Homework Statements

Biochemical Engineering

Spring 2016


Table of Contents

DueHomeworkSolutions
02/04Homework #1 See Homework Statement
02/11Homework #2  
02/18Homework #3  
02/25Homework #4  
03/03Homework #5  
03/10Homework #6  
03/24Homework #7  
03/31Homework #8  
04/07Homework #9  
04/14Homework #10  
04/21Homework #11  
04/28Homework #12  
05/05Homework #13  
    
  
 


Homework #1 (due 02/04)

  1. Read Chapter 1 of Shuler & Kargi (bioprocess engineer)

  2. Get textbook. Review computer methods. Make sure you know how to do regression analysis, solve numerically linear/nonlinear algebraic equations and ordinary differential equations (ODE) with a computer.

  3. List five biochemical engineering products. Pick out one of these products and (1~2 page) describe how it is manufactured and attach a process flow diagram.

Homework #2 (due 02/11)

  1. Read Chapter 2 of Shuler & Kargi (biological basics)

  2. Problem 2.11, 2.12, 2.13 of Shuler & Kargi

    Solution (Prob 2.11, 2.12, 2.13):

                       advantages                             disadvantages
      ------------------------------------------------------------------------------
      defined medium   more reproducible                      expensive
                       clean extracelluar product recovery    miss some micronutrient
      complex medium   inexpensive (some are expensive)       run-to-run variation
                       unknown nutrient requirement           complex product recovery
                       higher growth/yield
    
      Prob 2.12. produce antibiotics in large scale
        Complex medium.  Reason: inexpensive product, unknown nutritional requirement
      Prob 2.13. produce high-value protein
        Defined medium.  Reason: expensive product, need consistently reproducible results
    

  3. Problem 2.16 of Shuler & Kargi

    Solution (Prob 2.16):

       Gram -: two cell membranes (outer and inner) separated by periplasmic space;
               excess protein accumulates in periplasmic space
               structure: intracellular|cytoplasmic membrane|periplasmid space|1 peptidoglycan layer|outer membrane| extracellular
               typical example: E. coli
    
       Gram +: cell wall plus inner membrane;
               easier protein secretion
               structure: intracellular|cytoplasmic membrane|periplasmid space|cell wall=peptidoglycan layers| extracellular
               typical example: Bacillus subtilis
    
    Bioprocessing implication: protein product excretes more easily in gram + bacteria than gram - bacteria.


Homework #3 (due 02/18)

  1. Read Chapter 3.1-3.3 of Shuler & Kargi (enzyme kinetics)

  2. Problem 3.3 of Shuler & Kargi

    Solution:

  3. Problem 3.6 of Shuler & Kargi

    Solution:

    • Reaction Rate Parameter Estimation


Homework #4 (due 02/25)

  1. Read Chapter 3.4-3.7 of Shuler & Kargi (immobilized enzyme)

  2. Figure 6 of Bitan's paper describes a mechanism for the formation of amyloid, a culprit of Alzheimer's disease. The following is the gist of it.
        monomer (M) <--> paranuclei (N) <--> large oligomers (O) <--> protofibrils (P) --> fibril (F)
        monomer (M) --> protofibrils (P)
        paranuclei (N) --> protofibrils (P)
    
    You propose a means of reducing fibril by "mopping" away one of the precursors, say, protofibrils (P), with a monoclonal antibody (A).
                                     kap=1
      antibody (A) + protofibril (P) <--> AP
                                     kpa=1
    
    For this homework problem, let's simulate with the following kinetic parameters, where kij is the rate from I to J. (Units are omitted for simplicity.)
      kmn=1  knm=1
      kno=1  kon=1
      kop=1  kpo=0.5 (smaller, because Bitan's figure has a shorter arrow)
      kpf=1
      kmp=0.1  knp=0.1 (smaller, because Bitan's figure has dotted arrows)
    
    Translate this mechanism into a set of dynamic equations of the standard 1st-order form "dy/dt=f(t,y)". Integrate dy/dt from t=0 to t=10, starting with M=1 and A=0. Plot all species with time. Repeat by adding antibody (A=1 at t=0), and graphically compare this case with antibody to the case without antibody. What if you design an antibody that binds stronger and faster to P (kap=100 & kpa=0.1 instead)?

    Solution:

  3. Generate Figure 3.15 with the parameter values given in that figure's caption. Plot for a few exposure times: 5 min, 10 min, 15 min, 20 min, 25 min.

