AP Biology Lab Eight: Population Genetics

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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
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  AP Biology Lab 8: Population Genetics Introduction G.H Hardy and W. Weinberg developed a theory that evolution could be described as achange of the frequency of alleles in an entire population. In a diploid organism that hasgene 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 aallele. An example is, in a population of 100 organisms, if 45% of the alleles are A thenthe frequency is .45. The remaining alleles would be 55% or .55. This is the allelefrequency. An equation called the Hardy Weinberg equation for the allele frequencies of a population is p 2+ 2pq+ q 2 = 1. P represents the A allele frequency. The letter q representsthe a allele. Hardy and Weinberg also gave five conditions that would ensure the allelefrequencies of a population would remain constant. A. The breeding population is large. The effect of a change in allele frequencies isreduced.B. Mating is random. Organisms show no mating preference for a particulargenotype.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 ontheir 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 beable 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 HardyWeinberg 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 anda 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 (p 2 +2pq) is a found as a decimal bydividing the total number of tasters by the total number of students in the class. The  frequency of nontasters (q 2 ) 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 befound. q is found by taking the square root of q 2 . p is found by using the equation 1-q=p.Also calculate the frequencies of the North American population. Finally find 2pq thatrepresents the percentage of the heterozygous tasters in the class. Record the results intable 8.1 ResultsTable 8.1 Phenotypic Proportions of Tasters and Nontasters and Frequencies of theDetermining Alleles   Phenotypes Allele FrequenciesTasters P 2 + 2pq   Nontasters Q 2   p QClassPopulation# % # %7 77.78 2 22.22.53 .47NorthAmericanPopulation55 45 .33 .671.   What is the % of heterozygous tasters 2pq in your class? 49.82% 2.What % of the North American population is heterozygous for the tastertrait? 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 frequencyof .5 for the dominant allele A and the recessive allele a and genotype frequencies of .25AA, .50 Aa and .25 aa. Your initial genotype is Aa. Record this on the data page. Eachmember 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 setof chromosomes to the next generation.  Hypothesis By conducting the experiment under ideal conditions we will be able to show an idealHardy Weinberg population. Materials The materials used in this experiment are as follows: cards labeled A and a, a pencil anda piece of paper. Methods Begin the experiment by turning over the four cards so the letters are not showing, shufflethem, 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 thealleles of the first offspring. One of you should record the genotype of this offspring inthe 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 thisgeneration is over. Now you and your partner need to assume the genotypes of the twonew offspring. Next, the students should obtain the cards requires to assume their newgenotype. Each person should then randomly pick out another person to mate with onorder to produce the offspring of the next generation. Follow the same mating methodsused to produce offspring of the first generation. Record your data. Remember toassume your new genotype after each generation. The teacher will collect class data after each generation. ResultsCase I   AA Aa aaF 1 1 5 2F 2 2 4 2F 3 1 6 1F 4 1 5 2F 5 1 5 2Number of offspring with genotype AA =6x2= 12 A alleles  Number of offspring with genotype Aa = 25x1 = 25 A allelesTotal = 37 A allelesP= .46Number of offspring with genotype aa = 9x2 = 18 allelesNumber of offspring with genotype Aa = 25Total = 43Q = .542pq=.49.22+.49+..291.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 istoo 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 HardyWeinberg equation to determine the frequencies of the alleles. Introduction In this case you will modify the simulation to make it more realistic. In the naturalenvironment, not all genotypes have the same rate of survival; that is, the environmentmight favor some genotypes while selecting against others. An example is the humancondition, 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 thehomozygous recessive individuals never survive, and that heterozygous and homozygousdominant 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 thegenotype aa is produced it dies. The parents must continue to reproduce until two fertile
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