1. adsorption from an infinitely large volume · 2018. 8. 16. · 1. adsorption from an infinitely...
TRANSCRIPT
Adsorption in biosensors
1. Adsorption from an infinitely large volume
1.1 Thermodynamic aspects In most biosensors, the recognition event such as the antigen – antibody reaction or the oligonucleotide hybridisation occurs between a target molecule in the sample solution and the receptor adsorbed on a solid support, which can be a microtiter plate, a microbead or a microarray chip. To understand, which step controls the binding reaction, we shall consider first the simplest case where the recognition process is slow compared to the transport in solution, and where the sample volume is large enough to consider that the reactant concentration is constant. To treat, the adsorption process, it is important to define a relation between the target surface concentration, noted Γ in mol·m–2, and the species bulk concentration, c in mol·dm–3, called the adsorption isotherm. The isotherm mostly used in biosensor research is the Langmuir isotherm based on the following hypotheses:
- The surface is supposed to be defined by a number of adsorption sites given by the surface concentration is Γmax.
- Equilibrium between adsorbed and bulk target species - No lateral interactions between the adsorbed target species
Fig. 1. Adsorption according to the Langmuir isotherm
The reaction can be written as
Empty site + Target Species ! Adsorbate
In this way, we can define the adsorption equilibrium constant as
K = konkoff
= !!max " !( ) cbulk
= !1"!( ) cbulk
(1)
where θ is the surface coverage defined as the ration Γ/Γmax. The adsorption equilibrium is expressed in M–1. The surface coverage equilibrium !eq can be written as
!eq = Kcbulk
1+ Kcbulk= "
1+" (2)
2
The dimensionless parameter ψ is useful to define the adsorption process. If ψ is much smaller than unity, then we have an adsorption from a dilute solution where the surface coverage is small and directly proportional to the bulk concentration; we then speak of the linearized Langmuir isotherm. If ψ is much greater than unity, then the coverage tends to a full monolayer. When ψ is equal to unity, the equilibrium coverage is half of the full coverage. In the case of biosensing, we deal generally with dilute solutions and therefore the equilibrium surface concentration is directly proportional to the bulk concentration.
!eq = !maxKcbulk (3)
Exercise : For all the calculations, we shall take K=109 M–1, kon=106 M–1·s–1, a bulk analyte
concentration of 0.1 pM unless specified otherwise, and !max " 10#9 mol·m#2 .
Calculate the area of an adsorption site, the equilibrium surface coverage and ψ .
1.2 Kinetic aspects The rate law for a binding reaction is just given by the difference between the adsorption process considered as a second order reaction between the reactant and a free site and the desorption process simply considered as a first order reaction, and we can write
d!dt
= koncbulk (1!! )! koff! (4)
This differential equation can be re-arranged as
d!dt
+! koncbulk + koff!" #$ = konc
bulk (5)
and the solution is then given by
! = koncbulk
koncbulk + koff
1! exp ! koncbulk + koff( )t"
#$%
"#&
$%' = "
1+"1! exp !t /#[ ]"# $% (6)
where τ is the time constant for the adsorption defined as
! = 1konc
bulk + koff= td
1+" (7)
where td is the desorption time td =1/ koff( ) . This equation clearly shows that the adsorption kinetics from dilute solutions is controlled by the desorption time. The adsorption kinetics is illustrated schematically in Figure 2 for different values of ψ
Fig. 2. Time dependence of the surface coverage
3
Figures 3 & 4 illustrate the dependence of τ on kon and koff.
5
4
3
2
1
0
-1
-2
log
( /
s )
-15 -10 -5 0
log (c /M)
kon = 105
koff = 10–4
koff = 10–3
koff = 10–2
koff = 10–1
Fig. 3. Influence of desorption time koff in s–1.
5
4
3
2
1
0
-1
-2
log
( /
s )
-15 -10 -5 0
log (c /M)
kon = 106, 105, 104, 103
koff = 10–4
Fig. 4. Influence of adsorption time kon in M–1·s–1.
At low concentrations, the adsorption time constant is dominated by the desorption rate whilst it is dominated by the adsorption rate at high concentrations. The transition point on Figures 3 & 4 corresponds to a concentration of K–1. We can also define the adsorption time ta =1/ konc
bulk( ) such that
! !1 = ta!1 + td
!1 (8)
At low bulk concentrations ! <<1( ) , from eq. (3) & (6), the surface concentration of target species is given by
!(t) = !eq 1" exp "t / td( )#$ %& (9)
At short times, we can linearize the exponential term to show that the surface coverage is proportional to time
! = koncbulkt = t
ta= " t
td (10)
or
4
! = !eqttd
(11)
Exercise : Calculate ta , td , and τ. Please comment. Plot Γ(t)/Γeq. To give some orders of magnitude, Table 1 gives some kinetic values for some classical systems.
kon / M–1·s–1 koff / s
–1 IgE 106 10–3
IgG4 106 10–2 Avidin-Biotin 108 10–7
ss-DNA 6·104 5·10–5 Exercise : Please comment on the different values
1.3 Adsorption measurement To monitor the adsorption of target species onto a substrate, one can distinguish label-free techniques and techniques relying on a chemical labelling step. Label free techniques include wire conductivity measurements, capacitance measurements, quartz microbalance measurements, resonators and surface plasmon resonance measurements. Chemical labelling is used for immunoassays and protein arrays using often a secondary antibody labelled with an enzyme. Enzyme activity is then monitored by addition of a substrate. In the case of DNA, the complementary oligo sequence is itself labelled often with a fluorophore. Label-free techniques are very useful to measure the kinetics of binding. For example, a flow system combined to SPR can be used to measure a sensorgram. The measurement is done in two steps. First, a buffer solution is passed on the detector, and at time t=0, a solution of target species is injected, the adsorption is then monitored. At a given time, the flowing solution is switched to the eluent and the desorption is monitored as shown below.
