Lab4_Worksheet_BIO120Fall2021-topost

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Feb 20, 2024

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Lab 4: Population Genetics Worksheet What you will need to complete Lab 4: This Worksheet, Chapter 4 and Appendix A (pages 15 to 25) of the BIO120 Lab Manual (PDFs on Quercus) A way to record your data results (either in this Worksheet or on printed paper) Your Lab 4 Genotype Card : download your assigned Genotype Card from the “Lab 4: Worksheet and Genotype Cards” page on Quercus before lab. Use your Species Number (posted under Grades) to determine which Genotype Card you should download. See Quercus for more details. Note: There are two graded assessments due by the start time of Lab 4: Proposal Lab 4 Quiz Case 1: Random Mating, No Selection 1. Obtain your fish’s genotype from your Genotype Card (initial frequencies: approx. 25% AA, 50% Aa, 25% aa) 2. Your TA will split you into randomly assigned pairs. 3. Flip a coin* (heads = Allele 1 , tails = Allele 2 ) and pass this allele on to your fish’s offspring. Your partner’s fish will contribute the other allele. 4. Produce two offspring (to keep population size consistent) (i.e., complete Step 3 two times). 5. Your parent fish has now died! One partner takes on the genotype of one offspring, and the other partner takes the other offspring’s genotype. 6. Press the Leave button in the Zoom toolbar to return to the Main room 7. Your TA will move everyone to a new random pairing and you will repeat the process for five generations. Record your genotypes in Table 4-1. * If you do not have a physical coin, you can use a virtual coin by searching “coin flip” in Google BIO120 Fall 2021 Page 1
Table 4-1 . Data table for Case 1: Random Mating, No Selection Your initial genotype: AA Generation Offspring 1 Offspring 2 Your genotype for the next generation F 1 Aa Aa Aa F 2 Aa Aa Aa F 3 Aa Aa Aa F 4 Aa Aa Aa F 5 Aa aa Aa Your final genotype: Aa Table 4-2 . Class result for Case 1: Random Mating, No Selection Sum of entire population (class) F 5 offspring: AA: 4 Aa: 10 aa: 5 Genotype frequencies: AA: 0.21 Aa: 0.53 aa: 0.26 Allele frequencies: A: 0.47 a: 0.53 Table 4-3 . Calculation of expected genotypes for Case 1 Genotype AA Aa aa Hardy-Weinberg p 2 2pq q 2 Expected genotype frequency 0.22 0.50 0.28 Expected number of individuals 4.26 9.47 5.26 BIO120 Fall 2021 Page 2
Table 4-4 . Chi-square test for Case 1 Genotype AA homozygotes Aa heterozygotes aa homozygotes Total Observed ( O ) 4 10 5 19 Expected ( E ) 4.26 9.47 5.26 19 O E -0.26 0.53 0.26 (O E) 2 0.0676 0.2809 0.0676 (O E) 2 / E 0.0159 0.0297 0.0159 2 = 0.06 Null hypothesis: The fish population is at Hardy-Weinberg equilibrium Degrees of freedom ( df ) = 1 At P = 0.05 and df = 1, the critical 2 value is 3.84 Do we reject or not reject the null hypothesis? We do not reject the null hypothesis. Is this fish population in Hardy-Weinberg equilibrium at this locus? Yes Is this the result we expect? Explain why or why not. Yes BIO120 Fall 2021 Page 3
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Case 2: Selection For Gene 2, there is 100% negative selection against the homozygous recessive genotype ( bb ). Table 4-5 . Results for Case 2: Selection Sum of entire population: BB: 1645 Bb: 355 bb: 0 Genotype frequencies: BB: 0.8225 Bb: 0.1775 bb: 0 Allele frequencies: B: 0.91125 b: 0.08875 Table 4-6 . Calculation of expected genotypes for Case 2 Genotype BB Bb bb Hardy-Weinberg p 2 2pq q 2 Expected genotype frequency 0.83 0.16 0.01 Expected number of individuals 1660.75 323.49 15.75 Table 4-7 . Chi-square test for Case 2 Genotype BB homozygotes Bb heterozygotes bb homozygotes Total Observed ( O ) 1645 355 0 2000 Expected ( E ) 1660.75 323.49 15.75 2000 O E -15.75 31.51 -15.75 (O E) 2 248.16 992.64 248.16 (O E) 2 / E 0.15 3.07 15.75 2 = 18,97 BIO120 Fall 2021 Page 4
Null hypothesis: The gene 2 of the lab fish population is at H-W equilibrium. Degrees of freedom ( df ) = 1 At P = 0.05 and df = 1, the critical 2 value is 3.84 Do we reject or not reject the null hypothesis? We reject the null hypothesis. Is this fish population in Hardy-Weinberg equilibrium at this locus? No Is this the result we expect? Explain why or why not. Yes because bb under heavy selection and will die. _____________________________________________________________________________________ Case 3: Genetic drift You can use the dice simulation on Google by searching “roll dice”. Then click multiple times on the pink 10-sided dice on the bottom, until you can see 10 die in the window. (To delete a dice, click on it when it is in the window.) BIO120 Fall 2021 If you need to delete a dice, click on it when it is in this window Click on the 10-sided die here to add 10 die to the window above Page 5
Table 4-8 . Data table for Case 3: Genetic Drift Gen Frequency of A allele # of A alleles # of a alleles Gen Frequency of A allele # of A alleles # of a alleles Parent 0.5 1 to 5 6 to 10 11 0.7 1 to 7 8 to 10 1 0.6 1 to 6 7 to 10 12 0.2 1 to 2 3 to 10 2 0.3 1 to 3 4 to 10 13 0.5 1 to 5 6 to 10 3 0.2 1 to 2 3 to 10 14 0.5 1 to 5 6 to 10 4 0.4 1 to 4 5 to 10 15 0.5 1 to 5 6 to 10 5 0.3 1 to 3 4 to 10 16 0.3 1 to 3 4 to 10 6 0.5 1 to 5 6 to 10 17 0.2 1 to 2 3 to 10 7 0.4 1 to 4 5 to 10 18 0.1 1 2 to 10 8 0.7 1 to 7 8 to 10 19 0.1 1 2 to 10 9 0.8 1 to 8 9 to 10 20 0 10 0.9 1 to 9 10 21 Table 4-9 . Class data for Case 3: Genetic Drift Enter in each cell below (one per student) the number of generations it took for one of the alleles to disappear 7 5 11 8 4 8 7 8 7 7 27 20 23 Mean = Do you predict that the mean number of generations it takes for an allele to disappear would be higher or lower if the population was larger (e.g., if you rolled 100 dice)? BIO120 Fall 2021 Page 6
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