What should be the frequency of a Single Nucleotide variation to be called as pathogenic variant for a rare disease?

Question

[garimav89]

In a rare disease having a small population size, direct sequencing data has given some single nucleotide mutations present in diseased individuals (Samples = 500 , controls = 500)for example). But the novel point mutation which are found is say only in 1/500. Can this mutation be called as novel and snp?

Secondly if this mutation is also present in healthy individuals, then it should be considered or not?

Some point mutations found in affected individuals (say 4/500), but also in controls (3/500), then can it be called as a mutation?

Thanks for your kind help

Answer 1

[Devon Ryan]

If a variant is only found in one person then it is, by definition, a novel SNP. DO NOT REFER TO THIS AS A MUTATION WITHOUT FURTHER EVIDENCE! You will find many variants, few of them will be mutations.

If a variant is found in 0.2% of patients and a similar rate in controls then it's not causative. That should be obvious. There is no statistical difference between 4/500 and 3/500 (the p-value is 1 if you do a Fisher's test).

Why don't you summarize by genes rather than variants. If you have a bunch of variants in one gene (or a group of genes in a particular complex or pathway) then do the summarization and statistics in that way.

[garimav89]

I have bunch of variants in one gene but the information regarding their presence in same patient or not is not provided in the research papers, I went through.

[Devon Ryan]

Ah, so this isn't your own data. You'll have to contact the authors and ask them.

Answer 2

[Carlo Yague]

So your question is: Is that SNP associated with the disease?

               mutated        normal
diseased             4           496
healthy              3           497

In that case, it's obvious that your SNP is not significantly more present in diseased samples.

To make sure, you can make a Fisher's test (here in R):

fisher.test(matrix(c(4,496,3,497),nrow=2,ncol=2,),alternative="greater")

The results tells you that your SNP is not significantly more present in diseased samples:

Fisher's Exact Test for Count Data

data:  matrix(c(4, 496, 3, 497), nrow = 2, ncol = 2, )
p-value = 0.5
-----------^^
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval: 0.2899126   ->  Inf
odds ratio = 1.335632

[Carlo Yague]

ps : pvalue = 1 as Devon Ryan said if you do a two-sided test, i.e., without adding

alternative="greater"

[Devon Ryan]

I can't believe I forget alternative="greater"! Good catch!