| Table of Contents |
| Due | Homework | Solutions |
|---|---|---|
| 02/04 | Homework #1 | See Homework Statement |
| 02/11 | Homework #2 | |
| 02/18 | Homework #3 | |
| 02/25 | Homework #4 | |
| 03/03 | Homework #5 | |
| 03/10 | Homework #6 | |
| 03/24 | Homework #7 | |
| 03/31 | Homework #8 | |
| 04/07 | Homework #9 | |
| 04/14 | Homework #10 | |
| 04/21 | Homework #11 | |
| 04/28 | Homework #12 | |
| 05/05 | Homework #13 | |
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
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.
Solution:
Solution:
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:
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.
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
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.
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.
Solution:
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
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.)
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"
APPRLICDSRVLQRYLL
EAKEAENITTGCAEHCSLNENITVPDTKVNFYAWKRMEVGQQAVEVWQGLALLSEAVL
RGQALLVNSSQPWEPLQLHVDKAVSGLRSLTTLLRALGAQKEAISPPDAASAAPLRTI
TADTFRKLFRVYSNFLRGKLKLYTGEACRTGDR
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
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
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.
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.
host chromosome:
thymidine
plasmid: thymidine + gene of interest (insulin)
host + plasmid = complete
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.
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 cagct
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.
ATGAC 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'
BamHI 20-mer primers from above
forward (+) primer: 5'-xxxGGATCC-ATGGGGGTGCACGAATGTCC-3'
reverse (-) primer: 5'-xxxGGATCC-agctgtttcctgtgtgaaat-3'
Solution:
Ammonium sulfate required:
(30g/L)(1000L)(12% N)(mole a.s./28g N)(132g a.s./mole a.s.)=17.0kg
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.
Solution:
Solution:
Solution:
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