Re: MMPBSA Question

From: McGuire, Kelly (
Date: Sat Oct 01 2022 - 20:12:39 CDT

I made sure I'm using independent samples for the standard error. I increased from 10 ps decorrelation to 20 ps, 50 ps, and 100 ps, and the standard errors were all below 1 kcal/mol, but large SD. It seems that the standard deviation is not a useful measurement in MMPBSA. So, report the standard error instead...? Just want to report honest results.

Dr. Kelly L. McGuire

PhD Biophysics

Department of Physiology and Developmental Biology

Brigham Young University

LSB 3050

Provo, UT 84602

From: <> on behalf of Josh Vermaas <>
Sent: Thursday, September 22, 2022 1:15 PM
To: <>; Mcguire, Kelly <klmcguire_at_UCSD.EDU>
Subject: Re: namd-l: MMPBSA Question

Hi Kelly,

Would you expect your standard deviation to go down with more frames/data points? The standard deviation measures in some general sense how far away a typical measurement is away from the mean of a measurement. If you took 1000 samples from a gaussian distribution, the standard deviation would be basically the same as if you took 1,000,000 samples. The standard error of the mean does decrease with increasing number of samples.

The better question I think is how many actually independent samples you have in your 5000 DCD frames. CaFE is assuming something like a ~100ps decorrelation time between frames. Is that enough? No idea, but technically you want independent samples when calculating a standard error.


On 9/22/22 1:56 PM, Mcguire, Kelly wrote:
Question about standard deviations with MMPBSA. Should I use millions of snapshots for better standard deviations. I ran a 10 nanosecond simulation of my protein-peptide complex (2 fs timesteps, saved to dcd every 1000 steps). That would be 5,000 frames total in the DCD. I used CaFE and NAMD to get my MMPBSA result (single trajectory method). For CaFE I used stride 5 for the trajectory so that I would use a frame every 10 ps, thus a total of 1,000 frames for the MMPBSA calculation. My result is in the attached file. I get -29.3 kcal/mol with a standard deviation of 11.5. That's a large standard deviation. But, my standard error seems decent (11.5/sqrt(1,000) = 0.37), which appears to be what others report for these calculations instead of the standard deviation. So is this a good result for a single trajectory only using 1,000 frames for the calculation, or do I really still need to use millions of frames? Thanks!


Dr. Kelly McGuire
Herzik Lab - Postdoc
Chemistry/Biochemistry Department
Natural Science Building, 4104A, 4106A, 4017

Josh Vermaas<>
Assistant Professor, Plant Research Laboratory and Biochemistry and Molecular Biology
Michigan State University

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