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Unformatted text preview: Objectives Biology 1M03 Tutorial 1 Hardy‐Weinberg Student Package !"Understand Hardy‐Weinberg theory and apply to in‐class simulation data !"Apply Chi square test to population data to determine whether populations are in Hardy‐Weinberg equilibrium ! Read the Introduction in this lab manual ! Read section 25.1 (pp. 504‐508) of Freeman’s Biological Science, 3rd Ed. ! Complete the ELM pre‐lab quiz Preparation Introduction Chapter 24 in your text described how evolution by natural selection causes changes in the characteristics of populations. But how do we know if a population is changing? The Hardy‐Weinberg equation allows us to calculate allelic or genotypic frequencies for a population. A population is in Hardy‐Weinberg equilibrium if these frequencies are constant from one generation to the next. If the genetic frequencies are in equilibrium, we can say the population is not evolving. When we observe a change in genetic frequencies, the population is not in equilibrium, and evolution is thus occurring. Hardy‐Weinberg equilibrium theory states that a population will remain in equilibrium unless at least one of the following acts on the population: 1. 2. 3. 4. 5. Genetic drift – chance fluctuations in gene frequency especially in small population Gene flow – transfer of alleles into and out of a population due to migration Mutation – new mutations cause changes in the frequency of alleles Non‐random mating (sexual selection) ‐ mate selection eliminates random mixing Natural selection – differential survival and reproductive success of certain alleles increases their frequency As you might suspect, it’s highly unlikely that a natural population will satisfy all of the above conditions; infinitely large population size, no migration, no mutation, and no natural or sexual selection. In other words, most populations violate at least one of these assumptions and are therefore evolving. Thus the Hardy‐Weinberg theorem is a useful tool for establishing a null hypothesis when we want to test whether evolution is occurring in a population. To do this, we consider the two possible form of a specific allele, and assign the letter p to the dominant allele (A) and the letter q to the recessive allele (a). Since all individuals have two alleles (they may be AA, Aa, or aa) we can express the population’s gene frequencies in terms of p and q: Table 1: genotypic frequencies for all individuals in a population Allele 2 p q 2 p pq p Allele 1 q pq q2 Assuming the values in the above table represent all of the alleles in a population, we can assume the genotype frequencies (because pp, pq, and qq represent genotypes) found in the table can be summed to equal 1.00: p2 + 2pq + q2 = 1 We can also assume that the allelic frequencies (because p and q represent alleles) can be summed to equal 1.00: p + q = 1 We can derive at least two uses from these equations: 1. Any change in allelic OR genotypic frequencies (from one generation to the next) suggests that evolution is occurring and one of the 5 conditions required for equilibrium has been violated 2. If we know one allelic or genotypic parameter of a population, we may use it to solve for other, unknown parameters in the population. For example: If we know the percentage of individuals in a population (that is in equilibrium) who show a recessive phenotype, we can calculate genotypic and allelic frequencies. Suppose you have determined 16% of a population possesses a continuous hairline, a recessive trait (q2). The alternative phenotype, a widow’s peak hairline, therefore has a prevalence of 84% in the population (p2+2pq). We can use the Hardy‐ Weinberg equation to determine how many individuals are homozygous dominant and how many are heterozygous. Since p + q =1 and q2= 0.16 or 16 %, p = 1 ‐ 0.4, or p = 0.6. This means p2 =(0.6)2 = 0.36. Finally, 2pq = 2(0.6)(0.4) = 0.48. Thus, the values for the Hardy‐Weinberg Equation are: p2 (homozygous dominant) = 0.36 or 36% 2pq (heterozygous) = 0.48 or 48% 2 q (homozygous recessive) = 0.16 or 16% We can verify that: p2 + 2pq + q2 = 1: Since: 0.36 + 0.48 + 0.16 = 1.00 Chi square test: Review Suppose you have a normal, six‐sided die and you would like to test the hypothesis that each of the six sides may turn up with equal probability (i.e. 1/6). Thus, your null hypothesis is that there is no underlying factor causing any one side turn up more