ࡱ>  JBbjbj,, (NNJ:TCT"vvv!!!!!!!#Y&!vrvvz!!~V!v!P>J$!0"&&vvv :   IB Biology Name _______________________________________ Population Genetics & Evolution Period ________ Date ___________________ BACKGROUND: Natural selection is the mechanism of evolution. Natural Selection assumes that there is variation within a population for traits controlled by pairs of genes called alleles. Allele frequency (also sometimes called gene frequency) is the percent occurrence of an allele. When Natural Selection is occurring, one allele is more successful than its partner in the population. We say that the successful allele is "being selected for" and the less successful allele is "being selected against". When scientists are examining a population they need a tool to determine whether or not selection is taking place. In 1908 G.H. Hardy and W Weinberg independently suggested a scheme whereby evolution could be viewed as changes in the frequency of alleles in a population. They reasoned that 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 a population of 100 individuals (each with two loci) in which 40% of the loci are A, p would be 0.40. The rest of the loci (60%) would be a, and q would equal 0.60 (i.e., p + q = 1.0). These are referred to as allele frequencies also called gene frequencies. If certain conditions are met the frequency of the possible diploid combinations of these alleles (AA, Aa, aa ) should equal p2 + 2pq +q2 = 1.0. Hardy and Weinberg also argued that if five conditions are met, the population's allele and genotype frequencies will remain constant from generation to generation. These conditions are as follows: 1. The population is very large. The effects of chance on changes in allele frequencies is thereby greatly reduced. 2. Individuals show no mating preference for A or a , i.e. mating is random 3. There is no mutation of the alleles. 4. No differential migration occurs 5. All genotypes have an equal chance of surviving and reproducing, i.e. there is no selection Basically, the Hardy-Weinberg equation describes the status quo. If the five conditions are met, then no change will occur in either allele or genotype frequencies in the population. Of what value is such a rule? It provides a yardstick by which changes in allele frequency, and therefore evolution, can be measured. One can look at a population and ask: Is evolution occurring with respect to one particular gene locus? The purpose of this laboratory simulation is to examine the conditions necessary for maintaining the status quo and see how selection changes allele frequency. PART 1: Estimating Allele Frequencies for a Specific Trait within a Sample Population Using the class as a sample population, an allele frequency of a gene for which there are two alleles that are expressed in the population. Some pairs of alleles that could be used are shown below: dominant / recessive PTC Taster / non-taster tongue rolling / non-rolling tongue dangling earlobes / attached earlobes hitch-hiker thumb / straight thumb . In this activity the dominant allele will be indicated by A and the recessive allele will be indicated with a. A person who has the dominant characteristic may have a genotype that is either homozygous (AA) or heterozygous (Aa). To estimate the frequency of the dominant allele in the population, one must find p. To find p, one must first determine q (the frequency of the recessive allele), because only the genotype of the homozygous recessive individuals is known for sure. Dumas adaptation of AP Biology Lab/2011 Procedure: 1. Choose one trait to survey among class members. Survey the population and record totals in Data Table 1. 2. A decimal number representing the frequency of dominant genotypes (p2 + 2pq) should be calculated by dividing the number of dominants in the class by the total number of students in the class. A decimal number representing the frequency of recessives (q2) can be obtained by dividing the number of recessives by the total number of students. Record these numbers in Data Table 1. 3. Use the Hardy-Weinberg equation to determine the frequencies (p and q) of the two alleles. The frequency 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 Data Table 1 for the class. Attach a worksheet showing your calculations in order. 4. Write a title for Data Table 1 on the line provided. Data Table 1: __________________________________________________ Survey  ResultsPhenotypesAllele  FrequencyPopulation Size  Dominants Recessives % dominants p2 + 2pq  % recessives q2 p  q Questions: 1. What is the percentage of heterozygous individuals (2pq) in the class? 2. How might knowledge of gene frequencies in different sub-populations of people be useful? CASE 1: A Test of an Ideal Hardy-Weinberg Population--No Selection It will be assumed that the class is initially a population of randomly mating heterozygous individuals, each of genotype Aa. Therefore, the initial allele frequency is 0.5 for the dominant allele A and for the recessive allele a, that is, p = 0.5 and q = 0.5. Each member of the class will receive four cards. Two cards will have the letter A printed on them and the other two the letter a. A represents the dominant allele and a is a recessive allele. The four cards represent the products of meiosis. Each student should work with a partner. Each student (parent) contributes a haploid set of chromosomes to the next generation. So for each trait, each partner donates one allele to each offspring. Procedure: 1. Turn four cards over so the letters are not showing, shuffle them, and take the card on the top to contribute to the production of the first offspring. Your partner should do the same. Put the two cards together. You are now the proud parents of the first offspring. One of you should record the genotype of this offspring in the Generation 1 row of Data Table 2. Each student pair must produce two offspring, so all four cards must be reshuffled and the process repeated to produce a second offspring. 2. Your partner should then record the genotype of the second offspring in his or her Data Table 2. The very short reproductive career of this generation is over. You and your partner now become the next generation by assuming the genotypes of the two offspring. That is, student 1 assumes the genotype of the first offspring and student 2 assumes the genotype of the second offspring. In the example diagrammed below, for the first generation, a student with genotype Aa mates with another student with a genotype Aa and produces a first offspring with a genotype Aa and a second offspring with genotype aa . 3. Each student should obtain, if necessary, new cards representing the alleles in his or her respective gametes after the process of meiosis. In the example, student 1 becomes genotype Aa and obtains cards A, A, a, a; student 2 becomes aa and obtains cards a, a, a, a. Each participant should randomly seek out another person with whom to mate in order to produce the offspring of the next generation. 