Transcript Eukaryotic Chromosome Mapping

Eukaryotic Chromosome
Mapping
Using Genetic Recombination to
Estimate Distances Between
Genes
Linked Genes
Mendel’s
experiments
Linked Genes
Gene
location
Genes on separate
chromosomes
Genes on the same
chromosome
Gamete
types
Equal numbers of all
possible allele
combinations
More parental
combinations than
recombinant
combinations
Independent Assortment
vs. Gene Linkage
Example from Drosophila
Red eyes, x Pink eyes
Beige body
Ebony body
RRBB
rrbb
F1:
Red eyes, Beige body
RrBb
Independent Assortment
vs. Gene Linkage
Testcross: cross to individual of
known genotype
F1:Red eyes X Pink eyes
Beige body
Ebony body
RrBb
rrbb
Independent Assortment
vs. Gene Linkage
F2 phenotype
Red eyes
Beige body
Number of
Offspring
398
Expected for
Unlinked Genes
250
Pink eyes
Ebony body
Red eyes
Ebony body
382
250
108
250
Pink eyes
Beige body
112
250
Independent Assortment
vs. Gene Linkage
F1:Red eyes X Pink eyes
Beige body
Ebony body
RrBb
rrbb
rb
RB
Rb
RrBb
Red
Beige
Rrbb
Red
Ebony
rB
rrBb
Pink
Beige
rb
rrbb
Pink
Ebony
Independent Assortment
vs. Gene Linkage
If genes are linked:
Red eyes, x Pink eyes
Beige body
Ebony body
F1:
R B
r b
r b
R B
Red eyes, Beige body
R B
Coupling
r
b
or Cis
Configuration
Independent Assortment
vs. Gene Linkage
F1: Red eyes, Beige body
R
B
X
r
b
Four types of gametes are produced
Parental
Recombinant
R
B
R
b
r
b
r
B
Independent Assortment
vs. Gene Linkage
F1:Red eyes X Pink eyes
Beige body
Ebony body
R B
r b
r b
r b
R B
r b
R B
r b
r b
R b
r B
r b
r b
R b
r b
r B
r b
Independent Assortment
vs. Gene Linkage
F2 phenotype
Red eyes
Beige body
Pink eyes
Ebony body
Red eyes
Ebony body
Pink eyes
Beige body
Number of
Offspring
398
382
108
112
Chromosome
arrangement
RB//rb
Parental
rb//rb
Parental
Rb//rb
Recombinant
rB//rb
Recombinant
Genetic Map Units
1% recombination = 1 map unit
= 1 centimorgan
108 + 112 x 100 = 22%
1000
These genes are located 22 map units
apart on the same chromosome.
Limits of Genetic Mapping
Genes that are 50 map units apart will
appear to assort independently.
The calculated distance between any
TWO genes on the same chromosome
should be less than 50 map units.
Predicting Gamete Frequencies
for Linked Genes
Red eyes, x Pink eyes
Ebony body
Beige body
F1:
R b
r B
r B
R b
Red eyes, Beige body
R
b
r
B
Repulsion
or Trans
Configuration
Predicting Gamete Frequencies
for Linked Genes
F1: Red eyes
Beige body
R b
r B
The genes are 22 map units apart, therefore we
expect 22% recombinant gametes and 78%
parental gametes.
0.11
R B
0.39
R b
0.11
r b
0.39
r B
0.22 recombinants
0.78 parentals
Using a Three-point Testcross to
Determine Genetic Distance
• A cross between two parental strains is
used to produce a tri-hybrid
(heterozygous for three genes).
• The tri-hybrid is crossed to an organism
that is homozygous recessive for all
three genes.
• Eight classes of offspring are analyzed
to determine recombination frequencies.
Problem 1, Page 2-1
In corn, a strain homozygous for the recessive
alleles a (green), d (dwarf) and rg (normal
leaves) was crossed to a strain homozygous
for the dominant alleles of each of these
genes, namely A (red), D (tall) and Rg
(ragged leaves). Offspring of this cross were
then crossed to plants that were green, dwarf
and had normal leaves. The following
phenotypic classes were observed.