Adsorption Desorption
Time Exercise : Please explain the physical principles of a label-free technique (min 1 A4 page). Derive the equation and draw a sensorgram for the system used above, namely K=109 M–1, kon=106 M–1·s–1, a bulk analyte concentration of 0.1 pM.
5
2. Adsorption from a finite volume
2.1 Thermodynamic aspects In microsystems, the surface to volume ratio can be very large and the bulk concentration can decrease when the target molecules adsorb. At equilibrium, the bulk concentration and therefore the equilibrium surface coverage will be different than for large systems having a small S/V ratio.
t = 0 c0
Equilibrium ceq < c0
Adsorption
Fig. 5. Adsorption in a microchannel
Another way to increase even further the surface to volume ratio is to fill the channel with magnetic beads coated by antibodies. In this case, the area corresponds to the total bead area, and the volume corresponds to the void volume between the beads. Let’s consider the injection of target molecules in a microchannel, as illustrated in Figure 5. The number of molecules injected is nin = cinV , where V is the volume of the solution. These molecules distribute themselves between the walls and the bulk, and if we consider a linearized Langmuir isotherm, we have:
nin = nwall + nsolution = !eqµ A + ceqV = K!maxceqA + ceqV (12)
where !eqµ is the equilibrium surface concentration in the microchannel, which is related to the
sample concentration by
!eqµ = K!maxV
V + K!maxA"#$
%&'cin =
!eq1+ K!maxA /V
=! !eq1+!
(13)
with ! a dimensionless geometric factor, given by
! = VAK!max
(14)
We can see that when the volume is large, we recover eq.(3). We can also define a wall collection efficiency as
! = nwallnin
=A!eq
µ
Vcin= K!maxA
V + K!maxA= 1
1+" (15)
We can see that α tends to unity for large surfaces and to zero for large volumes.
6
Exercise: Calculate α and ! for a microchannel 1cm long x100µm wide x 50µm deep, and for a microtiter plate V = 100µL, A = 1 cm2. Example : IgG-Antigen
K=109 M
–1
kon=106 M
–1·s
–1
cbulk
= 0.1 pM
D=10–10
m2·s
–1
! = 100 "m
#max = 10–9
mol·m–2
100!m
50!m
1cm
V = 50nL
A = 3mm2
!= 0.01667
" = 0.9836
# eq $ 10–6
DiagnoSwiss Immunochip
With one fill, we have only 0.0001% of a monolayer for a 0.1pM solution
Enzyme Linked ImmunoSorbent Assay ELISA
Y Y Y Y
YYYY
YY
Capture antibody adsorptionWash
Surface blockingWash
Antigen adsorptionWash
Y Y
Y
Counter antibody adsorption
Addition of substrateDetection
A market of more than 20 billion $...
Total well volume = 300 L
100 L = 1cm2
Enzyme Linked ImmunoSorbent Assay ELISA
Y Y Y Y
YYYY
YY
Capture antibody adsorptionWash
Surface blockingWash
Antigen adsorptionWash
Y Y
Y
Counter antibody adsorption
Addition of substrateDetection
A market of more than 20 billion $...
Total well volume = 300 L
100 L = 1cm2
2.2 Comparison between a microtiter plate and a microchannel Exercise: Please fill the following table using the numerical values listed above: Microtiter well Microchannel Volume 100μL Initial number of antigens in solution Surface area 100 mm2 Maximum antibody sites Adsorbed antigens (Maximum)
2.3 Kinetic aspects To estimate the kinetics of adsorption in a microsystem, we can write eq.(4) as
dnwallnsdt
= konnin ! nwall
V"#$
%&'ns ! nwall
ns
"#$
%&'! koff
nwallns
"#$
%&'
(16)
where ns is the total number of sites available on the walls of the microsystem. Implicitly, we assume that the bulk concentration is homogeneous, i.e. we neglect any mass transport contribution. By neglecting the second order term, we can re-write this differential equation as
dnwalldt
+kon nin + ns( )
V+ koff
!"#
$%&nwall = konninns
V (17)
The solution of eq.(17) is
nwall = konninns
Vkon nin + ns( )
V+ koff
!"#
$%&
1' exp 'kon nin + ns( )
V+ koff
!"#
$%&t
(
)*
+
,-
(
)**
+
,--
(18)
We can re-arrange this equation introducing the geometric factor ! to read
nwall = nin1+! 1+ Kcin( ) 1! exp !
kon nin + ns( )V
+ koff"#$
%&'t
(
)*
+
,-
(
)**
+
,--
(19)
The adsorption time constant in a microsystem τµ can be compared to that of a planar surface in a large volume as when nin is much greater than ns the two expressions converge.
7
!µ = 1kon nin + ns( )
V+ koff
= 1kon cin + (A /V )!max( )+ koff
= " td1+ Kcin" +"
(20)
If we still define ψ as equal to Kcin , then when ψ <<1, eq.(19) reduces to
nwall = !nin 1! exp ! t!"td
"
#$
%
&'
"
#$
%
&' (21)
Exercise: Calculate the time constant for the same microchannel and microtiter plate as above
2.4 Comparison between a microtiter plate and a microchannel Exercise: Please fill the following table using the numerical values listed above assuming that the kinetics of the binding is the rate-limiting step: Microtiter well Microchannel Volume 100μL Initial number of antigens in solution Adsorbed antigens (Maximum)
Adsorbed antigens After 3 min XXXXX
Adsorbed antigens After 15 min XXXXX
Adsorbed antigens After 10 s XXXXX
Adsorbed antigens After 30 s XXXXX
8
3. Adsorption on a planar substrate with semi infinite linear diffusion
3.1 General equations When the rate of adsorption is fast compared to the time scale of diffusion, the kinetics of the binding may depend on the rate of arrival of the target species to the surface. To know in which case we operate, it is useful to define a diffusion time as tdiff = !
