BIOL2400 Final Exam Questions With Correct Answers Latest 2024 (GRADED)

BIOL2400 Final Exam Questions With Correct Answers Latest 2024 (GRADED) Estimate the frequency of the A1 and A2 alleles (separately) for each population and write it in the last two columns. a) If we use the formula on the equation sheet that uses the number of alleles (which is double the number of perch): p Eramosa = (2*125+550)/2000 = 0.4 q Eramosa = 1-p = 0.6 p Speed River = (2*270+500)/2000 = 0.52 q Speed River = 1-p = 0.48 [b] Are these populations approximately in Hardy-Weinberg equilibrium? If they are not in HWE then which assumptions are likely being violated? For Eramosa under HWE expected genotype frequencies would be: p 2 2pq q 2 (0.4) 2 2(0.4)(0.6) (0.6) 2 0.16 0.48 0.36 We multiply these by the number of fish 1000 to get the number expected of each genotype: For the Eramosa River*: Chi squared, χ2 = 21.267 with 2 degrees of freedom. The two-tailed P value 0.0001. Very highly significant! Could there be selection for heterozygotes for this locus in the Eramosa River? Genotype CC CG GG Observed numbers Expected number 160.0 480.0 360.0 For the Speed River: Chi squared, χ2 = equals 0.003 with 2 degrees of freedom. The two-tailed P value equals 0.99870 with 2 degrees of freedom. Not significant. Genotype CC CG GG Observed numbers Expected number 270.4 499.2 230.4 For Guelph Lake: P value and statistical significance: Chi squared, χ2 = 40.000 with 2 degrees of freedom. The two-tailed P value is less than 0.0001. Very highly significant! Could there be inbreeding in the Guelph Lake population? Genotype CC CG GG Observed numbers Expected numbers 562.5 375.0 62.5 *Chi square calculator at: [c] Do you think the perch are swimming between the Eramosa and Speed Rivers and Guelph Lake? Why or why not? No. If there were high amounts of migration between the two rivers and between the Speed River and Guelph Lake then the allele frequencies would be very similar at all three locations. 2) [10 marks] You are studying a population of deer mice Peromyscus maniculus in Algonquin Park. You are interested in whether there is a heritable component to body weight and to running speed for a sprint of 30 metres. You trap wild mice for two generations and estimate that for this population: Table 1. Phenotypic statistics and heritability for a deer mouse population. Trait mean variance h 2 N body weight (g) 30.0 10.7 0.37a 400 Sprint speed (ms-1) 10 4.0 0.61 a 400 a Narrow sense heritability from the slope of a linear regression with the average offspring value as the y value and the average value for their parents as the x value. a) Which trait has the larger phenotypic variance? b) Calculate which trait has a larger additive genetic variance. (Assume that the phenotypic variances are similar in the laboratory and in the field and that the trait has the same genetic basis in both environments). Recall: h2 = VA / V P For body weight: From Table: mean body weight = 30 g, phenotypic variance, V P, = 10.7, h2 = 0.37 Therefore : VA = 3.959 For sprint speed mean sprint speed = 10 ms-1 , V P = 4.0, h2 = 0.61 Therefore: VA = 2.44 Body weight has a larger additive (=heritable) genetic variance than sprint speed. c) Your summer’s trapping data shows that there was intense selection for decreased body size (S, the selection differential is −1.271) likely because of changes in predator abundance. Use this estimate of S and the data from the Table 1 above to predict what the mean body size would be after one generation of directional selection. Recall: R = h2 S. Note that the selection differential, S = 1.271 − (It is negative because the mice are getting lighter). After one generation: R = (0.37)( 1.271) − R = 0.47 − g per generation Also recall that: