Chromosome mapping by recombination Meiosis is the basis of transmission genetics The recombination that occurs during meiosis in heterozygotes generates data that are a useful tool for making linkage maps ID: 527152
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Slide1
Linkage
Chromosome mapping by recombination
Meiosis is the basis of transmission genetics
The recombination that occurs during meiosis (in heterozygotes) generates data that are a useful tool for making linkage maps
Linkage maps assist in understanding evolution, synteny, and selection response
Slide2
F1
Parent 1
Parent 2
Populations for Linkage AnalysisSlide3
Doubled Haploids
Parent 1
Parent 2
F1
GametesSlide4
×
P1
P2
P2
Independent
8 parental: 8 non-
parental
Two loci:
-Spike color
- Plant height
Linked
14 parental: 2 non-parental
What is Linkage?Slide5Slide6
Linkage
The association of two or more phenotypic characters in inheritance because the genes controlling these characters are located in the same chromosome.
The closer the gene locations, the stronger the association
Genes carried by the same chromosome are members of the same linkage group
T
he number of linkage groups corresponds, therefore, to the basic number of chromosomes in the organism in question. Slide7
Chromosomes
Plant
2n
= _X = _
Genes
Arabidposis thaliana 2n = 2x = 10
27,000Fragaria vesca
2n = 2x = 14
35,000Theobroma cacao
2n = 2x = 2029,000
Zea
mays
2n = 2x = 20
40,000Slide8
Linkage
Crossing over refers to the physical exchange between homologous chromosomes. Meiotic crossing over is a potent source of genetic variability.
Recombination is the genetic result of crossing over and is detected by new combinations of alleles at two or more loci. Slide9
Prophase I
(5 stages)
1.
Leptonema
: Longitudinal duality of chromosomes not discernible.
2. Zygonema: Pairing of homologous chromosomes. Formation of
synaptonemal
complex and zygotene
DNA synthesis.
3. Pachynema: Pairing persists:
synaptonemal
complex + crossovers. Bivalent = 2 homologous chromosomes = 2 sets of 2 chromatids. Crossing over occurs (
chiasma
;
chiasmata
).
4.
Diplonema
:
Synaptonemal
complex dissolves; visualize longitudinal duality.
5.
Diakinesis
: Continued bivalent contraction.
MeiosisSlide10
How do nonsister chromatids ensure that crossing over between them will occur without the loss or gain of a single nucleotide? One plausible mechanism for which there is considerable laboratory evidence postulates the following events.
www.accessexcellence.org
Crossing OverSlide11
Linkage
Complete:
Pairs of loci are so close together that crossing over rarely occurs and recombinant types are generally not recovered.
0 10
0 crossovers between loci 1,2,3 and 10 crossovers between the cluster of loci 1/2/3 and locus 100 (in a population of 100 individuals)
Partial:
Pairs of loci are sufficiently far apart that some recombinant types are recovered. The "distance" between genes ranges from a few percent recombination to 50% recombination
.
1,2,3……..100Slide12
Linkage
Terms describing the allelic condition at linked loci
cis
aka
coupling
trans
aka
repulsion
A……..B
a……..b
A……..b
a……..BSlide13
Linkage
At Pachynema, there can be breakage of chromatids, followed by their fusion with sister chromatids, or with non-sister chromatids. As long as orientation corresponds, reciprocal exchanges between sister chromatids will not be detected and will not lead to genetic variability. Slide14
Keys points in non-sister chromatid exchange