2 /D , where δ is the diffusion layer thickness and D the diffusion coefficient. The diffusion layer thickness varies with time in unstirred solutions, but can be fixed by controlling the hydrodynamics e.g. flow cells or the geometry e.g. microbeads (steady-state diffusion). To be able to treat kinetic and mass transport, we shall use the two-layer model, where we define a volumic surface concentration csurf that differs from the bulk value cbulk such that the diffusion flux is driven by the gradient (csurf – cbulk)/δ . Indeed, a diffusion flux results from a gradient of concentration. By rewriting eq.(4) in terms of the surface concentration Γ and a volumic concentration at the surface csurf , we have
d!(t)dt
= koncsurf !max " !(t)( )" koff!(t) (22)
Two layer model
Bulk
cbulk
Surface layer
zcsurf
Adsorbed layer
!
Fig.6. Two-layer model with the adsorbed and the surface layers.
The diffusion equation is given by Fick’s second law, which for a linear geometry reads
!c(x, t)!t
= D ! 2c(x, t)! x2
(23)
where D is the diffusion coefficient (m2·s–1). The variation of the surface concentration of the target species with time is equal to the flux of arrival at the surface:
d!(t)dt
= D !c(x, t)! x
"#$
%&' x=0
= ( J(t) (24)
9
Unfortunately, these equations cannot be solved analytically in the general case, but specific solutions can be obtained for example assuming that the diffusion process is steady-state.
3.2 Steady-state approximation for kinetic-diffusion control at short times Here, we consider the simple case based on neglecting the presence of occupied sites, i.e. ! <<!eq . This condition is fulfilled at very short times for all values of ψ. In this case, eq.(22) reduces to
d!(t)dt
= koncsurf!max (25)
If we consider furthermore, that the diffusion layer thickness is steady-state, e.g a well-stirred system or a spot in a protein array, we have
d!(t)dt
= koncsurf!max = " J = D
!cbulk " csurf( ) (26)
We can calculate the volumic surface concentration as
csurf = Dcbulk
kon!max! + D (27)
and eq.(26) gives
d!(t)dt
= konDcbulk!max
kon!max! + D (28)
As a result, the surface concentration increases linearly with time
!(t) = koncbulk!max
1+ kon!max!D
"
#
$$$
%
&
'''t =
!eq1+ Da"#$
%&'ttd
(29)
When D is very large, eq.(29) tends to eq.(10). We can see that when the diffusion layer thicknesses are thin or for large values of D (i.e. small molecules) then the mass transfer is very
fast. The dimensionless ratio kon!max!D
is called the Damköhler number, Da. The Damköhler
number represents the ratio of the characteristic adsorption rate to the characteristic diffusion rate, both rates being in m·s–1. When Da is large then the diffusion plays an important role, and inversely if Da is small, then the diffusion is not limiting. For Da <<1 , we recover eq,(11), whereas for Da >>1 we have a pure diffusion control
!(t) = Dcbulk
!t (30)
In a general manner, it is not possible to solve the time dependence of an adsorption-diffusion process. Nonetheless, we can apply a perturbation approach to eq.(29) and consider that the diffusion layer thickness grows with time according to Einstein’s equation such that ! = 2Dt . As a result, eq.(29) now reads
10
!(t) = koncbulk!max
1+ kon!max2tD
"
#
$$$
%
&
'''t = !eq
ttd
"#$
%&'
1
1+ kon!max2tD
(31)
Figure 7 shows a surface concentration-time plot showing the influence of the diffusion in stirred and unstirred solutions. Exercise : Please plot Γ(t)/Γeq using D = 10–10 m2·s–1 and δ = 100µm. Please also calculate the Damköhler number.
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Γ /Γ
eq
10008006004002000
t /s
Kinetic controlKinetic-Steady state diffusionKinetic-Time dependent
diffusion layer thickness
Fig. 7. Influence of the diffusion on the adsorption. ( K=109 M–1, kon=106 M–1·s–1, cbulk = 0.1 pM,
D=10–10 m2·s–1, δ = 100 µm and !max "10#9 mol·m#2 ). θ << θeq at short times.
The top curve represents the ideal stirring condition where the volumic surface concentration is always equal to the bulk concentration. In this case, the adsorption is limited by the kinetics of the binding reaction (kinetic control). The dashed line represents a stirring condition that fixes the diffusion layer thickness to 100 µm. As the Damköhker number is equal to one, the slope is equal to half that of the initial tangent of the top curve. In this case, the adsorption process is controlled both by the kinetics and the diffusion. The bottom curve represents a non-stirred system where the diffusion layer thickness grows with square root of time. We see that stirring only plays a role after 1 minute. Indeed, at short times the transient Damköhler number
Da =kon!max!
D= kon!max
2tD
is very small.
Exercise : Calculate the diffusion layer thickness and the Damköhler number at t = 100s and t = 1000s for unstirred conditions.
11
3.3 Steady-state approximation for kinetic-diffusion control in dilute solutions When the surface concentration is proportional to the volumic surface concentration ! <<1( ) , both the surface and the volumic surface concentrations vary with time in conjunction with the flux of oncoming target species
d!(t)dt
+ z dcsurf (t)dt
= " J = D!
cbulk " csurf( ) (32)
where z is the arbitrary thickness of the volumic surface layer
d!(t)dt
1+ zK!max
"#$
%&'
= D!
cbulk ( !(t)K!max
"#$
%&'
(33)
We then obtain the following differential equation
d!(t)dt
1+ zK!max
"#$
%&'+ D
!K!max
"#$
%&'!(t) = D
!cbulk (34)
The solution is eq.(34) is
!(t) = K!maxcbulk 1" exp " Dt
! z + K!max( )#
$%
&
'(
#
$%%
&
'((
(35)
if z << KΓmax, then eq.(35) reduces to
!(t) = !eq 1" exp " 1Da
ttd
#$%
&'(
)
*+
,
-.
)
*++
,
-..
(36)
Eq.(35) describes the kinetic of surface adsorption under steady-state diffusion control. Again, the use of the Damköhler number permits to know if the binding process is diffusion control or not. If Da>1, the diffusion slows the adsorption process. Now, again we can apply the perturbation approach and introduce a time dependent diffusion layer thickness ! = 2Dt to obtain
!(t) = K!maxcbulk 1" exp " D / 2t
K!max
#$%
&'(t
)
*++
,
-..
