Mapping populations Controlled crosses between two parents –two alleles/locus, gene frequencies = 0.5 –gametic phase disequilibrium is due to linkage, not other causes Examples –Backcross (BC 1 or BC 2 ) –F 2 or F 2:3 –Recombinant inbred lines (RIL) –Doubled haploid (DH)

Recombinant Inbred Lines (RILs)

GenerationAAAaaa F10100%0 F225%50%25% F337.5%25%37.5% F443.75%12.5%43.75% F %6.25%46.875% F %3.125% % F1049.9%0.2%49.9% A (1/2) a (1/2) A (1/2) AA (1/4) Aa (1/4) a (1/2) aA (1/4) Aa (1/4) ♀ ♂

expected frequency r = 0r = Recombinant Inbred Lines (RILs)

R R R R RILs

Doubled Haploids

expected frequency r = 0r = Doubled Haploids (DHs)

DOUBLED HAPLOIDS RR R R RR R R R R

F2 Population

Expected Genotypic Frequencies for F 2 Progeny when r = 0 or r = 0.5 Between Two Loci in Coupling (AB/ab) Configuration GenotypeExpected Frequency r = 0r = 0.5 AB/ABp 1 = 0.25(1 - r) 2 1/4 = 0.251/16 = AB/aBp 2 = 0.50r(1 - r) 0.02/16 = AB/Abp 3 = 0.50r(1 - r) 0.02/16 = AB/abp 4 = 0.50(1 - r) 2 1/2 = 0.52/16 = Ab/aBp 5 = 0.50r /16 = Ab/Abp 6 = 0.25r /16 = Ab/abp 7 = 0.50r(1 - r) 0.02/16 = aB/aBp 8 = 0.25r /16 = aB/abp 9 = 0.50r(1 - r) 0.02/16 = ab/abp 10 = 0.25(1 - r) 2 1/4 = 0.251/16 =

Expected and Observed Genotypic Frequencies Coupling (AB/ab) and Repulsion (Ab/aB) F 2 Progeny GenotypeObserved Frequency CouplingRepulsion AB/ABp1p1 p 1 = 0.25(1 - r) 2 p 1 = 0.25r 2 AB/aBp2p2 p 2 = 0.50r(1 - r) AB/Abp3p3 p 3 = 0.50r(1 - r) AB/abp4p4 p 4 = 0.50(1 - r) 2 p 4 = 0.50r 2 Ab/aBp5p5 p 5 = 0.50r 2 p 5 = 0.50(1 – r) 2 Ab/Abp6p6 p 6 = 0.25r 2 p 6 = 0.25(1 – r) 2 Ab/abp7p7 p 7 = 0.50r(1 - r) aB/aBp8p8 p 8 = 0.25r 2 p 8 = 0.25(1 – r) 2 aB/abp9p9 p 9 = 0.50r(1 - r) ab/abp 10 p 10 = 0.25(1 - r) 2 p 10 = 0.25r 2 Co-dominant Fully classified double hets. Locus A = A and a Locus B = B and b r = recombination frequency between locus A and B

Expected and Observed Genotypic Frequencies Coupling (AB/ab) F 2 Progeny GenotypeObserved Frequency Coupling AB/ABq1q1 q 1 = 0.25(1 - r) 2 AB/aBq2q2 q 2 = 0.50r(1 - r) AB/Abq3q3 q 3 = 0.50r(1 - r) AB/ab + Ab/aBq4q4 q 4 = p 4 + p 5 = 0.50[(1 - r) 2 +r 2 ] Ab/Abq5q5 q 5 = 0.25r 2 Ab/abq6q6 q 6 = 0.50r(1 - r) aB/aBq7q7 q 7 = 0.25r 2 aB/abq8q8 q 8 = 0.50r(1 - r) ab/abq9q9 q 9 = 0.25(1 - r) 2 Co-dominant Unclassified double heterozygotes Locus A = A and a Locus B = B and b r = recombination frequency between locus A and B

Expected and Observed Genotypic Frequencies Coupling (AB/ab) and Repulsion (Ab/aB) F 2 Progeny GenotypeObserved Frequency CouplingRepulsion A_B_f1f1 f 1 = 0.25(3 - 2r + r 2 )f 1 = 0.25(2 + r 2 ) A_bbf2f2 f 2 = 0.25(2r – r 2 )f 2 = 0.25(1 – r 2 ) aaB_f3f3 f 3 = 0.25(2r – r 2 )f 3 = 0.25(1 – r 2 ) aabbf4f4 f 4 = 0.25(1 - r) 2 f 4 = 0.25r 2 Dominant Locus A = A and a Locus B = B and b r = recombination frequency between locus A and B