:
Crossing over
does
not involve loss or addition of
chromatinOnly two chromatids are involved in any single crossover
event
There may be multiple crossovers between non-sister chromatids
Any
combination of crossover configurations can occur, and the outcome of such configurations can be radically different
Linkage Slide15
Linkage
Factors affecting meiotic crossing over
Sex
chromosomesSlide16
Dioecy
Evolution of sex chromosomes from autosomes
Accumulation of sex-determining genes on a single chromosome with no homolog
prevents
recombination between sex-determining genes
Linkage Slide17
Linkage
Factors affecting meiotic crossing over
Position on chromosomeSlide18
Linkage
Types of Crossovers
Single crossover
2-strand double crossover
3-strand double crossover
4-strand double crossover Slide19
Oregon Wolfe Barley Map
187 classical lociSlide20
Locus abbreviations and alleles
Vrs1 = V: V, v
GBSS = W:
W,w
Nud
= N:
N,n
LKs2 = L:
L,l
N
L
N
L
W
W
n
l
n
l
w
w
V
V
v
v
Linkage Relationships
Oregon Wolfe Barley Parents
OWB Dominant
2-row; normal starch; hulled grain; long awns
VV WW NN LL
OWB Recessive
6
-row; waxy starch; naked grain; short awns
vv
ww
nn
llSlide21
N
L
N
L
W
W
n
l
n
l
w
w
V
V
v
v
An example of how independent assortment and recombination could lead to an entirely new combination of traits
: 2-row; waxy starch; hulled grain; short awn
OWB Dominant
2-row; normal starch; hulled grain; long awns
VV WW NN LL
OWB Recessive
6
-row; waxy starch; naked grain; short awns
vv
ww
nn
ll
x
n
l
N
L
w
W
v
V
F1Slide22
n
l
N
L
w
W
v
V
n
l
w
v
N
L
W
V
n
l
w
v
N
L
W
V
Meiosis 1 & Crossing Over – Cell 1
After
DNA replication
n
l
w
v
n
l
W
V
N
L
w
v
N
L
W
V
Pairing and
crossing over
n
l
w
v
n
l
W
V
N
L
w
v
N
L
W
V
Anaphase I
Pre-meiotic
Heterozygous parent
Ch
2
Ch
7Slide23
Meiosis 2 & Crossing Over – Cell 1
n
l
w
v
n
l
W
V
N
L
w
v
N
L
W
V
Ch
2
Ch
7
Ch
2
Ch
7
Anaphase II
Ch
2
Ch
7
N
L
W
V
Non-Recombinant
2row, non-waxy, hulled, long awn
N
L
w
v
Recombinant
6row, waxy, hulled, long awn
n
l
W
V
Recombinant
2row, non-waxy, naked, short awn
n
l
w
v
Non-Recombinant
6row, waxy, naked, short awnSlide24
n
l
N
L
w
W
v
V
n
l
w
N
L
W
n
l
w
N
L
W
v
V
v
V
Meiosis 1 & Crossing Over – Cell 2
After
DNA replication
n
l
w
n
l
W
N
L
w
N
L
W
v
V
v
V
Pairing and
crossing over
n
l
w
n
l
W
N
L
w
N
L
W
v
V
v
V
Anaphase I
Pre-meiotic
Heterozygous parent
Ch
2
Ch
7Slide25
Meiosis 2 & Crossing Over – Cell 2
N
L
W
v
N
L
w
V
n
l
W
v
n
l
w
V
Ch
2
Ch
7
Ch
2
Ch
7
Anaphase II
Ch
2
Ch
7
Non-Recombinant
2row, waxy, naked, short awn
n
l
w
V
6row, non-waxy, naked, short awn
Recombinant
n
l
W
v
Recombinant
2row, waxy, hulled, short awn
N
L
w
V
Non-Recombinant
6row, non-waxy, hulled, long awn
N
L
W
vSlide26
N
L
w
V
N
L
W
v
Non-Recombinant
Non-Recombinant
Recombinant
Recombinant
Doubled Haploid Phenotypes
2row, waxy, naked, short awn
6row, non-waxy, hulled, long awn
6row, non-waxy, naked, short awn
2row, waxy, hulled, short awn
n
l
w
V
n
l
W
v
N
L
w
V
N
L
W
v
n
l
w
V
n
l
W
v
n
l
W
V
n
l
w
v
N
L
W
V
N
L
w
v
n
l
W
V
n
l
w
v
Non-Recombinant
Non-Recombinant
Recombinant
Recombinant
2row, non-waxy, hulled, long awn
6row, waxy, naked, short awn
6row, waxy, hulled, long awn
2row, non-waxy, naked, short awn
N
L
W
V
N
L
w
v
Ch
2
Ch
7Slide27
Gametes
VW
Vw
vW
vw
DH
VVWW
VVww
vvWW
vvww
Class
Non-recombinant
Recombinant
Recombinant
Non-Recombinant
1
1
1
1
0.25
0.25
0.25
0.25
Dihybrid
Ratio for Doubled Haploids
(2/4 loci)
