Practical 1 Key - Association Testing

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Last updated: 2024-06-13

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---

Before you begin:

• Make sure that R is installed on your computer
• For this lab, we will use a few R libraries:

library(data.table)

Case-Control Association Testing

Introduction

We will be using the LHON dataset covered in the lecture notes for this portion of the exercises. The LHON dataset is from a case-control study and includes both phenotype and genotype data for a candidate gene.

Let’s first load the LHON data file into the R session. We need to define the path to the file (if you have it downloaded on your machine, change the path to the file location).

LHON_FILE <- "https://raw.githubusercontent.com/joellembatchou/SISG2024_Association_Mapping/master/data/LHON.txt" 

We can now read the file

LHON <- fread(LHON_FILE, header=TRUE)

Helpful suggestions for R

There are many ways to obtain summary information for a dataset. Here are some short examples:

• Get information on number of rows/columns as well as the variables present in the data set

str(LHON)

Classes 'data.table' and 'data.frame':  328 obs. of  3 variables:
 $ IID  : chr  "ID1" "ID2" "ID3" "ID4" ...
 $ GENO : chr  "TT" "CT" "TT" "CT" ...
 $ PHENO: chr  "CONTROL" "CONTROL" "CASE" "CONTROL" ...
 - attr(*, ".internal.selfref")=<externalptr> 

• Get counts for a specific variable in the table (use $ to access a variable)

table(LHON$GENO)

 CC  CT  TT 
 16  74 238 

# cross tabulation for two variables
table(LHON$GENO, LHON$PHENO)

    
     CASE CONTROL
  CC    6      10
  CT    8      66
  TT   75     163

• Functions like as.numeric() and factor() will be useful to convert between numeric and categorical variables.

LHON$GENO[1:5] # see the first 5 entries

[1] "TT" "CT" "TT" "CT" "TT"

as.numeric(factor(LHON$GENO, levels = c("CC", "CT", "TT")))[1:5] # convert to numeric specifying the order of the labels

[1] 3 2 3 2 3

• Note: For any R function you don’t know the input syntax, you can get that information using ?<function_name>, e.g. ?table

Exercises

Here are some things to look at:

1. Examine the variables in the dataset
   • How many observations? (use str function)
   • How many cases/controls? (use table function)
   • What are the genotypes present in the variable GENO? (use table function)
     • To visualize the counts, you can use barplot(table(LHON$GENO))
   • What is the distribution of the genotypes across cases/controls? (use table function)

barplot(table(LHON$GENO))

Past versions of unnamed-chunk-4-1.png

Version	Author	Date
de1eb44	Joelle Mbatchou	2024-06-12

# Visualize the distribution of the genotypes across cases/controls
barplot(table(LHON$PHENO, LHON$GENO), col = c("blue","red"))
legend("topleft", legend = c("CASE", "CONTROL"), fill = c("blue","red"))

Past versions of unnamed-chunk-4-2.png

Version	Author	Date
de1eb44	Joelle Mbatchou	2024-06-12

# Compute allele frequency for allele 'T'
(1 * sum(LHON$GENO == "CT") + 2 * sum(LHON$GENO == "TT")) / (2 * nrow(LHON))

[1] 0.8384146

2. Perform a logistic regression analysis for this data with CC as the reference genotype using the glm() function.

• First convert the GENO variable to a factor

GENO_factor <- factor(LHON$GENO, levels = c("CC", "CT", "TT")) # convert to numeric specifying the order of the labels

• Convert the phenotype to a 0/1 variable

pheno_binary <- 1 * (LHON$PHENO == "CASE")

• Check that the entries in pheno_binary with 1 correspond to PHENO='CASE'

table(pheno_binary, LHON$PHENO)

            
pheno_binary CASE CONTROL
           0    0     239
           1   89       0

• Run logistic regression using the glm function

logistic_model_LHON <- glm(pheno_binary ~ GENO_factor, family = binomial(link = "logit"))

You can get information about the model fit and parameter estimates (i.e. coefficients):

summary(logistic_model_LHON)

Call:
glm(formula = pheno_binary ~ GENO_factor, family = binomial(link = "logit"))

Coefficients:
              Estimate Std. Error z value Pr(>|z|)  
(Intercept)    -0.5108     0.5164  -0.989   0.3226  
GENO_factorCT  -1.5994     0.6378  -2.508   0.0122 *
GENO_factorTT  -0.2654     0.5349  -0.496   0.6197  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 383.49  on 327  degrees of freedom
Residual deviance: 368.48  on 325  degrees of freedom
AIC: 374.48

Number of Fisher Scoring iterations: 4

3. Obtain odds ratios and confidence intervals (CI) for the CT and TT genotypes relative to the CC reference genotype. Interpret.
   • use the lecture notes to obtain odds ratios & CI from estimates and standard errors.

