Download ChE 170B Midterm: Growth Models, Protein Structure, and Microbial Coexistence and more Exams Engineering Chemistry in PDF only on Docsity!
ChE 170B. Midterm Examination 1 6 March 2009
- Growth Models a) Using a Monod growth model, a line was fit to E. coli growth data as shown below in Figure 1. It was found that μmax was 1.2. The same strain of E. coli was grown in a similar experiment using 20% more glucose. What would the new growth curve look like? Draw your estimate on the plot below given that μmax was 2.
Answer: Higher final growth rate (not quite double), and steeper slope during exponential phase since reaches half max growth at a lower substrate concentration than the original. (3 points)
b) Below is a typical growth curve for an organism undergoing substrate inhibition. Set up mass and substrate balances based on this data. You do not need to solve. Hint: The substrate concentration that yields the highest growth rate is Scritical =.
What are μmax, Ki, and Ks?
Growth rate
μ
[Substrate]
Mass Balance (3 points):
Values (4 points): μmax, is 0. Ks is about 30 (many things will do so long as this is shown)
Scritical =. = 180
Ki is about 1080
- Cell line development. Your company has generated a hybridoma cell line that produces the beta-globulin antibody at 3 pg/cell/day. How might you generate a cell line with improved production of your protein?
Two possible answers, each requiring a sentence or two of explanation: DHFR, stringent selection (5 points)
Once you have generated your improved cell line, how might you determine where in the genome it has inserted?
Sequence out using a primer designed from your gene of interest. Sequence in both directions and identify genome elements closely associated. (5 points)
Design a way to direct recombinant genes for future cell lines to integrate at this site. Describe similarities and differences of this process to knocking out a gene in bacteria. Diagrams are necessary.
The sequence you determined in the previous section is important here. This is similar to making a gene knockout in bacteria in that the gene you want is embedded in sequence homologous to the region you want to insert into. Instead of a marker gene (e.g. kan resistance) you are inserting your gene of interest. (10 points: 4 for diagram, 3 for the use of flanking DNA, 3 for describing similarities)
plasmid
DNA replication occurs on crossover rather than native DNA
Occasional crossover allows for more rapid integration of your gene into the transcriptional hot spot Your gene flanked by sequence for the transcriptional hot spot.
Your gene
- Microbial Coexistence in a Chemostat
2 2 2
1 1 1
0
2 2 2 2
1 1 1
1
X
Y
X
Y
DS S
dt
dS
DX S X
dt
dX
DX S X
dt
dX
where: X 1 = Biomass of species 1 X 2 = Biomass of species 2 S= Substrate concentration μ 1 = specific growth rate of species 1 μ 2 = specific growth rate of species 2 D = dilution rate
Solve the mass balance equations for the steady state substrate concentration Sss, and Dc the dilution rate for coexistence. Show all steps! What major assumption do you have to make?
Major assumption: Set all rates to 0 because you are at steady state. (5 points)
Solve this with the quadratic formula
where a = ; b = ; c = 0
There are two roots (5 points, no 0 is OK):
Thus (5 points):
- Protein folding. Your company has scaled up a new process to manufacture a therapeutic protein for clinical trials. Refractile bodies from a 1,000 L E. coli fermentation are dissolved in 8M urea, 20mM dithiothreitol; and then refolded by dilution into a 10,000 L tank, 2M urea, 4mM glutathione, pH 9. After refolding, the solution is held in the tank awaiting chromatography for final purification. Because of limitations in the capacity of the first chromatography column, only 1 / 5 of the tank can be processed at one time, so the material is processed in 5 batches which are purified on succeeding days. Analysis of material purified on the first and fifth days shows the following differences:
- Reversed phase HPLC – at day 5, an additional peak is present (about 5% as large as the main product peak).
- SDS-PAGE – no differences day 1 to 5.
- Isoelectric focusing – at day 5, an additional band is present with more acidic isoelectric point.
- Peptide map – at day 5, a small additional peak is present.
- Mass spec – at day 5, an additional species with a mass increase of 43da is present.
What is the contaminant which is present at day 5? How do you know?
The storage conditions and analytical results are consistent with carbamylation of the desired product. Carbamylation results in an increase in mass of 43da (detected by mass spec, but too small a change to be noticed on SDS-PAGE) and makes the protein more acidic (IEF shift). Carbamylation should have been anticipated as an adverse result of prolonged exposure to urea at high pH. (5 points)
While maintaining the same production schedule (refolding at 10,000 L scale, and purifying 5 batches over 5 days) suggest at least two changes to minimize formation of the contaminant.
