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I hope this question is suitable for this site. I am concerned about the Chip experiment part so I think it should be okay. I am a Applied Math student starting to get into bioinformatics and so I've been looking at Chip Seq data. But to make sense of the Chip Seq data I wish to understand how the experiment is first performed.

Basically my question relates to step 1 of the following diagram: enter image description here

Now my main question is how come the reads are overlapped? Where did we get so many of the (almost the) same reads?

My "understanding" is that:

1) They take a DNA sequence and crosslink the protien of interest. Then they get rid of the DNA sequence surrounding this area of interest so now we have a "small" sequence of DNA and in this small DNA is somewhere where our TF binds. Now we make copies of this small DNA seq and run it through a sequencing machine. Is this correct?

2) They take a bunch of cells = bunch of DNA sequences. Then they do the same procedure above (by crosslinking and getting a "smaller" DNA sequence of interest). Since they had many cells to begin with, this means they had many DNA seq to begin with. Now they shear the DNA seq and we have fragments. Then we align these up with the reference genome.

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2 Answers 2

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The basic steps of ChIP-Seq are:

  1. Crosslinking proteins to DNA - this fixes the proteins in their natural positions
  2. Nuclease digestion - this removes regions that are unbound to protein; nucleases are sterically hindered from digesting protein-bound DNA
  3. Immunoprecipitation - this allows isolation of the target protein by binding it to a selective antibody
  4. Sequencing to determine the bound DNA sequence
  5. Mapping the reads to the reference genome

Why so many reads? They are taking many reads from different cells. Why are the reads so short? Many modern next-gen sequencing methods are limited to ~30 bp reads. Thus, the reads you see are not actually the entire protected sequence but rather just a shorter fragment from the 5' end (where sequencing starts from). The why behind this is explained by Jiang and Pugh, 2009 (Nucleosome positioning and gene regulation: advances through genomics):

Three DNA sequencing technologies have been used to map nucleosomes. Pyrosequencing using the Roche 454 GS20/FLX sequences nucleosomel DNA end-to-end, allowing both nucleosome borders to be linked in a single sequence. This method provides the greatest mapping accuracy, particularly in genomic regions of low complexity. By contrast, other platforms, such as those provided by the Illumina-Solexa Genome Analyzer and Applied Biosystems SOLiD, generate only 25-35 by sequence tags, requiring both nucleosome borders to be inferred. Nonetheless, these short-read technologies produce >100 times the number of sequence tags at a similar cost as the long-read technologies, and so the short-read technology is currently the only practical technology for mapping nucleosomes in large genomes. The higher tag count of the short-read technology enhances mapping accuracy and thus provides a practical way of mapping nucleosomes.

Why are the reads overlapped? I can think of a few explanation for that. For starters, Protein-DNA binding is variable in its specificity. In other words, a given protein might bind at the exact same site every time or it may bind 'around' a site or it may bind randomly. A protein that binds an exact sequence would be expected to produce reads that are more in phase that one that binds more variably. Next, the nuclease digestion process is random (or at least expected to be). Thus the DNA cleavage point relative to the protein is variable and your reads start from different points. Again, from Jiang and Pugh:

At any given genomic locus, the preferential positioning of nucleosomes--called phasing--can be described. At most loci, there is an approximately Gaussian (normal) distribution of nucleosome positions around particular genomic coordinates, ranging from ~30 by for highly phased nucleosomes to a random continuous distribution throughout an array. How much of this variation is due to genuine positional heterogeneity and how much is an artefact that is caused by overtrimming or undertrimming of the DNA at nucleosome borders by micrococcal nuclease during sample preparation remains to be determined.

Does that answer your question? Basically the reads are overlapped because there are many of them and they may differ either because of actual variance in the DNA binding site (which is biologically significant) or because of variance in the DNA digestion site (which is a methodological flaw). As with any biological experiment, you must perform many replicates to statistically determine the source of the variance and this is done by taking many reads. I recommend you read the article I quoted from, it is a good overview of histone mapping with ChIP and is likely also applicable to what you're studying.

Also, do you know why there are two sets of reads (red and blue)? I would venture a guess that they are taken from opposite strands. In other words, they are reads of the same DNA region taken from different directions. That might help clarify what's going on.

To conclude:

  • DNA is not extracted from a single cell. In usual procedures it is always from a collection of cells or from tissues.
  • The overlap in TF associated reads arises because of the TF bound sequence coming from different cells in this collection.
  • Each DNA piece is amplified and clusters are formed in the NGS sequencers (for the ease of detection);
  • In a homogeneous cell collection i.e. all cells are of same type, the variability in the read position usually arises because of flexibility of binding sites (i.e TF wont exactly bind to at a certain locus). Also, as previously mentioned the variability can arise during the technical procedures.
  • In case of heterogeneous tissues the variability arises because of different biochemical status of different cells. Strictly speaking there is always some heterogeneity even in an apparently homogeneous tissue but tissues such as brain are highly heterogeneous
  • In the case of nucleosomes the positioning is not exact because the sequence contribution to the position is very minimal. Certain nucleosome positions are tightly maintained for e.g. around the TSS (transcription start sites) whereas in other regions the positioning is fuzzy because of the dynamic nature of the chromatin.
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I added conclusions to summarize your answer. –  WYSIWYG Apr 21 at 9:13
    