)
*++
,
-..
= !eq 1" exp " D / 2tK!max
#$%
&'(t
)
*++
,
-..
)
*++
,
-..
(37)
In the case a linearized Langmuir isotherm, the time dependence of the surface concentration in the case of a diffusion control binding (very fast kinetics) can be obtained analytically as shown in Annexe 1.
!(t) = !eq 1" exp(a2Dt)erfc(a Dt )#$ %& (38)
with
a = 1K!max
(39)
12
1.0
0.8
0.6
0.4
0.2
0.0
Γ /Γ
eq
1000080006000400020000
t /s
Kinetic-Steady state diffusion Diffusion control Kinetic-Time dependent
diffusion layer thickness
Fig. 8. Comparison between stirred and unstirred diffusion controlled reactions . ( K=109 M–1,
kon=106 M–1·s–1, cbulk = 0.1 pM, D=10–10 m2·s–1, δ = 100 µm and !max "10#9 mol·m#2 ).
This figure shows that the perturbation theory gives a rather good approximation to the exact solution eq.(38).
3.4 Steady-state approximation for kinetic-diffusion control - General case Eq.(26) can be more generally written as
d!(t)dt
= koncsurf!max 1"!( )" koff!max! = " J = D
!cbulk " csurf( ) (40)
or in terms of surface coverage
d! (t)dt
= koncsurf 1!!( )! koff! = D
!"maxcbulk ! csurf( ) (41)
from which we can as before obtain an expression for the surface concentration
csurf =cbulk + Da! /K( )1+ Da 1!!( ) (42)
The differential equation (41) now reads
d! (t)dt
=konc
bulk + Dakoff !( ) 1!!( )1+ Da 1!!( ) ! koff! =
koncbulk !! konc
bulk + koff( )1+ Da 1!!( )
or
1+ Da 1!!( )! !" ! +1( )"
#$
%
&'d! = dt
td (43)
We can integrate this expression explicitly assuming no initial coverage to get
Da! ! 1+ Da" +1
"#$
%&'ln 1!! " +1
""#$
%&'
"#$
%&'
=t " +1( )td
(44)
13
We can either plot the time as a function of the surface coverage or see different limiting cases to recover the previous approximations.
1.0
0.8
0.6
0.4
0.2
0.0
Γ /Γ
eq
300025002000150010005000
t /s
ψ =0.1ψ =1ψ =10
Figure 9. Comparison between kinetic control under ideal stirring (full line) and with some
diffusion influence according to eq.(44) (dotted line). Da = 1, td=1000s.
We define a diffusion time as tdiff = !2 /D , where δ is the diffusion layer thickness and D the
diffusion coefficient. Exercise: If δ = 100 µm and D = 10–10 m2·s–1, calculate tdiff and compare with ta and td. Compare this general solution with the two approximations, namely at short times and for dilute solutions.
14
4. Adsorption-diffusion in microsystems
4.1 Steady-state approximation Let us consider a microchannel of height 2h.
L
2h
l
By symmetry, the diffusion layer thickness is h as the top molecules diffuse to the top plate and the bottom ones to the bottom plates. We can therefore write the diffusion equation as in eq.(26) as
d!(t)dt
= koncsurf!max = " J = D
!cbulk " csurf( ) (45)
To take in to account the depletion in the center of the channel upon adsorption as in equation (16) reproduced below
dnwallnsdt
= konnin ! nwall
V"#$
%&'ns ! nwall
ns
"#$
%&'! koff
nwallns
"#$
%&'
(46)
and the diffusion equation reads
d!(t)dt
= dnwallAdt
= konnsurfV
"#$
%&' !max =
D!
nin ( nwall ( nsurfV
"#$
%&' (47)
As in equation (27), we can express nsurf as
nsurf =nin ! nwallkon"max!
D+1
= nin ! nwallDa +1
(48)
We can now solve the differential equation
dnwalldt
= Akon!maxV
nin " nwallDa +1
#$%
&'( ==
1!td
nin " nwallDa +1
#$%
&'(
(49)
The solution of this equation is
nwall = nin 1! exp ! t!td Da +1( )
"#$
%&'
(
)**
+
,--
(50)
When Da>>1, we are diffusion controlled with a fast kinetic, we have
nwall = nin 1! exp ! t!tdDa
"#$
%&'
(
)*
+
,- (51)
15
and when the mass transport is fast, we recover nearly eq.(21)
nwall = nin 1! exp ! t!td
"#$
%&'
(
)*
+
,- (52)
Exercise : Calculate the number of antigens adsorbed in the same microchannel as above after 30 s and compared to the number one would reach at equilibrium. Please comment.
4.2 Comparison between a microtiter plate and a microchannel Discuss the differences between a microtiter plate and a microchannel. Please fill the following table using the numerical values listed above taking into account the diffusion: Microtiter well Microchannel Volume 100μL Initial number of antigens in solution Adsorbed antigens (Maximum)
Adsorbed antigens After 3 min XXXXX
Adsorbed antigens After 15 min XXXXX
Adsorbed antigens After 10 s XXXXX
Adsorbed antigens After 30 s XXXXX
16
5. Coating a microchannel
5.1 Iterative stop-flow with adsorption equilibrium One way to coat a microchannel is the stop-flow method that consists in filling the channel quickly and to allow the adsorption to take place. This process is then repeated n-times until the surface concentration is high enough to allow a detection using, for example, a sandwich assay. During the iterative process, the total number of molecules present after the nth fill of the microchannel channel is equal to what was adsorb on the wall at the (n-1) fill plus the number of molecules injected during the nth fill
ntot,N = nwall, N –1 + nin (53)
In this way, we can write for the successive iterations. First fill:
ntot,1 = nin (54)
nwall, 1 = !nin (55)
Second fill:
ntot,2 = nwall, 1 + nin = 1+!( )nin (56)
ntot,2 = nwall,2 + nsolution = !eq,2µ A + ceqV = K!maxceqA + ceqV (57)
!eq,2µ = K!max
V + K!maxA"#$
%&'ntot,2 (58)
nwall,2 = !eq,2µ A = K!maxA
V + K!maxA"#$
%&'ntot,2 = ! 1+!( )nin (59)
Then, at the n-th fill we have
ntot,n = nwall, n!1 + nin = 1+! +! 2 + ... +! n–1( )nin (60)
and
nwall,n = ! +! 2 + ... +! n–1( ) · nin (61)
Considering the properties of geometric series
nwall,Nnin
= = ! ! i
i=0
N–1
! =! 1"! N( )1"!
