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

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

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

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

It would be easier to use an example to illustrate:

If you start with a sample of 100ug/ml

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

And your sample start with 1ml of the starting material,
= 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

Bravo!! J_K!! I would like to ask my PI to invite you as postdoc in our lab Noboby never told me so precisely how and why. Although in all fairness, I wonder they know it this well!! Thank you!!

jiro_killua on Feb 5 2009, 05:24 PM said:

thank you for breaking it down, that was extremely helpful

is it strongly recommended to run a dilution assay on the input material against the primers beforehand, what information is this giving you?

what amounts of dna should go into the reactions, is 5ng to little, how is this determined?

thanks

dna_nerd on Feb 6 2009, 12:34 AM said:

Maybe you mean doing a standard curve with a dissociation curve?

This is important because it tells you whether you can use that pair of primers

It provides you with 2 pieces of information:

1. With the dissociation curve, you will know whether you get a single product from realtime PCR
If it has a single peak, then it's a single product
If it has more than one peak, that means your primers either amplify multiple product from the genome, or, it has primer dimer

To distinguish the two, you can tell from the well that has no DNA added (the blank), if it's from the genome, there should be no CT, if it's primer dimer, then there will be a CT (usually comparable to your most dilute standard, like around 32-36)

2. the standard curve will tell you whether your primer is efficient.
With the standard curve, you will get a slope and a y-intercept
If, for example, you do a 10-fold serial dilution of the input DNA and do a standard curve, then ideally each dilution should have a cycle difference of 3.321928

This is because log10 to the base 2 is 3.321928

For example, the ideal case will be like this:

S1: 1000ug CT: 20
S2: 100ug CT: 23.32
S3: 10ug CT: 26.64
S4: 1ug CT: 29.96
NTC: 0ug CT: undetectable

This is ideal, and the slope will be exactly -3.321928

If you have primer dimer, the slope will be smaller, like 3.0 (then you will also see from dissociation curve)
If your slope is 3.6, then maybe the primer is not very efficient (doesn't mean you cannot use it, but just not as good)

dna_nerd on Feb 6 2009, 12:34 AM 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

jiro_killua

You should write the next real time for dummies!!

You break it down so well, and it is instantly understandable. I have been doing real time PCR, and doing melting and standard and looking for -3.2 slope and etc, but without as lucid understanding as you have. Perhaps I should learn to learn things like you do

Now I know who to ask for real time troubles. What else is your expertise?

TanyHark on Feb 6 2009, 10:08 PM said:

haha...thanks for your appreciation

I just try to tell what I know

I'm relatively active in the Methylation and ChIP forum just because I'm doing this recently

This is an interesting and useful forum to find information and I still got so much to learn!!

you are a wealth of knowledge

i have more questions

so the standard curve (using genomic dna) tells you the efficiency of the primers, but also to determine the proper amount of dna to use in the analysis?

if you see that 10ng of genomic dna gives a Ct of around 20, then you know to add 10ng each for input, Ab and no-Ab samples?

dna_nerd on Feb 7 2009, 02:08 AM said:

Do not add 10ng of input, Ab and IgG control to do the real-time

instead of normalizing them with DNA concentration, you add equal volume to the realtime PCR

For example, you can try to resuspend all the sample pellet in 30ul, and use 1ul in each well

If as I said before, the starting volume of Input is 100ul and sample is 1ml, and at the end, you resuspend them both in 30ul, you will have an adjustment factor of 10

So, after absolute quantification,

Input: 80ng
Sample: 40ng
(Assume the CT of Input is 21 and Sample is 22)

Then %Input = 40ng/80ng = 50%

This is before adjustment

But because the Input starting material is only 1/10 of the sample (remember you only take 100ul for input and used 1ml for sample), you will have to divide the % Input by 10

Therefore, actually % Input = 50% / 10 = 5%

dna_nerd on Feb 5 2009, 04:48 PM said:

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.