Midterm I Answer Key

# Midterm I Answer Key - Name 1 ID a b 1:1:1:1 =(164-150...

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Name: ______________________________ ID: _______________________________ 1 1. a. 1:1:1:1 b. ± 2 = (164-150 ) 2 + (172-150) 2 + (130-150) 2 + (134-150) 2 = 150 150 150 150 1.31 + 3.23 + 2.67 + 1.71 = 8.92 degrees of freedom = 3 c. The genes are unlinked. d. Reject; the genes appear to be linked. X 2 > expected for unlinked genes, i.e. P < 0.05. e. Linked. Distance between calculates to 44 map units. total recombinants/total progeny = 264/600 X 100 = 44 % 2. a. 119+86+10+15 = 230 = 23 x 100% = 23% 1000 1000 b. 63+82+10+15 = 170 = 0.17 x 100% = 17% 1000 1000 c. Add 23% and 17% to get 40%. Or, calculate recombination from the crossovers. Add the double crossovers twice to get the correct answer. 119+86+63+82+10+15+10+15 = 400 = 0.40 x 100% = 40% 1000 1000 d. cn ------------23m.u.------------------c---------17m.u.-----------px e. Yes. Observed double crossovers 25 Expected double crossovers 0.23 x 0.17 x 1000 = 39.1 Coefficient of coincidence 25/39.1 = 0.64 Interference = 1 – 0.64 = 0.36

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Midterm I Answer Key - Name 1 ID a b 1:1:1:1 =(164-150...

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