Polylactic acid (PLA) Degradation Module

Introduction

The efficiency of Polylactic acid is important for the performance of MAPLE. Therefore, we chose a strong enzyme Proteinase K which is suggested by many literature as an efficient PLA degrading enzyme. We performed several assays in order to determine the kinetic properties of Proteinase K. The PLA degradation model was then built based on the experimental results.

Objective and Design

1. With the defined kinetic properties of Proteinase K. The model can predict the time needed to degrade a certain amount of PLA. The information is important as it estimates the efficiency of the MAPLE system when degrading the large amount of plastic.Therefore, further improvements can be made if the system is not efficient.

2. The model also involves the gene expression model of the Proteinase K with an inducible promoter. Therefore, gene expression can be regulated by adjusting the inducer concentration. The overall plastic degradation can be regulated by the gene expression.

3. The secretion model is also contained in the model which assumes the efficiency of enzyme secretion to the culture.

Specifications of the model

Polylactic acid (PLA) degradation involves proteinase K:

The overall PLA degradation model is shown as below:

There are two compartments which represents cells and the culture from left to right. The "cell" compartment contains the gene expression module whereas the "culture" compartment contains the degradation module. The "secretion" block that connects two compartments is the secretion module.

Parameters and assumptions

Gene expression module of PudA

1.Derivation of the maximal expression rate,β

• Average molecular weight (Mw) of a base pair = 660g/mol[http://www.geneinfinity.org/sp/sp_dnaprop.html][http://www.lifetechnologies.com/uk/en/home/references/ambion-tech-support/rna-tools-and-calculators/dna-and-rna-molecular-weights-and-conversions.html]
• Average mass of a base pair = 660g/mol x 1.66x10^-24 = 1.1x10^-21g
• Volume of an E.coli cell = 1µm³[http://kirschner.med.harvard.edu/files/bionumbers/fundamentalBioNumbersHandout.pdf] = 1x10^-15L
  • ∴Mass concentration =
  • ∴Molar concentration of 1 base pair in the volume of E.coli = = 1.66x10^-6 mM
• BioBrick assembly plasmid pSB1C3 is a high copy number plasmid (100-300 copies per cell)[http://parts.igem.org/Part:pSB1C3?title=Part:pSB1C3]
  • assume 200 copies per cell
• ∴ concentration of the gene per cell = N x 200 x 1.66x10^-6mM, where N = number of base pairs
  • ∴ concentration of the gene BDH2 (N = 768) in the volume of an E.coli cell is = 0.25mM
• Transcription rate in E.coli = 80bp/s[http://kirschner.med.harvard.edu/files/bionumbers/fundamentalBioNumbersHandout.pdf] = 80 x 1.66x10^-6mM/s = 80 x 1.66x10^-6 x 60mM/min = 7.97x10^-3mM/min
• ∴ Rate of mRNA_Proteinase K production under the control of pBAD = 7.97x10^-3 ÷ 0.25 = 0.032/min

• Average molecular weight(Mw) of an amino acid(aa)= 110g/mol[http://www.genscript.com/conversion.html][http://www.promega.com/~/media/Files/Resources/Technical%20References/Amino%20Acid%20Abbreviations%20and%20Molecular%20Weights.pdf]
• Average mass of an amino acid = 110g/mol x 1.66x10^-24=1.83x10^-22g/L
  • ∴Mass concentration of one aa in the volume of an E.coli = = 1.83x10^-6g/L
  • ∴Molar concentration of one aa = = 1.66x10^-5mM
• Translation rate = 20aa/s = (20 x 1.66x10^-5 x 60)mM/min = 0.020mM/min
• BDH2 comprises of 256aa[http://www.uniprot.org/uniprot/Q2PEN2&format=html]
  • ∴concentration of BDH2's aa in the volume of an E.coli = 1.66x10^-5mM x 256 = 4.25x10^-3mM
• ∴ Rate of protein production = 0.020 ÷ 4.25x10^-3 = 4.7/min