# Odds ratios for CT
exp(-1.5994)

[1] 0.2020177

# Odds ratios for TT
exp(-0.2654)

[1] 0.7668991

# Other way for all genotypes at once
exp(coef(logistic_model_LHON))

  (Intercept) GENO_factorCT GENO_factorTT 
    0.6000000     0.2020202     0.7668712 

# CI for CT
exp( -1.5994 + c(-1,1) * 1.96 * 0.6378)

[1] 0.05787394 0.70517308

# CI for TT
exp( -0.2654 + c(-1,1) * 1.96 * 0.5349)

[1] 0.2687956 2.1880353

# Using R function `confint.default()`
exp(confint.default(logistic_model_LHON, level = 0.95))

                   2.5 %   97.5 %
(Intercept)   0.21806837 1.650858
GENO_factorCT 0.05787424 0.705187
GENO_factorTT 0.26878265 2.187981

4. Is there evidence of differences in odds of being a case for the CT and TT genotypes (compared to CC)?

Check the p-values.

Extra

5. Perform the logistic regression analysis with an additive genotype coding (e.g. counting the number of ‘T’ alleles).
   • Hint: To convert to numerical, create a new variable with values 0/1/2 based on the genotypes (you can then use table() function to make sure the new variable was defined correctly).

# create additive coding variable
GENO_additive_T <- 0 + 1 * (LHON$GENO == "CT") + 2 * (LHON$GENO == "TT")
# check it is correct by comparing with the original variable
table(GENO_additive_T, LHON$GENO)

               
GENO_additive_T  CC  CT  TT
              0  16   0   0
              1   0  74   0
              2   0   0 238

# fit the logistic model
logistic_model_additive_LHON <- glm(pheno_binary ~ GENO_additive_T, family = binomial(link = "logit"))
summary(logistic_model_additive_LHON)

Call:
glm(formula = pheno_binary ~ GENO_additive_T, family = binomial(link = "logit"))

Coefficients:
                Estimate Std. Error z value Pr(>|z|)    
(Intercept)      -1.8077     0.4554  -3.970  7.2e-05 ***
GENO_additive_T   0.4787     0.2505   1.911   0.0559 .  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 383.49  on 327  degrees of freedom
Residual deviance: 379.47  on 326  degrees of freedom
AIC: 383.47

Number of Fisher Scoring iterations: 4

6. Obtain odds ratios and confidence intervals. Is there evidence of an association? How does it compare with the 2-parameter model?

exp(coef(logistic_model_additive_LHON))

    (Intercept) GENO_additive_T 
      0.1640322       1.6140439 

exp(confint.default(logistic_model_additive_LHON, level = 0.95))

                     2.5 %    97.5 %
(Intercept)     0.06718883 0.4004616
GENO_additive_T 0.98792490 2.6369796

Association Testing with Quantitative Traits

Introduction

We will be using the Blood Pressure dataset for this portion of the exercises. This dataset contains diastolic and systolic blood pressure measurements for 1000 individuals, and genotype data at 11 SNPs in a candidate gene for blood pressure. Covariates such as gender (sex) and body mass index (bmi) are included as well.

Let’s first load the file into R. We need to define the path to the file (if you have it downloaded on your machine, change the path to the file location).

BP_FILE <- "https://raw.githubusercontent.com/joellembatchou/SISG2024_Association_Mapping/master/data/bpdata.csv" 

Use the following command to read it into R:

BP <- fread(BP_FILE, header=TRUE)

• Get a snippet of the data:

head(BP, 2)

   V1    sex sbp dbp snp1 snp2 snp3 snp4 snp5 snp6 snp7 snp8 snp9 snp10 snp11
1:  1 FEMALE 171  89   CC   TT   TT   TT   CC   GG   AA   TT   TT    CC    TT
2:  2   MALE 160  99   TT   TT   CC <NA>   CC   AG   AT   CC   CT    CC    CT
   bmi
1:  25
2:  35

Exercises

Here are some things to try:

1. Perform a linear regression of systolic blood pressure (sbp) on SNP3 using the lm() function.

linear_model_BP <- lm(sbp ~ snp3, data = BP)

You can get information about the model fit and parameter estimates (i.e. coefficients):

summary(linear_model_BP)

Call:
lm(formula = sbp ~ snp3, data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-55.931 -12.428  -0.931  10.572  60.572 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 140.4283     0.7361 190.773   <2e-16 ***
snp3TC        2.5026     1.2840   1.949   0.0516 .  
snp3TT        5.2859     3.1868   1.659   0.0975 .  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.34 on 957 degrees of freedom
  (40 observations deleted due to missingness)
Multiple R-squared:  0.006019,  Adjusted R-squared:  0.003942 
F-statistic: 2.898 on 2 and 957 DF,  p-value: 0.05563

2. Is there any evidence of an effect of the SNP on systolic blood pressure?

Check the p-values.

