Statistics Genomics Module #3

 

1Quest

Statistics Genomics Module #3

cheet sheet.without code.

1. Load the example SNP data with the following code:

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library(snpStats)

library(broom)

data(for.exercise)

use <- seq(1, ncol(snps.10), 10)

sub.10 <- snps.10[,use]

snpdata = sub.10@.Data

status = subject.support$cc

Fit a linear model and a logistic regression model to the data for the 3rd SNP. What are the coefficients for the SNP variable? How are they interpreted? (Hint: Don't forget to recode the 0 values to NA for the SNP data)

1 / 1 point

Linear Model = -0.16

Logistic Model = -0.04

Both models are fit using a dominance model. So in the linear model case, the coefficient is the decrease in probability associated with one copy of the minor allele. In the logistic regression case, it is the decrease in the log odds ratio associated with one copy of the minor allele.

Linear Model = 0.54

Logistic Model = 0.18

Both models are fit on the additive scale. So in both cases the coefficient is the decrease in probability associated with each additional copy of the minor allele.

Linear Model = -0.04

Logistic Model = -0.16

Both models are fit on the additive scale. So in the linear model case, the coefficient is the decrease in probability associated with each additional copy of the minor allele. In the logistic regression case, it is the decrease in the log odds ratio associated with each additional copy of the minor allele.

Linear Model = -0.04

Logistic Model = -0.16

Both models are fit on the additive scale. So in both cases the coefficient is the decrease in probability associated with each additional copy of the minor allele.

Correct

2.

Question 2

In the previous question why might the choice of logistic regression be better than the choice of linear regression?

1 / 1 point

The log odds is always more interpretable than a change in probability on the additive scale.

The logistic regression model is the natural model for GWAS data.

It is customary to use logistic regression for case-control data like those obtained from genome-wide association studies.

If you included more variables it would be possible to get negative estimates for the probability of being a case from the linear model, but this would be prevented with the logistic regression model.

Correct

3.

Question 3

Load the example SNP data with the following code:

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library(snpStats)

library(broom)

data(for.exercise)

use <- seq(1, ncol(snps.10), 10)

sub.10 <- snps.10[,use]

snpdata = sub.10@.Data

status = subject.support$cc

Fit a logistic regression model on a recessive (need 2 copies of minor allele to confer risk) and additive scale for the 10th SNP. Make a table of the fitted values versus the case/control status. Does one model fit better than the other?

1 / 1 point

The additive model fits much better since there are fewer parameters to fit and the effect size is so large.

The recessive model fits much better since there are more parameters to fit and the effect size is so large.

The additive model shows a strong effect, but the recessive model shows no difference so the additive model is better.

No, in all cases, the fitted values are near 0.5 and there are about an equal number of cases and controls in each group. This is true regardless of whether you fit a recessive or additive model.

Correct

4.

Question 4

Load the example SNP data with the following code:

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data(for.exercise)

use <- seq(1, ncol(snps.10), 10)

sub.10 <- snps.10[,use]

snpdata = sub.10@.Data

status = subject.support$cc

Fit an additive logistic regression model to each SNP. What is the average effect size? What is the max? What is the minimum?

1 / 1 point

Average effect size = 0.007, minimum = -4.25, maximum = 3.90

Average effect size = -0.02, minimum =-3.59 , maximum = 4.16

Average effect size = 1.35, minimum =-6.26 , maximum = 6.26

Average effect size = 0.02, minimum = -0.88, maximum = 0.88

Correct

5.

Question 5

Load the example SNP data with the following code:

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library(snpStats)

library(broom)

data(for.exercise)

use <- seq(1, ncol(snps.10), 10)

sub.10 <- snps.10[,use]

snpdata = sub.10@.Data

status = subject.support$cc

Fit an additive logistic regression model to each SNP and square the coefficients. What is the correlation with the results from using $\mathtt{\text{snp.rhs.tests}}$ and $\mathtt{\text{chi.squared}}$? Why does this make sense?

1 / 1 point

0.81 They are both testing for the same association using the same additive regression model on the logistic scale. But it doesn't make sense since they should be perfectly correlated.

