Lecture 18: Diffusion: Fick’s First and Second Laws

1 Physical Principles in BiologyBiology 3550 Fall 2021 Lecture 18: Diffusion: fick s First and Second LawsMonday, 4 October 2021 David P. GoldenbergUniversity of Idealized Macroscopic diffusion Experiment0xC(x)0xHow will plot ofC(x)versusxchange with time? fick s First Law of DiffusionJ= 2x2fidCdxSymbols: J=flux of molecules (or moles) per unit area per unit time: x=RMS displacement along thex-direction. fi=average duration of random steps. dCdx= derivative of concentration withx, concentration gradient. If concentration increases withx, flux is in the diffusion Coefficient,DConsider the term from fick s First law: 2x2fiBoth xandfiare parameters describing the random walk of molecules(or larger particles) undergoing diffusion , and are constant for a giventype of particle under defined solution a new parameter, the diffusion coefficient,D:D= 2x2fiThe usual form of fick s First law:J= DdCdxDcan be experimentally determined, without knowing anything aboutthe microscopic random walk of Molecules Across a Cellular MembraneHigh concentrationLow concentrationMembranePoreSome plausible values: diffusion coefficient for a small molecule.

2 10 10m2=s(to be considered later)Thickness of a biological membrane : 3 nmDiameter of a molecular pore: 1 nmConcentrations: 50 mM and 5 mMApplying fick s First LawApproximate the concentration gradient:dCdx concentration differencemembrane thickness=50 mM 5 mM3 nm= 15 mM=nmConvert this to units with moles and meters:15 mM=nm =0:015 moles=L10 9m 103Lm3= 1:5 1010moles=m4 Flux from fick s First law:J= DdCdx= 10 10m2=s 1:5 1010moles=m4= 1:5 moles=(s m2)Negative sign indicates that diffusion is in the direction opposite to theconcentration through a membrane PoreThe pore:Diameter = 1 nmArea = r2= (0:5 nm)2= 7:8 10 19m2 Flux:J= 1:5 moles=(s m2)Moles per Second through the pore:1:5 moles=(s m2) 7:8 10 19m2= 1:2 10 18moles=sDiffusion through a membrane PoreMoles per Second through the pore:1:5 moles=(s m2) 7:8 10 19m2= 1:2 10 18moles=sMolecules per Second through the pore:1:2 10 18moles=s 6:02 1023molecules=mole = 7 105molecules=sHow Many Molecules are in the Pore?

3 Pore volume:V= r2 l= (0:5 nm)2 3 nm= 2:4 nm3 (10 9m=nm)3= 2:4 10 27m3= 2:4 10 27m3 103L=m3= 2:4 10 24 LAssume that the concentration in the pore is the mean of theconcentrations on the two sides:C= (50 mM 5 mM)=2 = 22 mM = 0:022 moles=LHow Many Molecules are in the Pore?Pore volume: 10 24 LConcentration: moles/LMoles in pore:0:022 moles=L 2:4 10 24L = 5:3 10 26molesMolecules in pore:5:3 10 26moles 6:02 1023molecules=mole 0:03 moleculesMost of the time, the pore is empty !An Idealized Macroscopic diffusion Experiment0xC(x)0xHow will plot ofC(x)versusxchange with time?Clicker Question #1 Where will the flux,J, be greatest?0xC(x)0xCBACBAAt point B, where the concentration gradient is , the molecules move at the same rate everywhere!

4 fick s Second Law of DiffusionHow does the concentration at a given pointchange with time?Net number of molecules moving to the rightat two sides of a slice, during intervaldt:Nx=JxAdtNx+ x=Jx+ xAdtChange in number of molecules in the slice:dN=Nx Nx+ x=AJxdt AJx+ xdt=Adt`Jx Jx+ x fick s Second Law of DiffusionChange in concentration in the slice:dC=dNA x=Adt`Jx Jx+ x A x= dtJx+ x Jx xIn the limit of smalldtand small x:dCdt= Jx+ x Jx x= dJdxHow doesJchange withx? fick s Second Law of DiffusionFick s First law:J= DdCdxDerivative ofJwith respect tox:dJdx= Dd2 Cdx2 fick s Second law:dCdt= dJdx=Dd2 Cdx2 Also called the diffusion is it good for? How do we use it? fick s First and Second Laws of DiffusionxC(x)xFirst law:J= DdCdxFlux,J, at positionxisproportional to the concentrationgradient at that law:dCdt=Dd2 Cdx2 Rate of change in concentration atpositionxis proportional to thederivative of the Question #2 Where will the concentration increase most rapidly?

5 XC(x)xCBACBAAt point A, where the concentration gradient increases most rapidly withrespect s Second Law of DiffusiondCdt=Dd2 Cdx2A Second -order differential equation .The solution to the equation is a function:C=f(x;t)that satisfies the equation:df(x;t)dt=Dd2f(x;t)dx2 The trick is to findC=f(x;t).The solution depends on the shape of thevolume and the initial concentrations, theboundary from a Sharp Boundary0xC(x)0xAtt= 0 Forx <0:C(x) = 0,dCdx= 0,d2 Cdx2= 0 Atx= 0:dCdx Forx 0:C(x) = 1,dCdx= 0,d2 Cdx2= 0 diffusion from a Sharp Boundary0xC(x)0xaaConsider molecules at a positionx=a >0:Molecules will begin to diffuse via a random will the molecules initially at positionabe distributed after a time,t?

6 diffusion from a Sharp BoundaryDistribution of molecules originally atpositionx=ax0aRelative Concentrationp(x) =1p2 n 2x e (x a)2=(2n 2x )n= number of steps in random walk 2x = mean-square step distancealongx-axisDiffusion from a Sharp BoundaryDistribution of molecules originally atpositionx=ax0aRelative Concentrationp(x) =1p2 n 2x e (x a)2=(2n 2x ) diffusion coefficient,D= 2x2fifi=average time of each RW stepAfter time,t,n=t=fin 2x =t 2x fi= 2 DtDiffusion from a Sharp BoundaryDistribution of molecules originally atpositionx=ax0aRelative Concentrationp(x) =1 4 Dte (x a)2=(4Dt) diffusion from a Sharp BoundaryDistribution of molecules from all starting points,a ConcentrationAt positionx, concentration is the sum of molecules that have diffusedfroma 0C(x;t) =1 4 DtZ 0e (x a)2=(4Dt)da