Class 17: Lab Session

Laura Lu (PID: A17844089)

Section 1. Proportion og G/G in a population

Downloaded a CSV file from Ensemble < https://useast.ensembl.org/Homo_sapiens/Variation/Sample?db=core;r=17:39894595-39895595;v=rs8067378;vdb=variation;vf=959672880#373531_tablePanel >

Here we read this CSV file

mxl <- read.csv("373531-SampleGenotypes-Homo_sapiens_Variation_Sample_rs8067378.csv")
head(mxl) 

  Sample..Male.Female.Unknown. Genotype..forward.strand. Population.s. Father
1                  NA19648 (F)                       A|A ALL, AMR, MXL      -
2                  NA19649 (M)                       G|G ALL, AMR, MXL      -
3                  NA19651 (F)                       A|A ALL, AMR, MXL      -
4                  NA19652 (M)                       G|G ALL, AMR, MXL      -
5                  NA19654 (F)                       G|G ALL, AMR, MXL      -
6                  NA19655 (M)                       A|G ALL, AMR, MXL      -
  Mother
1      -
2      -
3      -
4      -
5      -
6      -

table(mxl$Genotype..forward.strand.) 

A|A A|G G|A G|G 
 22  21  12   9 

table(mxl$Genotype..forward.strand.) / nrow(mxl) * 100 

    A|A     A|G     G|A     G|G 
34.3750 32.8125 18.7500 14.0625 

Now let’s look at a different population. I picked the GBR.

gbr <- read.csv("373522-SampleGenotypes-Homo_sapiens_Variation_Sample_rs8067378.csv")

Find the proportion of G|Gv

round(table(gbr$Genotype..forward.strand.) / nrow(gbr) * 100, 2) 

  A|A   A|G   G|A   G|G 
25.27 18.68 26.37 29.67 

This variant that is associated with childhood asthma is more frequent in GBR population than the MKL population.

Let’s now dig into this further.

Section 4: Population Scale Analysis

One sample is obivously not enough to know what is happening in a population. You are interested in assessin genetic differences on a population scale.

How many samples do we have?

expr <- read.table("rs8067378_ENSG00000172057.6.txt")
head(expr) 

   sample geno      exp
1 HG00367  A/G 28.96038
2 NA20768  A/G 20.24449
3 HG00361  A/A 31.32628
4 HG00135  A/A 34.11169
5 NA18870  G/G 18.25141
6 NA11993  A/A 32.89721

nrow(expr) 

[1] 462

table(expr$geno) 

A/A A/G G/G 
108 233 121 

library(ggplot2)

Let’s make a boxplot

ggplot(expr) + aes(geno, exp, fill=geno) + 
  geom_boxplot(notch=TRUE)

> Q13. Read this file into R and determine the sample size for each genotype and their corresponding median expression levels for each of these genotypes

data <- read.table("rs8067378_ENSG00000172057.6.txt")
  header = TRUE

head(data)

   sample geno      exp
1 HG00367  A/G 28.96038
2 NA20768  A/G 20.24449
3 HG00361  A/A 31.32628
4 HG00135  A/A 34.11169
5 NA18870  G/G 18.25141
6 NA11993  A/A 32.89721

summary(data)

    sample              geno                exp        
 Length:462         Length:462         Min.   : 6.675  
 Class :character   Class :character   1st Qu.:20.004  
 Mode  :character   Mode  :character   Median :25.116  
                                       Mean   :25.640  
                                       3rd Qu.:30.779  
                                       Max.   :51.518  

table(data$geno)

A/A A/G G/G 
108 233 121 

tapply(data$exp, data$geno, median)

     A/A      A/G      G/G 
31.24847 25.06486 20.07363 

aggregate(exp ~ geno, data, median)

  geno      exp
1  A/A 31.24847
2  A/G 25.06486
3  G/G 20.07363

> Q14. Generate a boxplot with a box per genotype, what could you infer from the relative expression value between A/A and G/G displayed in this plot? Does the SNP effect the expression of ORMDL3?

boxplot(exp ~ geno, 
        data = data,
        xlab = "Genotype",
        ylab = "ORMDL3 Expression",
        main = "ORMDL3 Expression by rs8067378 Genotype")