> 0.99. They are both testing for the same association using the same additive regression model on the logistic scale but using slightly different tests.

> 0.99. They are both testing for the same association using the same additive regression model on the logistic scale. But it doesn't make sense since they should be perfectly correlated.

0.002 It doesn't make sense since they are both testing for the same association using the same additive regression model on the logistic scale. But it doesn't make sense since they should be perfectly correlated.

Correct

6.

Question 6

Load the Montgomery and Pickrell eSet:

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con =url("http://bowtie-bio.sourceforge.net/recount/ExpressionSets/montpick_eset.RData")

load(file=con)

close(con)

mp = montpick.eset

pdata=pData(mp)

edata=as.data.frame(exprs(mp))

fdata = fData(mp)

Do the log2(data + 1) transform and fit calculate F-statistics for the difference between studies/populations using genefilter:rowFtests and using genefilter:rowttests. Do you get the same statistic? Do you get the same p-value?

1 / 1 point

You get the same p-values and statistics. This is because the F-statistic and t-statistic are the exact same in this case.

You get different p-values and statistics. The F-statistic and t-statistic are testing the same thing but do it totally differently.

You get different p-values but the same statistic. This is because the F-statistic and t-statistic test the same thing when doing a two group test and one is a transform of the other.

You get the same p-value but different statistics. This is because the F-statistic and t-statistic test the same thing when doing a two group test and one is a transform of the other.

Correct

7.

Question 7

Load the Montgomery and Pickrell eSet:

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con =url("http://bowtie-bio.sourceforge.net/recount/ExpressionSets/montpick_eset.RData")

load(file=con)

close(con)

mp = montpick.eset

pdata=pData(mp)

edata=as.data.frame(exprs(mp))

edata = edata[rowMeans(edata) > 100,]

fdata = fData(mp)

First test for differences between the studies using the $\mathtt{\text{DESeq2}}$ package using the $\mathtt{\text{DESeq}}$ function. Then do the log2(data + 1) transform and do the test for differences between studies using the $\mathtt{\text{limma}}$ package and the $\mathtt{\text{lmFit}}$, $\mathtt{\text{ebayes}}$ and $\mathtt{\text{topTable}}$ functions. What is the correlation in the statistics between the two analyses? Are there more differences for the large statistics or the small statistics (hint: Make an MA-plot).

1 / 1 point

0.85. There are more differences for the large statistics.

0.63. There are more differences for the large statistics.

0.93. There are more differences for the large statistics.

0.93. There are more differences for the small statistics.

Correct

8.

Question 8

Apply the Benjamni-Hochberg correction to the P-values from the two previous analyses. How many results are statistically significant at an FDR of 0.05 in each analysis?

1 / 1 point

DESeq = 1119 significant;

limma = 2328 significant

DESeq = 12 significant;

limma = 3significant

DESeq = 1995 significant;

limma = 2807 significant

DESeq = 2807 significant;

limma = 1995 significant

Correct

9.

Question 9

Is the number of significant differences surprising for the analysis comparing studies from Question 8? Why or why not?

1 / 1 point

No. There are very few genes different between studies and that is what we would expect.

Yes and no. It is surprising because there very few genes that are significantly different, but it isn't that surprising because we would expect that when comparing measurements from very different batches.

Yes. This is a very large number of genes different between studies and we don't have a good explanation.

Yes and no. It is surprising because there is a large fraction of the genes that are significantly different, but it isn't that surprising because we would expect that when comparing measurements from very different batches.

Correct

10.

Question 10

Suppose you observed the following P-values from the comparison of differences between studies. Why might you be suspicious of the analysis?

1 / 1 point

There are too many small p-values so there are too may statistically significant results.

The p-values should have a spike near zero (the significant results) and be flat to the right hand side (the null results) so the distribution pushed toward one suggests conservative p-value calculation.

This p-value histogram appears correct in the case where there is a large number of statistically significant results.

The p-values should be uniformly distributed, but here they are not uniformly distributed.

Correct