Diana,

This belongs in the Discussion on line, not in a personal e-mail to me.  I am going to transfer this to that location.  The length of DNA versus the diameter of the cell nucleus determine the minimal number of foldings the DNA must undergo as they are the largest available dimensions for each element. (You might also argue that the DNA length versus the circumference is the appropriate dimension of comparison.  Here the number of minimal foldings will be about 3-fold -- pi-fold-- less.)  One meter of DNA is folded into a volume with the longest dimension equal to 5 x 10 ^ -6 meters, so the minimal number of foldings is 1/(5x10^-6) = 0.2 x 10^6 or 200,000 times.  But it actually must be more than that as the minimal intranuclear dimension available once a single fold has taken place will be less than 5 um.  Even using circumference, the number of folds should exceed 70,000.  The point is there must be many, many folds even in short lengths of the DNA, 1% of the DNA still contains at least 700-2000 folds at minimum.  And to transcribe, replicate, or repair that DNA it needs to be unfolded.  If an immune cell undergoes somatic gene rearrangements, DNA has to undergo non-replicative DNA cleavage and repair involving both single and double stranded breaks that will involve many folds, especially if a lot of DNA is being removed in the process.  This may result in mistakes that ultimately lead to the production of transformed cells capable of escaping apoptotic cell elimination pathways and establishing clones that produce antiself antibodies.

Note, there are a couple of minor computational errors I see in what is presented (my comments in red, bold type).

Hopefully more back in the swing now that Phoenix is behind me,
Ken Campbell 

Diana Esshaki wrote:

Hi Dr. Campbell,

I am confused about the question from the last session about DNA size.
The answers posted seem a bit conflicted....This is how I worked through it:

One bp of DNA occupies a cylinder 0.33 nm tall with a diameter 2.37 nm.
One complete copy of DNA = 3x10^9bp

Length of 1 complete copy of DNA:
(3.3x10^-10m/bp)(3x10^9bp) = 1m

Volume of a typical cell nucleus:
Assume that the internal space of a nucleus in a cell is described by a sphere---
= 4/3(pi)(r^3)
= 4/3(3.14...)(5.0x10^-6m)^3
= 4/3(3.14...)(1.25x10^-16m)
= 5.24x10^-16m^3

Volume of DNA:
Assume that the volume of DNA is similar to the volume of a cylinder---
= (pi)r^2h
= (3.14...)(1.185x10^-9m)^2(1m)
= 4.41x10^-18m^3

So, to take things a little further...
Volume of 1bp:
= (3.14...)(1.185x10^-9m)^2(0.33x10^-9m)
= 1.46x10^-27m^3

Volume of 1chromosome: = total volume of DNA (computed above) / number of chromosomes, for humans = 46 (22, XX or 22, XY) = 4.41x10^-18m^3/ 46 = 9.59x10^-20m^3
=(1.46x10^-27m)(3x10^9bp)
=4.38x10^-18m^3/ chromo

Fraction of the nucleus:
= (4.38x10^-18)/(5.24x10^-16)
= 0.0084
= 0.8%
Therefore the DNA occupies less than 1% the internal volume of the nucleus.

Now, how often does DNA fold to fit into the volume of the nucleus?
First of all, I was confused about how Erin and Michael got their answers-- around 600,000 times.
The answer was obtained by dividing the volume of DNA by the volume of the nucleus. This is a fractional calculation, not the number of folds I thought.

I followed Dina's idea somewhat. BUT, I looked at it as though the length of the fold is the diameter of the sphere (not the radius).
I calculated the volume of a cube inside a circle (nucleus) as well as outside the circle. (The answer must be somewhere between.)

a.)Volume of a cube INSIDE the circle:
the diagnol is the diameter of the circle (10x10^-6m) making the sides of a circle 7.07x10^-6m (a2+b2 = c2).
Therefore, 1m/7.07x10^-6m = 144,272times
b.) Volume of a cube OUTSIDE the circle:
1m/(1x10^-5m) = 102,000times

Average of the two answers: 123,136 times.

Thanks for your help, :)
Diana








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