(62)
As illustrated in Figure 10, we have an additive filling for the values of α close to unity.
17
15
10
5
0
n wall / n i
n
20151050
N
! = 0.99
! = 0.9
! = 0.8! = 0.7
Fig. 10. Iterative filling
The advantage of the stop-flow method is to guarantee a homogeneous coating of the wall. Exercise: With previous parameters, calculate how many fills are required to coat a microchannel with 100’000 antigens and estimate the time required, the sample volume consumption and the number of analyte molecules wasted.
5.2 Iterative stop-flow with kinetic control To estimate the kinetics of adsorption in a microsystem, we can write
dnwallnsdt
= konnin ! nwall ! n0( )
V"#$
%&'ns ! nwall
ns
"#$
%&'! koff
nwallns
"#$
%&'
(63)
where ns is the total number of sites available on the walls of the microsystem, and n0 the number of molecules adsorbed from the previous step. Implicitly, we assume that the bulk concentration is constant, i.e. we neglect any mass transport contribution. We can re-write this differential equation as
dnwalldt
+kon nin + ns + n0( )
V+ koff
!"#
$%&nwall =
kon nin + n0( )nsV
(64)
which gives that
nwall = n0 exp !kon nin + ns + n0( )
V+ koff
"#$
%&'t
(
)*
+
,-
+kon nin + n0( )ns
Vkon nin + ns + n0( )
V+ koff
"#$
%&'
1! exp !kon nin + ns + n0( )
V+ koff
"#$
%&'t
(
)*
+
,-
(
)**
+
,--
(65)
We can re-arrange this equation introducing the geometric factor ! to read
18
nwall = n0 exp !kon nin + ns + n0( )
V+ koff
"#$
%&'t
(
)*
+
,-
+ nin + n01+! + K! cin + c0( ) 1! exp !
kon nin + ns + n0( )V
+ koff"#$
%&'t
(
)*
+
,-
(
)**
+
,--
(66)
The adsorption time constant in a microsystem τµ can be compared to that of a planar surface in a large volume as when nin is much greater than ns the two expressions converge.
!µ = 1kon nin + ns + n0( )
V+ koff
= td1+!!1 + K cin + c0( )
= td!0
(67)
If we still define ψ as equal to Kcin , then when ψ <<1, this equation reduces to
nwall = n0 exp ! !0ttd
"
#$
%
&' +
nin + n0"!0
1! exp ! !0ttd
"
#$
%
&'
"
#$
%
&'
= nin + n0"!0
+ n0 !nin + n0"!0
()*
+,-
exp ! !0ttd
"
#$
%
&'
(68)
First fill, we have
nwall,1 = nin!"0
1! exp ! "0ttd
"
#$
%
&'
"
#$
%
&' (69)
Second fill,
nwall,2 = nwall,1 exp ! !1ttd
"
#$
%
&' +
nin + nwall,1"!1
1! exp ! !1ttd
"
#$
%
&'
"
#$
%
&' (70)
Third fill,
nwall,3 = nwall,2 exp ! !2ttd
"
#$
%
&' +
nin + nwall,2"!2
1! exp ! !2ttd
"
#$
%
&'
"
#$
%
&' (71)
etc… Exercise: With previous parameters, calculate how many fills are required to coat a microchannel with 100’000 antigens and estimate the time required, the sample volume consumption and the number of analyte molecules wasted.
5.3 Continuous flow coating We can coat the microchannel by flowing a solution of target species through. If the binding kinetics is fast compared to the flow rate, we shall coat first the entrance of the channel and then
19
progressively along the channel. If the binding kinetics is slow, the coating will be more homogeneous but it is important not to flow to fast, not too waste too many target species. As developed in Annexe 2, under pure diffusion control the diffusion layer thickness in a flow channel varies with the flow rate according to
! =
3a2Dx!
30
l∫ dx
0
l∫ dx
=
0.759Dhx< v>
30
l∫ dx
0
l∫ dx
=
340.759Dh< v>
3 l4/3
l (72)
The flux of the target species to the wall of a microchannel of length l, of width L, of half height h where the average liquid velocity is <v> is expressed by
J = 0.8131 cbLD2/3l2/3 < v >1/3 h!1/3 (73) Please fill the following table and comment
Linear speed cm/s
Flow rate nL/min δ/μm nadsorbed
during 1 min nin
during 1 min 0.01 0.1 1 10
Which flow rate is required to obtain a homogeneous coating along the channel? Compared how long it would take to immobilize 100’000 target species under pure kinetic control and pure diffusion control.
5.4 How to fill a microchannel? What is the best way to coat a microchannel, with continuous flow or by iterative filling?
20
6. Adsorption-diffusion on a spherical substrate Nowadays, most immunoassays take place on magnetic beads as illustrated in Figure 11.
Fig. 11. Immunoassay on a magnetic bead
An interesting property of spherical diffusion is to provide a finite diffusion layer thickness. This property is used in electrochemistry with microelectrodes that can generate steady-state currents in the absence of any hydrodynamics. The diffusion layer thickness is equal to the radius R of the substrate. Exercise : Please demonstrate For ψ >1, we can use eq.(29), which shows that the surface concentration increases linearly with time
!(t) = koncbulk!max
1+ kon!maxRD
"
#
$$$
%
&
'''t =
!eq1+ Da"#$
%&'ttd
(74)
The smaller the beads, the less important the diffusion. It is then only controlled by the kinetics. For ψ <<1, we can use eq.(35) and obtain an exponential decay
!(t) = K!maxcbulk 1" exp " D
RK!max
#$%
&'(t
)
*+
,
-.
