Question

You’re provided a lab protocol to perform a very short digestion of eukaryotic chromatin with micrococcal nuclease, which should result in DNA fragments of 200 base pairs. You setup the experiment but get called away suddenly and are gone for a length of time. When you return to complete the experiment and run the gel to visualize the samples, it shows 146 base pair fragments. Explain what occurred.

Answer 1

DNA present in chromosome in form of nucleosome, coild around histone protein. Each nucleosome consist of core histone and DNA present around them. 200bp of DNA associated with one nucleosome , out of this 200 bp 145 bp is tightly associated with nucleosome ( as it present around the histone as coiled form ) while rest of the part present as pinker DNA form present in between the two nucleosome.

Linker DNA is easily accessible to micrococcal nuclease.

Short time incubation of DNA sample with micrococcal nuclease result in digestion of most convenient sites ( linker DNA) . So form 200 bp fragments.

But in the question student has left the solution for a longer time, so here digestion is very deep and here nuclease almost cleave all the linker DNA. But as 146 bp is present around the nucleosome, so it is not accessible to nuclease sobthis part is not cleaved . That's why upon long term treatment of DNA with micrococcal nuclease cleave linker DNA and expected bands are of 146 base pair .