    Solution:

    The following shows a wider spread in time than what was asked for, but it basically demonstrates the same point. As a chemical engineer, you would normally want to find what temperature to carry out the reaction and for how long to maximize your profit. You are welcome to try it for fun.


Homework #5 (due 03/03)

  1. Prepare for Mid-term exam #1.

  2. Problem 3.19 of Shuler & Kargi. In addition, set up the ODEs (dP3/dt=?, dS/dt=?, etc) & the associated initial and boundary conditions, that, when solved, yield P3 as a function of time in a batch reactor; provide other information (e.g., suface to volume ratio) that are needed.

    Solution:

    Chemical engineers do not stop at deriving reaction rate v. In fact, reaction rate v is the starting point for designing a reactor/process. The following material balance equations govern what happens with time to the various species in a batch bioreactor.
      ε=void fraction=(fluid volume)/(fluid volume+bead volume)
      Given J(s)=v1(s)+v2(s) → find|solve for: s=s(sb)
      dSb/dt =-A/(ε*V)*(v1(s(sb))+v2(s(sb))   Sb(0) =Sb0
      dP1b/dt= A/(ε*V)*v1(s(sb))              P1b(0)=P1b0=0
      dP2b/dt= A/(ε*V)*v2(s(sb))              P2b(0)=P2b0=0
      P3=lesser of: P1b & P2b
    


Homework #6 (due 03/10)

  1. Read Chapter 4 of Shuler & Kargi (central dogma of biology)

  2. Problem 4.12 of Shuler & Kargi

    Solution:

    Glycosylation is the addition of a glycan|carbohydrate to a protein after translation (post-translational modification). N-linked glycosylation: point of attachment to a protein is the side groups of asparagine (Asn) or arginine (Arg); the sequence is Asn-Any-Ser|Thr.

    1. Where (cellular location, organelles): initially in endoplasmic reticulum (ER), subsequently continues in Golgi apparatus. Since prokaryotes do not have these organelles, glycosylation occurs only in eukaryotes.
    2. What. In these organelles, a series of enzymes catalyze the removal of sugar residues to form simple glycoform, then the addition of other sugar residues to complex glycoform. The composition of the enzymes (rather than a DNA|RNA template), which is cell compartment dependent, determines the eventual glycosylation pattern: where|site on protein, structure (branching pattern), glycan composition (sugar types, length).
    3. Simple glycoform (in yeast, insect cells, etc): terminal residue = mannose.
    4. Complex glycoform (in mammalian cells): terminal residue = N-acetyl glucosamine, galactose, and/or sialic acid.
    5. Why (functions): modulation of a protein's function, protein folding, cell-cell interaction (including immuno responses, modulation of antigens, modulation of antibodies, immuno-typing), cell-ECM interaction.

  3. Facilitated Transport and Active Transport. Given a substrate that moves across a cell membrane by facilitated transport, derive the transport rate expression. In facilitated transport, a carrier C binds to the substrate S to form a carrier-substrate complex CS at the outer side of the cell membrane. The complex diffuses from the outer side of the membrane toward the inner side of the membrane, where the substrate is released into the cell cytoplasm and the free carrier C is regenerated. The free carrier shuttles back, again by Fick's diffusion, toward the outer side of the membrane, where it picks up another substrate; and the cycle is repeated. The following "graphical reaction mechanism" describes this process, where the subscripts "e" and "i" stand for external and internal species, respectively.
              K                                              K
      Ce + Se <--> CSe --> diffusion across membrane --> CSi <--> Ci + Si
                   Ce <-- diffusion across membrane <-- Ci
    
    Derive the flux. That is the rate of transport of S across the cell membrane (an expression like JA in Equation 4.2 of Shuler & Kargi. Depending on your assumption (for example, equilibrium or quasi-steady state for the dissociation/association of CS), you may or may not reach a form that matches Equation 4.2 of Shuler and Kargi. Finally, plot flux versus Se (for simplicity, assign 1 to all model parameters, e.g., JA.max=1, K?=1, CA.i=1, ... etc). What part of the model needs to be modified (or what model parameters do you need to assign) to lead to active transport? Implement that needed change(s) in your model to demonstrate active transport. (See the comments at the bottom of p125 of Shuler & Kargi on how active transport is different from facilitated transport.)

