## NS FL4 B/B 53

erichsuh94
Posts: 13
Joined: Wed May 23, 2018 4:16 am

### NS FL4 B/B 53

I am having very difficult time visualizing the explanation of the solution.

As for the question, I understand that

Histone diameter: 11nm
Length of one nucleotide: 0.3nm

and that 200 basepairs coil around one histone.

So the length of unwound DNA (200 basepairs) which would be wrapped around a single histone is 0.3 nm X 200 = 60 nm.

But I am not quite clear where we are going with

" Once coiled around the protein, we can assume (without getting into unnecessary surface area calculations) that a single layer of DNA strand will be wrapped around the protein (strand thickness = 2 nm), meaning we now have the strand condensed to a ~ 4-nm layer (2 nm on each side) around the 11-nm-diameter protein. Thus, about 60 nm of DNA is replaced by 15 nm, meaning there is approximately a 45-nm decrease in length. 45/60 = 0.75 = a 75% decrease in length."

as in the explanation.

NS_Tutor_Katelyn
Posts: 136
Joined: Wed May 30, 2018 11:51 am
Location: Minnesota
Contact:

### Re: NS FL4 B/B 53

erichsuh94 wrote:I am having very difficult time visualizing the explanation of the solution.

As for the question, I understand that

Histone diameter: 11nm
Length of one nucleotide: 0.3nm

and that 200 basepairs coil around one histone.

So the length of unwound DNA (200 basepairs) which would be wrapped around a single histone is 0.3 nm X 200 = 60 nm.

But I am not quite clear where we are going with

" Once coiled around the protein, we can assume (without getting into unnecessary surface area calculations) that a single layer of DNA strand will be wrapped around the protein (strand thickness = 2 nm), meaning we now have the strand condensed to a ~ 4-nm layer (2 nm on each side) around the 11-nm-diameter protein. Thus, about 60 nm of DNA is replaced by 15 nm, meaning there is approximately a 45-nm decrease in length. 45/60 = 0.75 = a 75% decrease in length."

as in the explanation.