3. Provide a plot illustrating the relationship between sbp and the three genotypes at SNP3.

• How does it compare with the linear model fitted in question (1)?

# show the mean in the boxplots (by default, only the median is shown)
with(BP, {
  # Draw the boxplots
  boxplot(sbp ~ snp3)
  # Calculate means
  means <- tapply(sbp, snp3, mean)
  # Add means to the boxplots
  points(x = 1:length(means), y = means, col = "red", pch = 18)
})

Past versions of unnamed-chunk-16-1.png

Version	Author	Date
de1eb44	Joelle Mbatchou	2024-06-12

4. By default, the 2-parameter model is used since the SNP is stored in the data as categorical. Contrast the parameter estimates, p-values and confidence intervals obtained between this model and using:

   • additive (linear) model (counting the T allele)
   • dominant model
   • recessive model

Hint: for each case, generate the appropriate allele coding variable and pass it to the lm() function. For example with additive coding:

SNP3_additive <- 0 + 1 * (BP$snp3 == "TC") + 2 * (BP$snp3 == "TT")
linear_model_BP_additive <- lm(sbp ~ SNP3_additive, data = BP)
summary(linear_model_BP_additive)

Call:
lm(formula = sbp ~ SNP3_additive, data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-55.974 -12.418  -0.974  10.582  60.582 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)   140.4179     0.7219 194.506   <2e-16 ***
SNP3_additive   2.5556     1.0615   2.407   0.0163 *  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.33 on 958 degrees of freedom
  (40 observations deleted due to missingness)
Multiple R-squared:  0.006014,  Adjusted R-squared:  0.004976 
F-statistic: 5.796 on 1 and 958 DF,  p-value: 0.01625

SNP3_dominant <- 0 + 1 * (BP$snp3 == "TC") + 1 * (BP$snp3 == "TT")
linear_model_BP_dominant <- lm(sbp ~ SNP3_dominant, data = BP)
summary(linear_model_BP_dominant)

Call:
lm(formula = sbp ~ SNP3_dominant, data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-56.218 -12.428  -0.823  10.572  60.572 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)    140.428      0.736 190.801   <2e-16 ***
SNP3_dominant    2.790      1.238   2.253   0.0245 *  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.34 on 958 degrees of freedom
  (40 observations deleted due to missingness)
Multiple R-squared:  0.005269,  Adjusted R-squared:  0.00423 
F-statistic: 5.074 on 1 and 958 DF,  p-value: 0.02451

SNP3_recessive <- 0 + 0 * (BP$snp3 == "TC") + 1 * (BP$snp3 == "TT")
linear_model_BP_recessive <- lm(sbp ~ SNP3_recessive, data = BP)
summary(linear_model_BP_recessive)

Call:
lm(formula = sbp ~ SNP3_recessive, data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-54.251 -12.501  -1.251  10.749  59.749 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)     141.251      0.604 233.854   <2e-16 ***
SNP3_recessive    4.463      3.163   1.411    0.159    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.37 on 958 degrees of freedom
  (40 observations deleted due to missingness)
Multiple R-squared:  0.002074,  Adjusted R-squared:  0.001032 
F-statistic: 1.991 on 1 and 958 DF,  p-value: 0.1586

For question 5 and 6 below, R also has a ‘formula’ syntax, frequently used when specifying regression models with many predictors. To regress an outcome y on several covariates, the syntax is:

lm(y ~ covariate1 + covariate2 + covariate3)

5. Now redo the linear regression analysis of sbp from question 4 for the additive model, but this time adjust for sex and bmi. Do the results change?

linear_model_BP_cov_adj <- lm(sbp ~ sex + bmi + SNP3_additive, data = BP)
summary(linear_model_BP_cov_adj)

Call:
lm(formula = sbp ~ sex + bmi + SNP3_additive, data = BP)

