Transcript Meiosis, Recombination, and Mapping (Lisa Harper) ()

Meiosis, Recombination and
Mapping
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Meiosis
Why bother to map your gene?
Mapping a mutant defining a single gene
Mapping a sequence
Material for you:
•This presentation
•PDFs of important papers
•Review of basic Genetics
•Problem sets with answers
•Definitions of IBM map words
Contact me anytime!
Dempsy. Maize Handbook
Meiosis takes about 5
days, then there is
about 8-10 more days
to shed
Maize meiosis
Three major processes in
meiotic prophase
Leptotene
1. Chromosomes become visable as threads
2. The Axial Element is installed
Leptotene-Zygotene Transition
1. The bouquet is formed
2. Dramtic transient change in chromatin structure
3. It all begins…
Zygotene
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Pairing
Synapsis
Recombination
Pachytene
1. Chromosomes are completely synapsed
2. Recombination is completed
3. Interference probably works here
MITOSIS
METAPHASE
ANAPHASE
MEIOSIS
METAPHASE I
ANAPHASE I
METAPHASE II
ANAPHASE II
Sister Chromatid cohesion and chiasmata holds bivalents
together until the Metaphase to Anaphase to transition
Diplotene- Diakinesis
1. SC falls apart
2. Chromosomes further condense
3. Homologs held together by chiasmata
Homologous
Chromosome Pairing
Synapsis
Axial element
Lateral Element
Synapsis starts at the chromosome ends
Anderson and Stack, Chr. Res 2002
Complete Synapsis
Anderson et al, Genetics 2003
Recombination
5’
3’
Lichten
Number of RAD51 foci
Leptotene
Zygotene
500
400
300
200
Early pachytene
Diplotene
Pachytene
Late zygotene
Mid zygotene
0
Early zygotene
100
Leptotene
Distribution of
RAD51
recombination
protein in normal
maize meiosis
600
Late pachytene
Wojtek Pawlowski
Early Recombination Nodules in TEM
Anderson, et al Genetics 2001
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Allers and Lichten Cell 2001
Patterns of
distribution of
RAD51
recombination
protein in
normal maize
meiosis.
Organisms rely on crossovers to segregate chromosomes…
… but most organisms make very few crossovers
per chromosome pair
species
S.c.
N.c.
C.e.
A.t.
D.m .
M.m .
H.s.
Z.m.
haploid
16
chromosome
number (n)
7
6
5
4
20
23
10
genetic map
length (m.u.)
4300
1040
300
520
285
1600
3300
1600
haploid
genome size
(bp)
1.2 x 107 2.7 x 107 1.0 x 108 1.0 x 108 1.4 x 108 3.0 x 109 3.3 x 109 5.4 x 109
length of DNA/ 2.8
m.u. (kb/m.u.)
26
330
200
490
1900
1000
3400
crossovers
per
chromosome
2-5
1
2-3
0-2*
1-2
1-7
2-4
3-8
Anne Villeneuve
Inna Golubovskaya
Break
The Genetic
Maps of Maize
There 1176
genetic maps of
maize. We will
only use the old
fashion “genetic”
map and the
IBM2 map
Where is your gene?
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Relative to other genes?
One the chromosome?
Why do you care?
How do you find out?
Your gene relative to other Genes
• The genetic map is based on an abstract
principle: the percent recombination
between two genes (or markers).
• The amount of recombination is not uniform
along the length of the chromosome- but
that does not matter for mapping.
f1 an1
rhl
d8
150 170
184
197
Your gene on the chromosome
Why do you care?
• It makes to whole map better.
• Maybe there is already a sequence mapped to
your gene!
• Maybe there will be soon
• Maybe you think that the mutant of your
gene is just like a mutant in yeast. Using the
yeast sequence, get some maize GSSs, map,
do they map to your gene?
• Map-based cloning may (will) be possible
How do you map a gene?
