Population Genetics Workshop: Breeding Bunnies
Pre-workshop Exercise : Population Genetics
Complete Parts I and II of this assignment and bring them with you to workshop. The following questions are a review of basic probability. Most elementary texts in mathematics will contain this material. Part I. Probability Review Here, we provide a brief review of how to calculate a probability in a simple system, such as a gumball machine. Consider a bubblegum machine with four colors of bubblegum balls: red, green, blue, and yellow. We begin with some definitions: Sample space: all possible outcomes of (in this case) choosing a gumball from the gumball machine Experiment: action whose possible outcomes can be recorded, such as choosing a gumball from a gumball machine Outcome: one possible output of an experiment, such as choosing a red gumball from the gumball machine Event: a subset of the sample space With these definitions in mind, try the following problems 1. What is the sample space of choosing a gumball from the gumball machine if we know that there are 4 red gumballs, 3 green gumballs, 2 blue gumballs and 3 yellow gumballs? 2. What is the sample space of choosing two gumballs at a time in the above scenario?
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3. An example of an event of the experiment is choosing a red gumball. The probability of choosing a red gumball is the number of red gumballs divided by the total number of gumballs. For the sample space described in problem 1:
a. What is the probability of randomly choosing a red gumball (give your answer to the nearest hundredth decimal place)?
b. What is the probability of randomly drawing a yellow gumball? 4. The probability of drawing a gumball of a given color at random from this population of gumballs is equal to the frequency of the ball color in the population. What is the probability of green colored balls in this group of balls? 5. Suppose the red and blue gumballs are removed, and there are only green and yellow gumballs remaining in the gumball machine;
a. What is the new sample space?
b. What is the probability of randomly choosing a blue gumball?
c. What is the frequency of yellow gumballs in the new population of gumballs?
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Part II: Allele Frequencies and Hardy Weinberg Equilibrium The Hardy-Weinberg Principle was developed to describe the genetic characteristics of populations with no evolutionary forces acting on it (mating is random, there is no selection or migration, the population size is infinite and no new mutations occur). As a result, for populations in Hardy-Weinberg equilibrium the frequency of alleles and genotypes does not change over time (across generations). This also means that for any locus with two alleles, A and a, if we know their frequencies in the population (freq(A) = p, freq(a) = q) the diploid genotype frequencies can be predicted by the allele frequencies as follows: freq(AA) = p2 for the AA homozygotes in the population, freq(aa) = q2 for the aa homozygotes, and freq(Aa) = 2pq for the heterozygotes. So, consider a gene with two alleles A and a, with frequency p = 0.7 and q = 0.3, respectively. 7. Using this information answer the following questions a. Assume the population we are studying is a species of plant and we wish to determine allele frequencies. If we randomly sampled 100 individuals from the population, how many would we expect to have the A allele?
b. Fill in the following table of expected genotype frequencies in this population:
Genotypes AA Aa aa Expected Frequency
8. Now assume that we actually obtain the diploid genotype from 100 members of the population and find them in the following numbers: AA: 20 Aa: 70 aa: 10 a. What are the observed diploid genotype frequencies? AA: Aa: aa:
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b. Do you think the observed genotype frequencies deviate significantly from those expected under Hardy-Weinberg? If so, what might be the cause? Explain your answer. 9. To determine if the observed genotype frequencies are statistically significantly different we can use a goodness of fit (chi-square) test. Chi-square analyses are used to test for differences between expected and observed values and are used when the observed data fall into categories or classes (e.g., counts of events in categories like the number of males wearing blue vs. red shirts, not continuous measurements like weight or height – such measurements are analyzed using different statistical procedures like t- tests). The formula for the Chi-square test statistic is: ?2 = Si (Observedi-Expectedi)2 / (Expectedi) Where Observedi is the count of individuals in a particular category i (e.g., the number of boys wearing red shirts) and Expectedi is the expected count of individuals in category i. The numerator is squared so that negative and positive values contribute equally to the test statistic. To get the chi-square value, add up all of the resulting values for each category i. To initiate the analysis of data using this test enter the number of AA, Aa, and aa individuals observed in your sample from the population in the table below. Next calculate the number of expected AA, Aa, and aa individuals in a theoretical sample of the same size (100 individuals) if the population was in HWE. To do this, multiply 100 x the expected frequency for each genotype under HWE using the values you entered in the table for 6b.
