencies of p and q compare to the final allele frequencies? 11. In a large population, do you think the recessive allele will be eliminated in this example? Explain your answer. CASE FOUR: GENETIC DRIFT 12. How do the initial frequencies of p and q compare to the final allele frequencies? 13. What do your results indicate about the importance of population size as an evolutionary force? 5 HARDY-WEINBERG PROBLEMS 14. In Drosophila, the allele for normal length wings is dominant over the allele for vestigial wings. In a population of 1,000 individuals, 360 flies have the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait? 15. The allele for the ability to roll one’s tongue is dominant over the allele for the lack of this ability. In a population of 500 individuals, 25% are nonrollers. How many individuals would you expect to be homozygous dominant and heterozygous for this trait? 6 16. The allele for the hair pattern called widow’s peak is dominant over the allele for no widow’s peak. In a population of 1,000 individuals, 510 have widow’s peaks. How many individuals would you expect of each of the three possible genotypes? 17. In the US, about 16% of the population is Rh negative. The Rh negative allele is recessive to the Rh positive allele. If a student population of high school in the US is 2000, how many students would you expect for each of the three possible genotypes? 7 18. In certain African countries, 4% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population of 1,000 newborn babies in these countries, how many would you expect for each of the three possible genotypes? 19. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele? 8 CALCULATION CHARTS Part One: Initial Allele Frequencies Genotypes AA Aa Number of each genotype Multiply by… Part One: Final Allele Frequencies Number of A's Genotypes AA Aa Total number of A's Genotypes Aa aa Number of each genotype Multiply by… Number of each genotype Multiply by… Number of A's Total number of A's Number of a's Genotypes Aa aa Total number of a's Number of each genotype Multiply by… Number of a's Total number of a's Total Number of genes p q Total Number of genes p q Part Two: Initial Allele Frequencies Part Two: Final Allele Frequencies Genotypes AA Aa Number of each genotype Multiply by… Number of A's Genotypes AA Aa Total number of A's Genotypes Aa aa Number of each genotype Multiply by… Number of each genotype Multiply by… Number of A's Total number of A's Number of a's Genotypes Aa aa Total number of a's Number of each genotype Multiply by… Total number of a's Total Number of genes p q Total Number of genes p q 9 Number of a's Part Three: Initial Allele Frequencies Genotypes AA Aa Number of each genotype Multiply by… Part Three: Final Allele Frequencies Number of A's Genotypes AA Aa Total number of A's Genotypes Aa aa Number of each genotype Multiply by… Number of a's Genotypes Aa aa Genotypes Aa aa Multiply by… Multiply by… Number of a's Part Four: Final Allele Frequencies Number of A's Genotypes AA Aa Total number of A's Number of each genotype Number of each genotype Total Number of genes p q Part Four: Initial Allele Frequencies Multiply by… Number of A's Total number of a's Total Number of genes p q Number of each genotype Multiply by… Total number of A's Total number of a's Genotypes AA Aa Number of each genotype Number of each genotype Multiply by… Number of A's Total number of A's Number of a's Genotypes Aa aa Total number of a's Number of each genotype Multiply by… Total number of a's Total Number of genes p q Total Number of genes p q 10 Number of a's