    Solution

      Short answer:
        eqns 1 & 2: flux=DCS/δ*(CSe-CSi)=DC/δ*(Ci-Ce)
        eqns 3 & 4: K=Ce*Se/CSe=Ci*Si/CSi
        eqn 5:      Ctotal=(CSe+CSi+Ce+Ci)/2
        solve the above 5 equations for 5 unknowns: CSe, CSi, Ce, Ci, & flux
      Longer answer: more detailed explanation below.
    


Homework #7 (due 03/24)

  1. Read Chapter 5 of Shuler & Kargi (metabolic pathways)

  2. Consider human erythropoietin. We will practice transcription, translation, and post-transcriptional modification.

    1. Briefly, what is its function?
    2. Find the gene sequence from NCBI.
    3. Nucleotide sequence and # of bases in the primary transcript. (transcription)
    4. Nucleotide sequence and # of bases in the mature transcript. (post-transcriptional modification)
    5. Peptide sequence and # of peptides in erythropoietin. (translation)
    6. Position of N-glycosylation sites in erythropoietin. (post-translational modification)

    Solution:

    1. Erythropoietin is a protein secreted by the kidneys; it stimulates red blood cell production in bone marrow, treats anemia, and is an alternative to blood transfusion. It is one of the drugs that Lance Armstrong took.
    2. Go to NCBI. Pick from the pull-down menu "Nucleotide"; enter into the search field "human erythropoitin"; click "Search". NCBI returns ~4k hits; choose from the first few on the 1st page. A couple of DNA examples: M11319.1 and X02158.1. An mRNA example: NM_000799.2.
    3. Hereforth, we answer based on X02158.1 or epo.seq.
        primary transcript:  mRNA 397..3327 → # of bases=3327-397+1=2931
        (Capitalized portion is primary transcript)
            361 cgacccccgg ccagagccgc agagtccctg ggccacCCCG GCCGCTCGCT GCGCTGCGCC
             :
           3301 ACCTCATTGA CAAGAACTGA AACCACCaat atgactcttg gcttttctgt tttctgggaa
      
    4.   mature mRNA  join(397..627,1194..1339,1596..1682,2294..2473,2608..3327)
                   range   # bases
        --------------------------
        exon 1   397.. 627  231
        exon 2  1194..1339  146
        exon 3  1596..1682   87
        exon 4  2294..2473  180
        exon 5  2608..3327  720
        --------------------------
        total              1364
        epo.sq2 or epo.sq3  (Capitalized portion is CDS.)
      
    5. Petpitde sqquence (primary structure of protein)
        CDS  join(615..627,1194..1339,1596..1682,2294..2473,2608..2763)
        Total # of bases in the above range = 582
        Total # of bases that are codons (not counting the stop codon "tga") = 579
        Total # of primary peptides (including the signal peptide) = 579/3 = 193
                     range   # bases
          --------------------------
          exon 1   615.. 627   13
          exon 2  1194..1339  146
          exon 3  1596..1682   87
          exon 4  2294..2473  180
          exon 5  2608..2763  156
          --------------------------
          total               582
      
            601 ggccaggcgc ggagATGGGG GTGCACGgtg agtactcgcg ggctgggcgc tcccgcccgc
                               -M--G- -V-...
             :
           1261 CGCCCCACCA CGCCTCATCT GTGACAGCCG AGTCCTGCAG AGGTACCTCT TGGAGGCCAA
                 -A--P- ... (start of mature peptide)
             :
           2701 AATTTCCTCC GGGGAAAGCT GAAGCTGTAC ACAGGGGAGG CCTGCAGGAC AGGGGACAGA
                                                                     ...-G--D--R-
           2761 tgaccaggtg tgtccacctg ggcatatcca ccacctccct caccaacatt gcttgtgcca
                stop
      
