Molecules in Space

Some Continuum Problems

AKH 4, 27/03/03

Fick’s law, simplified

dcJdx

D

The flux J is proportional to a concentration gradientand travels “down” the gradient.

Tissue Necrosis

r + dr

r

R

2 2 2

32 2 2

1

22 2

2

4 4 4 0

( ) ; ;3 3

; ;3 6 6

r r dr

R

J r J r m r dr

d mr mrr J mr r J C Jdrdc mr mr mc C c c r R rdr

D D D

MAX 2

6 RcmR

D

Example: Tissue Metabolism A tissue is reported to consume 5 microliters O2/(kg·sec). The

bathing fluid has a dissolved oxygen concentration of 700 microliters O2/kg. (These are reasonable number for metabolism and dissolved oxygen.) If the diffusion coefficient in the tissue is 6 10-6 cm2/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

60.070 cmRR

m

Dc

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

Necrosis – first-order variant

2 2 2

32 2 2

1

4 4 4 0

( ) ; ;3 3

3

r r drJ r J r m r dr

rd mrr J mr m c

m

r J C Jdrdc rdr

c

D

Interpreting Receptor EquilibriaThe problem: It is easy to measure receptor-binding equilibria. These yield typical saturation kinetics:

D

cK c

However KD is the ratio of an “on” rate coefficient kON and an “off” rate coefficient kOFF.. 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 kOFF? Possible answer, estimate kON and determine kOFF as KD x kON

A note about units:

The "on" rate per receptor is [=] molecules/time. [=] molecules/volume. Thus [=] volume/time

The "off" rate per receptor is [=] molecules/time

moleculesThus [=] volume

on

on

off

offD

on

k cc k

k

kK

k

Diffusion-Limited kON

2 2 21

2

22

( ) 0 (no metabolism);

But 2 [flux area of receptor]

Substituting Fick's law of diffusion:

;2 2

1 1( ) ( )2

When diffu

R

R on

on on

on

d r J r J C R Jdr

R J k c

k c k cdc drr dcdr r

k cc R c

R

D D

Dsion-limited, ( ) 0, and

volumetime

2 [=] on

c R

k DR

Simple Convective Diffusion(“lumped” transverse transport)

0

0

0

( )

See the figure to understand the sign .

ln

exp

e L

e

e L

e

dcQ P c cdz

c c PL PAQ Qc c

c c PAQc c

Q c | z

Q c | z+dz

(inlet)

(outlet)

inlet - outlet = Q dc/dz

Details (1):

is the perimeter.P is the (local) permeabilityL is the contact areaQ is the flowrateThe ambient concentration is a constant,

ce.

Details (2):

1 2LM

1

2

0 L

0

L

0

yln

c c

cln

c

( )

e e

e

e

L LM

y yyy

c c

cc

N Q c c PA c

Mathematical Definition (Problem - independent)

Log -mean concentration difference :

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)

Transport can always be calculated with the log mean

1 out in 2 out in1 2( ) ( )

BUT the two c's are GEOMETRICLM

N Q c c Q c c

N PA c

Steady-State Re-examinedSteady-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.

0

expe

e

tc c PAc Vc

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

, implying a constant

Diffusion distance in a direction to the flow.

Px

D

General convective-diffusion problems (steady state)

2

2

or often:

v c c

c cvz r

2D

D

Loschmidt Diffusion Problem