Note: The “recombinant “ types in this case (between the V and W loci) are not due to
crossovers. They are due to random alignment of paired homologous chromosomes at
Metaphase I Slide28
Gametes
VW
Vw
vW
vw
DH
VVWW
VVww
vvWW
vvww
Non-recombinant
Recombinant
Recombinant
Non-Recombinant
0.25
0.25
0.25
0.25
Expected
20.5
20.5
20.5
20.5
Observed
20
15
25
22
(O-E)
2
/E
0.012
1.476
0.988
0.110
2
[3]
= 2.586; P=0.460; Null hypothesis of no linkage accepted
Independent Assortment : Loci are on different chromosomes
Random alignment of chromosomes at Metaphase 1 (slide 22 vs. slide 24)
Linkage Test – 2/6row v waxy/nonSlide29
Gametes
WN
Wn
wN
Wn
DH
WWNN
WWnn
wwNN
wwnn
Non-recombinant
Recombinant
Recombinant
Non-Recombinant
0.25
0.25
0.25
0.25
Expected
20.5
20.5
20.5
20.5
Observed
27
18
13
24
(O-E)
2
/E
2.061
0.305
2.744
0.598
2
[3]
= 5.707; P=0.128; Null hypothesis of no linkage accepted
Independent Assortment:
Sufficient crossovers between loci far enough apart on same
chromosome
Linkage Test – waxy/non v hulled/nakedSlide30
Gametes
WL
Wl
wL
wl
DH
WWLL
WWll
wwLL
wwll
Non-recombinant
Recombinant
Recombinant
Non-Recombinant
0.25
0.25
0.25
0.25
Expected
20.5
20.5
20.5
20.5
Observed
26
19
20
17
(O-E)
2
/E
1.476
0.110
0.012
0.598
2
[3]
= 2.195; P=0.533; Null hypothesis of no linkage accepted
Independent Assortment:
Sufficient crossovers
between loci far enough apart on same chromosome
Linkage Test – waxy/non v long/short awnSlide31
Gametes
NL
Nl
nL
nl
DH
NNLL
NNll
nnLL
nnll
Non-recombinant
Recombinant
Recombinant
Non-Recombinant
0.25
0.25
0.25
0.25
Expected
20.5
20.5
20.5
20.5
Observed
37
3
9
33
(O-E)
2
/E
13.280
14.939
6.451
7.622
2
[3]
= 42.293; P=<0.001; Null hypothesis of no linkage rejected
Not
Independent Assortment = linkage
Few crossovers
between
loci close together on
same chromosome
Link Test – naked/hulled v long/short awnSlide32
Gametes
4
a
b
Non Recombinant n4
3
a
B
Recombinant n3
2
A
b
Recombinant n2
1
A
B
Non Recombinant n1
Total N
Recombination frequency
Parent
2
1
4
3
A
B
A
B
a
b
a
b
RecombinationSlide33
Parentals
(Non-recombinants) >> Recombinants = linkage
Linkage is calculated as the total Recombinant types/ total population size
Nl
(n2) = 3;
nL
(n3) = 9; N = 82
r = (n2+n3)/N; r= 12/82 = 14.6%
N and L are approximately 14.6 map units apart
They are linked
“The
association of two or more phenotypic characters in inheritance because the genes controlling these characters are located in the same
chromosome”
Recombination FrequencySlide34
Three Point Linkage Test
W
N
L
r
WN
=0.39
r
NL
=0.14
r
WL
=0.48
The recombination frequency over the interval
W
–
L
(
r
WL
) is greater than either
r
WN
or
r
NL
therefore W and L are the two loci furthest apart and N is likely to lie between them. N is likely to be closer
t
o L as
r
WN
>
r
NL
.Slide35
% recombination values are biased due to double crossovers
Double crossovers
can give parental combinations of alleles and therefore underestimate the % recombination at values of ~ 10% recombination and higher
At values of ~ 10% recombination and less, double crossovers occur less often than expected
% recombination values are not additive: the maximum of recombination is 50%Therefore the centiMorgan – a computed value derived from
the observed % recombination Recombination and Map DistanceSlide36
Non-
Additivity
of recombination frequencies
A
B
C
r
AB
r
BC
r
AC
The recombination frequency over the interval A – C (
r
AC
) is less than the sum of
r
AB
and
r
BC
:
r
AC
<
r
AB
+
r
BC
.