4. After all matings have been recorded in column 2 of the Data Table, class data will be obtained. Data Table 2: GenerationOffspring's GenotypeClass Totals for each Genotype(AA, Aa, or aa )AAAaaa1 X2345Analysis of Results: 1. Genotype Frequency: From the genotypes recorded for Generation 5 in Data Table 2, compute the gene frequencies genotype frequencies for the class. Show sample calculations Frequency of AA ___________ of Aa _____________ of aa ______________ before generation 1 calculation Frequency of AA ___________ of Aa _____________ of aa ______________ after generation 5 What was the Genotype frequency at the beginning of the simulation (generation 1)____________________ Is the Genotype frequency the same as it was before Generation 1?___________ 2. Allele Frequency: The allele frequencies, p and q , should be calculated for the population after five generations of random mating. This is done to find out the answer to the question: What was the allele frequency before Generation 1? p = ___________ q = ______________ What was the allele frequency after Generation 5? p = ___________ q = ______________ Conclusions: 3. What does the Hardy-Weinberg equation predict for the new p and q? 4. Do the results you obtained in this simulation agree?________ 5. What major assumption(s) were not strictly followed in this simulation? CASE 2: A Test of a Hardy-Weinberg Population-- with Selection In humans, several genetic diseases have been well characterized. Some of these diseases are controlled by a single allele where the homozygous recessive genotype has a high probability of not reaching reproductive maturity, but both the homozygous dominant and the heterozygous individual survive. At this point, both homozygous dominant (AA) individuals and the heterozygous individuals (Aa) will be considered to be phenotypically identical. The selection will be made against the homozygous recessive individuals (aa) 100% of the time. Question: Are the allele frequencies before selection the same as they are after selection? 6. Based upon the procedure, predict what will happen to the allele frequencies. Procedure: 1. The procedure is similar to that in the preceding exercise. You should start with initial genotype (Aa) and, with another student, determine the genotype of your two offspring. This time, however, there is one important difference. Every time an offspring is aa, it dies. To maintain the constant population size, the two parents of an aa offspring must try again until they produce two surviving offspring (AA or Aa). 2. The process should proceed through five generations with selection against the homozygous recessive offspring. It is important to remember that an aa individual cannot mate. After each generation, you should record the genotype that you assume on Data Table 3. 3. Class data should be collected for each generation and recorded in Data Table 3. Data Table 3: Generation Offspring's Genotype Class Totals for each Genotype(AA, Aa, or aa )AAAa1 X2345Analysis of Results: 7. Genotype Frequency: From the genotypes recorded for Generation 5 in Data Table 2, compute the genotype frequencies for the class. Show sample calculations Frequency of AA ___________ of Aa _____________ of aa _______________ before generation 1 Frequency of AA ___________ of Aa _____________ of aa _______________ after generation 5 What was the Genotype frequency at the beginning of the simulation (generation 1)____________________ Is the Genotype frequency the same as it was before Generation 1?___________ 8. Allele Frequency: The allele frequencies, p and q , should be calculated for the population after five generations of random mating. This is done to find out the answer to the question: What was the allele frequency before Generation 1? p = ___________ q = ______________ What was the allele frequency after Generation 5? p = ___________ q = ______________ Conclusions: 9. How has the allele frequency of the population changed in Case 2? Describe. 10. Why has this happened? 11. Why might your results not conform to your predicted results? 12. Predict what would happen to frequencies of p and q if you simulated another five generations in Case 2? 13. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Would it be probable? Explain. CASE 3: A Test of a Hardy-Weinberg Population-- with Heterozygous Advantage From Case 2, it is easy to see that the lethal recessive allele rapidly decreases in the population. However, data from many human populations show an unexpectedly high frequency of the sickle-cell allele in some populations. In other words, our simulation does not accurately reflect the real situation; this is because individuals who are heterozygous are slightly more resistant to a deadly form of malaria than homozygous dominant individuals. In other words, there is a slight selection against homozygous dominant individuals as compared to heterozygous. This fact is easily incorporated into our simulation. In this round, keep everything the same as in Case 2, except that if your offspring is AA, flip a coin. If heads, the individual does not survive; if tails the individual does survive to reproduce. Simulate five generations, starting again with the initial genotype from Case 1. The genotype aa never survives and homozygous dominant individuals only survive if the coin toss comes up tails. Total the class genotypes and calculate the p and q frequencies. Starting with the F5 genotype, go through five more generations and again total the genotypes and calculate the frequencies of p and q. 14. Based upon the procedure for case 3, predict what will happen to the allele frequencies. Data Table 4: Generation Offspring's Genotype Class Totals for each Genotype(AA, Aa, or aa )AAAa1 X2345 Generation Offspring's Genotype Class Totals for each Genotype(AA, Aa, or aa )AAAa6 X78910 Analysis of Results: 15. Genotype Frequency: From the genotypes recorded for Generation 5 in Data Table 2, compute the gene frequencies genotype frequencies for the class. Show sample calculations Frequency of AA ___________ of Aa _____________ of aa _______________ before generation 1 Frequency of AA ___________ of Aa _____________ of aa _______________ after generation 5 What was the Genotype frequency at the beginning of the simulation (generation 1)____________________ Is the Genotype frequency the same as it was before Generation 1?___________ 16. Allele Frequency: The allele frequencies, p and q , should be calculated for the population after five generations of random mating. This is done to find out the answer to the question: What was the allele frequency before Generation 1? p = ___________ q = ______________ What was the allele frequency after Generation 5? p = ___________ q = ______________ Conclusions: 17. Explain how the changes in p and q frequencies in Case 2 compare with Case 1 and Case 3. 18. Do you think the recessive allele will be completely eliminated in either Case 2 or Case 3? Explain.    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