1. Analysis of single SNPs
A polymorphism
occurs when different individuals have many genetic variants at the same
location (loci) of their genomes. Every possible variant is called allele,
and if only one nucleotide has changed, the variant is named SNP (Single Nucleotide Polymorphism).
In this case, we usually have two possibilities for each loci,
for example changing T by C (denoted T→C). Then we say this loci is biallelic. Every individual carries
two alleles of every loci, one for each of the 22
autosomic chromosomes inherited independently from his parents. A genotype
is an observed couple of individual alleles for some loci. In the
former example, there are three potential genotypes: T/T, T/C and C/C. If the
two alleles are identical (T/T or C/C) we say the individual is homozygous. Otherwise, the individual will be heterozygous.
We can establish the
variant genotype as the less frequent, but it can differ amog
different populations.
1.1 Descriptive analysis
The
statistical analysis of a polymorphism is based on estimating the population
prevalence of each allele, and by doing it for each possible genotype we can
then estimate the genotype and allele frequencies. To estimate the genotype
frequencies (_{}) we calculate the observed rate of genotypes. Let be N the
sample size, then:
_{}
To
see an application, consider the test sample data included in the SNPstats
software (dataset 1). This dataset has information about 5 biallelic locus for 706 individuals. We choose one specific SNP (SNP2)
and the software performs the genotype frequencies (proportion") for all
the subjects and separately for every status subset, as shown in Figure 1:
Genotype frequencies (n=706) 


All subjects 
STATUS=0Control 
STATUS=1Case 

Genotype 
Count 
Proportion 
Count 
Proportion 
Count 
Proportion 
T/T 
255 
0.4 
123 
0.41 
132 
0.4 
T/C 
264 
0.42 
126 
0.42 
138 
0.41 
C/C 
116 
0.18 
52 
0.17 
64 
0.19 
NA 
71 
 
28 
 
43 
 
Figure 1. Note that the proportions are
calculated considering the 635 individuals with genotype information.
To calculate allele frequencies
(_{}) we double the sample chromosomes and count every allele
rate:
_{}
The results
of applying this formula to SNP2 are shown in the next table:
Allele frequencies (n=635) 


All subjects 
STATUS=0Control 
STATUS=1Case 

Allele 
Count 
Proportion 
Count 
Proportion 
Count 
Proportion 
T 
774 
0.61 
372 
0.62 
402 
0.6 
C 
496 
0.39 
230 
0.38 
266 
0.4 
Figure 2. Allele frequencies.
1.1.2 HardyWeinberg Equilibrium
First thing
we have to do before beginning an association analysis between genetic
polymorphisms and diseases is a test for HardyWeinberg equilibrium. This is
useful to assess independence between the alleles inherited from the parents.
The statistic test
HW=_{}
compares
the observed allele frequency with the expected one under the assumption of
independence, and has a chisquare distribution with one degree of freedom.
We consider
again the polymorphic biallelic loci SNP2 with a sample of size N=706 (71
missings) and two possible nucleotides, T and C, with probabilities p=0.61 and
q=0.39 respectively. The expected frequencies for each genotype are N_{}(=236,28) for T/T, 2Npq (=302.13) for T/C and N_{} (=96.58) for C/C which are compared with the observed
genotype frequencies shown in Figure 1.
The next
table shows the pvalues calculated via an exact test:
Exact
test for HardyWeinberg equilibrium (n=635) 


N11 
N12 
N22 
N1 
N2 
Pvalue 
All subjects 
255 
264 
116 
774 
496 
0.0015 
STATUS=0Control 
123 
126 
52 
372 
230 
0.051 
STATUS=1Case 
132 
138 
64 
402 
266 
0.012 
Figure 3. HardyWeinberg equilibrium.
If the
deviation was significative, first check the genotyping method to discard a
bias. At last the individuals could be
dependents due to the sampler method.
1.2 Analysis of association between polymorphisms and disease
The
statistic point of view lead us to describe a
polymorphism like a categorical variable with one level for each possible
genotype. The reference category often is the homozygous form.
To assess
association between one polymorphism and disease we make the contingency table
and then apply a chisquare test. The estimation of the OR (odds ratio) for each genotype respect to the reference genotype
will give us a notion about the measure of the association.
In case we
need to adjust the model by confounders variables, it
would rather use logistic regression models due to their versatility. These
models easily allow to assess interaction between the
polymorphism and the other factors.
Let us to
describe the logistic regression model. Let be p the case probability, G the
categorical variable with the polymorphisms codified and Z the variables to
adjust the model. The next equation defines the logistic model:
_{}
This
equation involves three parameters _{}that must be
estimated.
Now let be G
a SNP with a variant allele C which modifies the risk of be case. Like mentioned above, every individual
genotype is formed by two alleles. Now, the risk of every genotype depends on
the number of C copies carried (one or two).
According to the number of copies needed in order to modify the risk,
there are five inheritance models we can define:
Codominant
model: Is the most general model and it allows every
genotype to give a different and non additive risk. This model compares
heterozygous T/C (He) and homozygous for the variant allele C/C (Va) genotypes to the homozygous for the most frequent allele
T/T:
_{}
This model
estimates two ORs, one for He and one for
Dominant
model: A single copy of C is enough to modify the risk,
then heterozygous and homozygous genotypes have the same risk. We could compare
a combination of these two possible genotypes T/C+C/C (Do) to the homozygous
T/T.
_{}
Recessive
model: Two copies of C are necessary to change the risk.
Hence, T/C and T/T genotypes have the same effect. A combination of both
T/T+T/C (Re) is compared to the variant allele homozygous genotype C/C
_{}
Overdominant
model: Heterozigous are compared to a pool of both allele
homozygous, it is T/C (He) is compared
versus T/T+C/C.
_{}
Additive
model: Each copy of C modifies the risk in an additive
form, it is the homozygous C/C have double risk than heterozygous T/C. Now,
compare a combination of the two
genotypes with weights 2 and 1 respectively 2C/C+T/C (Ad), to T/T:
_{}
Look
below the models performed for SNP2 and
the ORs for each comparison:
SNP association with disease (crude analysis) 

