Linkage,Recombination
and Genetic Mapping
Linkage,Alleles of two genes segregate
together during meiosis,
+ +
w y
w
w y
y
+
+
+
+
w
w y
y
+
+
+
++ +
w y
Complete Linkage Partial Linkage No Linkage
1/2
1/2
1/4
1/4
1/4
1/4
>1/4
>1/4
<1/4
<1/4
During meiosis crossing over between homologous
chromosomes produces recombinant chromosomes
A
a
b
B
A
a
B
b
The frequency of recombinant gametes is
1/2 the frequency of crossing over
Each crossover event gives rise to two recombinant
and two parental chromosomes,
A crossover frequency of 100% gives a recombinant
chromosome frequency of 50%.
a b
a b
A B
A B
a b
a B
A b
A B
Tests for Linkage,Chi Square
The chi square test is used to test the goodness of
fit of observed data to expectations,This test can
be used to test whether two genes are linked.
Example,Flies heterozygous for black (b) and
vestiial (vg) were testcrossed,Are these two
genes linked?
progeny phenotypes number
grey normal 283
grey vestigial 1294
black normal 1418
black vestigial 241
Chi Square Test
Hypothesis,the genes are not linked,
Prediction,the progeny should be in a 1:1:1:1 ratio.
2 test:
obs exp (obs-exp)2/exp
283 809 342.0
1294 809 290.8
1418 809 458.4
241 809 398.8
Total 3236 3236 1490.0
Degrees of freedom = 4 -1 = 3 (one less than the number
of progeny classes)
Chi Square Test
From the calculated?2 value and the known degrees of
freedom,the?2 table reveals a probability value that
the deviation from predicted would occur randomly
For this example,probability <0.001
This means that the deviation from predicted is expected to
occur randomly less than 1 in 1000 times
Conclusion? The hypothesis is not correct,The genes
do not show independent assortment.
A.H,Sturtevant and linkage mapping
Sturtevant,strength of linkage is relative to physical
distance,
(1) Recombination is randomly distributed in the chromosome.
(2) Each gene has a unique location (locus) on the
chromosome.
(3) The frequency of recombination is proportional to the
distance between two loci on the chromosome.
ie,A - B is smaller than B - C
A B C
Linkage mapping
Linkage mapping is done using test crosses.
a b / A B X a b / a b
a b 100 Parental
a B 25 Recombinant
A b 25 Recombinant
A B 100 Parental
R = # recombinant progeny/ #Total progeny X 100 = cM
R = 50/250 X 100 = 20 cM
Three Point Cross
What can we learn from a three point
testcross?
The parental genotypes.
The order of genes on the chromosome.
The linkage map distance between the genes.
Whether there is interference,and how much.
Problem for three point mapping cross
Individuals heterozygous for three linked genes are test
crossed:
sc + cv/ + ec + X sc ec cv / sc ec cv
wild type + + + / sc ec cv 1
scute echinus crossveinless sc ec cv/ sc ec cv 4
scute sc + + / sc ec cv 997
echinus crossveinless + ec cv/ sc ec cv 1002
scute echinus sc ec +/ sc ec cv 681
crossveinless + + cv / sc ec cv 716
echinus + ec + / sc ec cv 8576
scute crossveinless sc + cv / sc ec cv 8808
Reminder,the cross is sc + cv/ + ec + X sc ec cv / sc ec cv
The progeny are scored for all three genes.
Phenotype Genotype Number
Parental types and gene order
Parental types are
always in excess:
Double crossovers
are the least
frequent:
Gene order =
wild type + + + / sc ec cv 1
scute echinus crossveinless sc ec cv/ sc ec cv 4
scute sc + + / sc ec cv 997
echinus crossveinless + ec cv/ sc ec cv 1002
scute echinus sc ec +/ sc ec cv 681
crossveinless + + cv / sc ec cv 716
echinus + ec + / sc ec cv 8576
scute crossveinless sc + cv / sc ec cv 8808
1 + + + 1 + ec + Double
2 sc ec cv 4 sc + cv crossover
3 sc + + 997 + ec + Region lI
4 + ec cv 1002 sc + cv
5 sc ec + 681 + ec + Region I
6 + + cv 716 sc + cv
7 + ec + 8576 + ec + Parental
8 sc + cv 8808 sc + cv
The map distances between the genes are calculated,and
combined to produce the map.
R sc ec = (681 + 716 + 4 + 1) / 20785 X 100 = 6.74
R ec cv = (997 + 1002 + 4 + 1) / 20785 X 100 = 9.65
sc ec cv
6.74 9.65
Linkage maps can also be used to predict the results of
crosses,
What phenotypes and how many of each do you expect
from the following cross,if there are 10,000 progeny?
sc ec cv/ + + + X sc ec cv/ sc ec cv
sc ec cv
6.74 9.65
Interference and Coincidence
The occurrence of a chiasma between two chromatids
may physically impede the occurrence of a second
chiasma nearby,a phenomenon called chiasma
interference (or chromosomal interference).
interference=1-coefficient of coincidence
Coefficient of coincidence=Observed double crossover
frequency divided by expected double crossover
frequency
Problem
In a cross
PJR/pjr X pjr/pjr
progenies can be divided into these classes,
PJR and pjr 179 and 173
pJR and Pjr 52 and 46
pjR and PJr 22 and 22
PjR and pJr 4 and 2
What is the coefficient of coincidence?