• You need a DIFFERENCE to map- two
alleles
• First localize in a general region
• Next fine structure mapping
How do you map two genes
relative to each other?
y10 lg2
y10 lg2
Y10 Lg2
y10 lg2
Gametes:
Parental
Recombinant
{
{
y10 lg2
y10 lg2
Y10 Lg2
y10 Lg2
Y10 lg2
Test Cross: Heterozygote cross multiply recessive (or co-dominant) tester line. The genotype
of the gametes of the “tested” parent is also the phenotype of the progeny.
y10 lg2
y10 lg2
Y10 Lg2
y10 lg2
Genotype
Parentals
Recombinants
{
{
Phenotype
Number of Progeny
y10 lg2
y10 lg2
Yellow, liguleless
48
Y10 Lg2
y10 lg2
Wildtype
42
y10 Lg2
y10 lg2
Yellow
6
liguleless
4
Y10 lg2
y10 lg2
y10 lg2
y10 lg2
Y10 Lg2
y10 lg2
Number of Progeny
48
P:
Recombinants
X 100 = cM
Total Progeny
42
6
R:
4
6+4
X 100 = 10% = 10cM
48+42+6+4
10 cM between y10 and lg2
How can you determine gene
order??
Three (or more) factor cross
mwy
MWY
MWY X mwy
F1
MWY
mwy X
True Breeding X
wild type
mwy
mwy
Phenotype
Min, white, yellow
Yellow
White eyes
Min
Yellow, white eyes
White eyes, min
Min, yellow
Wild type
True breeding
miniature, white-eyed,
yellow bodied fly
Test cross
(Genotype)
mwy
y++
w++
m++
yw+
wm+
my+
+++
Number
3501
25
3
1720
1734
35
6
3471
Put the 8 progeny types into 4 reciprocal classes.
The number of progeny per class is a clue.
P
m w y 3501
Parental
+ + + 3471
mwy
Parental:
+++
m + + 1720
SR y w + 1734
Single recombinants
w m + 35
SR
y + + 25
Single recombinants
DR
w++ 3
my+ 6
Double Recombinants
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
DR
+w+ 3
m+y 6
Recombinants
X 100 = cM
Total Progeny
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
DR
+w+ 3
m+y 6
Recombinants
X 100 = cM
Total Progeny
Recombination between m and w:
(1720 + 1734 + 3 + 6) / 10495 = 33cM
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
DR
+w+ 3
m+y 6
Recombinants
X 100 = cM
Total Progeny
Recombination between m and w:
(1720 + 1734 + 3 + 6) / 10495 = 33cM
Recombination between w and y:
(35 + 25 + 3 + 6) / 10495 = 0.7cM
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
DR
+w+ 3
m+y 6
Recombinants
X 100 = cM
Total Progeny
Recombination between m and w:
(1720 + 1734 + 3 + 6) / 10495 = 33cM
Recombination between w and y:
(35 + 25 + 3 + 6) / 10495 = 0.7cM
Recombination between m and y:
(1720 + 1734 + 35 + 625) / 10495 = 33.5cM
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
DR
+w+ 3
m+y 6
Recombinants
X 100 = cM
Total Progeny
Recombination between m and w:
(1720 + 1734 + 3 + 6) / 10495 = 33cM
Recombination between w and y:
(35 + 25 + 3 + 6) / 10495 = 0.7cM
Recombination between m and y:
(1720 + 1734 + 35 + 625) / 10495 = 33.5cM
W is in the middle, because it took two
crossovers to separate it from the other
allele
M w y
Parental:
+ + +
P
m w y 3501
+ + + 3471
m + + 1720
SR + w y 1734
w m + 35
SR
+ + y 25
+w+ 3
DR
m+y 6
Recombinants
X 100 = cM
Total Progeny
Recombination between m and w:
(1720 + 1734 + 3 + 6) / 10495 = 33cM
Recombination between w and y:
(35 + 25 + 3 + 6) / 10495 = 0.7cM
Recombination between m and y:
(1720 + 1734 + 35 + 625) / 10495 = 33.5cM
33 cM
m
0.7cM
w y
Crossover Interference:
Crossover in one region of a chromosome
reduces the probability of a second crossover nearby.