than 1/6 of the time. You decide to experiment, and you roll your die 30 times. Your record your results in a table: Die shows # times occured 6 4 4 7 4 5 You can use the chi‐square equation to determine whether your hypothesis that each side has a 1/6 probability of turning up is statistically significant. Your observed values are recorded in the table above, and your expected value for each of the six “outcomes” is 5, since 1/6 x 30 =5. !2 = ! (O‐ E)2 = ( 6 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 7 ‐ 5 )2 + ( 4 ‐ 5 )2 + ( 5 ‐ 5 )2 E 5 5 5 5 5 5 2 = (1) +(‐1)2 +(‐1)2 + (2)2 + (‐1)2 + (0)2 5 = 1.6 So, your chi square statistic is 1.6. In order to reject the null hypothesis, the chi‐square statistic must exceed the critical value found on the chi square distribution chart corresponding to the degrees of freedom (Df) in our experiment and the desired level of statistical significance. The degrees of freedom in a chi square test is equal to the number of possible outcomes minus one. So here Df =6‐1 = 5. The desired level of statistical significance in this case is 0.05, which means there that, if we were to obtain a chi square value larger than the critical value, there is 95% certainty that the observed difference is real, and not due to chance. Statistical significance Df 0.5 0.01 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.210 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.070 13.388 15.086 20.517 Here we see the value is 11.070. Since 1.6 < 11.070, we cannot reject the null hypothesis; and conclude that our data do not provide any evidence that the dice are unfair. However, when the chi square statistic exceeds the critical value, we reject the null hypothesis. Tutorial 1  Tutorial Exercise HIV and Human Evolution HIV/AIDS has infected over 40 million people worldwide. Many infected individuals die before reproducing, willingly abstain from sexual reproduction, or are less able to raise healthy offspring. As such, it is plausible that HIV is exerting selective pressure on the human population. One potential gene on which such pressure may act is that which codes for the CCR5 receptor, a peripheral protein that allows the HIV virus to gain entry to human immune cells. Some individuals possess a 32‐basepair mutation in this gene (ccr5‐!"#$%&Those with two copies (homozygotes) lack a functional CCR5 receptor and are thus immune to HIV infection. Additionally, heterozygotes tend to have slower disease progression, living longer are acquiring fewer secondary infections. "#$%&'$()*$+,$%.%/#$%012%3$*$/4+%()'4$5%643$*7%)8+9%human sub‐populations, with higher rates among Europeans whose ancestors survived previous disease epidemics including the bubonic plague and smallpox. It has been suggested that the pathogens causing these epidemics were also reliant on the CCR5 receptor for cellular $+/'7:%)+3%/#)/%4+34(43;)*5%64/#%/#$%012%3$*$/4+%6$'$%5484*)'*7%488;+$%3;'4+9%/#$%<;<+4,%&*)9;$%)+3%58)**%&=% epidemics. Many researchers are now interested in the CCR5 gene and its implication in human evolution in the context of AIDS. One group of researchers, Muxel et al (2008), has recently surveyed 121 Caucasians in Brazil, determining their genotypic status at the CCR5 locus. The researchers found the following: CCR5/CCR5 = 108 ''()*!"# = 12 !"#*!"#&& = 1 Is the study population in Hardy‐Weinberg equilibrium for this gene? Or is the population evolving? With a partner, apply the Hardy Weinberg equation to determine the expected frequencies and compare these to the observed genotypic frequencies. Create this chart on the board. Fill in the values when you take the exercise up with the class: 2 2 p 2pq q frequencies OBSERVED EXPECTED In‐tutorial Exercise In this tutorial, we will demonstrate a how genes in a population may fluctuate from one generation to the next. We will use the Hardy Weinberg equation and the chi‐square test to determine whether changes in genotypic or allelic frequencies are significantly different from previous generations. Scenario 1: No Selection The class will serve as a sample population, where each student will be assigned a genotype and become a member of the parent generation. Your TA will summarize your class’ genotype distribution on the board. You will need to copy this data into Table 2A, and calculate the genotypic and allelic frequencies. Once you are assigned a genotype, you need a mate and a coin. Any mate and any coin will do, random