BIOL10003 Lecture Notes - Lecture 26: Null Hypothesis, Contingency Table, Zygosity

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Lecture 26 - autosomal linkage mapping 3 markers. Finding the number of genes divide total number of offspring in the f2 by 4 if the resultant number is related to the phenotype ratios then it is one gene. If not, divide by 16, if number relates to the phenotype = 2 genes. Test cross heterozygous recessive male (no recombination in drosophila) Re(cid:272)o(cid:373)(cid:271)i(cid:374)a(cid:374)ts ge(cid:374)ot(cid:455)pes (cid:272)ha(cid:374)ge depe(cid:374)di(cid:374)g o(cid:374) (cid:272)is or tra(cid:374)s arra(cid:374)ge(cid:373)e(cid:374)t of pare(cid:374)t(cid:859)s chromosomes. Non recombinants (i. e. same genotype as parents) = highest proportion. Double recombinants (only middle locus swapped) = lowest proportions. A positive value for interference - observed drs are less than the expected drs. A crossover in one region inhibits a crossover in the adjacent region. A negative value for interference - observed drs are more than the expected drs. A crossover in one region promotes a crossover in the adjacent region.

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