Three-Point Test-Cross and Interference

I = 0.6 D ↔ E = 15 cM = 0.15 recombination frequency. E ↔ F = 25 cM = 0.25 recombination frequency. Expected DCO frequency = 0.15 x 0.25 = 0.0375 Expected DCO = 0.0375 x 726 = 27.225 The problem states that 11 double crossovers were observed in the test cross. c = observed DCO (11) / expected DCO (27.22) = 0.404 I = 1-C = 1 - 0.404 = 0.6

A three-point test cross revealed a total of 11 double crossovers among a total of 726 test-cross progeny. Using the genetic map provided, calculate the coefficient of Interference (I). I = 0.0375 I = 0.015 I = 0.4 I = 0.6 I = 1

20 cM total recombinants 49 + 46 + 2 + 97 +3 + 98 = 295 total offspring 353 + 352 + 295 = 1000 Recombination frequency = 295 / 1000 = 0.295 Map distance between A and B 0.295 x 100 = 29.5cm

Based on the data provided, what is the distance between genes A and B? PhenotypeOffspringPhenotypeOffspring A B C 353 A B c 49 a b c 352 a b C 46 a B C 2 A b C 97 A b c 3 a B c 98 0.02 cM 0.2 cM 2 cM 20 cM 200 cM

10cm total recombinants 49 + 46 + 2 + 3 = 100 total offspring 353 + 352 + 295 = 1000 Recombination frequency = 100 / 1000 = 0.10 Map distance between C and A 0.10 x 100 = 10 cm

Based on the data provided, what is the distance between genes C and A? PhenotypeOffspringPhenotypeOffspring A B C 353 A B c 49 a b c 352 a b C 46 a B C 2 . A b C 97 A b c 3 a B c 98 0.01 cM 0.1 cM 1 cM 10 cM 100 cM

3.75% Double Crossover Frequency = 0.15 x 0.25 = 0.0375 0.0375 x 100 = 3.75%

Given the genetic map of genes D, E and F. What is the expected frequency (probability) of double crossovers? 1.5% 2.0% 3.75% 5% 7.5%

Map B 33 + 37 / 462 + 468 +33 +37 = 0.07

Rr Tt x rr tt Testcross Phenotype Counts R T 462 r t 468 R t 33 r T 37 Which map is correct for this data?