1
$\begingroup$

Given my work on ICeChIP (a biological assay that measures the binding of one, two, or more proteins along the genome)

https://github.com/shah-rohan/icechip/blob/master/Scripts/computeHMDandError

I would appreciate to have the answer to the question :

given two proteins A and B, that bind together in 100000 locations, the binding strength of each protein A and B on in location (i) is denoted by values a(i) or b(i). The assay also provides a calibration number "cal".

enter image description here

Why the 95 confidence interval of binding strength of protein A has been calculated as :

100 / cal * sqrt (( a/ (b^2) + (a^2) / (b ^3)) * 1.96

Bogdan

$\endgroup$
3

1 Answer 1

1
$\begingroup$

The formula in question is a simple application of the approximate formula for the variance of a ratio of uncorrelated variables. In this case, it's the ratio of immunoprecipitated (IP'd) DNA to the total input DNA over specific regions of the genome, in chromatin immunoprecipitation (ChIP) experiments.

This answer provides a general approximate form for the variance of the ratio of two uncorrelated variables $X$ and $Y$:

$$\sigma_{X/Y}^2 \approx \sigma_{X}^2 \cdot \frac{1}{\mu_Y^2} + \sigma_{Y}^2 \cdot \frac{\mu_X^2}{\mu_Y^4}.$$

where $\mu_X$, $\mu_Y$, $\sigma_{X}^2$, $\sigma_{Y}^2$ are their means and variances.

In ChIP experiments one effectively counts the number of input and IP'd DNA fragments, so it's reasonable to assume that both are Poisson-distributed, with variances equal to their means. With $X$ and $Y$ both Poisson-distributed, the above simplifies to:

$$\sigma_{X/Y}^2 \approx \mu_{X} \cdot \frac{1}{\mu_Y^2} + \mu_{Y} \cdot \frac{\mu_X^2}{\mu_Y^4} = \frac{\mu_{X}}{\mu_Y^2} + \frac{\mu_X^2}{\mu_Y^3}.$$

The square root of that is the (approximate) standard deviation of the ratio, and 1.96 is the factor for converting the standard deviation to 95% confidence intervals under a normal approximation. Substitute your $a$ and $b$ for $\mu_X$ and $\mu_Y$ and we're almost there.

The remaining cal value comes from the internal calibration standard used in the ICeChIP method, to correct for antibody affinity. That gives a "Barcode IP enrichment"; those who developed the method "assume that standard deviation of Barcode IP enrichment is negligible" and thus "Barcode IP enrichment" can simply appear as a multiplicative factor in the CI calculation. The cal factor in your equation is the inverse of that "Barcode IP enrichment."

A few more details on ICeChIP

ChIP examines the fractions of different DNA regions associated with specific forms of proteins involved in gene regulation. DNA is broken down into short pieces and an antibody against a specific form of a protein is used to "pull down" associated DNA. The specific immunoprecipitated (IP'd) DNA is identified by polymerase chain reaction or by high-throughput DNA sequencing.

One would like to know the actual fraction of the input DNA that is associated with the protein in question. The IP/input ratio provides a first approximation. The antibody, however, is not completely effective at "pulling down" all of the associated DNA. The ICeChIP method uses internal standards to correct for the antibody's "pull down" effectiveness. The cal factor in the OP's formula is that correction factor. It's the inverse of the "Barcode IP enrichment" in a formula on the 21st page of Supplemental Material for the ICeChIP paper.


Adrian T. Grzybowski, Zhonglei Chen, Alexander J. Ruthenburg, "Calibrating ChIP-Seq with Nucleosomal Internal Standards to Measure Histone Modification Density Genome Wide." Mol Cell 58: 886-899, 2015.

$\endgroup$
1
  • $\begingroup$ This is a very profound explanation ! Thank you for sharing your extensive knowledge ! $\endgroup$
    – Bogdan
    Commented Sep 5, 2022 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.