Problem 1, Page 2-1
Offspring Resulting from
Three-Point Testcross
red, tall, ragged
265
green, dwarf, normal
275
red, tall, normal
24
green, dwarf, ragged
16
red, dwarf, normal
90
green, tall, ragged
70
red, dwarf, ragged
120
green, tall, normal
140
Problem 1, Page 2-1
A
Red
a
Green
D
Tall
d
Dwarf
Rg
Ragged leaves
rg
Normal leaves
Problem 1, Page 2-1
With Arbitrary Gene Order
a d rg
a d rg
A D Rg
A D Rg
X
F1
A D Rg
a d rg
Testcross
A D Rg
X
a d rg
a d rg
a d rg
F2
Problem 1, Page 2-1
With Arbitrary Gene Order
A D Rg
X
a d rg
a d rg
a d rg
Parentals: A D Rg
a d rg
Recombinants: A d rg
a D Rg
A D rg
a d Rg
A d Rg
a D rg
F2
a d rg
Used as a genetic
background to see
the contribution from
the tri-hybrid.
Problem 1, Page 2-1
•Determine which classes are parentals
The two parental classes will represent the
largest number of offspring in the F2
generation.
Information on the parents may be given in
the problem description itself.
Parentals: red, tall, ragged
green, dwarf, normal
Problem 1, Page 2-1
• Determine which classes are double
recombinants
Double recombinants have two
crossovers: one between the first and
middle gene and one between the
middle and third gene
These will be the two smallest classes.
Double Recombinants: red, tall, normal
green, dwarf, ragged
Problem 1, Page 2-1
•Determine the gene order
The middle gene is the one that changes
places in the double recombinants when
compared to the parental combinations.
A
a
Rg
X
rg
X
D
A
d
a
rg
D
Red, tall, normal
Rg
d
Green, dwarf, ragged
Problem 1, Page 2-1
This shows why other gene orders are incorrect.
A
D
X
a
Rg
X
d
rg
A
d
Rg
Red, dwarf, ragged
a
D
rg
Green, tall, normal
D
A
X
d
Rg
X
a
rg
D
a
Rg
Green, tall, ragged
d
A
rg
Red, dwarf, normal
Problem 1, Page 2-1
•Assign genotypes to all classes
Use correct gene order
Contribution of
F1 parent
red, tall, ragged
P
A Rg D
265
green, dwarf, normal
P
a rg d
275
red, tall, normal
DC
A rg D
24
green, dwarf, ragged
DC
a Rg d
16
red, dwarf, normal
A Rg
A rg d
90
green, tall, ragged
A Rg
a Rg D
70
red, dwarf, ragged
Rg D
A Rg d
120
green, tall, normal
Rg D
a rg D
140
Problem 1, Page 2-1
Recombination between A and Rg
Single Crossovers
A
X
a
Rg
D
A
rg
d
a
rg
d
Red, dwarf, normal
Rg
D
Green, tall, ragged
90
70
Double Crossovers
A
a
Rg
X
rg
X
D
A
d
a
rg
D
Red, tall, normal
24
Rg
d 16
Green, dwarf, ragged
Total = 200
Recombination = (200/1000) x 100 = 20 %
Problem 1, Page 2-1
Recombination between Rg and D
Single Crossovers
A
Rg
a
rg
X
D
A
d
a
Rg
d 120
Red, dwarf, ragged
rg
D
Green, tall, normal
140
Double Crossovers
A
a
Rg
X
rg
X
D
A
d
a
rg
D
Red, tall, normal
Rg
d
24
16
Total = 300
Green, dwarf, ragged
Recombination = (300/1000) x 100 = 30 %
Problem 1, Page 2-1
Two maps are possible:
A
Rg
20 map units
D
30 map units
or
D
Rg
30 map units
A
20 map units
Interference
• Interference: crossover in one region
inhibits crossover in an adjacent region
• Interference = 1 – (coefficient of coincidence)
• Coefficient of coincidence =
Observed double crossovers
Expected double crossovers
Calculating Interference
• Coefficient of coincidence =
Observed double crossovers =
Expected double crossovers
24 + 16
= 40 = 0.667
0.2 x 0.3 x 1000 60
• Interference =
1–(coefficient of coincidence) =
1- 0.667 = 0.333