This is because (rare) double recombination events (a recombination in both A - B and B - C) do not contribute to recombination between A and C.Slide37
Distance (
cM
)
Haldane
=-0.5*LN(1-2*r)
Distance (cM) Kosambi
=0.25*LN((1+2*r)/(1-2*r))
The centiMorgan is a unit of distance without a physical basis. This is readily apparent in organisms with large genomes, where the ratio of cM to base pairs can vary enormously across the
genome
Map Distance MeasuresSlide38
Mapping Functions
Mapping Function & RecombinationSlide39
Oregon Wolfe Barley Map
187 classical loci
BCD1434
0
DsT-66
7
Act8A
12
RbgMD
17
MWG837B
22
scind00046
ABC165C
25
Bmac0399
28
GBM1007
30
BCD098
36
GBM1042
49
Bmag0211
59
BG369940
64
ABC160
79
JS10C
109
MWG706A
119
KFP170
126
MWG2028
150
ABC261
151
KFP257B
152
WMC1E8
160
MWG912
162
ABG387A
scssr04163
166
1H
ABG058
0
scind02622
1
ABG008
13
scssr10226
41
scssr07759
45
GBM1066
47
Pox
50
scssr03381
62
Hot1
68
Ebmac0684
scssr02236
69
scssr00334
72
ABG356
75
GBM1023
77
scsnp03343
90
vrs1
96
Bmag0125
102
DsT-41
105
MWG503
110
GBM1062
111
KFP203
112
MWG882A
115
ABG072
133
Ebmc0415
150
cnx1
152
Zeo1
161
GBM1019
173
Aglu5
Aglu4
176
MWG720
179
GBM1012
185
MWG949A
scssr08447
193
2H
GBM1074
0
MWG798B
6
Dst-27
9
BCD706
12
alm
37
Bmac0209
41
ABC325
45
DsT-67
48
scssr25691
63
ABG377
65
Bmag0225
73
ABG499
98
scsnp00940
123
ABG004
147
scind02281
160
MWG883
165
HVM62
183
ABC805
191
ABC172
202
3H
MWG634
0
MWG077
20
HVM40
23
CDO542
29
scsnp01648
30
CDO122
31
hvknox3
35
ABC303
scssr20569
41
CDO795
44
DST-46
scind03751
scind04312a
50
Tef2
scind01728
52
GBM1020
60
Bmag0353
61
scind10455
68
scssr14079
80
ABG472
81
GBM1059
84
KFP221
93
Ebmac0701
95
MWG652B
96
GBM1048
102
Hsh
113
HVM67
114
KFP241.1
117
ABG601
127
4H
scssr02306
0
MWG618
7
ABC483
12
ABG610
13
ABG395
41
scssr02503
48
NRG045A
61
scsnp04260
62
Ale
63
ABC302
91
scind16991
94
scssr15334
96
scsnp06144
100
srh
111
RSB001A
135
scsnp00177
142
0SU-STS1
148
ABG003B
153
Tef3
184
MWG877
188
ABG496
203
E10757A
scsnp02109
224
ABG391
228
JS10B
229
ABC622
236
DsT-33
239
MWG602A
253
scssr03906
255
5H
Bmac0316
0
MWG602B
36
JS10A
48
GBM1021
55
GBM1068
67
BG299297
70
HVM31
73
rob
75
scssr02093
Bmag0009
76
ABG474
87
ABG388
97
scsnp21226
103
MWG820
107
GBM1008
134
scind04312b
142
GBM1022
145
Bmac0040
153
DsT-32B
157
DsT-28
170
DsT-74
171
MWG798A
MWG514
173
6H
ABG704
0
GBSS-I
16
AW982580
24
CDO475
MWG089
31
ABG380
33
scsnp00460
72
ABC255
73
ABC165D
74
scssr15864
87
GBM1030
91
MWG808
DAK642
94
scsnp00703
103
MWG2031
104
nud
108
lks2
124
ABC1024
128
Bmag0120
136
DsT-30
137
WG380B
138
ABC310B
149
Ris44
151
ABC253
169
ABG461A
179
WG380A
DsT-69
184
GBM1065
191
KFP255
215
ThA1
218
7H
124
–
108
=
16 cM on this map vs
.
