Difference between revisions of "Team:UMaryland/BCmodel"

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Bacterial Cellulose Modeling

How fast can PETase CBD degrade a bottle with cellulose binding?

A 500mL PET water bottle from a common brand weighs 4 grams, is 23 centimeters tall, and has a diameter of 7 cm. If we flatten this bottle out into a sheet, its surface area will be roughly 1000 square centimeters. This sheet will be covered by sheets of bacterial cellulose on either side.

Bacterial cellulose creates a "sandwich" to maximize PETase contact with the surface

Cellulose Binding Domain (CBD) has a dissociation constant of Kd= 0.038uM. Its maximum binding capacity, Cmax, is 0.43uMol per gram cellulose.

If cellulose has a surface area of 1000 square centimeters, working thickness of 0.1mm, and a density of 1.5 grams per cubic centimeter, the total mass of cellulose is about 15g. Thus, in a 1 L container, the maximum CBD protein binding concentration is 0.43uMol/g*15g/L = about 6.5uM protein

The degradation process involves heterogenous catalysis: solid PET substrate is interacting with dissolved PETase enzyme. We will assume that the concentration of PET vastly exceeds that of PETase. We will also assume a zero order reaction.

Equations from the assumption of zero order reaction

We can determine the rate constant of PETase degrading PET with values from Yoshida's 2016 paper. PET was incubated with 50nM PETase at pH 7.0 for 18 hours at 30C. 0.015mM of degradation product was observed from high-crystalline PET, the type found in a bottle. Plugging these parameters into the equation above, we obtained K = 0.00463 s^-1 for high crystalline PET exposed to PEtase.

PET is a long polymer, but we will assume complete degradation to MHET (MW = 210.185 g/mol). For a 4g bottle in a 1L container, that comes out to 0.019M

So, with K = 0.00463 s^-1, [PETase] = 6.5uM, and [product] = 0.019M, degradation based on a zero order reaction is expected to take roughly two days.

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-					CBD has a dissociation constant of Kd= 0.038uM. Thus, its maximum binding capacity, Cmax, is 0.43uMol per gram cellulose.
+					Cellulose Binding Domain (CBD) has a dissociation constant of Kd= 0.038uM. Its maximum binding capacity, Cmax, is 0.43uMol per gram cellulose.
 					<div>
 					<div>
 					If cellulose has a surface area of 1000 square centimeters,  working thickness of 0.1mm, and a density of 1.5 grams per cubic centimeter, the total mass of cellulose is about 15g. Thus, in a 1 L container, the maximum CBD protein binding concentration is 0.43uMol/g*15g/L = about 6.5uM protein
 					<div>
+					<div class="meatMeat">
+The degradation process involves heterogenous catalysis: solid PET substrate is interacting with dissolved PETase enzyme. We will assume that the concentration of PET vastly exceeds that of PETase. We will also assume a zero order reaction.
+					</div>
 <div class="imageBox">
 						<img src="https://static.igem.org/mediawiki/2018/7/74/T--UMaryland--heterogenous1.png" style="max-width: 100%" alt="Waluigi Time!">
-						<div class="imageBoxDescription">Bacterial cellulose creates a "sandwich" to maximize PETase contact with the surface</div>
+						<div class="imageBoxDescription">Equations from the assumption of zero order reaction</div>
 					</div>
 					<div class="meatMeat">
-The degradation process involves heterogenous catalysis: solid PET substrate is interacting with dissolved PETase enzyme. We will assume that the concentration of PET vastly exceeds that of PETase. We will also assume a zero order reaction.
+We can determine the rate constant of PETase degrading PET with values from Yoshida's 2016 paper. PET was incubated with 50nM PETase at pH 7.0 for 18 hours at 30C. 0.015mM of degradation product was observed from high-crystalline PET, the type found in a bottle. Plugging these parameters into the equation above, we obtained K = 0.00463 s^-1 for high crystalline PET exposed to PEtase.
+					</div>
+					</div>
+PET is a long polymer, but we will assume complete degradation to MHET (MW = 210.185 g/mol). For a 4g bottle in a 1L container, that comes out to 0.019M
+					</div>
+					</div>
+So, with K = 0.00463 s^-1, [PETase] = 6.5uM, and [product] = 0.019M, degradation based on a zero order reaction is expected to take roughly two days.
 					</div>
 			</div>