If no interference:
Freq. of d.c.o. involving interval x and y = (RF x) (RF y)
x
y
33 cM
m
0.7cM
w y
Exp dco=2.3%
Obs dco =0.08
“coefficient of coincidence” C = obs d.c.o. / exp d.c.o.
C = 1 intervals are independent
C = 0 complete interference
C=0.08/2.3
How can you map your MUTANT?
In most cases, you must do crosses
First localize it to a chromosome arm by:
-Crossing to many TB or T-waxy translocation lines and
scoring. This takes two (TB) to three (T-waxy) generations.
Then, make crosses to close markers, and map. A possible
discussion topic.
And/Or
-Make a mapping population, and use molecular markers to
localize, and to fine map
Molecular Markers for mapping:
RFLPs
SSRs
AFLPs
Indels
SNPs
RFLP: Restriction Fragment
Length Polymorphism
Allele A
Allele B
GCCGCATTCTATAAAG
CGGCGTAAGATATTTC
GCCGAATTCTATAAAG
CGGCTTAAGATATTTC
Detect by Southern blot:
A/A
A/B
B/B
SSR: simple sequence repeat
(length polymorphism)
PCR Primer
(CA)n
(CA)m
variation between strains in number of repeats
at a given locus
PCR yields products of different size:
Inserted Elements as Genetic
Markers
e.g. presence or absence of mapped
transposon
Product
No Product
SNP: single nucleotide
polymorphisms
 Outgrowth of sequencing projects
 If they affect a restriction enzyme site:
snip-SNP
 Detection of SNPs can be done without gels:
highly automated/high throughput and/or
highly parallel (simultaneous scoring of
MANY markers)
“Snip-SNP”
Single nucleotide polymorphism affecting
a restriction endonuclease site
Step 1)
PCR
Step 2)
cut
A
B
Indels: Insertions- deletions
Through sequencing efforts, small insertions or
deletions (indels)have been discovered. These
can be detected by PCR with unique primers.
A
B
Each chromosome is divided into bins
These bins are about 20 cM
Bins are defined by molecular markers
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
The 90 core bin markers are at
http://www.maizegdb.org/cgi-bin/bin_viewer.cgi
Map your mutant relative to a molecular marker
First, make a mapping population:
New mutant: lg5. This mutant was found in an MTM
(Mutator) population. You introgressed into B73
lg5
In a B73 background
lg5
B1 B2 lg5 B3 B4
B1 B2 lg5 B3 B4
F1
x
x
Mo17
M1 M2 Lg5 M3 M4
M1 M2 Lg5 M3 M4
B1 B2 lg5 B3 B4
M1 M2 Lg5 M3 M4
F1
B1 B2 lg5 B3 B4
M1 M2 Lg5 M3 M4
x
B1 B2 lg5 B3 B4
B1 B2 lg5 B3 B4
Mapping population
This population with segregate 1:1 for lg5. You
will make DNA from some number of lg plants,
and an equal number of wild type plants.
B73
Mo17
lg 1
lg 2
lg 3
lg 4
lg 5
lg 6
lg 7
lg 8
lg 9
lg 10
wt 1
wt 2
wt 3
wt 4
wt 5
wt 6
wt 7
wt 8
wt 9
wt10
DNA extracted
Digested with EcoR1/HindIII
Southern Probed with tub1
P: mutant allele of lg5 and B73 tub1 allele:
R: mutant allele of lg5 and Mo17 tub1 allele:
R: Wt allele of Lg5 and B73 tub1 allele:
P: Wt allele of Lg5 and Mo17 tub1 allele:
5+5
20
= 50% UNLinked!
5
5
5
5
B73
Mo17
lg 1
lg 2
lg 3
lg 4
lg 5
lg 6
lg 7
lg 8
lg 9
lg 10
wt 1
wt 2
wt 3
wt 4
wt 5
wt 6
wt 7
wt 8
wt 9
wt10
DNA extracted
Digested with Pst1
Southern Probed with umc128
P: mutant allele of lg5 and B73 tub1 allele:
R: mutant allele of lg5 and Mo17 tub1 allele:
R: Wt allele of Lg5 and B73 tub1 allele:
P: Wt allele of Lg5 and Mo17 tub1 allele:
1+1
20
9
1
1
9
= 10% = 10 cM Linked!