Residuals:
   Min     1Q Median     3Q    Max 
-58.83 -12.81  -0.82  11.58  57.80 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)   145.85380    3.00271  48.574  < 2e-16 ***
sexMALE        -4.77580    1.17642  -4.060 5.32e-05 ***
bmi            -0.09837    0.09481  -1.038   0.2997    
SNP3_additive   2.63566    1.05434   2.500   0.0126 *  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.19 on 955 degrees of freedom
  (41 observations deleted due to missingness)
Multiple R-squared:  0.02402,   Adjusted R-squared:  0.02096 
F-statistic: 7.836 on 3 and 955 DF,  p-value: 3.608e-05

Extra

6. What proportion of the variance of sbp is explained by all 11 SNPs combined using categorical coding?

• Use the summary() function to see the model results (the proportion of variance is the “Multiple R-squared” quantity)

linear_model_BP_all_snps <- lm(sbp ~ snp1+snp2+snp3+snp4+snp5+snp6+snp7+snp8+snp9+snp10+snp11, data = BP)
summary(linear_model_BP_all_snps)

Call:
lm(formula = sbp ~ snp1 + snp2 + snp3 + snp4 + snp5 + snp6 + 
    snp7 + snp8 + snp9 + snp10 + snp11, data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-50.722 -11.967  -0.703  11.021  61.704 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 133.1726    12.4033  10.737   <2e-16 ***
snp1CT       -1.7048     4.5991  -0.371    0.711    
snp1TT        1.9319     8.2839   0.233    0.816    
snp2AT        0.7347     5.5923   0.131    0.896    
snp2TT       -0.5118     6.9317  -0.074    0.941    
snp3TC        4.7672     5.0211   0.949    0.343    
snp3TT        6.6913     9.7904   0.683    0.495    
snp4CT       -0.4778     3.5501  -0.135    0.893    
snp4TT        2.3431     6.4874   0.361    0.718    
snp5CT        1.1896     3.0462   0.391    0.696    
snp5TT       -2.2787     7.5490  -0.302    0.763    
snp6AG       -3.0266     2.0697  -1.462    0.144    
snp6GG        2.1230     4.6650   0.455    0.649    
snp7AT       -3.0873     3.9148  -0.789    0.431    
snp7TT       -2.6319     4.3146  -0.610    0.542    
snp8CT       -1.5509     3.6318  -0.427    0.669    
snp8TT       -2.5507     7.3228  -0.348    0.728    
snp9CT        6.0693     7.6170   0.797    0.426    
snp9TT        4.7385     7.4517   0.636    0.525    
snp10CT       1.4330     1.6466   0.870    0.384    
snp10TT       1.9810     2.0699   0.957    0.339    
snp11CT       4.8005     6.5175   0.737    0.462    
snp11TT       4.0226     9.2775   0.434    0.665    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.2 on 707 degrees of freedom
  (270 observations deleted due to missingness)
Multiple R-squared:  0.02633,   Adjusted R-squared:  -0.003965 
F-statistic: 0.8691 on 22 and 707 DF,  p-value: 0.6372

summary(linear_model_BP_all_snps)$r.sq

[1] 0.02633265

• How would it differ if an additive coding is used for the 11 SNPs?
  • use unique() to check the genotypes for each SNP, e.g. unique(BP$snp1)
  • count the T allele (or A allele if applicable)

unique(BP$snp1)

[1] "CC" "TT" "CT" NA  

SNP1_additive <- 0 + 1 * (BP$snp1 == "CT") + 2 * (BP$snp1 == "TT")
unique(BP$snp2)

[1] "TT" "AT" "AA" NA  

SNP2_additive <- 0 + 1 * (BP$snp2 == "AT") + 2 * (BP$snp2 == "TT")
unique(BP$snp3)

[1] "TT" "CC" "TC" NA  

SNP3_additive <- 0 + 1 * (BP$snp3 == "TC") + 2 * (BP$snp3 == "TT")
unique(BP$snp4)

[1] "TT" NA   "CT" "CC"

SNP4_additive <- 0 + 1 * (BP$snp4 == "CT") + 2 * (BP$snp4 == "TT")
unique(BP$snp5)

[1] "CC" NA   "CT" "TT"

SNP5_additive <- 0 + 1 * (BP$snp5 == "CT") + 2 * (BP$snp5 == "TT")
unique(BP$snp6)

[1] "GG" "AG" "AA" NA  

SNP6_additive <- 0 + 1 * (BP$snp6 == "AG") + 2 * (BP$snp6 == "AA")
unique(BP$snp7)

