Download 11/01/11 Mapping: By recombinant frequency. -

11/01/11
Mapping:
e Nucleotide Polymorphism (SNP) mapping:
1) By recombinant frequency.
er the following scenario: you have identified --method
a mutant strain that
carries a single counting visible phenotype.
of detection:
ve mutation (*). This
mutant
was identified
in the (SNP)
standard
laboratory N2
--Single
Nucleotide
Polymorphism
mapping:
Consider
the following
scenario:
you have identified
a mutant
ound strain (strain isolated
in Bristol,
England).
The Hawaiian
CB4856 strain
is strain that carries a single
recessive mutation (*). This mutant was identified in the standard laboratory N2
ld type but has polymorphisms
relative
to
N2.
Cross
mutant
(*)
hermaphrodite
background strain (strain isolated in Bristol, England). The Hawaiian CB4856 strain is
awaiian males. Assume
on chromosome
I. (chromosomes
IV, mutant (*) hermaphrodite
For example:
also mutation
wild type(*)
butis has
polymorphisms
relative to N2.III,Cross
X not shown for simplicity):
with Hawaiian males. Assume mutation (*) is on chromosome I. (chromosomes III, IV,
V, and X not shown for simplicity):
SNP1
*
*
II
I
X
*
*
I
II
SNP1
F1 CROSS PROGENY:
OSS PROGENY:
I
*
SNP2
II
II
SNP2
m/m
F1
Phenotype: 100% wild type
50% hermaphrodites,
males
SNP150%F2:
SNP2
ype: 100% wild type
ermaphrodites, 50% males
X
X
SNP1
I
SNP2
II
SNP1
a/a
SNP2
m-recessive mutation
a-recessive, visible phenotype when homozygous
I
*
II
m/+;
a/+,
allow
to self-fertilize
SNP1
SNP2
pick individual hermaphrodite cross progeny
separate plates, allow self fertilization th
 if monto
and
a are not linked 1/16 of offspring will be a/a;m/m double-mutants
pick individual hermaphrodite
cross
 if m and aprogeny
are linked <1/16th of offspring will be a/a;m/m double-mutants
separate
allow100%
self fertilization
F2 Phenotypes: !onto
wild
type, "plates,
mutants.
hermaphrodites.
Step 1: determine which chromosome This
carries the
mutation: requires being able to detect difference between 1/16 and <1/16, which requires
method
Pick individual F2 progeny with mutant phenotype. What are the probabilities of seeing
notypes: ! wild type,SNPs
" mutants.
100% hermaphrodites.
examining
enough
F2associated
offspring
on chromosomes
unassociated with
mutation? On
chromosome
with to get statistical significance.
mutation?
To
find
the
exact
chromosomal
location of your gene, one needs to use “mapping strain” that has multiple
NON-RECOMBINANT
determine which chromosome carries the mutation:*
F1
F
*
markers
on multiple chromosomes, so that numerous genetic regions can be tested simultaneously.
dividual F2 progeny with
mutant
phenotype.2 What*are theRECOMBINANT
probabilities of seeing
I
*
Because
of
on chromosomes unassociated with mutation? On* chromosome associated their
with frequency and distribution in the genome molecular polymorphisms are the linkage
markers used for mapping.
on?
F2
*
*
F1
*
II *
*
NON-RECOMBINANT
F2
RECOMBINANT
25%
50%
25%
1
F2
25%
50%
2) With molecular markers:
--method of detection: genotyping
--Single nucleotide polymorphism (SNP) and simple sequence length polymorphisms (SSLP)
Detecting SNPs by:
-sequencing
-restriction enzyme target site
Detecting SSLP by:
-PCR
Most of contemporary mapping are done using molecular variants (SNP, SSLP), rather than with loci that give
visible phenotypes.
Compared to other markers that have been used for genetic mapping, SNPs have two distinct advantages:
1) Unlike conventional marker mutations that cause visible phenotypes, SNPs in general have no associated
phenotype. Thus, mutant phenotypes that are masked by conventional marker mutations, such as those with subtle
behavioral defects, can be mapped using SNPs.
2) SNPs are far denser than other markers, including both visible
Method that uses SNPs to determine rough position of mutation is called SNIP SNP mapping.
SNIP SNP detection requires PCR amplification of the SNP region, digestion with the appropriate restriction enzyme,
and gel electrophoresis.
SNP mapping is usually done in two phases:
2
-The first phase, chromosome mapping, which seeks to identify the relevant chromosome and rough position of the
gene of interest.
-The second phase, interval mapping, seeks to place the gene of interest in an interval between two SNPs, and can be
used iteratively to fine map the gene.
(PMID: 16156901): SNIP SNP mapping
1) Identified all DraI SNPs in a custom database that incorporated all SNPs
2) From among these select eight candidate DraI SNPs on each
3) Next, to enable simultaneous amplification of all selected SNPs in a 96-well format, chose primer pairs with similar
annealing temperatures and product length. Each primer would contain each selected DraI SNP.