Expected double crossover:
Map distance 1 X Map distance 2 = 20.8% X 10%
=0.0208
Observed:
6/500 = 0.012
Thus,coefficient of coincidence is equal to
0.012/0.0208=0.577
Genetic fine structure analysis
Fine structure analysis is genetic analysis of the internal
structure of genes.
Mapping analyses demonstrated that recombination can
occur between alleles within a gene.
Complementation Tests
Two independent recessive mutations cause a similar
phenotype:
they could be alleles of the same gene
they could be mutations of different genes,each of which
causes the same phenotype.
Complementation tests distinguish these possibilities
To test,a cross is made between the two individuals
that possess the same phenotype.
For example,aaBB and AAbb are both albinos,but
when crossed they give normal progeny (which are
AaBb,thus these two mutations complement each
other,
If they were in the same gene,they would not.
If the mutations are in different genes,then each will be
heterozygous and the wild type phenotype will occur
this is referred to as complementation (A)
If the mutations are in the same gene,then each copy of
the gene will carry a recessive mutation,and the mutant
phenotype will occur
this is referred to as noncomplementation (B)
Mutations that fail to complement usually are alleles of the
same gene.
G e n e 1 G e n e 2 G ene 1
A B
+
+
Linkage mapping produced the,beads on a string”
model of chromosomes.
Each gene has one location (locus).
Each gene has one genetic function that can mutate.
All alleles of one gene fail to complement,ie,a1/a2 = mutant
Recombination takes place between (but not within) genes.
New discoveries changed this model.
GENES
lozenge
Crossing over within a gene
P lzg X lzs
lzg lzs
F1 testcross lzg X lzg
lzs lzg
F2 phenotype +
lzg
(Very rare event)
Recombination between mutant lesions can generate
wild type alleles,
lzg lesion
lzs lesion
lozenge
locus
g
s+
+
+ +g s
g/s allele wild type
allele
Fig 6.27
RF = (NPD + 1/2T)/total tetrads X 100%
Two unusual phenomena
Mitotic Crossing-over
Gene Conversion
and Genetic Mapping
Linkage,Alleles of two genes segregate
together during meiosis,
+ +
w y
w
w y
y
+
+
+
+
w
w y
y
+
+
+
++ +
w y
Complete Linkage Partial Linkage No Linkage
1/2
1/2
1/4
1/4
1/4
1/4
>1/4
>1/4
<1/4
<1/4
During meiosis crossing over between homologous
chromosomes produces recombinant chromosomes
A
a
b
B
A
a
B
b
The frequency of recombinant gametes is
1/2 the frequency of crossing over
Each crossover event gives rise to two recombinant
and two parental chromosomes,
A crossover frequency of 100% gives a recombinant
chromosome frequency of 50%.
a b
a b
A B
A B
a b
a B
A b
A B
Tests for Linkage,Chi Square
The chi square test is used to test the goodness of
fit of observed data to expectations,This test can
be used to test whether two genes are linked.
Example,Flies heterozygous for black (b) and
vestiial (vg) were testcrossed,Are these two
genes linked?
progeny phenotypes number
grey normal 283
grey vestigial 1294
black normal 1418
black vestigial 241
Chi Square Test
Hypothesis,the genes are not linked,
Prediction,the progeny should be in a 1:1:1:1 ratio.
2 test:
obs exp (obs-exp)2/exp
283 809 342.0
1294 809 290.8
1418 809 458.4
241 809 398.8
Total 3236 3236 1490.0
Degrees of freedom = 4 -1 = 3 (one less than the number
of progeny classes)
Chi Square Test
From the calculated?2 value and the known degrees of
freedom,the?2 table reveals a probability value that
the deviation from predicted would occur randomly
For this example,probability <0.001
This means that the deviation from predicted is expected to
occur randomly less than 1 in 1000 times
Conclusion? The hypothesis is not correct,The genes
do not show independent assortment.
A.H,Sturtevant and linkage mapping
Sturtevant,strength of linkage is relative to physical
distance,
(1) Recombination is randomly distributed in the chromosome.
(2) Each gene has a unique location (locus) on the
chromosome.
(3) The frequency of recombination is proportional to the
distance between two loci on the chromosome.
ie,A - B is smaller than B - C
A B C
Linkage mapping
Linkage mapping is done using test crosses.
a b / A B X a b / a b
a b 100 Parental
a B 25 Recombinant
A b 25 Recombinant
A B 100 Parental
R = # recombinant progeny/ #Total progeny X 100 = cM
R = 50/250 X 100 = 20 cM
Three Point Cross
What can we learn from a three point
testcross?
