Chapter 10. Mass Transport and Biochemical Interactions BMOLE 425-625 Biomolecular Engineering Engineering in the Life Sciences Era BMOLE 452-689 – Transport Chapter 10. Mass Transport and Biochemical Interactions Text Book: Transport Phenomena in Biological Systems Authors: Truskey, Yuan, Katz Focus on what is presented in class and problems… Dr. Corey J. Bishop Assistant Professor of Biomedical Engineering Principal Investigator of the Pharmacoengineering Laboratory: pharmacoengineering.com Dwight Look College of Engineering Texas A&M University Emerging Technologies Building Room 5016 College Station, TX 77843 cbishop@tamu.edu Francis Crick and neuroscience? Why is E = mc2 and not ½ mc^2… Name all of the energy equations you know? Integral => 1/2 © 2008 Prof. Anthony Guiseppi-Elie; guiseppi@clemson.edu; T: 001 864 656 1712 F: 001 864 656 1713

Reaction Rates N in moles 𝑹 𝒊 = 1 𝑽 𝒅 𝑵 𝒊 𝒅𝒕 = 𝒅 𝑪 𝒊 𝒅𝒕 Ri is the reaction rate per unit volume Moles/unit volume/unit time 𝑹 𝒊 = 1 𝑽 𝒅 𝑵 𝒊 𝒅𝒕 = 𝒅 𝑪 𝒊 𝒅𝒕 𝑹 𝒊 𝒔 = 1 𝑺 𝒅 𝑵 𝒊 𝒅𝒕 Ris is the reaction rate per unit surface area Moles/unit area/unit time 𝒗 1 𝑨+ 𝒗 2 𝑩⇌𝑪 Reaction stoichiometry 𝑹 𝒄 = −1/ 𝑣 1 𝑹 𝑨 =−1/ 𝑣 2 𝑹 𝑩 Q = volumetric flow rate V = volume of reactor V/Q is = residence time − 𝑹 𝒊 = 𝑸 𝑽 𝑪 𝒊𝒐 − 𝑪 𝒊 Not often used with proteins, peptides, and nucleotides because of the limited supply (not steady flow).

Reaction mechanisms 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 Generally, Rates = constant * reactants * products * catalysts and a, b, c represent Reaction orders of individual reactants or products or catalysts and a +b +c = overall reaction order Is k actually a constant?

Reaction mechanisms 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 Generally, Rates = constant * reactants * products * catalysts and a, b, c represent Reaction orders of individual reactants or products or catalysts and a +b +c = overall reaction order Is k actually a constant? What affects k the most?

Reaction mechanisms 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 𝑹=𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 Generally, Rates = constant * reactants * products * catalysts and a, b, c represent Reaction orders of individual reactants or products or catalysts and a +b +c = overall reaction order Is k actually a constant? What affects k the most? Arrhenius equation 𝒌=𝑨 𝒆 − 𝑬 𝒂 𝑲 𝑩 𝑻 Ea = energy of activation (Joules) A = a constant for each chemical reaction defining the frequency of collisions in correct orientation Units of exponent? Units of k?

Reaction mechanisms Generally, Rates = constant * reactants * products * catalysts and a, b, c represent If disappearing or appearing dictates the sign Reaction orders of individual reactants or products or catalysts and a +b +c = overall reaction order 𝑹=𝒔𝒊𝒈𝒏 ∗𝒌 𝑪 𝑨 𝒂 𝑪 𝑩 𝒃 𝑪 𝑪 𝒄 Is k actually a constant? What affects k the most? Arrhenius equation 𝒌=𝑨 𝒆 − 𝑬 𝒂 𝑲 𝑩 𝑻 Ea = energy of activation (Joules) A = a constant for each chemical reaction defining the frequency of collisions in correct orientation Units of exponent? Units of k? How do you know the units of k?