4) Hawaiian males are crossed into the mutant strain (that was backcrossed 4 times) to produce heterozygous F1
animals.
5) Homozygous F2 animals from the heterozygous F1 animals are identified based on their mutant phenotype. At the
Polymorphism
(SNP) mapping:
same time, animals with a non- mutant phenotype, which are enriched--Single
for Nucleotide
Hawaiian
sequences
at the locus of
Consider the following scenario: you have identified a mutant strain that carries a single
recessive mutation (*). This mutant was identified in the standard laboratory N2
interest, are also isolated
BMC Genomics 2005, 6:118
http://www.biomedcentral.com/1471-2164/6/118
Chromosome mapping:
background strain (strain isolated in Bristol, England). The Hawaiian CB4856 strain is
also wild type but has polymorphisms relative to N2. Cross mutant (*) hermaphrodite
with Hawaiian males. Assume mutation (*) is on chromosome I. (chromosomes III, IV,
V, and X not shown for simplicity):
SNP1
I
A Chromosome mapping procedure
pick 30
mutant
pick 30
non-mutant
mutant/Hawaiian
Hawaiian
SNP2
SNP1
SNP2
Phenotype: 100% wild type
50% hermaphrodites, 50% males
PCR mix
(no primers)
pick individual hermaphrodite cross progeny
-The lysate is then added to a PCR mix lacking
primers, and the
onto separate plates, allow self fertilization
mix is aliquoted into every other row of a 96 well plate.
F2 Phenotypes: ! wild type, " mutants. 100% hermaphrodites.
LGI
LGII
LGIII
LGIV
LGV
LGX
left
SNP2
II
1
primers

I
SNP1
genomic PCR mix 8-channel
DNA (no primers) pipettor
F2
X
II
-Typically 30 mutant animals (homozygous Bristol) and 30 wildtype animals (heterozygous Bristol/Hawaiian or homozygous
CROSS PROGENY:
I
*
II
Hawaiian DNA) are Flysed
in 20 μL lysis
buffer.
mutant
mutant
mutant F
Hawaiian 1
*
*
right
PCR mix
(no primers)
PCR
Step 1: determine which chromosome carries the mutation:
-Primers are added by
replication
from
a master
Pick pin
individual
F progeny with mutant
phenotype.
What are theplate
probabilities of seeing
2
8-channel
pipettor
DraI
digest
pin replicate
LGI
LGII
LGIII
LGIV
LGV
LGX
digested PCR
products
SNPs on chromosomes unassociated with mutation? On chromosome associated with
mutation?
*
NON-RECOMBINANT
F1
F2
*
I
*
*
RECOMBINANT
*
F1
LGI
agarose gel
F2
25%
II
50%
25%
B Chromosome mapping primer set
9
3
Interval mapping
After determining the rough position of a mutation on a chromosome using
chromosome mapping, mutations can be quickly mapped to a genetic interval using
the same efficiencies of the 96-well format employed in chromosome mapping.
Interval mapping differs from chromosome mapping in that the genotype of individual
mutant animals, rather than the genotype of pooled animals, must be determined. Also,
it is necessary to assay these mutant DNAs for many SNPs within the interval for
which linkage has been established.
--Single Nucleotide Polymorphism (SNP) mapping:
The
closer
the
gene and the marker are, the lower the proportion of
Consider the following scenario: you have identified
a mutant
strain
thatmutant
carries a single
recessive mutation (*). This mutant was identified
in
the
standard
laboratory
N2
recombinant mutant progeny
will have the marker locus. Thus, the proportion
background strain (strain isolated in Bristol, England). The Hawaiian CB4856 strain is
recombinants
is
a
clue
which
marker is the closest to the mutant allele.
also wild type but has polymorphisms relative to N2. Cross mutant (*) hermaphrodite
with Hawaiian males. Assume mutation (*) is on chromosome I. (chromosomes III, IV,
V, and X not shown for simplicity):
SNP1
I
*
*
X
II
I
SNP2
II
SNP1
F1 CROSS PROGENY:
I
*
SNP1
Phenotype: 100% wild type
50% hermaphrodites, 50% males
of
SNP2
II
SNP2
pick individual hermaphrodite cross progeny
onto separate plates, allow self fertilization
F2 Phenotypes: ! wild type, " mutants. 100% hermaphrodites.
Step 1: determine which chromosome carries the mutation:
Pick individual F2 progeny with mutant phenotype. What are the probabilities of seeing
SNPs on chromosomes unassociated with mutation? On chromosome associated with
mutation?
*
NON-RECOMBINANT
F1
F2
*
I
*
*
RECOMBINANT
*
F1
4
F2
25%
II
50%