% INPUT - ChIP - (Feb/05/2009 )

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Thanks for explaining the calculation so well. I have a naive question about error bar calculation. I know how to determine the % input of ChIP'd DNA but I am not sure how to calculate errors for plotting error bars in my graph. I just got a stardard deviation from my triplicate PCR reaction, then how to convert it into a positive and a negative error.

-M&M81-

M&M81 on Feb 27 2009, 01:29 AM said:

Thanks for explaining the calculation so well. I have a naive question about error bar calculation. I know how to determine the % input of ChIP'd DNA but I am not sure how to calculate errors for plotting error bars in my graph. I just got a stardard deviation from my triplicate PCR reaction, then how to convert it into a positive and a negative error.

Do you guys calculate the %input of the triplicate PCR individually and then get a SD from them? so the error bar is equal to the SD calculated?

-M&M81-

M&M81 on Feb 26 2009, 05:39 PM said:

M&M81 on Feb 27 2009, 01:29 AM said:

Thanks for explaining the calculation so well. I have a naive question about error bar calculation. I know how to determine the % input of ChIP'd DNA but I am not sure how to calculate errors for plotting error bars in my graph. I just got a stardard deviation from my triplicate PCR reaction, then how to convert it into a positive and a negative error.

Do you guys calculate the %input of the triplicate PCR individually and then get a SD from them? so the error bar is equal to the SD calculated?

you can either use SD for error bar, or you can use SEM, which is equal to SD / square root

for people working on animals, usually the variation is bigger and SEM would make more sense

-jiro_killua-

Hi jiro_killua,

I used your method to calculate % of input for my samples. The result I got was the protein i studies do not bind to promoter region after treatment!

But when I carried out normal PCR, I can see a much more intense band on my agarose gel after treatment. Can you tell me why?

I really dont know how to study all the figure I got from qPCR!

-giny-

you can use some genomic DNA to do the standard curve, and first measure the DNA concentration by Nanodrop,

Use about 100ng DNA as you most concentrated standard, and then 10 fold serial dilute it

Hopefully the CT will be around 20

Hi all, I dont understand why we can use genomic DNA to do the standard curve? Is the purpose of doing this is to see our primers can work best at which concentration?

It is because our ChIP samples usually contain less DNA (those precipitated by antibody). Even though we know how much concentration we have to use to let the CT be around 20, but we will never know the concentration in our ChIP samples? or anyone of you will measure the amount of DNA in ChIP samples using NanoDrop?

-giny-

giny on Aug 11 2009, 11:00 PM said:

you can use some genomic DNA to do the standard curve, and first measure the DNA concentration by Nanodrop,

Use about 100ng DNA as you most concentrated standard, and then 10 fold serial dilute it

Hopefully the CT will be around 20

Hi all, I dont understand why we can use genomic DNA to do the standard curve? Is the purpose of doing this is to see our primers can work best at which concentration?

It is because our ChIP samples usually contain less DNA (those precipitated by antibody). Even though we know how much concentration we have to use to let the CT be around 20, but we will never know the concentration in our ChIP samples? or anyone of you will measure the amount of DNA in ChIP samples using NanoDrop?

Don't use Nanodrop to measure the DNA concentration of ChIPped samples.

You won't get anything reliable since the amount is so small.

The purpose of standard curve is NOT to see the concentration at which the primer work the best, it IS about and only about the efficiency of primers used, unrelated to the amount of starting material (but of course you need to have some DNA as starting material)

-jiro_killua-

i need some serious help with some real data based on these discussions

we are doing chip on embryos , so material is extremely limited
we started with chromatin in 1ml (from more than 1000 embryos!!!!) and saved 100 ul for input
the rest was divided into 2 X 450ul aliquots: one ChIped for the ab of interest and one with IgG
in the end of the procedure
the input is now in 50 ul and has a concentration of 46.3 ng/ul
the Chiped specific ab is in 200ul and has a concentration of 2.7ng/ul (our protocol recommended dividing the samples for ChIP into 4 pools and then combining at the end so that is what we did)

i am at a loss as to do all the corrections for the qPCR since the final dnas are in different volumes
why, if we know the final concentration of dna in the input and chip samples, do we still need to do the adjustment for the fact that the input represented less of the total chromatin at the start

-chipnlearn-

I would second everything Jiro has said so far (very good explanation) but would add one thing about using the signal to noise calculation in parallel to % of input.

Some regions of the genome, in my experience and in the experience of others, may be less easily pulled down by ChIP, non-specifically. In other words, for some regions of the genome regardless of what antibody you use, you will get less pull down than other regions. This is only a problem when you are comparing two genomic regions (like comparing your region of interest to a negative control).

Calculating the signal to noise (IP/mock) will eliminate this bias, however, as Jiro says, the mock signal is always very noisy because it is so low. For this reason I only do the signal to noise calculation after I have a lot of data points built up and just take the average. Also, it's VERY IMPORTANT to remember you are only doing this calculation to compare two regions of the genome using the same chromatin sample. If you compare two samples using the signal to noise calculation then you can run into problems if the samples don't have the same level of chromatin input.

There's no reason to ever publish this calculation so you can always express your data as % of input UNLESS the signal to noise calculation contradicts what you see with the % of input calculation.