Carbamylation can be slowed by a) reducing the pH during storage to minimize formation of reactive cyanate, and by b) including ethylenediamine or another cyanate scavenger. (5 points for 2 reasons)
- Cloning and molecular biology. Below are the reactions, nucleotide sequences and translations of two enzymes catalyzing steps in the lignin biodegradation pathway:
O 2 N PO 4 -
PhoA O 2 N OH
p-nitrophenyl phosphate (colorless)
p-nitrophenol
(yellow)
b. Design primers using BioBricks and SLIC to insert the two pieces into the vector. Diagrams, rationale, and special considerations in primer design are required. Use the back of this page.
This one was graded on a case-by case basis. No primers at all, only explanations: minus 6 points. Attempts of only one BB or SLIC process: minus 3 points.
BioBricks primers (5 points): First, remove the BglII site in PhoA, e.g. (worth 2 of 5 points) CCGAGCTGAGGTCTGGCCTC and complement (need silent mutation)
And the EcoRI site in PnpA, e.g. GCGAG AACGAGTTCC TGCTG and complement (need silent mutation)
Then design some BioBricks primers using the canonical primer ends: phoAF: TTG GAATTC ATG GGATCC ATGATCAATA AGAC (change Lys codon to avoid to many A’s) phoAR: TTC CTCGAG CTA AGATCT TCAGAC AGCTTCGGCA (Lys codon again)
pnpAF: TTG GAATTC ATG GGATCC ATGACCATCA CCAAA pnpAR: TTC CTCGAG CTA AGATCT TCACT TGATCAGCGG Although the question prompt put BglII in front of BamHI, the reverse will be accepted for full credit as the diagram in the notes put the BioBricks sites in that order.
SLIC Primers (5 points): Easiest way to do this is just make a long version of the final sequence you want and select primers from that. Special considerations include making sure there are not too many overlapping regions. (2 points for some explanation)
CGACTTCTAG AGAATTCATG AGATCT ATGATCAATA AAGCGTACGA... ...TCGAT TGCCAAGGCT GTCTGA GGAGG phoA intervening
ATAAT ATGACCATCA CCCAAGAAGCT... ...GCGTG CCGTGATCA AGTGA ATAG GATCCTAACT CGAGACTGGC sequence pnpA
CGTCGTTTTA
Thus, primers might be: (3 points)
phoA-fwd: AGAATTCATG AGATCT ATGATCAATA AAGCGTACGA phoA-rev: TTGGGTGATGGTCAT ATTAT GGTCC TCAGACAGCCTTGGCA
pnpA-fwd: TGCCAAGGCT GTCTGA GGAGG ATAAT ATGACCATCACCCAA pnpA-rev: AGTTAGGAT CCTAT TCACT TGATCACGG
vector-fwd: CCGTGATCA AGTGA ATAG GATCCTAACT vector-rev: TCGTACGCTT TATTGATCAT AGATCT CATGAATTCT
c. A paper you just read has suggested that the active site is the catalytic triad of PnpA is Arg24, Asn38, Glu71. You are interested in improving catalytic efficiency of the enzyme. Describe a method to do this. Include clear, labeled diagrams and the sequences of any primers you might order.
Site directed mutagenesis. Diagrams necessary (1 point)
Arg24: CAGGAAGGCNNNTACGTAAA and TTTACGTANNNGCCTTCCTG Asn38: CTGGTGGCCNNNTACCTGGG and CCCAGGTANNNGGCCACCAG Glu71: GCGAGAACNNNTTCCTGCTG and CAGCAGGAANNNGTTCTCGC (4 points)
d. How might you screen for improvements to the enzyme? Using the method you described above, how many colonies will you have to screen?
Since your enzyme converts a colored (PNP) substrate to a colorless (beta- ketoadipate) substrate, a colorometric assay is ideal. One way might be to express the enzyme in cells in the presence of PNP and look for the colonies that convert PNP to beta-ketoadipate more rapidly than the non-engineered control. (3 points)
If you were to screen ALL of the possible mutants, you would have to screen 3(202020) = 24,000 cfu (will also accept 606060 =216,000). However, if you screen for each mutation separately and only select ONE best fit each sequentially, you can screen 60 + 60 + 60 = 180-colonies. However, you may miss combinatorial improvements (e.g. Glu24 might work better in combination with Asn38, but in combination with Arg38 it might not). (2 points)*
Sufficient dilutions and spreading across several plates may make this possible in a single experiment. With this many items to screen, a selection method would be most efficient.
e. A strain of B. subtilis has been known to use PNP as growth substrate. How might you use this strain to improve your screen or select for improvements in your enzyme? Include diagrams of any plasmids necessary.
Transform genes of interest into a plasmid that will replicate in B. subtilis (not pUC19) Perform mutagenesis, express genes Grow cells on PNPP as sole carbon source Cells that grow faster than the wild-type contain improvements. (5 points)