Example: http://www.dartmouth.edu/~chem6/9/RxnRatesEquil.pdf

Example: Solve for Feq 1 http://www.dartmouth.edu/~chem6/9/RxnRatesEquil.pdf

Example: Solve for Feq 2 1 http://www.dartmouth.edu/~chem6/9/RxnRatesEquil.pdf

Example: 2 1 3 solve for k Solve for Feq http://www.dartmouth.edu/~chem6/9/RxnRatesEquil.pdf

Example: Solve for Feq 2 1 3 http://www.dartmouth.edu/~chem6/9/RxnRatesEquil.pdf

Laplace Transformations http://mathworld.wolfram.com/LaplaceTransform.html

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

Solving differential equations via laplace Convert differential equation to F(s) using Laplace table Solve for F(s) Apply initial conditions of f(t=0) and f’(t=0), etc. Calculate any unknowns by plugging in fake s values (i.e., 0, 1, 2) to solve Take the inverse using the Laplace table Done Example: y’’-10y’+9y=5t Sometimes easier to think about using f(t) and F(s)…

𝒇 ′′ 𝒕 −𝟏𝟎 𝒇 ′ 𝒕 +𝟗𝒇 𝒕 =𝟓𝒕;𝒇 𝟎 =−𝟏 𝒂𝒏𝒅 𝒇 ′ 𝟎 =𝟐 𝒇 ′′ 𝒕 −𝟏𝟎 𝒇 ′ 𝒕 +𝟗𝒇 𝒕 =𝟓𝒕;𝒇 𝟎 =−𝟏 𝒂𝒏𝒅 𝒇 ′ 𝟎 =𝟐 𝑳.𝑻. :𝓛 𝒇 ′′ 𝒕 = 𝒔 𝟐 𝑭 𝒔 −𝒔𝒇 𝟎 −𝒇′(𝟎) 𝑳.𝑻. :𝓛 𝒇′ 𝒕 =𝒔𝑭 𝒔 −𝒇 𝟎 𝑳.𝑻. :𝓛 𝒇 𝒕 =𝑭 𝒔 𝑳.𝑻. :𝓛 𝒕 = 𝟏! 𝒔 𝟏+𝟏 ; note; 0!=1 and 1!=1 𝒔 𝟐 𝑭 𝒔 −𝒔𝒇 𝟎 − 𝒇 ′ 𝟎 −𝟏𝟎 𝒔𝑭 𝒔 −𝒇 𝟎 +𝟗𝑭 𝒔 = 𝟓 𝒔 𝟐 𝒔 𝟐 𝑭 𝒔 +𝒔−𝟐− 𝟏𝟎𝒔𝑭 𝒔 +𝟏𝟎 +𝟗𝑭 𝒔 = 𝟓 𝒔 𝟐 𝒔 𝟐 𝑭 𝒔 +𝒔−𝟏𝟐−𝟏𝟎𝒔𝑭 𝒔 +𝟗𝑭 𝒔 = 𝟓 𝒔 𝟐 (𝒔 𝟐 −𝟏𝟎𝒔+𝟗)𝑭 𝒔 +𝒔−𝟏𝟐= 𝟓 𝒔 𝟐 𝒔−𝟏 𝒔−𝟗 𝑭 𝒔 += 𝟓 𝒔 𝟐 − 𝒔−𝟏𝟐 𝑭 𝒔 += 𝟓 𝒔 𝟐 − 𝒔−𝟏𝟐 𝒔−𝟏 𝒔−𝟗 = 𝟓− 𝒔 𝟑 −𝟏𝟐 𝒔 𝟐 𝒔 𝟐 𝒔−𝟏 𝒔−𝟗 = 𝑨 𝒔 + 𝑩 𝒔 𝟐 + 𝑪 𝒔−𝟏 + 𝑫 𝒔−𝟗 ;𝒔= −𝟏,𝟎,𝟏,𝟐 ;𝟒 𝒆𝒒., 𝟒 𝒖𝒏𝒌. 𝑭 𝒔 = 𝟓𝟎 𝟖𝟏 𝒔 + 𝟓 𝟗 𝒔 𝟐 + 𝟑𝟏 𝟖𝟏 𝒔−𝟗 + −𝟐 𝒔−𝟏 𝓛 −𝟏 𝑭 𝒔 =𝒇 𝒕 = 𝟓𝟎 𝟖𝟏 + 𝟓 𝟗 𝒕+ 𝟑𝟏 𝟖𝟏 𝒆 𝟗𝒕 −𝟐 𝒆 𝒕

http://www.sciencedirect.com/science/article/pii/S1742706116301325

http://www.sciencedirect.com/science/article/pii/S1742706116301325

Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 P = plasmid # (note that volume is not taken into account); simply # of plasmids in a given compartment at any given time t. Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 kdeg1 and kdeg2 equal to 0.62 And 0.084 hr-1 Where A+B at t=2 hours is total plasmid # in cell. i.e., 6500 plasmids Why 2 degradation rates of plasmids (polyplex context)? http://www.sciencedirect.com/science/article/pii/S1742706116301325

Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 P = plasmid # (note that volume is not taken into account); simply # of plasmids in a given compartment at any given time t. Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 kdeg1 and kdeg2 equal to 0.62 And 0.084 hr-1 Where A+B at t=2 hours is total plasmid # in cell. i.e., 6500 plasmids Why 2 degradation rates of plasmids (polyplex context)? What is the rate limiting step? Why is it what it is and not perhaps the expected nuclear uptake step? Why a Heaviside step function (HSSF)? Be able to do this http://www.sciencedirect.com/science/article/pii/S1742706116301325

Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 P = plasmid # (note that volume is not taken into account); simply # of plasmids in a given compartment at any given time t. Use: kcell, kne, kni, kbd equal to 7.5e-4, 0.72, 1.1, and 1.8 hr-1 kdeg1 and kdeg2 equal to 0.62 And 0.084 hr-1 Where A+B at t=2 hours is total plasmid # in cell. i.e., 6500 plasmids Why 2 degradation rates of plasmids (polyplex context)? What is the rate limiting step? Why is it what it is and not perhaps the expected nuclear uptake step? Why a Heaviside step function (HSSF)? Be able to do this Be able to do this Apply Laplace to solve… How would you do this? How would you deal with the HSSF without Taking the laplace of the HSSF. http://www.sciencedirect.com/science/article/pii/S1742706116301325

http://www.sciencedirect.com/science/article/pii/S1742706116301325

Why is Laplace with an e-st. : Could you use cos(st) or sin(st) Why is Laplace with an e-st?: Could you use cos(st) or sin(st)? or any other ex form or another answer of a differential equation?

Why is Laplace with an e-st. : Could you use cos(st) or sin(st) Why is Laplace with an e-st?: Could you use cos(st) or sin(st)? or any other ex form or another answer of a differential equation? The laplace table is simplified to the greatest extent using e-st. The most basic and most common answer of differential equations is e-xt. How do you solve more complicated partial differential equations?

Why is Laplace with an e-st. : Could you use cos(st) or sin(st) Why is Laplace with an e-st?: Could you use cos(st) or sin(st)? or any other ex form or another answer of a differential equation? The laplace table is simplified to the greatest extent using e-st. The most basic and most common answer of differential equations is e-xt. How do you solve more complicated partial differential equations? Assume P(x,y) = f1(x)*f2(y) => similar principle… Laplace(s)=definite integral(f1(st)*f2(t)) and t goes away… 23 March 1749 – 5 March 1827 Kind compliments of Wikipedia.org

Matrix Math for Chemical Reactions Chapter 2 Multi-Step Reactions: The methods for Analytical Solving the Direct Problem The following slides are based on this chapter cited below… https://link.springer.com/chapter/10.1007%2F978-3-7091-0531-3_2#page-1

Short-hand

Another example: Can you do this? Try it…

Another example:

Another example: ?

Another example: Now what?

Another example: Multiple by dt Separate variables properly integrate

Another example: Note: k matrix is Always an nxn matrix Multiple by dt Separate variables properly integrate Is this the correct notation? Eigenvalues and vectors are for this… Alternative?