        The above is given in the gene sequence file (193 peptides):
                           /translation="MGVHECPAWLWLLLSLLSLPLGLPVLGAPPRLICDSRVLQRYLL
                           EAKEAENITTGCAEHCSLNENITVPDTKVNFYAWKRMEVGQQAVEVWQGLALLSEAVL
                           RGQALLVNSSQPWEPLQLHVDKAVSGLRSLTTLLRALGAQKEAISPPDAASAAPLRTI
                           TADTFRKLFRVYSNFLRGKLKLYTGEACRTGDR"
        Total # of mature peptides (secreted) = 498/3 = 166
          mat_peptide     join(1262..1339,1596..1682,2294..2473,2608..2760)
                     range   # bases
          --------------------------
          exon 2  1262..1339   78
          exon 3  1596..1682   87
          exon 4  2294..2473  180
          exon 5  2608..2760  153
          --------------------------
          total               498
                           /translation="                           APPRLICDSRVLQRYLL
                           EAKEAENITTGCAEHCSLNENITVPDTKVNFYAWKRMEVGQQAVEVWQGLALLSEAVL
                           RGQALLVNSSQPWEPLQLHVDKAVSGLRSLTTLLRALGAQKEAISPPDAASAAPLRTI
                           TADTFRKLFRVYSNFLRGKLKLYTGEACRTGDR"
      
    6. The N-glycosylation sites are (Problem 4.12 of Shuler & Kargi): N-any-S|T and R-any-S|T. These are marked red for EPO: N-24, N-38, and N-83 (and 1 O- glycosylation site (S-126).
                                   APPRLICDSRVLQRYLL
        EAKEAENITTGCAEHCSLNENITVPDTKVNFYAWKRMEVGQQAVEVWQGLALLSEAVL
        RGQALLVNSSQPWEPLQLHVDKAVSGLRSLTTLLRALGAQKEAISPPDAASAAPLRTI
        TADTFRKLFRVYSNFLRGKLKLYTGEACRTGDR
      
    
    

  3. Problem 5.8 of Shuler & Kargi

    Solution (Prob 5.8):

    Transamination: Reaction that converts one amino acid to another or to an organic acid. Eqn 5.11 in Shuler & Kargi provides an example.

      glutamic acid + oxaloacetic acid → α-keto glutaric acid + aspartic acid
    

  4. Problem 5.11 of Shuler & Kargi

    Solution

      nitrogen fixation: synthesis of NH3 from atmospheric N2.
      mechanism: nitrogenase catalyzes reduction of N2: N2 + 6 H+ + 6 e- → 2 NH3
                                                        N2 + 6 H (as 3 NADH2) → 2 NH3
      environment: reducing or very low oxygen level
      Anaerobic organisms: ok
      Aerobic organisms:   in compartments that are protected from oxygen exposure or in very low oxygen environment
    


Homework #8 (due 03/31)

  1. Read Chapter 8 of Shuler & Kargi (recombinant DNA)

  2. Problem 8.11 of Shuler & Kargi. (What is an auxotroph? How might a thymidine auxotrophic mutant of E. coli be useful as a host in genetic engineering?)

    Solution:

    Auxotroph = an organism that cannot synthesize an essential nutrient component. Thus, that nutrient must be provided for cell growth, unless the cell is provided with the missing gene via a plasmid that also contains the gene of interest (say, insulin). Basically, whatever is missing in the chromosome, we provide it in a plasmid.

      host chromosome: thymidine
      plasmid: thymidine + gene of interest (insulin)
      host + plasmid = complete
    

    Grow on petri dish with thymidine in nutrient agar; stamp on petri dish without thymidine in nutrient ager; & see which colony is missing. Or, simply grow in nutrient containing thymidine to apply selection pressure.

  3. Given the gene of human erythropoietin from last homework, you are tasked to insert the gene into a plasmid named pUC19 (or search for it at NCBI). (Summary of sites in pUC19) To simplify purification, keep the signal peptide. Outline the general laboratory steps, including PCR (in that case give nucleotide sequence of primers) and/or the restriction enzyme(s) needed to insert the erythropoietin gene into pUC19. Print out the nucleotide sequence from the intermediate steps. On the nucleotide sequence of the final recombinant plasmid, clearly mark: the promoter|operator, RBS, and the erythropoietin gene. Ignoring the post-translational steps such as proper folding and glycosylation for now, will a bacterial host like E. coli be able to produce the primary structure of erythropoietin from your recombinant plasmid?