mating is the point here. Once you find a partner, you will “mate” twice: flip your coin once for your first mating, heads means you will donate allele #1, tails means you donate allele #2. My genotype “Tails” “Heads” Allele 2 Allele 1 Once you and your partner have both donated an allele, you now have one offspring, with two of its own alleles. You each must flip again donating two new alleles to create your second offspring. Our simulation is zero‐population growth, which means each set of parents will have two children to replace themselves. Unfortunately, this means now that you have two children you are dead. You are now a member of the F1 generation, you and your partner should assume the genotypes of your children and part ways to go find a new mate. But first, fill in the genotypes of your two F1‐generation children in Table 2B. You will mate a total of five times, writing down the genotypes of the resulting offspring in your chart. Each time, you and your partner will assume the genotypes of your offspring. Circle the offspring whose genotype you assume in each generation. Aa individuals Genotypic frequencies Allelic frequencies Table 2A: Class data table for Scenario 1: Random Mating/No selection (Parental generation) Total Total number(n) of nAA= ____________ nAa= ____________ naa= ____________ n= ___ ___ 2 p = 2pq = 2 q = nAA /nTotal = __________ nAa / nTotal = ___________ naa / nTotal = _________ p =freq(A) = ______________ q = freq(a) = _______________ Table 2B: Individual data table for Scenario 1 My initial genotype:______________ Generation F1 F2 F3 F4 F5 Offspring 1 Offspring 2 individuals Genotypic frequencies Allelic frequencies My final (F5) genotype:_______________ Table 2C: Class data table for Scenario 1: Random Mating/ No selection (F5 generation) Total Total number of nAA= ____________ nAa= ____________ naa= ____________ n= ___ ___ 2 p = 2pq = 2 q = nAA /nTotal = __________ nAa / nTotal = ___________ naa / nTotal = _________ p =freq(A) = ______________ q = freq(a) = _______________ 1. Are the genotype frequencies in the F5 generation the same as those in the initial (parent) generation? 2. What were the allelic frequencies in the initial population? Are they the same as the allelic frequencies in the F5 generation? 3. Is it possible for the genotypic frequencies to change but not the allelic frequencies? b) What would it mean to the population if this occurred? Would the population be in equilibrium? 4. The allelic and genotypic frequencies may differ from the initial population to the F5 generation, but we want to know if there is a significant difference. If we allow the number of individuals from the parental population to be the expected values, and let the F5 genotypic frequencies equal the observed values, we can perform a chi square test on our data. Fill in the Table 2D to calculate the chi square statistic: Table 2D: Chi square test for Scenario 1 Null Hypothesis: Expected results for AA: _________________ Observed results for AA: _________________ Expected results for Aa: _________________ Observed results for Aa: _________________ Expected results for aa: _________________ Observed results for aa: _________________ Calculation of chi square statistic: (Show your work): ! = ! (O‐ E) = E 2 2 Degrees of freedom: Probability: Reject null hypothesis? (Y/N) State what these results mean, in your own words: Critical Value: Scenario 2: Selection In humans, many genetic diseases occur when an individual receives two copies of a mutated allele. Since two copies are required for expression of the disease phenotype, the mutation‐bearing alleles in this scenario are referred to as recessive. The recessive genotype is considered lethal, causing premature death during fetal development, early infancy, or some time before reproductive maturity. However, heterozygous andhomozygous dominant individuals survive. Diseases that show this type of inheritance include sickle‐cell anemia, cystic fibrosis, and Tay‐Sachs disease. In the Scenario 2, we will again assign initial genotypic frequencies. Record your population’s data in Table 3A. The double recessive genotype is lethal, so offspring who inherit the aa genotype will die. Since our population is to have zero‐population growth, parents who produce aa offspring must continue to mate until two viable offspring are produced. For the purposes of the exercise, “aa” individuals in the parental generation will not be considered