Model 
Genotype 
STATUS=0Control 
STATUS=1Case 
OR (CI 95%) 
Pvalue 
AIC 
Codominant 
T/T 
123 (40.9%) 
132 (39.5%) 
1.00 
0.82 
884.2 
T/C 
126 (41.9%) 
138 (41.3%) 
1.02 (0.721.44) 

C/C 
52 (17.3%) 
64 (19.2%) 
1.15 (0.741.78) 

Dominant 
T/T 
123 (40.9%) 
132 (39.5%) 
1.00 
0.73 
882.5 
T/CC/C 
178 (59.1%) 
202 (60.5%) 
1.06 (0.771.45) 

Recessive 
T/TT/C 
249 (82.7%) 
270 (80.8%) 
1.00 
0.54 
882.2 
C/C 
52 (17.3%) 
64 (19.2%) 
1.14 (0.761.70) 

Overdominant 
T/TC/C 
175 (58.1%) 
196 (58.7%) 
1.00 
0.89 
882.6 
T/C 
126 (41.9%) 
138 (41.3%) 
0.98 (0.711.34) 

Additive 
 
 
 
1.06 (0.861.31) 
0.58 
882.3 
Figure 4. The last two columns are
useful for choose the best model.
1.2.1 Choose the best model
Given a SNP,
a criterion is necessary to decide the best inheritance model. To compare every
model to the most general model (the codominant) we can select the likelihood
ratio test (LRT). The likelihood ratio test
is a statistical test of the goodnessoffit between two models: we
compare a relatively more complex model to a simpler model to asses if it fits
a dataset better.
The LRT compares
the likelihood scores of the two models:
LR
= 2*(lnL1lnL2)
The statistic follows a chisquare
distribution with degrees of freedom equal to the number of additional
parameters in the more complex model. Even so, sometimes this test is not enough to discard models, and criteria like
the Akaike
information (AIC) could be useful to choose the inheritance model that
best fits the data. It is, the model with the less akaike information value,
that corresponds to minimize the expected entropy :
_{}
Figure 4.
shows the pvalue of the likelihood
ratio test for SNP2 and the AIC value. The inheritance model with less AIC is
the recessive.
1.3 Analysis of interactions with covariates
For the
interaction models a new term is added: the product between the genotype and an
environment variable:
_{}
_{}
or between
two genotype variables:
_{}
_{}coefficients allow us to describe the association between
each polymorphisms and the disease by way of calculate the ORs. Therefore, the
corresponding 95% confidence intervals are also computable.
Let be Z the
categorical covariate SEX of the sample data. The next tables show ORs for
every SNP2s variant relative to the most frequent, calculated for each
covariate level (male/female):
Corner interaction table (crude
analysis) 


Female 
Male 


STATUS=0Control 
STATUS=1Case 
OR (95% CI) 
STATUS=0Control 
STATUS=1Case 
OR (95% CI) 
T/T 
59 
50 
1.00 
64 
82 
1.51 (0.922.49) 
T/C 
57 
55 
1.14 (0.671.93) 
69 
83 
1.42 (0.872.33) 
C/C 
22 
21 
1.13 (0.562.28) 
30 
43 
1.69 (0.933.08) 
Figure 5.
First of the
next two tables (Figure 6.) shows the ORs
classifying first by the SNP, and then by SEX. The second table, first
separates by SNP and then by SEX:
SEX within SNP2 (crude analysis) 

T/T 


T/C 


C/C 


Interaction pvalue: 0.84 Trend test: 0.87 
Figure 6. Two pvalues are provided,
one for the interaction and another for the trend.
SNP2 within SEX (crude analysis) 

