P value difference between FISHER exact test and GLM - r

Question

I have a dateset plink.raw and I am testing if a maker CN00020133 (has three level 0, 1, 2) is associated with phenotype5. I want to compare 0 vs 1 and 2 vs 1 using GLM or fisher extract test:
table(plink.raw$phenotype5,plink.raw$CN00020133)
1 0 2
0 3559 0 7
1 14806 54 123
tested using GLM, I can see the p value for 0 vs 1 is 0.912894.
plink.raw$CN00020133 <- factor(plink.raw$CN00020133, levels=c("1","0","2"))
univariate=glm(phenotype5 ~ relevel(CN00020133,ref ="1"), family = binomial, data = plink.raw)
summary(univariate)
Call:
glm(formula = phenotype5 ~ relevel(CN00020133, ref = "1"),
family = binomial, data = plink.raw)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.4173 0.6564 0.6564 0.6564 0.6564
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 1.42555 0.01867 76.361 < 2e-16 ***
relevel(CN00020133, ref = "1")0 13.14051 120.12616 0.109 0.912894
relevel(CN00020133, ref = "1")2 1.44072 0.38902 3.703 0.000213 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 18158 on 18548 degrees of freedom
Residual deviance: 18114 on 18546 degrees of freedom
AIC: 18120
Number of Fisher Scoring iterations: 13
But if I tested it using fisher exact test, p value for 0 vs 1 is 4.618e-06.
Convictions <- matrix(c(0, 54, 3559, 14806), nrow = 2,dimnames = list(c("control", "case"),c("del/dup", "normal_copy")))
fisher.test(Convictions, alternative = "less")
Fisher's Exact Test for Count Data
data: Convictions
p-value = 9.048e-06
alternative hypothesis: true odds ratio is less than 1
95 percent confidence interval:
0.0000000 0.2374411
sample estimates:
odds ratio 0

Answer 1

You have a case of complete separation, also known as The Hauck-Donner effect occurs. If you CN00020133==0, all of them have 1 as phenotype, and this makes it hard to estimate the standard error of the coefficient. There's a fair amount of material on it, for example Alexej's blog, post by Brian Ripley, Ben Bolker's notes.
If you need to test for significance of the effect of "1", one solution is to use the likelihood ratio test:
df = rbind(
data.frame(phenotype=rep(0:1,c(3559,14806)),CN00020133="1"),
data.frame(phenotype=rep(0:1,c(0,54)),CN00020133="0")
)
anova(glm(phenotype ~ CN00020133,data=df,family=binomial),test="Chisq")
Analysis of Deviance Table
Model: binomial, link: logit
Response: phenotype
Terms added sequentially (first to last)
Df Deviance Resid. Df Resid. Dev Pr(>Chi)
NULL 18418 18082
CN00020133 1 23.227 18417 18059 1.44e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
This gives you different p-value compared to fisher test because it's a different distribution (binomial vs hyper-geometric). But more or less you can conclude there is an added effect of "1" using "0" as reference.
There is a implementation of Firth's logistic regression in R, you can try and see but I must say I am not very familiar with this:
library("logistf")
logistf(phenotype ~ CN00020133,data=df,family=binomial)