lethal; instead we will assume the lethality emerges only in the F1 generation. NOTE: Individuals in the parental generation with the aa phenotype should ensure they do not mate with another aa individual. Again, you will mate five times, recording the genotypes of your offspring in Table 3B: Table 3A: Class data table for Scenario 2: Selection against recessive gene (Parental generation) Total Total number(n) of nAA= ____________ nAa= ____________ naa= ____________ n= ___ ___ individuals Genotypic frequencies Allelic frequencies 2 p = 2pq = 2 q = nAA /nTotal = __________ nAa / nTotal = ___________ naa / nTotal = _________ p =freq(A) = ______________ q = freq(a) = _______________ Table 3B: Individual data table for Scenario 2 My initial genotype:______________ Generation F1 F2 F3 F4 F5 Offspring 1 Offspring 2 My final (F5) genotype:_______________ individuals Genotypic frequencies Allelic frequencies Table 3C: Class data table for Scenario 2: Selection against recessive gene (F5 generation) Total Total number of nAA= ____________ nAa= ____________ naa= ____________ n= ___ ___ 2 p = 2pq = 2 q = nAA /nTotal = __________ nAa / nTotal = ___________ naa / nTotal = _________ p =freq(A) = ______________ q = freq(a) = _______________ Use the data from Scenario 2 to complete the Post‐lab Assignment (attached). Hand in your assignment to the 1MO3 drop boxes by tomorrow at 1pm. Late assignments must be signed and stamped by Alastair Tracey before being submitted to the dropboxes. Tutorial 1: Post‐lab Assignment Chi‐square analysis of Scenario 2 data Name:_______________________________________ Student #: _____________________________ TA name: ____________________________________ Perform the chi‐square test on the data for Scenario 2, and answer the questions that follow. Table 3D: Chi square data for Scenario 2 Lab Section: ________________________ Null Hypothesis: Expected results for AA: _________________ Observed results for AA: _________________ Expected results for Aa: _________________ Observed results for Aa: _________________ Expected results for aa: _________________ Observed results for aa: _________________ Calculation of chi square statistic: (Show your work): ! = ! (O‐ E) = E 2 2 Degrees of freedom: Probability: Reject null hypothesis? (Y/N) State what these results mean, in your own words: 1. Critical Value: In Scenario 2, the recessive allele is lethal when two copies are inherited (aa genotype). This is the manner in which the sickle cell anemia gene acts. However, the recessive allele in some genetic diseases, such as sickle cell anemia, may be advantageous when only one copy is inherited, compared to the homozygous dominant genotype. Suppose, beginning with the F5 offspring, malaria enters the population at an epidemic level. Knowing that heterozygotes (Aa) are less likely than homozygote dominant individuals (AA) to die if infected with the malaria parasite, predict what would happen to the genotypic frequencies in the Scenario 2 after several generations: a. What would happen to the frequency of aa individuals? +,./0120*30./0120*4&.5160$ b. What would happen to the frequency of Aa individuals? +,./0120*30./0120*4&.5160$ c. What would happen to the frequency of AA individuals? +,./0120*30./0120*4&.5160$ References Andersen, M. (1993). An introduction to population genetics. Pages 141‐152 in Tested Studies for Laboratory Teaching. Vol. 14 (C.A. Goldman, Editor). Proceedings of the 14th Workshop/Conference of the Association for Biology Laboratory Education (ABLE). Lucotte. (2005). Distribution of the CCR5 gene 32‐basepair deletion in West Europe. A hypothesis about the possible dispersion of the mutation by the vikings in historical times. Human Immunology, 7#&+8$9&933‐936. Martinson, Chapman, Rees, Liu & Clegg (1997). Global distribution of the CCR5 gene 32‐basepair deletion. Nature, 16, 100‐103 Muxel, S., Borelli, S., Amarante, M., Voltarelli, J., Aoki, M., de Oliveira, C., and Watanabe, M. (2008). Association study of CCR5 delta 32 polymorphism among the HLA‐DRB1 Caucasian population in Northern Parana, Brazil. Journal of Clinical Laboratory Analysis, 22, 229‐233. ...
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This note was uploaded on 04/13/2011 for the course BIO 1M03 taught by Professor Jonathanstone,jamesquinn during the Spring '11 term at McMaster University.
 Spring '11
 JonathanStone,JamesQuinn
 Biology

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