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A) Fill out the following table using data from questions 6 and 7 above and then complete the equation below to calculate the chi – squared statistic (?2): Observed (O) Expected (E) (O-E)2/E FF Ff ff ?2= B) Next calculate the degrees of freedom for this data set. The equation for calculating degrees of freedom for a chi-square goodness of fit test is:
d.f = k – 1 – m where k is the number of categories and m is the number of independent parameters that we needed to use to calculate the expected number of individuals in the different categories1. In this case our independent parameter was one of the observed allele frequencies (one of which is independent because if we know p then we must know q because both must add up to one – so only one allele frequency is independent). So, for our data the there are three categories (genotypes) minus 1 (because if we know two of the expected genotype frequencies we already know the third one as all frequencies must sum to one) minus 1 (because we used one independent allele frequency to calculate our expected genotype frequencies)
d.f. = 3 – 1 – 1 C) If the chi-square value we calculate from our data exceed a critical chi-square value with a certain degree of freedom at a probability of P <0.05, then we can reject the null hypothesis. If you want to read more about where these P values come from see any introductory statistics textbook or look online at a site you trust (http://en.wikipedia.org/wiki/P-value). Look at the table of critical chi-square values below and compare your calculated chi-square value to the critical value given for P = 0.05 with 1 d.f. If the null hypothesis is that this locus is in Hardy-Weinberg equilibrium, do you accept or reject your null hypothesis? Explain why you reached this conclusion compare this to the answer you gave for 7b. df ?2, P = 0.05 1 3.84 2 5.99 3 7.82 4 9.49
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Population Genetics Workshop: Breeding Bunnies
Kathleen Hoffman, Jeff Leips, and Sarah Leupen
University of Maryland Baltimore County Edited by Adrienne DuBois
Objective: Biological evolution is a change in the frequency of alleles in a population over a period of time. The type of evolutionary mechanisms operating, and the initial frequency of an allele in the population, can influence the amount of change in allele frequencies. In this activity, you will examine natural selection in a small population of wild rabbits. Breeders of rabbits have long been familiar with a variety of genetic traits that affect the survival of rabbits in the wild, as well as in breeding populations. One such trait is the trait for furless rabbits (naked bunnies). This trait was first discovered in England by W.E. Castle in 1933. The furless rabbit is rarely found in the wild because the cold English winters are a strong selective force against it.
Note: In this activity, the dominant allele for normal fur is represented by F and the recessive allele for no fur is represented by f. Bunnies that inherit two F alleles or one F and one f allele have fur, while bunnies that inherit two f alleles have no fur. In this worksheet you will review some important principles and terms applicable to genetics. This worksheet will ask questions about Mendelian genetics. You will also utilize the Chi-square statistical test to test for deviations from Hardy Weinberg equilibrium. Materials per group
• 1 bag of beans (Red and White) • 4 cups
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Procedure: 1. The red beans represent the allele for fur (F), and the white beans represent the allele
for no fur (f). The bag represents the habitat where the rabbits live, and randomly mate.
2. Label one cup FF for the homozygous dominant genotype. Label a second cup Ff for the heterozygous condition. Label the third cup ff for those rabbits with the homozygous recessive genotype.
3. Use the beans (alleles) in your bag (habitat) to count and record your starting allele frequencies. (Note: don’t trust the numbers written on the bags. Count them yourselves.) Record the number of red beans (F) in the “Generation 0” row in the column labeled “Number of F Alleles;” white beans in the column “Number of f Alleles.” Calculate the frequency of each allele by dividing the number of each allele by the total number of alleles. Then, place all beans (alleles) back in the bag and shake up (mate) the rabbits.
4. Without looking at the beans, select two at a time, and record the results on the data
table next to “Generation 1.” For instance, if you draw one red and one white bean, place a mark in the chart under “Number of Ff individuals.” Continue drawing pairs of beans and recording the results in your chart until all beans have been selected and sorted. Place the diploid “rabbits” into the appropriate cup: FF, Ff, or ff. (Remember that the total number of individuals will be half the total number of beans because each rabbit requires two alleles.)
5. The ff bunnies are born furless. They cannot survive to adulthood, so they die before
they get a chance to mate. Therefore, take all the beans out of the ff cup and place them aside (but remember where you put them! All beans must be returned to the TA to receive a grade for the lab).
6. Now count the F and f alleles (beans) that remain in each of the surviving bunny cups
(FF and Ff) and (i.e., you are counting the alleles of the surviving bunnies. These alleles will contribute to the next generation.) Enter the number of each allele and their frequencies in the right side of the table.
7. Place the alleles of the surviving rabbits (which have grown, survived and reached
reproductive age) back into the bag and mate them again to get the next generation.
8. Repeat steps four through seven to obtain generations two through five.
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Record your data here:
Parents Alleles contributing to next generation Gen. Number of
FF individuals
Number of Ff individuals
Number of ff individua ls
Number of F alleles
Number of f alleles
Frequen cy of F
Frequen cy of f
0
1
2
3
4
5
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Discussion questions: 1. Prepare a graph with allele frequency as a function of time (generation number) and
graph your allele frequencies. (Allele frequency is plotted along the y-axis and time [generation] is plotted on the x-axis.) Frequency should be represented in decimals. Plot all frequencies on one graph. Use a solid line for F and a dashed line for f. Draw the graph of the change in F on the white-board for your table. Clearly label the axes and note the starting frequency of F (in Generation 0) for your group. Compare the frequencies of the alleles in the 5th generation to the starting generation. Would you say this particular bunny population has evolved (i.e. allele frequency has changed more than 1%) over 5 generations? Why or why not?
