| Due | Homework | Solutions (Computer Files) |
|---|---|---|
| 9/11 | Homework #1 | |
| 9/18 | Homework #2 | |
| 9/25 | Homework #3 | |
| 10/02 | Homework #4 | |
| 10/09 | Homework #5 | |
| 10/16 | Homework #6 | |
| 10/23 | Homework #7 | |
| 10/30 | Homework #8 | |
| 11/06 | Homework #9 | |
| 11/18 | Homework #10 | |
| 11/25 | Homework #11 | |
| 12/04 | Homework #12 | |
| 12/11 | Homework #13 | |
An amino acid acts as a zwitterion, i.e., it can be either:
1) positively charged in an acid solution
2) negatively charged in an alkaline solution, or
3) neutral at the isoelectric point.
Zwitterion Reaction:
NH3+-CHR-COOH ----> NH3+-CHR-COO- + H+
NH3+-CHR-COO- ----> NH2-CHR-COO- + H+
which can be abbreviated as:
A+ ----> A + H+
A ----> A- + H+
The Dissociation constants for these steps are:
Dissociation constant #1: pK1=2.34=-log(K1)=-log([A][H+]/[A+]) eq. (1)
Dissociation constant #2: pK2=9.69=-log(K2)=-log([A-][H+]/[A]) eq. (2)
Add one more equation to find the fraction:
[A+] + [A] + [A-] = 1 eq. (3)
From three eqs (1)--(3), solve for three unknowns ([A+], [A], [A-]).
Plot these three fractions vs pH for pH=[0, 0.1, .., 12]
Solution:
See the "Zwitterion Behavior of Amino Acid"
link in the class web page.
dr/dt = a*r - b*r*f r(0)=3 ...Rabbits or bacteria (Prey)
df/dt =-c*f + d*r*f f(0)=2 ...Foxes or protozoa (Predator)
where
a = 2
b = 2
c = 1
d = 1
Find the steady state values of r and f. Plot r and f vs time t
for t=[0, 0.1, .., 10]
Solution:
Steady state is when d?/dt=0.
0 = a*rss - b*rss*fss
0 =-c*fss + d*rss*fss
Solving the above coupled set of algebraic equations yields
two separate steady states:
(rss, fss) = (0,0)
(rss, fss) = (c/d, a/b)
However, these steady states are unstable.
Solution:
Solution:
Solution:
Once the reaction rate expression v and its model parameters are
given, one can integrate ds/dt=-v to find s vs t.
Solution:
The assumption that led to the Michaelis-Menten rate expression
holds when E0<<s0, which is not the case
here. Thus, we go back to the original set of ODEs that were set up
before the equilibrium or quasi-steady state assumption was applied, and
we fit the model parameters at that point.
Solution:
Transformation of a nonlinear problem to a linear problem
distorts the problem. For example, in Lineweaver-Burk double
reciprocal plot (1/v vs 1/s), we attempt to minimize errors in
1/v rather than v. The problem thus gets totally inverted. A
small error in v or a small value of v is transformed to a large
error in 1/v or a large value of v. Forcing linear regression
on a nonlinear problem is like turning a mole hill into a
mountain and vice versa.
Solution:
reaction rate = dP/dt = -dS/dt = v = k2*S*E/(Keq+E)
where k2 = 1/hr
Keq = 1 g enzyme
The enzyme E deactivates with time in a first-order fashion:
dE/dt = -kd*E where kd= 0.01/hrThe price-cost structure for each gram of each species relative to 1 gram of P is given below.
species value $ relative to P
-------------------------------------------------------
P basis=1
S0 (fresh charge in bioreactor) α=0.2
E0 (fresh charge in bioreactor) β=10
S (residual in bioreactor) γ=-2α (i.e., negative value, meaning it costs $ to dispose residue)
E (residual in bioreactor) δ=0 (i.e., not recovered, thus having no salvage value)
Your boss asks you to maximize the rate at which profit is made.
profit in each batch = 1*P - α*S0 - β*E0 + γ*S(t) + δ*E(t) profit rate = profit/t
Solution:
Note that there are other profit measures. For example, one can
maximize the rate of return based on initial cost in substrate and enzyme.
Each objective function to be maximized results in different optima.
Solution:
Solution:
Given Data on E. coli:
diameter: 1um
length: 2um
water content: 75%
protein content: 60% of dry weight
rate of amino acid addition per ribosome: 20 amino acid per second
average molecular weight of amino acid: 126 g/mol = 2.09*10-22g
doubling time: 45 minutes
Find the number of ribosomes.
weight of E. coli = π(0.5*10-6m)2*(2*10-6m)*(106g/m3)
= 1.57*10-12g
number of amino acids = (weight of E. coli)*(25%)*(60%)/(2.09*10-22g)
= 1.127*109 AA
number of ribosomes = (1.127*109 AA)/(45min)/(20 AA/sec-ribosome)
= 2.1*104 ribosomes
Solution (Prob 4.6):
5'-ccg uau cga cuu gua aca acg cgc-3' ... mRNA
Pro Tyr Arg Leu Val Thr Thr Arg ... amino acids (three-letter)
P Y R L V T T R ... amino acids (one-letter)
5'-ccg tat cga ctt gta aca acg cgc-3' ... DNA (anti-sense strand)
3'-ggc ata gct gaa cat tgt tgc gcg-5' ... DNA (sense strand, template strand)
Strictly speaking, both strands make up the DNA. If you are to specify just
the template strand, you should give the sequence in the "standard" notation
below.