)
*++
,
-..
= !eq 1" exp " DRK!max
#$%
&'(t
)
*+
,
-.
)
*++
,
-..
= !eq 1" exp " 1Da
ttd
#$%
&'(
)
*+
,
-.
)
*++
,
-..
(75) Exercise: Calculate Da for a 1µm diameter bead. Calculate the number of antibodies per bead. Calculate the mass of beads (Ø =1µm), the time and the capture efficiency to bind 100’000 molecules from a 0.1pM solution. Sample volume = 10 µL Is it diffusion or kinetically controlled? In the case of commercial magnetic beads, the diameter is typically 1 micron and one mg contains about 109 beads.
21
7. ELISA assay (from www.interferonsource.com) “Enzyme Linked Immuno-Sorbent Assay (ELISA) is a powerful technique for detection and quantitation of biological substances such as proteins, peptides, antibodies, and hormones. By combining the specificity of antibodies with the sensitivity of simple enzyme assay, ELISA can provide a quick and useful measurement of the concentration of an unknown antigen or antibody. Currently, there are three major types of ELISA assays commonly used by researchers. They are: indirect ELISA, typically used for screening antibodies; sandwich ELISA (or antigen capture), for analysis of antigen present; and competitive ELISA, for antigen specificity”.
Figure 14. Elisa scheme http://www.interferonsource.com/New_InterferonSource/ELISA/ELISA_Description.html,
TetraMethylBenzidine for oxidation by HRP
“The "sandwich" technique is so called because the antigen being assayed is held between two different antibodies. In this method:
1. Plate is coated with a capture antibody. 2. Sample is then added, and antigen present binds to capture antibody. 3. The detecting antibody is then added and binds to a different region (epitope) of the
antigen. 4. Enzyme linked secondary antibody is added and binds to the detecting antibody. 5. The substrate is then added and the reaction between the substrate and the enzyme
product can be monitored optically or electrochemically. 6. The signal generated is directly proportional to the amount of antibody bound antigen.
Optimizing an ELISA assay requires the careful selection of antibodies and enzyme-substrate reporting system. Once optimized, sandwich ELISA technique is fast and accurate. If a purified antigen standard is available, this method can be used to detect the presence and to determine the quantity of antigen in an unknown sample. The sensitivity of the sandwich ELISA is dependent on 3 factors:
a) The number of molecules of the first antibody that are bound to the solid phase, namely, the microtiter plate.
b) The avidity of the antibodies (both capture and detection) for the antigen c) The specific activity of the detection antibody that is in part dependent on the number
and type of labelled moieties it contains. It is important to note that while an ELISA assay is a useful tool to detect the presence and the quantity of an antigen in the sample, it does not
22
provide information concerning the biological activity of the sample. ELISAs are not generally used to discriminate active or non-active forms of a protein. It may also detect degraded proteins that have intact epitopes”.
8. Protein array (from Arrait.com) “Antibody, Antigen, Peptide and other Protein Microarrays Custom Made to Order. Antibody microarrays to Cancer, Apoptosis, Growth Factors, Cytokines, Receptors and More... Users can to assay thousands of specific proteins with the ArrayIt Protein Microarrays. A wide variety of proteins representing a range of biological functions are detectable. ArrayIt microarrays allow users to measure differences in protein abundance using a two-color fluorescent labeling technique that can be provided with the microarrays. “Sandwich” detection schemes can also be implemented, please discuss your application goals with us. Detection is typically fluorescent-based using microarray scanners, but microarrays for other detection platforms can be provided. The protocols are simple, and the detection limits are extremely high.
Figure 15. Protein array scheme
http://www.arrayit.com/Products/Microarrays/
Figure 15. Two color antibody microarray. Compare two biological samples to measure the absolute and relative differences in protein expression. This procedure is fluorescence-based and compatible with all current microarray scanners. Extracted proteins are directly labelled (kit provided) and analysed. The covalently immobilized antibodies capture fluorescently labelled antigens during the reaction step. Normal and cancerous cells can be compared. The raw data provide a measure of proteins from samples A & B.
23
9. Redox detection in a microchannel Most immunoassays use an optical detection method based either on colorimetry, fluorescence or nowadays more and more chemiluminescence. In ELISA assays, an enzyme reaction takes place that produce often by cleavage an optical probe.
Fig 16 : Gravicell from DiagnoSwiss SA
The Michaelis-Menten equation for an enzyme reaction is rarely expressed in its explicit form that gives the product concentration as a function of time
–[P] + KM ln[S0] – [P][S0 ]
!
