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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).

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mapping populations
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 (BC1 or BC2)
    • F2 or F2:3
    • Recombinant inbred lines (RIL)
    • Doubled haploid (DH)
slide4

slide6

RILs

R

R

R

R

slide8

Doubled Haploids (DHs)

expected

frequency

slide12
Expected Genotypic Frequencies for F2 Progeny when r = 0 or r = 0.5 Between Two Loci in Coupling (AB/ab) Configuration
expected and observed genotypic frequencies coupling ab ab and repulsion ab ab f 2 progeny
Expected and Observed Genotypic FrequenciesCoupling (AB/ab) and Repulsion (Ab/aB) F2 Progeny
  • 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
Expected and Observed Genotypic FrequenciesCoupling (AB/ab) F2 Progeny
  • 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 progeny1
Expected and Observed Genotypic FrequenciesCoupling (AB/ab) and Repulsion (Ab/aB) F2 Progeny
  • Dominant
  • Locus A = A and a
  • Locus B = B and b
  • r = recombination frequency between locus A and B
analysis
Analysis
  • Single-locus analysis
  • Two-locus analysis
  • Detecting linkage and grouping
  • Ordering loci
  • Multi-point analysis
mendelian genetic 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 analysis1
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 analysis2
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.
one way or single locus tests
One-way or single-locus tests

Goodness of fit statistics

  • C2statistics
  • Log likelihood ratio statistics
  • (G-statistics)

Pr[C2 > 2df] = 

Pr[G > 2df] = 

i = ith genotype (or allele, or phenotype)

one way or single locus tests1
One-way or single-locus tests

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

i = ith genotype

k = 2 genotypic classes

Individual G-statistics for samples A and B

Pr[GA > 2k-1] =

Pr[14.8 > 21] = 0.0001

Pr[GB > 2k-1] =

Pr[6.88 > 21] = 0.0086

Null hypothesis is rejected for both samples

one way or single locus tests2
One-way or single-locus tests

Null hypothesis

1aa to 1Aa ratio for

pooled samples

Two backcross populations (A and B) genotyped for a co-dominant marker (Brandt and Knapp 1993)

i = ith genotype j = jth sample

k = genotypic classes

p = No. of samples (populations)

Pooled G-statistic across samples

Pr[GP > 2k-1] = Pr[20.7 > 21] = 0.0000054

Null hypothesis is rejected

one way or single locus tests3
One-way or single-locus tests

Two backcross populations (A and B) genotyped for a co-dominant marker (Brandt and Knapp 1993)

Null hypothesis

Samples A and B are homogenous

i = ith genotype

j = jth sample (population)

k = genotypic classes

p = No. of samples (populations)

n = Total No. of observations

The heterogeneity G-statistic is

Pr[GH > 2(k-1)(p-1)] = Pr[0.94 > 21] = 0.33 (N.S.)

one way or single locus tests4
One-way or single-locus tests

Relationship between G statistics

Pr[GT > 2p(k-1)] = Pr[21.7 > 22] = 0.00002

k = genotypic classes

p = No. of samples (populations)

one way or single locus tests5
One-way or single-locus tests

F2 progeny of Ae. cylindrica genotyped for the SSR marker barc98.

Null hypothesis

1:2:1 ratio of aa:Aa:AA

i = ith genotype

k = 3 genotypic classes

Individual G-statistics for samples A and B

Pr[G > 2k-1] = Pr[1.67 > 22] = 0.434

Null hypothesis is not rejected

calculating probability values for chi square distributions
Calculating probability values for Chi-square distributions

SAS program

data pv;

Input x df;

datalines;

3.75 2

;

data pvalue;

set pv;

pvalue = 1 – probchi (x, df);

output;

proc print;

run;

Output

Obs x df pvalue

1 3.75 2 0.15335

Excel formula

=CHIDIST(x , degrees_fredom)

=CHIDIST(3.75 , 2)

Output

0.15335