Chromosome damage

Thread Starter

boks

Joined Oct 10, 2008
218
Somebody tells you that a single ultraviolet photon carries an energy equivalent of about 10
electron volts (eV, see Appendix B). You propose a damage mechanism: A photon delivers that
energy into a volume the size of the cell nucleus and heats it up; then the increased thermal
motion knocks the chromosomes apart in some way. Is this a reasonable proposal? Why or why
not?


(Heat produced) = (mechanical energy input) * (0.24 cal/J)

so that

Heat produced = 3.84E-19 cal
 
Last edited:

jpanhalt

Joined Jan 18, 2008
11,087
That question is basically an exercise in dealing with exponents and knowing the definitions of eV, joules, and calories.

As a hint, consider a cell that is 8 microns in diameter (8X10^-6 meter). That would be comparable in size to the nucleus of a lymphocyte. If its specific gravity is 1.0, the mass is approximately 267X10^-12 g (assuming I did the exponents correctly). Your 10 eV is 16X10^-19 joule. From that you should be able to calculate the expected temperature rise. You can look up typical bond strengths and by inspection will realize that the temperature rise of the whole cell is way too little to break a bond in DNA.

The problem with the model you propose is that you are distributing the energy evenly to the whole mass of the cell. Cells are pretty big relative to the masses of individual atoms or molecules of water.

John
 

Thread Starter

boks

Joined Oct 10, 2008
218
I've learnt that the avarage size of a human cell nucleus is 1.7 micrometers. Assuming mass density equal to that of water, that will mean that the mass of the nucleus is ca. 4E-15 kg.

I'm not sure how to calculate the temperature rise. Can I simply use

\(
\Delta E_k = \frac{3}{2} k_B \Delta T
\)

as for gases?
 
Last edited:

jpanhalt

Joined Jan 18, 2008
11,087
Quote from Wikipedia:
The nucleus is the largest cellular organelle in animals.[4] In mammalian cells, the average diameter typically varies from 11 to 22 micrometers (μm) and occupies about 10% of the total volume.
Wikipedia is not always correct. I chose the lymphocyte in my example because it is a cell that is easily identified, relatively uniform in size, and has a nucleus about the same diameter as that of an RBC. Moreover, bone marrow and lymphoid tissue are composed of some of the most radiation sensitive cells in the body. Among those tissues, you will find the immature forms -- those cells for which radiation damage would have the greatest downstream effects -- have even larger nuclei.

For calculating the temperature rise, I assumed adiabatic conditions and water again. For whatever volume of cell you use and joules (or calories) added, you should be able to calculate temperature rise easily assuming those two conditions.

John
 
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