The parental genotypes.
The order of genes on the chromosome.
The linkage map distance between the genes.
Whether there is interference,and how much.
Problem for three point mapping cross
Individuals heterozygous for three linked genes are test
crossed:
sc + cv/ + ec + X sc ec cv / sc ec cv
wild type + + + / sc ec cv 1
scute echinus crossveinless sc ec cv/ sc ec cv 4
scute sc + + / sc ec cv 997
echinus crossveinless + ec cv/ sc ec cv 1002
scute echinus sc ec +/ sc ec cv 681
crossveinless + + cv / sc ec cv 716
echinus + ec + / sc ec cv 8576
scute crossveinless sc + cv / sc ec cv 8808
Reminder,the cross is sc + cv/ + ec + X sc ec cv / sc ec cv
The progeny are scored for all three genes.
Phenotype Genotype Number
Parental types and gene order
Parental types are
always in excess:
Double crossovers
are the least
frequent:
Gene order =
wild type + + + / sc ec cv 1
scute echinus crossveinless sc ec cv/ sc ec cv 4
scute sc + + / sc ec cv 997
echinus crossveinless + ec cv/ sc ec cv 1002
scute echinus sc ec +/ sc ec cv 681
crossveinless + + cv / sc ec cv 716
echinus + ec + / sc ec cv 8576
scute crossveinless sc + cv / sc ec cv 8808
1 + + + 1 + ec + Double
2 sc ec cv 4 sc + cv crossover
3 sc + + 997 + ec + Region lI
4 + ec cv 1002 sc + cv
5 sc ec + 681 + ec + Region I
6 + + cv 716 sc + cv
7 + ec + 8576 + ec + Parental
8 sc + cv 8808 sc + cv
The map distances between the genes are calculated,and
combined to produce the map.
R sc ec = (681 + 716 + 4 + 1) / 20785 X 100 = 6.74
R ec cv = (997 + 1002 + 4 + 1) / 20785 X 100 = 9.65
sc ec cv
6.74 9.65
Linkage maps can also be used to predict the results of
crosses,
What phenotypes and how many of each do you expect
from the following cross,if there are 10,000 progeny?
sc ec cv/ + + + X sc ec cv/ sc ec cv
sc ec cv
6.74 9.65
Interference and Coincidence
The occurrence of a chiasma between two chromatids
may physically impede the occurrence of a second
chiasma nearby,a phenomenon called chiasma
interference (or chromosomal interference).
interference=1-coefficient of coincidence
Coefficient of coincidence=Observed double crossover
frequency divided by expected double crossover
frequency
Problem
In a cross
PJR/pjr X pjr/pjr
progenies can be divided into these classes,
PJR and pjr 179 and 173
pJR and Pjr 52 and 46
pjR and PJr 22 and 22
PjR and pJr 4 and 2
What is the coefficient of coincidence?
Expected double crossover:
Map distance 1 X Map distance 2 = 20.8% X 10%
=0.0208
Observed:
6/500 = 0.012
Thus,coefficient of coincidence is equal to
0.012/0.0208=0.577
Genetic fine structure analysis
Fine structure analysis is genetic analysis of the internal
structure of genes.
Mapping analyses demonstrated that recombination can
occur between alleles within a gene.
Complementation Tests
Two independent recessive mutations cause a similar
phenotype:
they could be alleles of the same gene
they could be mutations of different genes,each of which
causes the same phenotype.
Complementation tests distinguish these possibilities
To test,a cross is made between the two individuals
that possess the same phenotype.
For example,aaBB and AAbb are both albinos,but
when crossed they give normal progeny (which are
AaBb,thus these two mutations complement each
other,
If they were in the same gene,they would not.
If the mutations are in different genes,then each will be
heterozygous and the wild type phenotype will occur
this is referred to as complementation (A)
If the mutations are in the same gene,then each copy of
the gene will carry a recessive mutation,and the mutant
phenotype will occur
this is referred to as noncomplementation (B)
Mutations that fail to complement usually are alleles of the
same gene.
G e n e 1 G e n e 2 G ene 1
A B
+
+
Linkage mapping produced the,beads on a string”
model of chromosomes.
Each gene has one location (locus).
Each gene has one genetic function that can mutate.
All alleles of one gene fail to complement,ie,a1/a2 = mutant
Recombination takes place between (but not within) genes.
New discoveries changed this model.
GENES
lozenge
Crossing over within a gene
P lzg X lzs
lzg lzs
F1 testcross lzg X lzg
lzs lzg
F2 phenotype +
lzg
(Very rare event)
Recombination between mutant lesions can generate
wild type alleles,
lzg lesion
lzs lesion
lozenge
locus
g
s+
+
+ +g s
g/s allele wild type
allele
Fig 6.27
RF = (NPD + 1/2T)/total tetrads X 100%
Two unusual phenomena
Mitotic Crossing-over
Gene Conversion