More details… Remember? Same slide as presented in an earlier chapter… For low Reynold‘s #s, the drag force is (K is translation tensor; v = velocity): 𝑭 𝑫 =𝑲 ∗𝒗 K is a symmetric tensor and the components thereof are friction coefficients fij The principal friction coefficients are where within f? How do we compute it? Det(K – fI) = 0 (f1 =f2 =f3 etc for an isotropic body) Where does this come from… Av=λv… Eigenvectors are scalable and translatable… v(A- λ?)=0… v(A- λI)=0 These are the eigenvalues and are the friction coefficients. The only non-zero solutions are Calculated using the determinant… For an nxn matrix, the eigenvectors will be a what? nx1… I is a what? nxn identity matrix. fbar = average = the harmonic mean: 1/fbar = 1/3(1/f1 + 1/f2 + 1/f3 +…) Note that for a sphere f = 6ᴫμR… sound familiar? S.E.

v(A- λ?)=0… not mathematically possible v(A- λI)=0 Solve for λ and then using Av= λv solve for eigenvectors Remember they are scalable and translatable and usually they are normalized to have a magnitude of 1.

LHS= RHS=nx1 vector (rxc) Calculate eigenvalues and vectors of k: k = nxn matrix X<i> = nx1 vector λi = scalar LHS= RHS=nx1 vector (rxc) Calculate eigenvalues and vectors of k: X_barbar = matrix of eigenvectors (as columns) Multiple by dt Separate variables properly integrate Matrix diagonalization…

Matrix diagonalization A = PDP-1 P = eigenvector matrix D = diagonalization of Eigenvalues via Identity Matrix i.e., k= xΛx-1 So Capital_Lambda_barbar = lambda*Identity matrix

A = randn(3,3); [V,D]=eig(A); V = 3x3 and D = a diagonal 3x3 with eigenvalues on diag. A*V = V*D and A*V(:,1)=V(:,1)*D(1,1) If A = symmetric then there are 3 eigenvalues (not necessarily distinct) and 3 orthogonal eigenvectors… How to from A calculate eigenvalues and eigenvectors. To do manually: Ex: from A= [3 0 0; 0 3 0; 0 0 3] or any A… making it simple for example… A-lambda*I = 0; solve for lambdas; each root is a lambda i.e., (λ-3)*(λ-3)*(λ-3)=0; all 3 are 0 but of course don’t have to be… Then solve for V1 by A*V(:,1)=V(:,1)*D(1,1) Then solve for V2 by A*V(:,2)=V(:,2)*D(2,2) Then solve for V3 by A*V(:,3)=V(:,3)*D(3,3);

[V,D]=eig(A); V = 3x3 and D = a diagonal 3x3 with eigenvalues on diag. A = randn(3,3); [V,D]=eig(A); V = 3x3 and D = a diagonal 3x3 with eigenvalues on diag. A*V = V*D; but isn’t Av=(λ=D)V?... Why is it wrong in matlab? and A*V(:,1)= D(1,1) *V(:,1) If A = symmetric then there are 3 eigenvalues (not necessarily distinct) and 3 orthogonal eigenvectors… How to from A calculate eigenvalues and eigenvectors. To do manually: Ex: from A= [3 0 0; 0 3 0; 0 0 3] or any A… making it simple for example… A-lambda*I = 0; solve for lambdas; each root is a lambda i.e., (λ-3)*(λ-3)*(λ-3)=0; all 3 are 0 but of course don’t have to be… Then solve for V1 by A*V(:,1)=V(:,1)*D(1,1) Then solve for V2 by A*V(:,2)=V(:,2)*D(2,2) Then solve for V3 by A*V(:,3)=V(:,3)*D(3,3); Now let’s convince ourselves: Why?

Is this equal to exp(V. D. m. inv(D))Co in matlab Is this equal to exp(V*D*m*inv(D))Co in matlab? Is this equal to Coexp(Kt)? Matlab won’t let you use symbolic math with t; reserved for other things… MatLab has a hard time simplifying: V*exp(D*t)*inv(V)*Co

Clc k*C = dC/dt syms k1, k2, k3, k4 k*dt = 1/C dC exp(kt)+constant = C(t) exp(kt)*constant=C(t) syms k1, k2, k3, k4 How do you solve for V and D manually from K?

Given: K = [1 -1; 2 4]; what is D and V Caclulate compartments CA(t) and CB(t) using: You should know though that to make sense there should be rate constants (and/or numbers) in the matrix K but for Simplicity to show the math process K is a function of numbers only here Be able to get to this end stage from knowing a compartmental scheme… i.e., I want you to be able to know every step... To get to C(t) using matrix math using chemical/rate reactions.