If you have mapping data, submit a maize
newsletter article, and Mary Polacco will
incorporate map data. Soon you can import
mapping data directly to MaizeGDB, but still,
write the MNL article so everyone can see the
data!
How do you map a sequence?
High resolution maps: IBM2 and IBM2 neighbor map
Why the
huge
difference?
Making the Intermated B73 X MO17 mapping population
Mo17
B73
X
Single F2
plant was
selfed
X
Grow up 200 plants,
random mating
Genotype of 5 of those 200 plants
Select 100 ears, pick 5 kernals from each ear. Put in a
bag, shake, plant, more random matings (2nd generation).
Repeat, (repeat….)
From these lines, generate Recombinant Inbred lines by repeated selfing (5x or more)
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Cool, but how do you map with them??
Date: Tue, 2 Mar 2004 13:43:57 -0600
To: [email protected]
From: Ed Coe <[email protected]>
Reply-To: [email protected]
Subject: Want to know the map location of your gene?
Want to know the map location of your gene?
You can easily obtain maize map location with the Community IBM Map Data Entry (CIMDE)
utility.
Kits containing a 96-well plate with DNA of 94 IBM lines and the parents, with protocols, are
available:
http://www.maizemap.org/dna_kits.htm
(alternatively, seeds of the 94-member panel of IBM stocks can be requested from the Stock
Center)
Data submission:
http://www.maizemap.org/CIMDE/cimde.html
Maps are generated and reported-back promptly after the data are submitted. The data, if you so
choose, will be incorporated into the combined community map, cIBM (see MaizeGDB for recent
posting of maps cIBM 1 through 10).
Ed Coe
Karen Cone
Georgia Davis
Mary Polacco
$240
There are 288 lines in the IBM population. They send you DNA from 94
representatives. You use primers from your gene of interest, that amplify a
different length fragment in Mo17 vs B73. Use these pimers on all 94 lines:
B73
Mo17
PCR + EcoR1
B73
Mo17
Line1
Line2
Line3
Line4
Line5
Line6
Line7
Line8
Line9
Line10
EcoR1
So, you gene’s ‘code’ is: MMBBMMBBMB…
Then, the MMPers figure out, with their computers, where
that matches in the genome.
So, you gene’s ‘code’ is: MMBBMM …
Line 1
Line 2
Line 3
MMMMMB
Nope!
MBMBMM
Nope!
MMBBMM
YEAH!
Line 4
Line 5
Line 6
High resolution maps: IBM2 and IBM2 neighbor map
Why the
huge
difference?
IBM2 vs
IBM2
neighbors
On frame vs off frame
Oat x maize addition line
Genomic in situ hybridization of an addition line with fluorescein-labeled maize DNA. The pair of maize chromosomes are yellow-green, and the
oat chromosomes are counterstained red with propidium iodide.
Chromosome 1 addition line
Unusual characteristics noticed in chromosome 1 OMA lines:Erect leaf blade, photoperiod neutral response, sectoring among shoots for the maize chromosome
Chromosome 10 Addition Line
Characteristics of chromosome 10 OMA lines: grassy type and very sterile
Oat-corn chromosome 9 monosomic addition line
Oat line without a
corn chromosome
Oat line with a corn
chromosome piece transfer
Selfing
Oat line with a modified
corn chromosome
Mapping with OMA lines
primer33A
primer33C
primer34A
primer34C
primer34A: will amplify ESTs AW066336 and AW066011. These two sequences are very similar, e-172 and 335/343 (97%) and may
represent two alleles from one gene.
primer34C: was designed to amplify TC77059. This sequence is more distantly related to AW066336, e-112, 269/289 (93%).
The gene represented by ESTs AW066336 and AW066011 appears to lie on chromosome 5. The PCR primers, primer34C, detects the
related sequence on chromosome 4.
3-D Deconvolution Microscopy
raw data
deconvolved data