7 Linkage&ChiSquare
Transcript of 7 Linkage&ChiSquare
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Linkage & Mapping in
Eukaryotes
Part I
Linkage & Chi square for linkage
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These haploid cells contain a
combination of alleles found
in the original chromosomes
Non-recombinant meiotic cells
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These are
termed parental
ornon-
recombinant
cellsThese haploid cells contain
a combination of alleles
NOT found in the original
chromosomes
This new combination
of alleles is a result ofgenetic recombination
These are termed non-
parental orrecombinant
cells
Recombinant meiotic cells
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Amuch greater proportion
of the two types found inthe parental generation
Bateson and Punnett
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Linked Genes In Drosophila
Morgans experiment:prvs.pr+
(purple vs. redeyes) and vgvs vg+ (vestigial wings vs. normalwings)
Theprand vg loci are located on an autosomeautosome
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F1pr+vg+ /prvg Female X prvg / prvgMale
P0:prvg /prvg X pr+ vg+ /pr+ vg+
prvg
pr+ vg+ pr+ vg+/ prvg
prvg / prvgprvg
Red normal 1339
Purple vestigial 1195
Red vestigial 151
Purple normal 154
Instead inInstead in F2F2::
89.3% parental chromosomes
10.7% recombinant chromosomes
Expected from test cross:
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Frequency OfRecombination
The 10.7% recombinant chromosomes (non-
parental) called frequency of recombinationfrequency of recombination
Results from crossingcrossing--overoverof the chromosomesin meiosis in the F1 femalesF1 females in flies
The recombination frequency can range from 0%(complete linkagecomplete linkage) to 50% (unlinkedunlinked)
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Frequency OfRecombination
Genes with recombination frequencies of< 50%< 50%are present in the same chromosome (linkedlinked).
Two genes that undergo independentindependentassortmentassortment, indicated by a recombinationfrequency }} 50%50%,, either are on nonnon--homologoushomologouschromosomes (unlinked) orfar apartfar apart on the samechromosome.
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=w+ / + m
transtrans orrepulsionrepulsion
configuration
=wm / + +
ciscis orcouplingcoupling
configuration
Two Configuration For Heterozygous
Linked Genes
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Morgan used wand m in the transtrans configuration
He also looked at the cis configuration
Parents: pr+ vg /pr+ vg X prvg+ /prvg+
F2Red eyes vestigial wings 1067
Purple eyes normal wings 965
Red eyes normal wings 157
Purple eyes vestigial wings 146
87.0%
13.0%
F1pr+vg /prvg+ Female X prvg / prvgMale
Linked Genes Behave the Same in
Cis orTrans Configurations
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The Chi-Square Test for Linkage
If we want to determine
whether 2 genes are linked:
25% each class = unlinked
< 25% for 2 classes = linked
Can use a chi-square to test
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AN EXAMPLE
2 genes R & S - performed cross and scored progeny
Phenotype data:
y Phenotypic classes in progeny
y 395RS, 382 rs, 223 rS, 247 Rs
What are the parents of this cross?
Recombination frequency = 470/1247 = 0.377
RF= 0.377 between R and S genes
Does RF= 0.377 fits the hypothesis that it is the same
as RF= 0.5? 13
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1. Hypothesis:1. Hypothesis:
No linkageNo linkage = Our data represents a 1:1:1:1 ratio and thus
R and S are unlinked.
Why do we use no linkage as a hypothesis????
2. Prediction = 311.75 of each class (total =1247; for
each class predicted if it is a 1:1:1:1)
AN EXAMPLE: 7 steps to G2
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AN EXAMPLE: 7 steps to G2
3.3. GG22 AnalysisAnalysis
395 311.75=83.25 (83.25)2/311.75= 22.23382 311.75= 70.25 (70.25)2/311.75=15.83
223 311.75= -88.75 (-88.75)2/311.75= 25.27
247 311.75= -64.75 (-64.75)2/311.75=13.45
G2 = 76.75
4. We have 4 classes of data how many degrees of
freedom?
p> 0.995 0.975 0.900 0.500 0.100 0.050 0.025 0.010 0.005
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a homozygous in
one cell
b homozygous in
the other cell
undetectable
detectable
twin spot
After mitotic recombination,daughter cells can have 2
configurations
Twin spots caused by mitotic recombination
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