Analysis 1.Single-locus analysis 2.Two-locus analysis 3.Detecting linkage and grouping 4.Ordering loci 5.Multi-point analysis

Mendelian Genetic Analysis Phenotypic and Genotypic Distributions The expected segregation ratio of a gene is a function of the transmission probabilities If a gene produces a discrete phenotypic distribution, then an intrinsic hypothesis can be formulated to test whether the gene produces a phenotypic distribution consistent with a expected segregation ratio of the gene The heritability of a phenotypic trait that produces a Mendelian phenotypic distribution is ~1.0. Such traits are said to be fully penetrant The heritability of a DNA marker is theoretically ~1.0; however, it is affected by genotyping errors

Mendelian Genetic Analysis Hypothesis Tests The expected segregation ratio (null hypothesis) is specified on the basis of the observed phenotypic or genotypic distribution One-way tests are performed to test for normal segregation of individual phenotypic or DNA markers –If the observed segregation ratio does not fit the expected segregation ratio, then the null hypothesis is rejected. The expected segregation ratio is incorrect Selection may have operated on the locus The locus may not be fully penetrant A Type I error has been committed

Mendelian Genetic Analysis Hypothesis Tests Two-way tests are performed to test for independent assortment (null hypothesis - no linkage) between two phenotypic or DNA markers. –If two genes do not sort independently, then the null hypothesis is rejected The two genes are linked (r < 0.50) The expected segregation ratio is incorrect A Type I error has been committed.

Mendelian Genetics Analysis Null Hypothesis AcceptReject True No error 1 -  Type I error  False positive False Type II error  False negative No error 1 - 

One-way or single-locus tests   statistics Log likelihood ratio statistics (G-statistics) i = i th genotype (or allele, or phenotype) Pr[  2 >  2 df ] =  Pr[G >  2 df ] =  Goodness of fit statistics

One-way or single-locus tests GenotypeSample ASample BTotal aa Aa Total Two backcross populations (A and B) genotyped for a co-dominant marker (Brandt and Knapp 1993) Null hypothesis 1:1 ratio of aa to Aa Pr[G A >  2 k-1 ] = Pr[14.8 >  2 1 ] = Pr[G B >  2 k-1 ] = Pr[6.88 >  2 1 ] = Null hypothesis is rejected for both samples Individual G-statistics for samples A and B i = i th genotype k = 2 genotypic classes

One-way or single-locus tests GenotypeSample ASample BTotal aa Aa Total Two backcross populations (A and B) genotyped for a co-dominant marker (Brandt and Knapp 1993) Null hypothesis 1aa to 1Aa ratio for pooled samples Pr[G P >  2 k-1 ] = Pr[20.7 >  2 1 ] = Null hypothesis is rejected Pooled G-statistic across samples i = i th genotype j = j th sample k = genotypic classes p = No. of samples (populations)

One-way or single-locus tests GenotypeSample ASample BTotal aa Aa Total Two backcross populations (A and B) genotyped for a co- dominant marker (Brandt and Knapp 1993) Null hypothesis Samples A and B are homogenous Pr[G H >  2 (k-1)(p-1) ] = Pr[0.94 >  2 1 ] = 0.33 (N.S.) The heterogeneity G-statistic is i = i th genotype j = j th sample (population) k = genotypic classes p = No. of samples (populations) n = Total No. of observations

One-way or single-locus tests Pr[G T >  2 p(k-1) ] = Pr[21.7 >  2 2 ] = SourceGdfPr > G Sample A14.7k-1 = 2-1 = Sample B6.9k-1 = 2-1 = Total21.6p(k-1) = 2(2-1) = Pooled20.7k-1 = 2-1 = Heterogeneity0.9(k-1)(p-1) = (2-1)(2-1) = Total21.6p(k-1) = 2(2-1) = Relationship between G statistics k = genotypic classes p = No. of samples (populations)

One-way or single-locus tests Allelic constitutionGenotypeObservedExpected 120bp /120bpaa bp /124bpAa bp /124bpAA Total94 F 2 progeny of Ae. cylindrica genotyped for the SSR marker barc98. Null hypothesis 1:2:1 ratio of aa:Aa:AA Pr[G >  2 k-1 ] = Pr[1.67 >  2 2 ] = Null hypothesis is not rejected Individual G-statistics for samples A and B i = i th genotype k = 3 genotypic classes

Calculating probability values for Chi-square distributions SAS program data pv; Input x df; datalines; ; data pvalue; set pv; pvalue = 1 – probchi (x, df); output; proc print; run; Output Obs x df pvalue Excel formula =CHIDIST(x, degrees_fredom) =CHIDIST(3.75, 2) Output