Female 


Male 


Interaction pvalue: 0.84 Trend test: 0.84 
Figure 7.
2. Analysis of multiple SNPs
In many situations
the causal polymorphism is unknown. In order to locate it, more than one polymorphism are taking into account to be analyzed.
2.1 Linkage disequilibrium and haplotypes
The
statistic correlation observable between different polymorphisms closer located
in the same chromosome, is called Linkage disequilibrium. This
association has its origin in the meiosis, the process of cell division which
occurs in the maturation of sex cells. At the end of the process therell be
two gametes, each gamete with one copy of each couple of chromosomes. Take
notice these chromosomes wont be identical to the parental chromosomes due to
the recombination (fragments of different copies of the same couple of
chromosomes are interchanged during the meiosis). Thereby, the chromosomes inherited by the offspring
will be a combination between mother and father chromosomes.
Even so, the
probability of have a recombination between closer locus are very low, and
usually these locus are transmitted in block. So that, the polymorphisms next to the causal will be
in association with the disease too, and analyze sets of locus could be very
useful for locate the true causal polymorphism.
The set of
polymorphisms transmitted together in every resultant chromosome is denominated
haplotype.
Given the sample genotypes, every individual has two possible haplotypes, one
per chromosome. Undesirably, usually this genotypes
are given with non chromosomical location, because in many cases sophisticated
laboratory techniques are required to. Due to this lack of information, when an
individual has at last two heterozygous locus his couple of haplotypes is unknown and in practice,
estimation methods like the EM algorithm or Markov Chain Monte Carlo methods
are used.
To determine
the linkage disequilibrium we calculate the D statistic, it is the deviation between the expected haplotype
frequency (under the assumption of no association) and the observed frequency.
Let be _{}and _{}the probabilities of
two alleles, and _{}the observed
probability of the couple. Then,
D=_{}_{}_{}
SNPstats also
offers the D' statistic that is equal to D scaled in [1,1] range.
The
correlation coefficient between alleles is provided too.
Look at the
next table to see the results for two SNPs (SNP1 and SNP2) given by SNPstats:
D statistic 
D' statistic 
r
statistic 


Snp1 
Snp2 

Snp1 
Snp2 

Snp1 
Snp2 
Snp1 
. 
0.2157 
Snp1 
. 
0.9814 
Snp1 
. 
0.9206 
Snp2 
. 
. 
Snp2 
. 
. 
Snp2 
. 
. 
Figure 8.
2.2 Analysis of haplotypes
Some
individuals have uncertain haplotypes
due to the lack of chromosomical information mentioned above. Hence the
count of sample haplotypes isnt straightforward. Moreover, missing values in the data increase this uncertainty.
There are
many different methods to estimate the Haplotype frequency. One of the
most commonly used is the twostage
iterative method named EM algorithm (Expectation
Maximization algorithm).
First,
initial values for the haplotype frequencies are given. Then, the E pass
consists in recalculate the expected genotype frequency for the genotypes with
uncertainty haplotypes (under Hardy Weinberg equilibrium) using the haplotypes
frequency . Using the recalculated genotype frequencies, the M pass calculates
every haplotype frequency. It is, count the compatibles haplotypes for every
genotype. At last, the algorithm converges desirably to the haplotype
frequencies. It could be advisable to repeat the method with different start
point to avoid local maximums.
The
haplotype frequency estimated via the EM Algorithm for two SNPs (SNP1 and
SNP2) is shown in the next table:
Haplotype frequencies estimation 


SNP1 
SNP2 
Total 
group.0.Control 
group.1.Case 
CumFreq 
1 
C 
T 
0.6062 
0.6169 
0.5965 
0.6062 
2 
G 
C 
0.3556 
0.3614 
0.3504 
0.9618 
3 
C 
C 
0.0341 
0.02 
0.0469 
0.9959 
4 
G 
T 
0.0041 
0.0017 
0.0062 
1 
Figure 9. Haplotype frequencies are shown for each level of the status variable. The cumulative frequency
is useful to decide which haplotypes are rare.
2.2.1 Analysis of association between haplotypes and disease
If theres no
uncertainty, the individual has his own couple of haplotypes. The association
between haplotypes and disease can be analyzed via logistic regression. A
chromosome analysis is usually done instead of treating individuals. The sample
is duplicated and then each individual is doubly represented with his two
haplotypes. The risk for every haplotype will be compared respect to the
reference category, it is the most frequent haplotype.
If there is
uncertainty, methods like the EM algorithm lead us to have a rebuilt haplotype
sample. The uncertain individuals
contribute with more than two haplotypes, it is with the haplotypes compatibles
with his genotype. In this case, every haplotype take a different weight in the
logistic regression model.
Analyzing
the association between haplotypes related to SNP1 and SNP2, and disease, SNPstats the next table:
Haplotype association with disease (crude analysis) 


SNP1 
SNP2 
Freq 
OR (95% CI) 
Pvalue 

1 
C 
T 
0.6062 
1.00 
 

2 
G 
C 
0.3556 
0.99 (0.80  1.23) 
0.94 

3 
C 
C 
0.0342 
2.41 (1.22  4.75) 
0.011 

rare 
* 
* 
0.0041 
3.78 (0.42  34.24) 
0.24 

Figure 10. The significative OR is red
written