Download Linkage Mapping 2 3 – point linkage mapping One crossover Two

Recombination Frequency
Estimates
Linkage Mapping 2
– Increasing sample size
– Using codominant genes/markers
– Progeny testing F2:3s to better classify genotypes
with dominant markers (e.g., distinguish AA from Aa)
– Using F2 populations with codominant markers
– Using backcross populations with dominant markers
CS741
2009
Jim Holland
3 – point linkage mapping
Gene Pair
F - Rj1
F - Idh1
Rj1 - Idh1
Recombination Frequency
0.41 ± 0.05
0.22 ± 0.03
0.25 ± 0.03
F
Idh1
0.22
• They are ESTIMATES!
• Variance or standard error of the estimates can
be obtained from 2nd derivative (see Mather,
1951).
• You can get better estimates (smaller std.
deviations) by:
Rj1
0.25
Recombination frequencies are not
additive!
• The number of crossovers that occur
along a chromosome is additive, but there
is not a 1:1 relationship between crossover
events and recombination.
• What is the relationship between
crossovers and recombination?
• Remember that crossovers are the
CAUSE and recombinations are the
RESULT.
Notice: 0.22 + 0.25 > 0.41!
One crossover
A--------------- B
A--------------- B
a--------------- b
a--------------- b
→
A--------------- B
A--------------- b
a --------------- B
a --------------- b
Parental gamete
Recombinant gamete
Recombinant gamete
Parental gamete
50% recombinant gametes produced!
Two crossovers
A --------------- B
A --------------- B
a --------------- b
a --------------- b
→
A --------------- B
A --------------- B
a --------------- b
a --------------- b
Parental gamete
Parental gamete
Parental gamete
Parental gamete
A --------------- B
A --------------- B
a --------------- b
a --------------- b
→
A --------------- b
A --------------- b
a --------------- B
a --------------- B
Recombinant gamete
Recombinant gamete
Recombinant gamete
Recombinant gamete
A --------------- B
A --------------- B
a --------------- b
a --------------- b
→
A --------------- B
A --------------- b
a --------------- b
a --------------- B
Parental gamete
Recombinant gamete
Recombinant gamete
Parental gamete
A --------------- B
A --------------- B
a --------------- b
a --------------- b
→
A --------------- b
A --------------- B
a --------------- B
a --------------- b
Recombinant gamete
Parental gamete
Recombinant gamete
Parental gamete
Averaged over all four possibilities – 50% recombinant gametes are produced!
1
Crossovers and Recombination
Mapping Functions
• Any number of crossovers greater than zero will
produce 50% recombinant gametes on average.
• This is why recombination frequency is not a
linear function of the average number of
crossovers between two loci.
• If loci are widely separated on the chromosome,
several crossovers may occur between them
regularly at each meiosis, but they will still have
only a maximum recombination frequency of
50%.
• How to make a genetic map linear, so that map
distances of separate intervals can be added up
to equal distance of combined intervals?
• Mapping Functions! - genetic distances as a
function of the presumed number of crossovers
that occur between genes.
• An interval in which an average of one
crossover occurs every meiosis is defined to
have a genetic map distance of 50 cM.
Mapping Functions
Mapping Functions
• The problem is that we do not actually observe
crossovers, we observe the resulting recombinations.
• By making assumptions about the level of interference
(whether crossovers are really independent along
chromosome or if one crossover reduces the probability
of a nearby crossover), the number of crossovers can be
estimated from the observed recombination frequency.
• So, we are working backwards from the observed result
(recombinations) to the unobserved cause (crossovers).
• The expected number of crossovers across different
intervals then can be summed and treated as a linear
map distance.
Map Distance vs. Recombination
Frequency
⎛ 1 ⎞ ⎛ 1 + 2r ⎞
cM = ⎜ ⎟ ln⎜
⎟ × 100
⎝ 4 ⎠ ⎝ 1 − 2r ⎠
• Kosambi is usually preferred, as there is
evidence for interference, but in reality the level
of interference is unknown, and anyway seems
to vary across the genome (Sherman & Stack,
95)
Map Distance: Example
Gene Pair
F - Idh1
Rj1 - Idh1
F - Rj1
140
120
100
80
60
40
20
0.48
0.4
0.44
0.36
0.32
0.28
0.2
0.24
0.16
0.12
0.08
0
0
0.04
Kosambi cM
• Haldane mapping function assumes no
interference.
