Missing p values for SNPs that are perfectly associated with trait

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Jeff Groh

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Feb 20, 2024, 2:45:46 PMFeb 20
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Hi,

I have a set of SNPs that I know from independent evidence are perfectly associated with the trait of interest (they are fixed between alleles at a sex locus, and I have an even sample of the two sexes, hence the allele freq of 0.75 at these SNPs). Gemma seems to not report p values for these SNPs. Here's an example of the output:

chr rs       ps n_miss allele1 allele0    af    logl_H1        l_mle        p_lrt       N
chr7  . 32346390      0       C       T 0.750        NaN          NaN          NaN 3477683
chr7  . 32346409      2       A       G 0.738  48.724300 1.000000e+05 4.257681e-34 3477684
chr7  . 32346418      4       A       C 0.750  27.123910 1.000000e+05 1.212737e-24 3477685
chr7  . 32346551      0       T       G 0.750        NaN          NaN          NaN 3477686
chr7  . 32346603      0       C       T 0.750        NaN          NaN          NaN 3477687
chr7  . 32346636      0       A       G 0.750        NaN          NaN          NaN 3477688
chr7  . 32346654      0       T       A 0.750        NaN          NaN          NaN 3477689
chr7  . 32346674      0       A       G 0.750        NaN          NaN          NaN 3477690
chr7  . 32346681      1       C       A 0.756  94.225260 1.000000e+05 5.827659e-54 3477691
chr7  . 32346686      1       A       G 0.756  50.811570 1.000000e-05 5.208712e-35 3477692

I get the same result for either wald p values or lrt p values.

Why does GEMMA not report p values for these SNPs?

Thanks,
Jeff

Michael Nagle

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Feb 20, 2024, 4:05:34 PMFeb 20
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If I understand right, you have a situation where R^2 = 1 for the relationship between your trait and a given SNP.

Calculation of Wald p-values requires the variance of the regression coefficient, and the variance would be undefined in this case.

I'm not sure why exactly there would be no p-value for the LRT. I'm not with the GEMMA development team, but they should know the specific answer.

In general, if you have a situation where R^2 = 1 for y and x, linear regression is probably not appropriate.

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