LSCI 204 Study Guide - Final Guide: Factor Analysis, Centimorgan, N-Terminus
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Problem 1: we only need to look at markers c and d. 217 + 2 + 214 + 2 = 435 c + + d c d 10 + 28 + 14 + 29 = 81. = 15. 6 cm: the two true-breeding lines were crossed to give the f1 females that were used in this test cross to homozygous recessive males to give the data that we see. By analyzing the data to find what markers are linked and unlinked, we can determine the true-breeding parental phenotypes. To determine linkage we need to analyze the markers pairwise: a and b ab a+ All classes are represented equally, so conclude that a and b are unlinked: a and c. Two factor analysis shows that a and c are unlinked: a and d. Two factor analysis shows that a and d are unlinked. 10 + 217 + 14 + 214 = 455 parental.