    Solution:

      Step 0.  Start with pUC19 and epo.sq3: (Capitalized = primary structure: 219..797)
      Step 1a. Look for one restriction enzyme that cuts in:
               i) the 5'UTR of EPO, ii) 3'UTR of EPO, and iii) pUC19.
               There is no one single restriction enzyme that can make these 3 cuts.
               There are two enzymes that cut both the 5'UTR and 3'UTR of EPO:
                 PspOMI (gggcc/c) & ApaI (g/ggccc).
               Create one such site (gggccc) in pUC19 (within beta-galactosidase CDS, usually within mcs) via site-directed mutagenesis through PCR.
               Note that although both PspOMI and ApaI have the same recognition bases (gggccc), the resulting sticky ends are incompatible.
               Treat the site-mutated pUC19 and EPO gene (cDNA from mRNA) with either PspOMI or ApaI.
      Step 1b. Add one common restriction site in each of 5'UTR and 3'UTR of EPO as we create EPO cDNA from EPO mRNA.
               The sequence is chosen based on available restriction sites in pUC19 (within beta-galactosidase CDS, usually within mcs)
      Step 1c. Add two dintinct restriction sites in each of 5'UTR and 3'UTR of EPO as we create EPO cDNA from EPO mRNA (see Step 1d).
               The sequence is chosen based on available restriction sites in pUC19 (within beta-galactosidase CDS, usually within mcs)
      Step 1d. Look for two restriction enzymes that respectively cut
               i) the 5'UTR of EPO, ii) 3'UTR of EPO, and iii) a cut each in pUC19.
               We find several combinations: one pair are XmaI and SacI.
               Treat pUC19 and EPO gene (cDNA from mRNA) with XmaI and SacI.
      Step 2.  Ligate.  There is no frame shift.
    
    
     ==============================================================================
      There is no one single restriction enzyme that can make a cut in the 5'UTR and 3'UTR of EPO and a cut in pUC19.
      We can use two restriction enzymes XmaI and SacI, each making a cut outside the CDS of EPO:
      Two different restriction sequences provie directionality, resulting in a higher probability of inserting EPO in the correct direction.
      The translation product has 41 extra peptides preceeding the EPO signal peptide..
        enzyme sequence  EPO  pUC19
        ---------------------------
        XmaI   c/ccggg   148   268
        SacI   gagct/c  1085   278
    
           :                                        .  .   .  .  .   .  .  .  .
     pUC  181 tgtgagcgga taacaatttc acacaggaaa cagctATGAC CATGATTACG CCAAGCTTGC
                .  .  .   .  .  .   .  .  .
     pUC  241 ATGCCTGCAG GTCGACTCTA GAGGATCC .   .  .  .   .  .  .   .  .  .  .
     EPO  121                               cc gggatgaggg cccccggtgt ggtcacccgg
                .  .  .   .  .  .   .  .  .  .   .  .  .
     EPO  181 cgccccaggt cgctgaggga ccccggccag gcgcggagAT GGGGGTGCAC GAATGTCCTG
           :
     EPO 1081 gcctgagct
     pUC  241                                               CGAATTCA CTGGCCGTCG
    
     ==============================================================================
     Or, we can add any sequence via PCR to either side of the target gene.
     To demonstrate, we add to EPO BamHI (g/gatcc) and EcoRI (g/aattc) sites, which does not exist in EPO.
     We also add a stop codon "taa" to terminate the translation of beta-galactosidase and re-start the translation of EPO.
     Once the lacZ promoter is induced, there will be two translation products:
       the 18-peptide N-terminus fragment of beta-lactamase and EPO.
    
        enzyme sequence  EPO  pUC19
        ---------------------------
        BamHI  g/gatcc         263
        EcoRI  g/aattc         284
    
      Given DNA strand #1 (+)                    5'-ATGGGGGTGCACGAATGTCC...........GCAGGACAGGGGACAGAtga-3'
      PCR primer #2 (-,reverse)                                                 3'-CGTCCTGTCCCCTGTCTACTcttaagcgc-5'
      ruler                                                                        12345678901234567890-EcoRI
    
      ruler                               BamHI-sTer12345678901234567890
      PCR primer #1 (+,forward)     5'-cgcggatccataaATGGGGGTGCACGAATGTCC-3'
      Given DNA strand #2 (-)                    3'-TACCCCCACGTGCTTACAGG...........CGTCCTGTCCCCTGTCTACT-5'
    
      where s after the BamHI site ggatcc is a spacer to align the frame for the stop codon taa.
      We digest pUC and the PCR product with BamHI and EcoRI, then ligate.
           :                                        .  .   .  .  .   .  .  .  .
     pUC  181 tgtgagcgga taacaatttc acacaggaaa cagctATGAC CATGATTACG CCAAGCTTGC
                .  .  .   .  .  .   .
     pUC  241 ATGCCTGCAG GTCGACTCTA GAG       .  .  .  .   .  .  .   .  .  .  .
     EPO  181                                 gatccataaAT GGGGGTGCAC GAATGTCCTG
           :
     EPO  781 GCAGGACAGG GGACAGAtga g
     pUC  241                                                 AATTCA CTGGCCGTCG
    
    In each of the above cases, we transform E. coli with the recombinant plasmid. Plate on a petri dish containing X-Gal (5-bromo-4-chloro-3-indolyl galactopyranoside), which when digested by beta-galactosidase releases a blue product. After induction with and IPTG (isopropyl thiogalactopyranoside), we select white colonies.