" #
$
% & = ' vmt (76)
At short times, we can linearize this equation
–[P] ! KM[P][S0 ]
= ! vmt (77)
to obtain
[P]= vmt1+ KM / [S0 ] (78)
In the case of excess substrate concentration, this equation reduces to
[P]=vmt (79)
In the case of immune-assays with electrochemical detection, the enzyme reaction produces a redox active probe. In the example below, alkali phosphatase is used to cleave para-aminophenolphosphate (PAPP) to produce para-aminophenol (PAP) that can be oxidized to quinone-imine. By amperometry, the current is directly proportional to the PAP concentration
24
PAPP AP
NH2
OPO32-
NH2
OH
ALP,H2O
pH = 9+ HPO4 2-
AP QI
NH
O
NH2
OH
2e-
2H+
Ag Concentration
-1.0E-09
0.0E+00
1.0E-09
2.0E-09
3.0E-09
4.0E-09
5.0E-09
6.0E-09
7.0E-09
8.0E-09
0 200 400 600 800 1000 1200 1400 1600
Time [s]
I [A
]
10 pM
2 pM
1 pM0 pM
Fig 17 : Amperometric detection with PAPP, and PAP oxidation
In the case of an oxidation of microdisc electrode, the current is equal to 4nFDcr, where n is the number of electrons exchanged, F is the Faraday constant, D the diffusion coefficient of the redox probe, c its concentration and r the radius of the microdisc electrode. Exercise : Assume a Michaelis-Menten rate law, and calculate the number of PAP molecules produced as a function of time. Calculate the current one would get on a 25 µm Ø gold disc microelectrodes. How many electrodes can one fit in the microchannel? In the literature, we can find k2= 50 s–1
25
Annexe 1: ψ << 1. Diffusion control In the case of an unstirred solution where the diffusion time is greater than the adsorption process time, we have to solve Fick’s equation in the solution phase. Using the Laplace transform, we can show that the transform of the concentration is
c (x, s) = c(x, 0)s
+ A(s) exp! s
Dx
A1.1
Exercise : Derive eq.A1.1 Similarly, the transform of the flux is
J (s) = !D !c (x, s)! x
"#$
%&'x=0
= D A(s) sD
"
#$
%
&' A1.2
which gives
c (x, s) = c(x, 0)s
+ J (s) 1Ds
exp! s
Dx
A1.3
The Laplace transform of eq. can be written as
!J (s) = s"(s)! "(0) = s"(s) A1.4
assuming that the surface was initially free of adsorbed molecules. So, eq.(A1.3) can be written
c (x, s) = cbulk
s! "(s) s
Dexp
! sDx
A1.5
If we assume that the adsorption process is fast compared to the diffusion, and if assumes that we have a low surface coverage ! <<1( ) , then using eq.(3) taking the volumic concentration at the surface
!(t) = !maxKcsurf (t) A1.6
we have using eq.(A1.5) at x = 0,
!(s)K!max
= cbulk
s" !(s) s
D A1.7
or
!(s) = cbulk
s 1K!max
+ sD
"#$
%&'
A1.8
We can re-inject this equation in eq.(A1.5) to obtain
26
c (x, s) = cbulk
s! cbulk
1K"max
+ sD
#$%
&'(
1sD
exp! s
Dx
A1.9
In this way, the transform of the concentration at the surface is
c (0, s) = cbulk
s! cbulk
s DK"max
+ s#$%
&'(
A1.10
Knowing the following inverse transform
L exp(u2t)erfc(u t ){ } = 1s s + u( ) A1.11
we have
csurf (t) = cbulk 1! exp(a2Dt)erfc(a Dt )"# $%
A1.12 with
a = 1K!max
A1.13
or
!(t) = !eq 1" exp(a2Dt)erfc(a Dt )#$ %& A1.14
It is important to realise that the term in brackets depends only on the adimensional term u = a2Dt as shown below, which in terms depends only on the physical parameters K, Γmax and D, but not on the bulk concentration. The latter affects only Γeq.
Fig.8. Illustration of equation Error! Reference source not found.
The transform of the flux using eq.(A1.4) is
J (s) = !D !c (x, s)! x
"#$
%&' x=0
= c Da D + s( ) A1.15
27
which gives
J(t) = c D 1! t
! a D expa2Dt erfc(a Dt )"
#$%&'
A1.16
This expression of the flux can be compared to the Cottrell equation in chronoamperometry
JCottrell(t) = c D 1! t
!"#
$%&
A1.17
We can therefore see that for small values of a D , i.e. large K values or for a large specific surface area providing a large Γmax or even for small molecules with a small diffusion coefficient, the Cottrell equation is a good approximation. Exercise: Derive the equivalent of the Cottrell equation by the perturbation method of a steady-state flux.
28
Annexe 2: Laminar flow. The Grätz problem We shall consider here a solution flowing parallel to two plane onto which molecules are adsorbed on a section of length l as shown in Figure 13. The problem is analogous to the current obtained in a band electrode in a flow channel.
Fig. 13. Adsorption in a flow channel
The equation of flux conservation is given by
!c!t
= " divJ = D#2c " v·gradc A2.1
In steady state, for a 2D system where the x axis is parallel to the electrode and the y axis perpendicular to it, we have only
vx!c!x
+ vy!c!y
= D !2c!y2
+ !2c!x2
"
#$%
&'( D !2c
!y2 A2.2
by neglecting the longitudinal diffusion. The continuity equation for an incompressible fluid is classically given by
!vx!x
+!vy!y
= div(!v) = 0 A2.3
We can solve this equation by doing a series development on y, to obtain
vx = vx (y = 0)+ y!vx!y
"#$
%&' y=0
+ ... = y!(x) A2.4
and
vy = vy(y = 0)+ y!vy!y
"#$
%&' y=0
+ 12 y
2 !2vy!y2
"
#$
%
&'y=0
+ ...
= 12 y
2 !!y
!vy!y
"#$
%&' y=0
+ ... = ( 12 y
2 !!y
!vx!x
"#$
%&' y=0
+ ...
= ( 12 y
2 !!x
!vx!y
"#$
%&' y=0
+ ... = ( 12 y
2! '(x)+ ...
A2.5
with
29
!(x) = !vx!y
"#$
%&' y=0
A2.6
and since
!vy!y
"#$
%&' y=0
= ( !vx!x
"#$
%&' y=0
= ( y! '(x)( )y=0 = 0 A2.7
assuming that
vx (y = 0) = vy(y = 0) = 0 A2.8
The flux conservation equation (1.90) becomes
y!(x) !c!x
" y2! '(x)2
!c!y
= D !2c!y2
A2.9
If the flow is parallel to the wall such that vy is zero, then equation (1.98) reduces to
y!(x) !c!x
= D !2c!y2
A2.10
For a Poiseuille flow profile between two planes located at –h and h, the velocity profile is given by
vx = vmax 1!Y 2
h2"
#$%
&'= 3< v >
21! Y
2
h2"
#$%
&' A2.11
with Y = y ! h , and then β(x) is constant and equal to 3<v>/h.
