IMED2004 Study Guide - Final Guide: Mendelian Inheritance
Document Summary
In some crosses, genes are obviously linked as clearly more nr progeny than. Other crosses, the difference is not so obvious . 54 aabb; 56 aabb; 42 aabb and 48 aabb. Can test using chi-square test evaluate likelihood that chance alone is responsible for. Deviations between number of progeny observed and numbers expected. Yes then genes assorting independently; no, then genes linked. Calculate expected probability of each progeny type and evaluate whether observed numbers deviate significantly from expected numbers: With independent assortment we expect each phenotype: Aabb, aabb, aabb, aabb. This expected probability of each genotype is based on multiplication rule of probability: If probability of aa is and the probability of bb is , then probability of aabb = x = . In calculation we are making two assumptions: Probability of each single-locus genotype is . Genotypes at the two loci are inherited independently ie. x . Determining the probability that the difference between observed and.