Earlier… How?

A = 3x3 V = 3x3 D = 3x3 A*V = V*D and A*V(:,1)=V(:,1)*D(1,1)=D(1,1)*V(:,1) but A*V does not equal D*V

Principal Component Analysis Reduces complex data => linearly uncorrelated orthogonal components Typically: Av = λv

Principal Component Analysis Reduces complex data => linearly uncorrelated orthogonal components Typically: Av = λv (A – 11TA(1/n) = C)TC = M M = variance-covariance matrix Mv = λv; v = eigenvectors; λ = eigenvalues Mv = scores: inter-variable correlations v = coefficients rank variables’ contributions A = randn(3) B=cov(A) Stdev = sqrt(variance) Sqrt(B(1,1)=stdev(A(:,1))

PCA 1) rows of samples with all the variables as columns 2) cannot be missing any data 3) normalize all of the data 4) [coefs, scores, variances, X] = princomp(normalized_data); 5) Main plots Percentage of variance contribution Scores plot Loading plot Further analysis for trends to predict

Normalized Variables Physico-chemical properties: Cellular function Mn, Mw (MW*), PDI, DP, B:S, B, S, B+S Joback/Crippen’s Fragmentation method: BP*, MP*, CV*, GFE*, LogP*, MR*, HtF*, tPSA * = repeating unit (vs full polymer) Cellular function Viability, uptake, transfection

percent_exp = 100*variances/sum(variances)

Figure S1 plot(coefs(:,1:2)) Mention some variables line up with PC1 and other variables line up with PC2 and some are in between. plot(coefs(:,1:2)) Figure 1. Top: the variances are explained for the first5 PCs. The top 4 variables of the 27 are ranked by the degree to which they contribute to the PCs according to their associated coefficients and listed above the variances. The “(-)” indicates a negative contribution; Bottom: The loading plot of all of the variables showing relative correlations.

Variables driving function Ranking of variables: Acos(Θ) = correlation strength E number index was most neutral in all cases: 0.02, 0.01, -0.01.

Scores Plot plot(scores(:,1),scores(:,2)) Mn Mw “Self-assembled” B+S

Red = highest transfection level “Self-assembled” The scores plot versus transfection efficacy with red being the highest level of transfection; B: region B of 3A with a B+S (sum of carbons in backbone and sidechain) value of 7; C: region C of 3A with a B+S value of 8; D: region D of 3A with a B+S value of 9.

I developed an algorithm to assess whether the aortic valve is opening in rotary VADs which can be implemented in the software for RPM auto-regulation RPMs are set to a value based on echocardiography infrequently Problems with not regulating RPMs: 1) LV collapse events which inhibits blood flow 2) Calcification of aortic valve if no blood traverses normal path (aortic valve stenosis 3) Stenotic valve can result in thromboembolism and subsequent cerebrovascular accident/stroke Software would inhibition of thrombosis formation and aortic valve calcification (bridge to recovery)

I developed an algorithm to assess whether the aortic valve is opening in rotary VADs which can be implemented in the software for RPM auto-regulation Fouriér Transformation Beginnings of PCA Resembles Arterial Pressure Line with Dichrotic Notch

I developed an algorithm to assess whether the aortic valve is opening in rotary VADs which can be implemented in the software for RPM auto-regulation

Principal Component Analysis Dot product of PC1 and electrical signal in time Confirmed with echocardiography

PCA: Predicting seizures from electroencephalograms Predicting cardiac arrhythmias from pacemakers It can be used as a learning algorithm You can use facial recognition with it You can breakdown someone’s voice and look at the strongest frequencies contributing to it and do voice recognition

More PK/PD (sometimes written PKPD) ADME

Common variables for PK/PD Cp=plasma concentration of drug DB, D0=drug amount in body, initial drug concentration Cl, ClR, ClH, ClT , ClNR=clearance, renal clearance, hepatic, total body, non-renal VD= volume distribution of drug R = rate of infusion into systemic circulation k, km, ke, ka=degradation rate, metabolism rate, excretion rate, absorption rate AUC = area under the curve (Cp(t)) F= bioavailability Du= drug amount in urine Ab = absorption