What does it mean when the signal to noise shows enrichment and why dont you publish this? because of different dilutions etc. we are having a hard time calculating % input but can definitely see that there are more amplicons in the specific ab sample than in the IgG sample?

-chipnlearn-

jiro_killua on Feb 5 2009, 07:24 PM said:

dna_nerd on Feb 5 2009, 05:48 PM said:

anybody feel like working me through the concept of % INPUT as it pertains to qPCR analysis after a ChIP experiment (or possibly point me in the right direction)?

I understand how to evaluate fold change in respect to signal over noise but can't quite grasp the concept of why % INPUT is a better way to present the data

thanks

% Input basically means the % of DNA being precipitated by your antibody

It would be easier to use an example to illustrate:

And you set aside 100ul of sample as Input, and used 1ml to do the ChIP

After the ChIP, you decrosslink and precipitate both you Input and ChIPped sample, and do realtimePCR

If, for simplicity, the CT for both your sample and input are cycle 20, then, the % Input will be 10%

Because your Input contain 100ul of the starting material,
= 100ul x (100ug/ml) = 10ug

= 1ml x (100ug/ml) = 100ug

So, if after ChIP, your sample and input have the sample CT (that means same quantity), that would mean from the 100ug, 10% was being precipitated by your antibody

To explain why % Input is better than Fold signal to noise, compare this:

After absolute quantification, the amount of DNA in different samples are as follow:
CTL Input: 10ug
CTL sample: 1ug
CTL IgG (noise): 0.05ug

Treatment Input: 8ug
Treatment sample: 0.2ug
Treatment IgG (noise): 0.01ug

If you use % Input:
CTL % Input = 1ug/10ug = 10%
Treatment % Input = 0.2ug/8ug = 4%

Conclusion: treatment decrease the binding

If you use signal to noise fold change:
CTL = 1ug/0.05ug = 20fold
Treatment = 0.2ug/0.01ug = 20fold

Conclusion: no change

% Input is useful in normalizing the starting material, such that you would not see a false positive due to using more DNA to start with in one sample

Problem with signal to noise fold change is that the noise varies so much that would heavily affect your results, and more importantly, a 0.05ug vs 0.01ug background really has no biological meaning, they are both low, and especially can be very inaccurate when the cycle number in realtime PCR gets higher

Hi,

In the protocol that our lab is following, we set aside 10% input (we use 200ug sample, 20ug input). On top of this, when we do PCR purification, we dilute the imput (but not the sample) 1:10. Due to the extra dilution step at the end, does that mean that the % input is now 1% rather than the original 10%?

If I chose to isolate 1% input at the beginning of the protocol (2ug), and then did not dilute 1:10 at the end, could I expect to get similar Ct values from this 1% input to the values I get when I isolate 10% input and then dilute the DNA 1:10?

Thanks a lot, I've asked people around and no one can give me a straight answer, and I am now really confused....

-Martina-

Huhh, I just have one question. Taken the example, shouldn't noise be taken into consideration as well? As treatment seems to be affecting IgG binding too? Wouldn't it make more sense using % input over noise?

I am currently trying to draw comparison among cell lines and apparently the stickiness of the antibodies vary among cells and in between IPs. How can I interpret the data while the backgrounds are all different from each other? Anyone has any idea about this?

jiro_killua on Fri Feb 6 00:24:58 2009 said:

anybody feel like working me through the concept of % INPUT as it pertains to qPCR analysis after a ChIP experiment (or possibly point me in the right direction)?

I understand how to evaluate fold change in respect to signal over noise but can't quite grasp the concept of why % INPUT is a better way to present the data

thanks

% Input basically means the % of DNA being precipitated by your antibody

It would be easier to use an example to illustrate:

And you set aside 100ul of sample as Input, and used 1ml to do the ChIP

After the ChIP, you decrosslink and precipitate both you Input and ChIPped sample, and do realtimePCR

If, for simplicity, the CT for both your sample and input are cycle 20, then, the % Input will be 10%

Because your Input contain 100ul of the starting material,
= 100ul x (100ug/ml) = 10ug

= 1ml x (100ug/ml) = 100ug

So, if after ChIP, your sample and input have the sample CT (that means same quantity), that would mean from the 100ug, 10% was being precipitated by your antibody

To explain why % Input is better than Fold signal to noise, compare this:

After absolute quantification, the amount of DNA in different samples are as follow:
CTL Input: 10ug
CTL sample: 1ug
CTL IgG (noise): 0.05ug

Treatment Input: 8ug
Treatment sample: 0.2ug
Treatment IgG (noise): 0.01ug

If you use % Input:
CTL % Input = 1ug/10ug = 10%
Treatment % Input = 0.2ug/8ug = 4%

Conclusion: treatment decrease the binding

If you use signal to noise fold change:
CTL = 1ug/0.05ug = 20fold
Treatment = 0.2ug/0.01ug = 20fold

Conclusion: no change

% Input is useful in normalizing the starting material, such that you would not see a false positive due to using more DNA to start with in one sample

Problem with signal to noise fold change is that the noise varies so much that would heavily affect your results, and more importantly, a 0.05ug vs 0.01ug background really has no biological meaning, they are both low, and especially can be very inaccurate when the cycle number in realtime PCR gets higher

-Shan-
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