14% two-locus recombination Slide40
Oregon Wolfe Barley Map
187 classical + 555 pilot-OPA1 + 467 pilot-OPA2 = 1209 loci
111- 96 = 15 cM on this map vs. 14% two-locus recombination Slide41
94.2 – 81.6 = 12.6 cM on this map vs.
14% two-locus recombination
cM < % recombination???
R
ecombination vs. Map DistanceSlide42
Determine order of loci in the chromosome
Determine how likely it is that non-parental combinations of alleles at linked loci will be obtained
Determine if trait relationships are due to linkage or pleiotropy
Create complete linkage maps (average ~ 100 cM/chromosome) where # chromosomes (n level) = number of linkage groups
Build a platform for cloning genesA framework for whole genome sequences
Identify genes controlling quantitative traits
Value of Genetic Linkage MapsSlide43
Visualise
the consequences of recombination Graphical
genotypes (haplotypes) of selected a mapping population
Graphical GenotypesSlide44
Linkage analysis is now an "automated"
procedure
D
ata on thousands, tens of thousands of loci
Key issues in linkage map construction are locus order and distance
The likelihood odds (LOD) score is a test statistic that is used to test the hypothesis that there is no linkage
LOD of 3.00 is approximately equal to P ~ .001LOD
> 3 then one concludes that two loci are indeed linked.
Making Linkage MapsSlide45
Locus ordering
No. of loci
k
Possible orders
No. of triplets
2
1
0
3
3
1
5
60
10
10
1,814,400
120
20
1.22 X 10
18
1,140
40
4.08 X 10
47
9,880
Number of orders among
k
loci
Number of triplets among
k
loci Slide46
Oregon Wolfe Barley Map
187 classical lociSlide47
Oregon Wolfe Barley Map
187 classical + 722
DArT
= 909 lociSlide48
Oregon Wolfe Barley Map
187 classical + 555 pilot-OPA1 + 467 pilot-OPA2 = 1209 lociSlide49
Oregon Wolfe Barley Map
187 classical + 555 pilot-OPA1 + 467 pilot-OPA2 + 722
DArT
= 1931 lociSlide50
Oregon Wolfe Barley Linkage
Map - (
2383 loci, Haldane
cM
)Slide51
Oregon Wolfe Barley Linkage
Map (2383 loci
+ 463RAD = 2846 loci)Slide52
Homoeology:
Chromosomes, or chromosome segments, which are similar in terms of the order and function of the genetic
loci.Homoeologous chromosomes may occur within a single allopolyploid individual (e.g. the A,B, and D
genomes in wheat)Or they may be found in related species (e.g. the 1A, 1B, 1D series in wheat
and the 1H of barley)
Map RelationshipsSlide53
Synteny
or
syntenic regions: Genetic
loci that are linked on the same chromosome Chromosomes and/or sections of chromosomes are reassembled in different configurations in related
organisms
Fragaria: Prunus
Genome Organisation
& MapsSlide54
Orthology
:
Genes in different species which are so similar in sequence that they are assumed to have originated from a single ancestral gene.
doi
/10.1371/journal.pcbi.1000703
Genome Organisation & MapsSlide55
Interference and MappingSlide56
Interference means that recombination events in adjacent intervals interfere. The occurrence of an event in a given interval may reduce or enhance the occurrence of an event in its
neighbourhood
.
Positive interference refers to the ‘suppression’ of recombination events in the
neighbourhood of a given one.
Negative interference refers to the opposite: enhancement of clusters of recombination events. Positive interference results in less double recombinants (over adjacent intervals) than expected on the basis of independence of recombination events.
Interference
r
AC
=rAB+r
BC
-2
C
r
AB
r
BCSlide57
Interference
C
= coefficient of coincidence
A
B
C
a
b
c
Interference
I
= 1 -
C
Coefficient
of coincidence
For N individuals, the expected number of double crossovers =
r
AB
r
BC
NSlide58
Observed
22
24
16
14
8
10
2
4
Coefficient of Coincidence (C)Slide59
Interference
No interference
C
= 1 and Interference = 1-C = 0
Complete interferenceC = 0 and Interference = 1-C = 1
Negative interferenceC > 1 and Interference = 1-C < 0Positive interference
C < 1 and Interference = 1-C > 0