Homework #9 (due 04/07)

  1. Read Chapter 7 of Shuler & Kargi (stoichiometry)

  2. One of the solutions to Problem #3 of Homework #8 uses two restriction enzymes XmaI and SacI. The protein that is expressed has 41 extra peptides preceeding the EPO signal peptide. You decide to perform the PCR step to cut out the codons that correspond to these 41 extra peptides, so that the translation product is the human erythropoitin (EPO), no more, no less. Provide the sequence of the forward and backward primers, say the standard 20 bases.

    Solution:

    Deleting bases means not copying these bases with PCR. The 123 bases that correspond to the 41 extra peptide are stricken out. The flanking 20 bases are in red. Showing work necessitates displaying the reverse primer in the nonstandard 3'→5' notation, but at the end, we should write in the standard notation, otherwise, there will be confusion and you will eventually get burned badly -- remember, I am "Nam Wang", not "Gnaw Man"; and "I love you" does not equal "you love me".

     pUC  181 tgtgagcgga taacaatttc acacaggaaa cagctATGAC CATGATTACG CCAAGCTTGC
                           3'-taaag tgtgtccttt gtcga-5' ... reverse primer (in nonstandard 3'→5' notation)
    
     pUC  241 ATGCCTGCAG GTCGACTCTA GAGGATCC
     EPO  121                               cc gggatgaggg cccccggtgt ggtcacccgg
    
     EPO  181 cgccccaggt cgctgaggga ccccggccag gcgcggagAT GGGGGTGCAC GAATGTCCTG
                                                    5'-AT GGGGGTGCAC GAATGTCC-3' ... forward primer
     Summary (in standrd 5'→3' notation)
       forward (+) primer: 5'-ATGGGGGTGCACGAATGTCC-3'
       reverse (-) primer: 5'-agctgtttcctgtgtgaaat-3'
    
    Comment. PCR creates linear blunt-ended, double-strand DNA. If we intend to re-circularize the PCR product (after we cut out the above bases), we might want to simultaneously introduce a common restriction site to both ends during PCR. Furthermore, we add a few extra bases "xxx" to help the restriction enzyme grab on to the end of to digest the linear PCR product to create sticky ends. Note that the restriction site for EcoRI (g/aattc) is no good for this case, because there is already an EcoRI site in pUC19 beyond the end of EPO CDS. The following pair of primers introduce the site for BamHI (g/gatcc). After PCR, we digest with BamHI, then circularize and ligate; the new circular plasmid produces exactly the same primary translation product as human EPO, no more, no less. The PCR approach in Problem #3 of Homework #8, which results in the premature termination of the beta-galactosidase fragment and restarting EPO, works in a prokaryote host, but not in an eukaryote host. Thus, in practice, we perform this PCR-digestion-recircularization step after the initial cloning step if we are to place this plasmid in an eukaryote host.
                                 BamHI  20-mer primers from above
       forward (+) primer: 5'-xxxGGATCC-ATGGGGGTGCACGAATGTCC-3'
       reverse (-) primer: 5'-xxxGGATCC-agctgtttcctgtgtgaaat-3'
    

  3. Problem 7.1 of Shuler & Kargi

    Solution:

    Ammonium sulfate required:
    (30g/L)(1000L)(12% N)(mole a.s./28g N)(132g a.s./mole a.s.)=17.0kg
    


Homework #10 (due 04/14)

  1. Read Chapter 6.1-6.2 of Shuler & Kargi (cell growth)

  2. Prepare for mid-term exam #2. No problem submission.


Homework #11 (due 04/21)

