AP Lab 8: Population Genetics and Evolution



AP Lab 8: Population Genetics and Evolution (Adapted from the 2001 Student Lab Manual)Purpose: In this lab, you will:learn about the Hardy-Weinberg law of genetic equilibriumstudy the relationship between evolution and changes in the allele frequency by using your class to represent a sample populationPrelab questions:What is the Hardy-Weinberg equation?The Hardy-Weinberg equation is “p2+2pq+q2=1” and it is used to calculate the genetic variation of a population at equilibrium by calculating the frequency of the alleles in the population. Which part of the equation represents: the homozygous dominant, the heterozygotes, and the homozygous recessive?The heterozygotes is represented by “2pq”The homozygous recessive is represented by “q2”The homozygous dominant is represented by “p2”How does the equation p+q=1 relate to Hardy-Weinberg?This relates to the Hardy-Weinberg equation because the sum of the alleles (dominant and recessive) must combine to form the genotype in 1 population, making the sum 1 (100%).What are the 5 conditions for Hardy-Weinberg equilibrium to exist?No mutationRandom matingNo natural selectionLarge population sizeMust be isolated; no migration or emigrationExercise A: Estimating Allele Frequencies for a Specific Trait Within a Sample PopulationFor this exercise, you will use the class as a sample population to find the allele frequencies for PTC tasting. For a person expressing the dominant allele, T, PTC has a bitter taste. If a person is homozygous recessive, s/he cannot taste PTC.Using the PTC taste-test papers, tear off a short strip and press it to your tongue tip. Tasters will sense a bitter taste.Calculate the frequency of the taster population by dividing the total number of tasters by the total number of students in the class. Calculate the frequency of the non-taster population by dividing the number of non-tasters by the number of students in the class. Use the Hardy-Weinberg equation to determine the frequencies (p and q) of the two alleles. Data for Exercise A:PhenotypesAllele Frequency Based on the H-W equation# and % Tasters (p2+2pq)# and % Non-Tasters (q2)PqClass Population2071%829%.54.46N. American Population55%45%.33.67Exercise B: Case StudiesCase 1-A Test of an Ideal Hardy-Weinberg PopulationThe class will represent a breeding population. In order to simulate random mating, choose another student at random. For this simulation, we will assume that gender an genotype are irrelevant to mate selection. The class is a population of randomly mating heterozygous individuals with an initial gene frequency of p=.5 and q=.5. For the phenotypes, AA=.25, Aa=.50, and aa=.25. Your initial genotype is therefore Aa. Record this on your data chart. You will be given 2 cards. These are your gametes. One card will have A and the other will have a. Each parent will contribute 1 gamete to each child.Turn the cards over so that the letters don’t show and shuffle them. The card on top will contribute to the production of your 1st offspring. Your partner will do the same.The product of your 2 cards will be the genotype of your 1st offspring. Record this on your data chart and produce another offspring with your partner. Be sure to reshuffle your cards as in step 1 before making your offspring. Your partner can record the 2nd child on their data chart.Each of you assumes the genotype of your child. You should obtain the necessary cards from the “gene pool” pile at the front of the room.You then need to find a new partner and repeat 1-3.You’ll do this for 6 Generations, recording the genotypes each timeData for Case 1Initial Class Frequencies:AA= .25Aa= .50aa= .25My initial Genotype= AaF1= aaF2= aaF3= AaF4= AaF5= AaUsing the following equations, calculate p and q:Number of A Alleles Present at the 5th generation=Total # of A alleles__________ Total number of alleles in classNumber of a Alleles Present at the 5th generation=Total # of a alleles___________ Total number of alleles in classp= .5q= .5Final Class Frequencies:(# of people who had it in class)AA=6Aa=16aa= 6AA=.25Aa=.50aa= .25 Questions for Case 1What does the H-W equation predict for p and q?It predicts that the new p and q are the same as the initial p and q. Do the results you obtained agree? Why or why not?My specific results do not agree to the simulation because the population size was too small to obtain true results in relation to the H-W rules of equilibrium. What major assumption(s) were not strictly followed in this simulation?In this lab the assumption of a large population size was not followed (one of the major rules for H-W equilibrium). Case 2: SelectionNow, we’re going to make this a little more realistic. In the natural environment, not all genotypes have the same rate of survival. For this simulation, you will assume that the homozygous recessive individuals never survive, while the others survive most of the time.Repeat the procedure as you did in Case 1, except that any aa offspring will die. Since we need a constant population size, you and your partner need to reproduce until you make 2 surviving offspring.Again, proceed through 5 generations.Data for Case 2Initial Class Frequencies:AA= .25Aa= .50aa= .25My initial Genotype= AaF1=AaF2=AAF3=AAF4=AAF5=AAp=.81q= .19Final