Practical 1 - Association Testing

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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:

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)
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'
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)

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

Extra

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).

GENO_additive_T <- 0 + 1 * (LHON$GENO == "CT") + 2 * (LHON$GENO == "TT")

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

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:

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)

Is there any evidence of an effect of the SNP on systolic blood pressure?
Provide a plot illustrating the relationship between sbp and the three genotypes at SNP3.

with(BP, boxplot(sbp ~ snp3))

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

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)

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)

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?

Extra

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)
- 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)

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):
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[13] htmltools_0.5.5  httpuv_1.6.11    sass_0.4.6       fansi_1.0.4     
[17] rmarkdown_2.22   jquerylib_0.1.4  evaluate_0.21    tibble_3.2.1    
[21] fastmap_1.1.1    yaml_2.3.7       lifecycle_1.0.3  whisker_0.4.1   
[25] stringr_1.5.0    compiler_4.3.0   fs_1.6.2         Rcpp_1.0.10     
[29] pkgconfig_2.0.3  rstudioapi_0.14  later_1.3.1      digest_0.6.31   
[33] R6_2.5.1         utf8_1.2.3       curl_5.0.1       pillar_1.9.0    
[37] magrittr_2.0.3   bslib_0.5.0      tools_4.3.0      cachem_1.0.8