  1. Read Chapter 6.3-6.5 of Shuler & Kargi (cell growth)

  2. Problem 6.2 of Shuler & Kargi. Continue with the following quetions.
    1. Repeat when the form of the specific rate μ is similar to Eqn. 3.42.
    2. With the latter form (Eqn, 3.42), simulate the batch growth until fermentation is almost over. Use the following model parameters: μm=1hr-1, Ks=1g/L, pK1=6, pK2=8, Yx/s=0.5(g biomass/g substrate), dH/dt=dp/dt=qp*x=α*μ*x+β*x (Eqn 6.18), α=10-5 (mole H/g biomass) β=10-7(mole H/hr/g biomass). I.C.: at t=0, x=0.1g/L, s=10g/L, p=H=10-9mole/L.
    3. Based on the simulated data of (x,s,pH) over fermentation time t, estimate the model parameters (μm Ks, K1, K2, Yx, α, β). Does it seem possible to estimate all or at least some of these model parameters from the dynamic data?
    4. If you control pH at 7, what is the product productivity (in g product/L/hr)? Compare the product productivity achieved with pH controlled at 7 to that achieved without pH control.

    Solution:

    The following file describes how to estimate model parameters by comparing the dynamic data (x,s,p); this is the most rigorous noise-tolerant method. Other ways of estimating model parameters are to first form μ=(dx/dt)/x, ds/dt, and dp/dt from the dynamic data (x,s,p); then regress μ to find μm, Ks, K1, & K2, either linearly or nonlinearly as described in Part b). Find Yx by regressing ds/dt against dx/dt=μ*x; the linear regression slope is 1/Yx. Find α & β by regressing dp/dt against two independent variables dx/dt=μ*x & x; the linear regression coefficients are α and β, respectively. This alternate method is less error-tolerant because the measurement errors are amplified when we calculate the derivatives (dx/dt, ds/dt, dp/dt, etc); we encounter this pitfall when we simulate with noisy data.


Homework #12 (due 04/28)

  1. Read Chapter 9 of Shuler & Kargi (bioreactor operation)

  2. Problem 6.15 of Shuler & Kargi

    Solution:

  3. Problem 9.2 of Shuler & Kargi. The motivation behind a two-stage chemostat is to increase product productivity. Find the combination of D1 & D2 that maximizes product productivity. Note that product productivity is defined as gram of product formed per hour per total reactor volume V=V1+V2. To help you visualize the effect of D1 & D2, you may want to plot product productivity versus D1 while D2=F/V2 is fixed at the given value; also plot product productivity versus D2 while D1=F/V1 is fixed at the given value; and/or do a 3-D surface|contour plot of productivity versus D1 & D2. Remember that a negative concentration makes no physical sense; concentrations are always constrained to be non-negative.

    Solution:

    • PDF Version (p9-2.pdf)
    • Mathcad File (p9-2.mcd) ... Maximize product productivity by changing D1 & D2
    • PDF Version (p9-2a.pdf)
    • Mathcad File (p9-2a.xmcd) ... More ways to maximize product productivity by changing D1 & D2


Homework #13 (due 05/05)

  1. Read Chapter 10 of Shuler & Kargi (Scale-up)

  2. Continue with Problem 9.2 of Shuler & Kargi, except that you carry out the two different stages of cell growth and product formation in a batch fermentor. Start the fermentor with x0=0.1g/L, s0=5g/L, and p0=0g/L Find the time tf1 to switch the first cell growth stage to the second product formation stage (say, induction of a recombinant organism) and the optimal harvest time tf2 so that the product productivity is maximized. In this case, if we ignore the downtime between batch runs, is the batch mode more productive than the continuous mode from the last homework assignment?

  3. Continue with Problem 2 where the product (say, EPO that you have previously worked on) is encoded in a plasmid. Upon cell division, there is a small probability P=0.001 of a plasmid-bearing cell producing a plasmid-free offspring. Start the batch fermentor with x0=0.1g/L, y0=0g/L, s0=5g/L, and p0=0g/L, where x & y are the concentrations of plasmid-bearing and plasmid-free cells, respectively. Induce product formation at 5 hour; run until substrate is exhausted; plot (x,y,s,p) versus time t and note the value of p and fraction of plasmid-free cells y/(x+y) at the end of the run. Repeat for induction at 10 hour. This problem attempts to demonstrate that, if we induce too early, non-productive plasmid-free cells take over the fermentor.

    Solution:


Homework #14 (due 05/??)

  1. Teaching evaluation for the instructor.
  2. Prepare for final exam (05/17/16, 1:30pm-3:30pm).


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Biochemical Engineering -- Homework Statements
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