[1] "AA" "AT" "TT" NA  

SNP7_additive <- 0 + 1 * (BP$snp7 == "AT") + 2 * (BP$snp7 == "AA")
unique(BP$snp8)

[1] "TT" "CC" "CT" NA  

SNP8_additive <- 0 + 1 * (BP$snp8 == "CT") + 2 * (BP$snp8 == "TT")
unique(BP$snp9)

[1] "TT" "CT" NA   "CC"

SNP9_additive <- 0 + 1 * (BP$snp9 == "CT") + 2 * (BP$snp9 == "TT")
unique(BP$snp10)

[1] "CC" "CT" "TT" NA  

SNP10_additive <- 0 + 1 * (BP$snp10 == "CT") + 2 * (BP$snp10 == "TT")
unique(BP$snp11)

[1] "TT" "CT" "CC" NA  

SNP11_additive <- 0 + 1 * (BP$snp11 == "CT") + 2 * (BP$snp11 == "TT")

linear_model_BP_all_snps_additive <- lm(sbp ~ SNP1_additive+SNP2_additive+SNP3_additive+SNP4_additive+SNP5_additive+SNP6_additive+SNP7_additive+SNP8_additive+SNP9_additive+SNP10_additive+SNP11_additive, data = BP)
summary(linear_model_BP_all_snps_additive)

Call:
lm(formula = sbp ~ SNP1_additive + SNP2_additive + SNP3_additive + 
    SNP4_additive + SNP5_additive + SNP6_additive + SNP7_additive + 
    SNP8_additive + SNP9_additive + SNP10_additive + SNP11_additive, 
    data = BP)

Residuals:
    Min      1Q  Median      3Q     Max 
-53.638 -12.849  -0.522  11.032  61.683 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)    134.43839    9.42384  14.266   <2e-16 ***
SNP1_additive    1.88456    4.03838   0.467    0.641    
SNP2_additive   -1.95639    2.96674  -0.659    0.510    
SNP3_additive    4.60730    4.65652   0.989    0.323    
SNP4_additive    0.05946    3.11138   0.019    0.985    
SNP5_additive   -0.26494    2.58719  -0.102    0.918    
SNP6_additive    1.17284    1.80185   0.651    0.515    
SNP7_additive    0.28939    1.78362   0.162    0.871    
SNP8_additive    0.70702    2.78030   0.254    0.799    
SNP9_additive    2.17197    2.54774   0.853    0.394    
SNP10_additive   0.60685    1.01229   0.599    0.549    
SNP11_additive  -0.39009    4.15347  -0.094    0.925    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 18.17 on 718 degrees of freedom
  (270 observations deleted due to missingness)
Multiple R-squared:  0.01418,   Adjusted R-squared:  -0.0009268 
F-statistic: 0.9386 on 11 and 718 DF,  p-value: 0.5022

summary(linear_model_BP_all_snps_additive)$r.sq

[1] 0.01417639

Session information

sessionInfo()

R version 4.3.0 (2023-04-21)
Platform: aarch64-apple-darwin20 (64-bit)
Running under: macOS 14.5

Matrix products: default
BLAS:   /Library/Frameworks/R.framework/Versions/4.3-arm64/Resources/lib/libRblas.0.dylib 
LAPACK: /Library/Frameworks/R.framework/Versions/4.3-arm64/Resources/lib/libRlapack.dylib;  LAPACK version 3.11.0

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

time zone: America/New_York
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] data.table_1.14.8

loaded via a namespace (and not attached):
 [1] vctrs_0.6.2      cli_3.6.1        knitr_1.43       rlang_1.1.1     
 [5] xfun_0.39        highr_0.10       stringi_1.7.12   promises_1.2.0.1
 [9] jsonlite_1.8.5   workflowr_1.7.0  glue_1.6.2       rprojroot_2.0.3 
[13] git2r_0.32.0     htmltools_0.5.5  httpuv_1.6.11    sass_0.4.6      
[17] fansi_1.0.4      rmarkdown_2.22   jquerylib_0.1.4  evaluate_0.21   
[21] tibble_3.2.1     fastmap_1.1.1    yaml_2.3.7       lifecycle_1.0.3 
[25] whisker_0.4.1    stringr_1.5.0    compiler_4.3.0   fs_1.6.2        
[29] Rcpp_1.0.10      pkgconfig_2.0.3  rstudioapi_0.14  later_1.3.1     
[33] digest_0.6.31    R6_2.5.1         utf8_1.2.3       curl_5.0.1      
[37] pillar_1.9.0     magrittr_2.0.3   bslib_0.5.0      tools_4.3.0     
[41] cachem_1.0.8