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Molecules in Space Some Continuum Problems AKH 4, 27/03/03.

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Presentation on theme: "Molecules in Space Some Continuum Problems AKH 4, 27/03/03."— Presentation transcript:

1 Molecules in Space Some Continuum Problems AKH 4, 27/03/03

2 Fick’s law, simplified The flux J is proportional to a concentration gradient and travels “down” the gradient.

3 Tissue Necrosis

4 Example: Tissue Metabolism A tissue is reported to consume 5 microliters O 2 /(kg·sec). The bathing fluid has a dissolved oxygen concentration of 700 microliters O 2 /kg. (These are reasonable number for metabolism and dissolved oxygen.) If the diffusion coefficient in the tissue is 6 10 -6 cm 2 /sec, what is maximum radius of a sphere of this tissue that does not have a necrotic core? We rearrange the formula for the critical metabolic rate, obtaining

5 Plot the oxygen concentration distribution for the same size of sphere when the oxygen consumtion rate is 200 and 400 microliters O2 /(kg·min). The red line shows the distribution for a lower metabolic rate. The blue line is for the critical rate, and the green line for the higher rate. Since concentration cannot be less than zero, all values for radii less than ~0.5 should be replaced by 0

6 Necrosis – first-order variant

7 Interpreting Receptor Equilibria The problem: It is easy to measure receptor-binding equilibria. These yield typical saturation kinetics: However K D is the ratio of an “on” rate coefficient k ON and an “off” rate coefficient k OFF.. The latter is the reciprocal of the mean lifetime of the ligand on the receptor and is an important determinant of what the ligand does to the cell. Problem: how to estimate k OFF ? Possible answer, estimate k ON and determine k OFF as K D x k ON

8 A note about units:

9 Diffusion-Limited k ON

10 Simple Convective Diffusion (“lumped” transverse transport)

11 Details (1):   is the perimeter.  P is the (local) permeability   L is the contact area  Q is the flowrate  The ambient concentration is a constant, c e.

12 Details (2):

13 Exchange between two liquids along a line.  Previous analysis implied the passage of a liquid along a line, through a ‘reservoir’.  The side-by-side flow of two liquids is interesting.  The flow may be concurrent (in the same direction), or countercurrent (in the opposite direction)

14 Transport can always be calculated with the log mean

15 Steady-State Re-examined  Steady-state depends on the ‘frame of reference’  The ‘Taystee Bread Factory’  This situation is like the unsteady compartment whose concentration varies with time. Compare time there with distance above.

16 What is the meaning of P, the permeability, in these formulae?

17 General convective-diffusion problems (steady state)

18 Loschmidt Diffusion Problem


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