Lecture 4 Guide

Lecture 4 Guide - Mendel’s breeding program that produced...

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Unformatted text preview: Mendel’s breeding program that produced a 9 : 3 : 3 : 1 ratio Punnett square illustrating the genotypes underlying a 9 : 3 : 3 : 1 ratio Independent assortment of chromosomes at meiosis explains Mendel’s ratio Independent assortment of chromosomes at meiosis explains Mendel’s ratio Independent assortment of chromosomes at meiosis explains Mendel’s ratio Stages of a Neurospora cross The linear meiosis of Neurospora Recombinants are meiotic output different from meiotic input In diploids, recombinants are best detected in a testcross Independent assortment produces 50 percent recombinants Linked alleles tend to be inherited together Crossing over produces new allelic combinations Chiasmata are the sites of crossing over Recombinants are produced by crossovers For linked genes, recombinant frequencies are less than 50 percent Hypothetical experiment: 1. Cross AB X ab ab ab 2. unlinked, expect Are the A and B genes linked? 1AB/ab : 1ab/ab : 1 Ab/ab : 1aB/ab 3. null hypothesis: the genes are not linked 100 progeny expect: 25 AB/ab – parental = 50 parental 25 ab/ab -- parental 25 aB/ab -- recomb = 50 recomb 25 Ab/ab -- recomb observe: Experiment 34 AB/ab – parental 30 ab/ab -- parental 15 aB/ab -- recomb 21 Ab/ab -- recomb 4. chi-square 2 Statistic to compare expected (exp) to observed (obs) actual result 2 = [(obs – exp)2/exp]parental + [(obs – exp)2/exp]recombinant 2 = [(34 – 25)2/25] + [(30 – 25)2/25] + [(15 – 25)2/25] + [(21 – 25)2/25] = 8.88 Degrees of freedom (Df) = # of classes – 1 Df = 4 – 1 = 3 Look up chi square value in Table, for the Degrees of freedom Chi square = 8.88 Df = 3 Critical Chi Square Values Cannot Reject Null Hypothesis 0.99 0.90 0.50 0.10 Null Hypothesis Rejected 0.05 0.01 0.001 p Values Degrees of Freedom 1 2 3 4 5 X2 Calculations – 0.02 0.11 0.30 0.55 0.02 0.21 0.58 1.06 1.61 0.45 1.39 2.37 3.36 4.35 2.71 4.61 6.25 7.78 9.24 3.84 5.99 7.81 9.49 11.07 6.64 9.21 11.35 13.28 15.09 10.83 13.82 16.27 18.47 20.52 P value is < 0.5 Therefore null hypothesis, i.e unlinked, is rejected. Conclusion is that A and B appear to be linked (not proven). 5. Recombination frequency… map distance…. between A and B 36 recombinants = 0.36 100 total progeny x 100% = 36 map units (mu) Map distances are generally additive Longer regions have more crossovers and thus higher recombinant frequencies Different gene orders give different double recombinants Mapping with a 3 point cross Example F1 test cross expect VBP/vbp VBP/vbp x vbp/vbp 1 VBP/vbp : 1 vbp/vbp progeny observed genotype VBP vbp Vbp vBP VbP vBp vbP VBp number 1779 1654 252 241 131 118 13 9 4197 class parental “ single crossover “ “ “ double crossover “ Double crossovers – the one that changes from parental – tells you which gene is in the middle. In this example, P is in the middle. gene order : V – P – B (or B – P – V) Calculating map units V to P recombination events involving V single cross-over : double cross-over : Vpb (252) vPb (13) vPB (241) VpB (9) 252 + 241 + 13 + 9 x 100% = 12.3% = 12.3 map units 4197 Calculating map units single cross-over : double cross-over : P to B VPb (131) vpB (118) vPb (13) VpB (9) 131 + 118 + 13 + 9 x 100 % = 6.4% = 6.4 map units 4197 Result: V-------- 12.3 mu-------P-------6.4 mu----------B Phenotypic ratios in progeny reveal the type of cross An RFLP linked to a mouse disease gene An RFLP linked to a mouse disease gene An RFLP linked to a mouse disease gene Some bands of a DNA fingerprint show linkage A microsatellite locus can show linkage to a disease gene ...
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This note was uploaded on 09/26/2009 for the course BIS 98659 taught by Professor Kimbrell during the Summer '09 term at UC Davis.

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