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2. Look at the graphs of the frequency of F over time for other pairs in your workshop group. Did any pairs have the same (or very similar) starting allele frequencies as you? In generation 5, are the allele frequencies the same between these groups? Explain why this might be.
3. Now compare the rates of change in the frequency of F across groups. In particular, look at your graph compared to a group with different starting allele frequencies. During the first 3 generations, does F change at the same rate for each group? Describe what you see and explain why this might be.
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The Hardy-Weinberg (HW) Principle Evolution is the change in allele frequency over time. In the absence of evolution, allele frequencies would stay the same in a population from generation to generation. The Hardy-Weinberg principle (HW) allows us to predict the genotype frequencies in the next generation of a population from the allele frequencies, provided that evolution is not occurring. In this way, HW provides a null hypothesis for evolution. If a population’s genotypic frequencies can be predicted by HW, then it is in HW equilibrium and evolution is not occurring. If, however, a population is not in HW equilibrium then evolution is occurring. HW principle:
• p = frequency of dominant allele (you have already calculated these values) • q = frequency of recessive allele (you have already calculated these values)
• p2 = predicted genotype frequency of homozygous dominant individuals (ex-AA) • 2pq = predicted genotype frequency of heterozygous individuals (ex-Aa) • q2 = predicted genotype frequency of homozygous recessive individuals (ex-aa)
• p2 + 2pq + q2 = 1
4. Calculate the expected (predicted) Hardy-Weinberg genotypic frequencies in the
starting generation for all three possible genotypes. (p = frequency of allele F, q = frequency of allele f. Hardy-Weinberg expected genotypes are FF = p2 Ff = 2pq ff = q2).
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5. Chi-square (?2) test: a) Now calculate the observed genotype frequencies in the last generation (i.e., what
percentage of diploid individuals were FF, Ff, and ff?)
b) Now you need to compare the observed genotype frequencies to the expected HW frequencies we will use a Chi-square test. Chi-square provides a statistical test for the hypothesis that the observed genotype frequencies are different than the expected HW genotype frequencies. First, use the expected frequencies you calculated in question 4 and the observed number of bunnies you had in generation 5 to figure out how many bunnies of each genotype you expected see to see in generation 5.
Expected # of FF bunnies = p2 x (total # of adult bunnies in generation 5) =
Expected # of Ff bunnies = 2pq x (total # of adult bunnies in generation 5) =
Expected # of ff bunnies = q2 x (total # of adult bunnies in generation 5) =
For Chi-square analysis, you will need to enter count data in the table below. For example, if you found an expected allele frequency of 0.60 or 60% for allele F, and you had a total of 50 adult bunnies in generation 5, your expected number of FF bunnies in generation 5 would be 18, because 0.602 x 50 = 18. Do this for each row of the “Expected” column. Next, fill out the “Observed” column with the actual genotype counts you saw in that final generation of bunnies – the actual numbers of FF, Ff and ff bunnies that were produced and survived to adulthood in generation 5. Note that the sum of your “Expected” column should be the same as the sum of your “Observed” column – if it isn’t, something is probably wrong with how you filled in your “Expected” column. Use these calculated count data to solve the equation (O-E)2 /E for each row, and sum those values to produce your Chi-square value.
Observed (O) Expected (E) (O-E)2/E
FF
Ff
ff
?2=
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c) Using the Chi-square table below, with 2 degrees of freedom (note- think about why you are using 2 degrees of freedom), what is the probability that you would observe the above outcome by random chance?
Degrees of Freedom
.99 .80 .50 .20 .10 .05 .01 .001
1 .00016 .064 .46 1.6 2.7 3.8 6.6 10.8 2 .20 .45 1.4 3.2 4.6 6.0 9.2 13.8 3 .12 1.0 2.4 4.6 6.3 7.8 11.3 16.3 4 .30 1.6 3.4 6.0 7.8 9.5 13.3 18.5
d) Remember that we learned last time that results are significant if the probability (p-
value) is less than 0.05. If the null hypothesis is that there was no effect of selection on genotype frequencies, do you accept or reject your null hypothesis? In other words, did natural selection significantly alter genotype frequencies in this scenario? Explain.
Literature cited: 1. Whitlock, M.C. and Schluter, D. 2009. The Analysis of Biological Data. Roberts and Co. Publishers. 2. Valenzano D. R., Kirschner J., Kamber R. A., Zhang E., Weber D., et al. , 2009. Mapping loci associated with tail color and sex determination in the short-lived fish Nothobranchius furzeri. Genetics 183: 1385–1395.