• Kosambi mapping function assumes a constant
and specific level of interference:
Recomb. Freq.
0.22 ± 0.03
0.25 ± 0.03
0.41 ± 0.05
cM
24
27
58
Notice that the map units are supposed to be more
additive than the original recombination units,
but in reality they are not always so! Differences
between the actual level of interference and the
assumptions in the mapping function cause this.
Recombination Frequency
2
LOD scores
LOD scores
• An alternative method of measuring the
evidence for linkage:
• “logarithm of odds”:
• From soybean example:
L(r$ = 0.41) =
250!
(.587025)144 (.162975) ( 44 + 39 ) (.087025) 23
144! 44! 39 ! 23!
⎛ (.587025)144 (.162975) ( 44 + 39 ) (.087025) 23 ⎞
⎛ L(r$ = 0.41) ⎞
LOD = log10 ⎜
⎟
⎟ = log10 ⎜
⎝ L(r = 0.5) ⎠
(9 / 16)144 (3 / 16) ( 44 + 39 ) (1 / 16) 23
⎠
⎝
⎛ L(r$MLE ) ⎞
LOD = log10 ⎜
⎟
⎝ L(r = 0.5) ⎠
⎛ 8.02425 × 10−124 ⎞
= log10 ⎜
⎟ = log10 (8.364) = 0.9224
⎝ 9.59378 × 10− 125 ⎠
LOD 1 means estimated recombination frequency is 10 times more likely than 50% recombination.
LOD 3 means estimated recombination frequency is 1000 times more likely than 50% recombination.
Typically in full genome linkage mapping, we consider genes with LOD 2.5 or more
linked. Why use such a stringent level of significance?
Multipoint Mapping in Mapmaker
Multipoint Mapping in Mapmaker
• First, must select the linear order of loci in the
linkage group.
• With many loci, it is not easy to know the correct
order (huge numbers of possible orders).
• Tightly linked loci make it even harder!
• So, usually start with a subset of loci to make
ordering simpler and a best guess of the order.
• Given the order, the maximum likelihood
estimate (MLE) of the recombination frequencies
is computed by extending the two locus
multinomial probability function to a multiple
locus probability function.
• Try different orders of loci.
• The LOD scores of the MLE for the different orders can
be compared, the highest LOD score is selected.
• Additional loci can be added one at a time to the
framework order to build up the map of the linkage
group.
• At each step, compute the MLE and LOD score and also
shuffle the order of loci a few times and compute LODs
for each new order.
• Keep the order with best LOD score.
• You are not guaranteed the most likely order! Two
different people can get two different maps from the
same data set!
Linkage Map Example
Portyanko et al., 2001
OT2
OT1 (KO22)
24
22
E
3
0.8
1.2
0.8
1.6
2.8
6.1
1.9
8.2
3.5
1.6
1.6
4.3
6.3
aa03.875
isu59a
adh2a
psr144
psr153
oisu1093
cdo795a [22,24;E]
rz474a
cdo590a [3,24;A]
e35m61-101.o
10.7
bcd1235d [25]
5.8
isu92
rz912
oisu1877b, cdo504 [E], rz630 bcd1734 [22;E]
cdo87 [E], cdo241a [E], oisu1961a [28]
e40m48-173.t bcd454 (=bcd450) [3;A,E,F,G], cdo1373b [E],
bcd1087 [E]
10.5
skdh [32]
E
5.6
2.6
oisu2191c
isu77d
bcd349 [G] cdo1373a [E]
rz474b
isu128b
10.3
bcd808d [24,28;E], cdo344c [30;F],
psr152a
cdo595 [6;D]
F
cdo216a* [3;A],, isu35c, plc, srgh8b**,
31.8
D
15.0
6
32
cdo407 [D]
cdo1090c [6,14,20,23,32]
4.8
cdo1081a [5;F]
12.4
cdo1090e [6,14,20,23,32]
bcd1840b [33,35], cdo216c [3;A]
isu107b**
8.2
72.5 cM
A
Markers in bold
are framework
markers,
positions are
certain with LOD
> 3.0
2.3
2.1
6.7
cdo216e [A]
cdo395 [32;A,B]
cdo216d [3;A]
pic20b
psr167
97.0 cM
Using common
The sum of interval
markers, you can
distances is used as total
compare maps across
populations and species map length for the linkage
group
Markers in italics
have positions
supported at LOD
< 3.0, so not used
to compute map
distances. They
may be in wrong
interval.
3