Thank you. I think was missing is that a TF may bind 'around' a binding site. I was thinking suppose you have a DNA seq ATGCGCTATATCGAT but you have a thousand of these since there are 1000 cells. TF binds to TATAT so when we crosslink and "digest" we get rid of everything surrouding it. Then we have TATAT left. Then we can break this up to TA, TAT since NGS readers can't read long seq. Now we have a bunch of TA, TAT floating around. If we align all of them up, we shouldnt see the picture I posted, but we should see a "ladder". This is not true, but do i have the process right? –  masfenix Apr 30 at 3:38
    
Your transcription factor may bind to TATAT but it may protect much more than that. For example, prokaryotic RNAP binds the core promoter at -10 and -35 (which is about 29 bp) but protects about 70 bp of DNA. Thus, you'd likely get more than just TATAT. Also, I expect that no NGS could actually sequence a fragment so short. Maybe I'm not understanding what you're getting at. –  canadianer Apr 30 at 4:53
    
The example seq I used was supposed to be a very long DNA sequence. The binding site TATAT was supposed to be 150 - 900bp long. Still too long for NGS, so we breakup TATAT into shorter fragments TA and TAT. And I wanted to see where my understanding had gone wrong. –  masfenix Apr 30 at 20:26

So that's not quite how ChIP-Seq (Chromatin ImmunoPrecipitation and Sequencing) works. First, you start out with a large number of cells per sample (preferably on the order of 107 or more - the more sample, the more immunoprecipitations you can do to different targets) and do whatever it is that you're going to do to induce transcription factor binding - expose cells to drugs, treat with cytokines, grow in different conditions, or simply compare two types of cell populations - normal vs. cancer, etc. After a certain period of time, determined by your experimental setup, you cross-link the cells with 1% formaldehyde, which fixes everything in place, including transcription factors bound to DNA. The cells are then broken apart by incubation in a detergent-containing buffer, and nuclei are harvested. After washing, the nuclei are either incubated with micrococcal nuclease (MNase) to digest DNA into strands of 1-5 nucleosomes in length (150-900 bp), or sheared by mechanical ultrasonication, hopefully leaving strands of the same length.

Once you have your DNA strands, you add antibodies to specific targets - either certain forms/post-translational modifications of the histone proteins that make up the nucleosome, or to specific transcription factors or TF families. The antibodies bind to their targets, and are enriched by incubating with (generally magnetic) beads that have the antibody-binding protein Protein G covalently attached to their surfaces. The beads are washed with low-salt and high-salt washes, then the formaldehyde cross-links are reversed and the chromatin is eluted from the beads in an elution buffer at high temperature (65°C) with vortexing in a thermomixer. Any remaining protein is removed by digesting with Proteinase K, the DNA is purified using a spin column, and you now have your enriched, purified DNA sample ready for sequencing.

ChIP method overview from CST Source

Antibody binding and enrichment of known sequences is often first analyzed by regular PCR, just to ensure that everything worked. Once your sample has been validated, it is then prepared for next-gen sequencing. We are now near Step 1 of your illustration above. The read lengths of some instruments may be shorter than the average length of the fragments you prepared during the ChIP step (150-900 bp), so during the library construction phase they are further broken down into shorter strands before having the common primer templates ligated to them (if that's the technology you're using). Even if the instrument allows you to use the full length of your DNA strands, you need to remember that the sample is the aggregate DNA from millions of cells, and they didn't all break after MNase digestion in exactly the same way.

After amplification and reading, you are then left a large number of short sequences, which need to be compared to a reference genome and aligned:

Mapping reads Source

Since we've already enriched for certain regions of DNA by using the antibodies which bind to DNA-bound or -associated proteins, there will be a larger number of similar overlapping reads in enriched areas, allowing us to tell where the histone, transcription factor, or other protein was bound when the sample was originally cross-linked.

Hopefully this all makes sense to you. I based my description of the ChIP protocol on Cell Signaling's SimpleChIP products, which I've found to be of quite good quality and reproducibility (no, I don't work for them!).

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Thank you. I can not express my sincerity and appreciation for typing that out for me. Just a couple of questions: 1) what does " enriched by incubating with..." mean? In other words, what does enriching the DNA do? –  masfenix Apr 20 at 5:12
    
1) When you introduce the antibodies to the sample, they bind the target protein but the complex is still in solution. The beads allow effective purification of the complexes by concentrating them on the beads, which are easily isolatable. 2) The NGS reads correspond to the red regions in the bottom image. Each 'line' is a single isolated DNA fragment which is sequenced from both ends to produce the 35 bp reads and then aligned to the genome. 3) That's essentially correct. –  canadianer Apr 20 at 5:12

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