Transcript pps - CS Technion

Mapping genes with
LOD score method
LOD score method


Aim: Determine , the recombinant
fraction (fraction of gametes that are
recombinant), using data from
relatively small families.
Reminder:  vary from 0 (2 genes
completely linked) to 0.5 (2 genes
are unlinked).
LOD score method (cont.)
There are 4 basic steps in the process:
1.
Determine the expected frequencies of F2 phenotypes for
every value of  from 0.01 to 0.5
2.
Determine the “likelihood” (L) that the family data
observed resulted from the given  value: the
maximum likelihood is the best estimate of  for given
data.
Determine the Odds Ratio and logarithm of the odds
ratio (lod score) by comparing the Likelihood for each
value of  to the Likelihood for unlinked genes (=0.5)
Add lod scores from different families to achieve an
acceptably high lod score so a specific most likely 
can be assigned.
3.
4.
LOD score method (cont.)
Lets see how it works on two genes showing complete
dominance:
P:
F1:
A
B
A
B
A
a
B
b
x
x
A_ B_
A_ bb
F2:
aa B_
aa bb
a
b
a
b
A
a
B
b
LOD score method (cont.)
Step 1: Calculate the expected frequency of offspring for values of  fro 0.01
to 0.5
Example:
Lets calculate expected offspring number for =0.2:
1.
P(Ab)=P(aB)=0.1 ; P(AB)=P(ab)=0.4
2.
AB/AB
0.16
Ab/AB
0.04
aB/AB
0.04
Ab/AB
0.16
AB/Ab
0.04
Ab/Ab
0.01
aB/Ab
0.01
Ab/Ab
0.04
AB/aB
0.04
Ab/aB
0.01
aB/aB
0.01
Ab/aB
0.04
AB/ab
0.16
Ab/ab
0.04
aB/ab
0.04
Ab/ab
0.16
3. F2 phenotype cell sums
expected freq
A_ B_
.16+.04+.04+.16+.04+.01+.04+.01+.16
0.66
A_ bb
0.01+0.04+0.04
0.09
aa B_
0.01+0.04+0.04
0.09
aa bb
0.16
0.16
LOD score method (cont.)
Step 2: Estimate the observed family data in light of the expected
distribution of offspring for each R value.
This is done by determining likelihood (L) of the observed family for each
value of R. The likelyhood is simply the probability of the observed family,
as determined by the multinomial theorem
(see http://mathworld.wolfram.com/MultinomialDistribution.html)
Lets define our terms for the observed family:

a = number of A_ B_ offspring

b = number of A_ bb offspring

c = number of aa B_ offspring

d = number of aa bb offspring

n = total offspring (a+b+c+d)
LOD score method (cont.)
…and terms for the expected family proportions (obtained fro Step1 above):

p = expected proportion of A_ B_ offspring

q = expected proportion of A_ bb offspring

r = expected proportion of aa B_ offspring

s = expected proportion of aa bb offspring
Then Likelihood will be calculated by the next formula:
n!
a b c d
L
p qr s
a!b!c!d!
LOD score method (cont.)
Example:
A family as in previous example has 5 children: 2 of A_ B_ phenotype, 1
with aa B_ and 2 with aa bb.
What is the likelihood of this family, given =0.2?
L=(5!/2!0!1!2!)(0.66)2(.09)0 (.09)1 (.16)2=0.0301
LOD score method (cont.)
Steps 3 and 4: Combining data from several families.
We want to be able to compare (and add) data from several different
families, to get a good estimate of R. To do this, the L values must be
standardized by calculating Odds Ratio (OR), which is the ratio of the L
for each  value divided by the L for =0.5 . Then, the logarithm of Odds
Ratio is taken; this is the lod score (Z).
Lod scores from different families can be added (this is equivalent to
multiplying the Odds Ratios, as in the AND rule for two events – family 1
and family 2 – both occurring).
A total lod score for some  value of 3.0 is considered proof of linkage
between two genes, which is not exactly right as will be explained futher…
Exclusion Mapping
In linkage analysis the main goal is localizing disease genes
relative to well-characterized marker loci (lod score > 3).
However with any given marker, the probability of finding
a positive test result is quite low as human genome is
quite large and most randomly selected markers are not
linked.
However, negative results are also results and may be used
for elimination of various chromosomal regions from
consideration…
Exclusion Mapping (cont.)
It’s important to remember that the likelihood ratio
test is a test of hypothesis of no linkage, such
that in the absence of a significant test result,
you fail to reject H0, meaning that there is no
significant evidence for linkage. However, this
does not mean that you accept H0 and have
proved by the failure to achieve a significant test
result that there is no linkage. It’s quite another
thing to prove the absence of linkage – a problem
that can be statistically very complicated…
Exclusion Mapping (cont.)
Morton has proposed (1955) that the test of linkage
be treated as a sequential likelihood ratio test
(LRT) of a simple hypothesis, = 1 .
He proposed that the new families continue to be
sampled until either the criterion Z(1)>3 is
fulfilled, in which case the hypothesis of no
linkage is rejected, or until Z(1)<-2, in which
case you would reject the hypothesis of linkage.
As long as -2<Z(1)<3, no conclusion may be
made.
Exclusion Mapping (cont.)
Chotai (1984) extended this concept to the general
case such that the positive test is considered
significant whenever Zmax>3; and the negative
test is considered significant on
{ | Z () < -2},
and the disease gene may be excluded from this
part of genome.
The same criteria may be applied for both two-point
and multipoint scores…
Model Errors and Exclusion
Mapping
It has bee shown that using incorrect model for the disease doesn’t
in general lead to an increased false-positive rate (ClergetDarpoux et al., 1986), as maximizing the lod score over models
does (Weeks et al. 1990a).
In other words, you are not more likely to obtain lod scores of 3 in
the absence of linkage under the wrong model than using the
correct one…
If there is linkage, however, there is lower power to detect it when
the model parameters are incorrectly specified…
Model Errors and Exclusion
Mapping (cont.)
Contrary to the lack of false-positives, the falsenegative rate may be astronomical when an
analysis is performed under incorrect model.
It’s quite easy to design an example where disease
gene will be mistakenly “excluded” from it’s
region by Z()<-2 criterion:
If there is a linkage in only 20% of families, then
summing the lod scores across the families can
easily lead us to spurious exclusions.
Model Errors and Exclusion
Mapping (cont.)
For this reason, doing a linkage analysis with a
complex disease, for which the model is not
accurately known, it’s not wise to use exclusion
analysis because the exclusion results obtained
apply only to that specific model. You can only
say that this region may be excluded only if the
analysis model is correct…