Class Frequencies:(# of people who had it in class)AA=17Aa=10aa= 0AA=.6561Aa=.3078aa= .0361 Questions for Case 2How do the new frequencies of p and q compare to the initial frequencies in Case 1?The frequency of p has increased whereas the frequency of q decreased. This is due to the fact that in this simulation natural selection has favored the dominant allele (p) for the fittest survival in Case 2, contributing to its increase in frequency over the 5 generations of reproductionWhat major assumptions were not strictly followed in this simulation?The major assumption of a large population was not followed in this simulation as we only had a population of 27, skewing the data we received. Predict what would happen to the frequencies of p and q if you simulated another five generations?If we simulated another 5 generations in case 2, the q frequency would continue to drop (inversely causing the p frequency to increase). However, the q frequency will never decrease to 0, as the heterozygous (the carriers of the recessive q allele) will continue to pass the q allele into the gene pool. In a large population, would it be possible to completely eliminate a deleterious (fatal) recessive allele? ExplainIt is not possible to completely eliminate a fatal recessive allele in a large population because the heterozygotes in the population can carry the “fatal allele” without dying themselves (as you only die from the allele if homozygous dominant). So they will continue to contribute the allele in the gene pool of the population; never actually “eliminating” the q allele. Case 3: Heterozygote AdvantageIt’s easy to see what happens to the lethal recessive allele in the population. However, the data for many human populations shows an unexpectantly high frequency of the sickle-cell allele in certain populations. Thus, our simulation does not accurately reflect the real situation because the heterozygous individuals are slightly more resistant to malaria than the homozygous dominant individuals. In this round, you’ll repeat your procedure as in Case 2. However, if your offspring is AA, flip a coin. If the result is heads, the individual does not survive. If the result is tails, the individual survives. Again, simulate 5 generations as before, keeping the population size the same by mating with your partner until you produce 2 surviving offspring.If time permits, start with the F5 genotype and simulate another 5 generations. Data for Case 3Initial Class Frequencies:AA=.25Aa=.50aa= .25My initial Genotype= AaF1=AaF2=AaF3=AaF4=AaF5=Aap=.72q= .28Final Class Frequencies (after F5):(# of people who had it in class)AA=12Aa=15aa= 0AA=.5184Aa=.4032aa= .0784 Final Class Frequencies (after F10): ***(we never went up to F10 in class)***AA=--Aa=--aa=--Questions:Explain how the changes in p and q frequencies in Case 2 compare with Case 1 and Case 3.The frequency of the p allele was higher in Case 2 than in Case 1 and 3 because the homozygous recessive in Case 2 was selected against; causing a major increase in the dominant allele (as only those with a dominant allele could live to contribute their genes in the gene pool). In turn, the natural selection working against the q allele in Case 2 causes the frequency of q to be lower in Case 2 than in Case 1 and 3. Do you think the recessive allele will be completely eliminated in either case 2 or Case 3?The recessive allele will never by completely eliminated in either case, because the heterozygous allele will remain in the population, contributing the recessive alleles in the gene pool. What is the importance of heterozygotes (the heterozygote advantage) in maintaining genetic variation in populations?The heterozygotes allow for there to be a greater genetic diversity, by carrying both of the alleles. This way, neither allele can be completely eliminated in a population. Case 4: Genetic DriftRecall that small populations have problems maintaining the genetic diversity of their gene poolFor this, you’ll be divided into a small population. You can only mate within your population.Repeat the mating procedure as you did for Case 1.Data for Case 4Initial Class Frequencies:AA= .25Aa=.50aa=.25My initial Genotype= Aa***I died in a forest fire, disabling me from reproducing***p=.46q= .54Final Class Frequencies:(# of people who had it in class)AA=2Aa=9aa= 3AA=.2116Aa=.4968aa= .2916 Questions for Case 4Explain how the initial genotypic frequencies of the population compare.The genotypic frequencies of the population in Case 4 have stayed generally around the same range. There was a slight decrease in homozygous dominant (.25 to .21) and a slight increase in homozygous recessive (.25 to .29); but no major changes in the heterozygous (.50 to .49). What do your results indicate about the importance of population size as an evolutionary force?My results indicate that having a smaller population size, contributes to an increase in genetic drift. As due to the few numbers of organisms, the population will be more heavily impacted than a population with a large number of organisms. Therefore evolutionary forces will have a heavier and faster impact when it is applied to a smaller population when compared to a larger population. ................
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