Fig. 14. Diffusion layer thickness in flow channels
Equation (1.99) can then be written
D !2c!y2
= 3< v >h
y !c!x
= 2vmaxh
y !c!x
A2.12
30
Diffusion layer thickness: Approximate solution To solve equation (1.99), we can consider the thickness of the diffusion layer δ (x) as illustrated above, and linearize the gradients and write
y! !c!x
= y! !c!ydydx
" y! #c#ydydx
= "! #c"d"dx
= !#c d"dx
A2.13
with y = ! (x) and !c = cb " c x = 0( ) . Similarly, we have
D !2c!y2
" D !!y
#c#y
= D #c! 2
A2.14
With these simplifications, eq.(1.99) reduces to
! d"dx
= D" 2
A2.15
that we can integrate to have
! 3 = 3Dx"
A2.16
The limiting current on the band electrode is then for the case illustrated in figure 13
I = nFDLcb dx! (x)0
l! = nFDLcb dx
3Dx"
3
= 3–1/3nFcbLD2/3"1/3l2/30
l! A2.17
By substituting the value of β equal to 3<v>/h.
I = nFcbLD2/3l2/3 < v >1/3 h!1/3 A2.18
or
I = 23
!"#
$%&1/3nFcbL vmaxD
2l2
h!
"#$
%&
1/3
= 0.8735nFcbL vmaxD2l2
h!
"#$
%&
1/3
A2.19
We can write this equation as a function of the volumetric flow rate FV given by
FV = 2 < v > hd A2.20
and obtain an equation.
I = 2!1/3nFcbL D2l2FVh2d
"
#$%
&'
1/3
= 0.7937 nFcbL D2l2FVh2d
"
#$%
&'
1/3
A2.21
Diffusion layer thickness: Exact solution We can solve analytically eq.(1.99), by using a similarity variable and write
c(x, y) = F y! (x)
!"#
$%&
= F !( ) A2.22
The boundary conditions are
31
c(0, y) = c(x,!) = cc(x, 0) = 0
A2.23
or
c(0, y) = F y! (0)
!"#
$%&
= F '( ) = c
c(x, 0) = F 0( ) = 0 A2.24 The partial derivatives are then
!c!x
= !!!xdFd!
= " y" 2d"dx
#$%
&'(dFd!
= " !"d"dxdFd!
A2.25
!c!y
= !!!ydFd!
= 1"dFd!
A2.26
!2c!y2
= !!y
!!!ydFd!
"#$
%&'
= 1"
!!y
dFd!
"#$
%&'
= 1"!!!y
dd!
dFd!
"#$
%&'
= 1" 2
d2Fd!2
"
#$%
&' A2.27
The two variables differential equation (1.99) becomes
d2Fd!2
!
"#$
%&+!2 "# 2
Dd#dx
!
"#$
%&dFd!
= 0 A2.28
The term in brackets must be constant with respect to x and we can write
!" 2
Dd"dx
!
"#$
%&= a2 A2.29
We should therefore solve the following equation
d2Fd!2
!
"#$
%&+ a2!2 dF
d!= 0 A2.30
A solution of this equation is
F(!) = c1 exp ! a2
3z3
"
#$%
&'dz + c20
!( A2.31
since
F '(!) = c1 exp ! a2
3!3
"
#$%
&' A2.32
and
F ''(!) = ! c1a2!2 exp ! a
2
3!3
"
#$%
&' A2.33
The integration constants of equation (1.120) must satisfy the boundary conditions (1.113), such that
32
c2 = 0 A2.34
and
c1 = c / exp ! a2
3z3
"
#$%
&'dz
0
() A2.35
Equation (1.120) becomes
F(!) = cexp ! a
2
3z3
"
#$%
&'dz
0
!(
exp ! a2
3z3
"
#$%
&'dz
0
)(
A2.36
The normalising integral is finite and equal to
exp ! a2
3z3
"
#$%
&'dz
0
() = 2
935/6
a2/3* 23
"#$
%&'
A2.37
The concentration profiles are therefore given by
c(x, y) = F y! (x)
!"#
$%&
= F(") =31/6a2/3c' 2
3!"#
$%&
2exp ( a
2
3z3
!
"#$
%&dz
0
") A2.38
To calculate the thickness of the diffusion layer, we can write
!c!y
"#$
%&' y=0
= !!!y
dFd!
"#$
%&'!=0
= 1"dFd!
"#$
%&'!=0
= (c"
= cb
" A2.39
and
dFd!
!"#
$%&!=0
=31/6a2/3' 2
3!"#
$%& c
b
2= 0.8131 a2/3cb = cb A2.40
This equation allows us to calculate a= 0.87116. By integration of eq.(1.118), we have
! =
3a2Dx!
30
l∫ dx
0
l∫ dx
=
0.759Dhx< v>
30
l∫ dx
0
l∫ dx
=
340.759Dh< v>
3 l4/3
l
A2.41 The limiting current on the band electrode is then for the case illustrated in figure 13
I = nFDLcb dx! (x)0
l! = nFDLcb dx
3a2Dx"
3
= 3–1/3a"2/3nFcbLD2/3"1/3l2/30
l!
33
A2.42
By substituting the value of β equal to 3<v>/h.
I = a!2/3nFcbLD2/3l2/3 < v >1/3 h!1/3 = 0.8131nFcbLD2/3l2/3 < v >1/3 h!1/3 A2.43
We can write this equation as a function of the volumetric flow rate FV given by eq.(1.109) and obtain an equation similar to equation (7.50) from the book “ Analytical & Physical electrochemistry”.
I = 2!1/3a!2/3nFcbL D2l2FVh2d
"
#$%
&'
1/3
= 0.6454 nFcbL D2l2FVh2d
"
#$%
&'
1/3
A2.44
As a function of the maximum velocity, we have
I = a!2/3nFcbLD2/3l2/3 2vmax3
"#$
%&'1/3h!1/3 = 2
3"#$
%&'1/3a!2/3nFcbL vmaxD
2l2
h"
#$%
&'
1/3
= 1.0744nFcbL vmaxD2l2
h"
#$%
&'
1/3 A2.45