5'-gcg cgt tgt tac aag tcg ata cgg-3' ... DNA (sense strand, template strand, in "standard" notation starting from 5')
Although the textbook says this is part of insulin, I do not find it in the
human insulin gene. Perhaps from another animal? BLAST fails to find the source.
MALWMRLLPLLALLALWGPDPAAAFVNQHLCGSHLVEALYLVCG ERGFFYTPKTRREAEDLQVGQVELGGGPGAGSLQPLALEGSLQKRGIVEQCCTSICSL YQLENYCN
Solution:
K K
C + Se <--> CSe --> diffusion across membrane --> CSi <--> Ci + Si
Ce <-- diffusion across membrane <-- Ci
Motivation: in traditional separation processes, a regular
membrane selects small size molecules because membrane pore size
and molecular size determine permeability. In comparison, in
facilitated transport, a liquid membrane selects molecules with
high solubility in the membrane. Thus, one can learn from bio
and do separation not based on size but on solubility, which can
be modified by adding carriers specific to the compound to be
selected.
Solution:
Depending on your assumption, you may or may not reach a form
that matches Equation 4.2 of Shuler and Kargi. Derive it
yourself and see.
The following files employ the equilibrium
assumption for the dissociation/association of CS.
The following files employ the quasi-steady state assumption
for the complex CS.
Solution:
VEGF helps angiogenesis (blood vessel growth). Thus, it helps
the formation of a baby, wound healing, and cardiovascular
diseases. However, it also helps formation/growth of tumor and
causes macular degeneration. Anti-VEGF is a common strategy in
fighting against cancer, for example, by mopping it up with a
monoclonal antibody, by blocking the integrin receptor site for
it with a blocker, by disrupting at any point in the
VEGF signal pathway that follows activation by VEGF, etc.
Since it, just like any other drugs, exerts both positive and
negative effects, an important practical engineering issue is
drug delivery to the right place, in the right amount, for the
right duration.
Solution:
Go to the NCBI
site; search "Gene" for "human VEGF A" yields the
complete gene sequence for VEGFA (download from UMCP)
or VEGFA (download from NCBI). It is located on chromosome 6.
Solution:
8 exons. Immediately after transcription, the introns are
interdispersed between the exons. In the sequence from the last
part, mRNA = (5001..21272) = 16272 bases
Solution:
There are 7 isoforms/variants due to different/alternative
splicing of exons as mRNA moves across the nucleus membrane.
The exons are:
Transcript Variant 1 Size
transcript variant 1 3665
transcript variant 2 3614
transcript variant 3 3596
transcript variant 4 3542
transcript variant 5 3507
transcript variant 6 3410
transcript variant 7 3476
mRNA in nucleus = (5001..21272) = 16272 bases
mature mRNA at ribosome = 3655 bases
exon # range in gene size
------------------------------
exon #1 5001.. 6097 1097
exon #2 9126.. 9177 52
exon #3 12254..12450 197
exon #4 13245..13321 77
exon #5 13674..13703 30
exon #6 15517..15639 123
exon #7 16741..16872 132
exon #8 19326..21272 1947
------------------------------
total in exon 3655
Below gives the mRNA sequence, where the codons are capitalized.
Transcript Variant 1 gives an identical spliced sequence
with 10 extra "a"s at the end.
exon #1 5001.. 6097 1097 bases
4981 ggcttggggc agccgggtag ctcggaggtc gtggcgctgg
5041 gggctagcac cagcgctctg tcgggaggcg cagcggttag gtggaccggt cagcggactc
5101 accggccagg gcgctcggtg ctggaatttg atattcattg atccgggttt tatccctctt
5161 cttttttctt aaacattttt ttttaaaact gtattgtttc tcgttttaat ttatttttgc
5221 ttgccattcc ccacttgaat cgggccgacg gcttggggag attgctctac ttccccaaat
5281 cactgtggat tttggaaacc agcagaaaga ggaaagaggt agcaagagct ccagagagaa
5341 gtcgaggaag agagagacgg ggtcagagag agcgcgcggg cgtgcgagca gcgaaagcga
5401 caggggcaaa gtgagtgacc tgcttttggg ggtgaccgcc ggagcgcggc gtgagccctc
5461 ccccttggga tcccgcagct gaccagtcgc gCTGACGGAC AGACAGACAG ACACCGCCCC
5521 CAGCCCCAGC TACCACCTCC TCCCCGGCCG GCGGCGGACA GTGGACGCGG CGGCGAGCCG
5581 CGGGCAGGGG CCGGAGCCCG CGCCCGGAGG CGGGGTGGAG GGGGTCGGGG CTCGCGGCGT
5641 CGCACTGAAA CTTTTCGTCC AACTTCTGGG CTGTTCTCGC TTCGGAGGAG CCGTGGTCCG
5701 CGCGGGGGAA GCCGAGCCGA GCGGAGCCGC GAGAAGTGCT AGCTCGGGCC GGGAGGAGCC
5761 GCAGCCGGAG GAGGGGGAGG AGGAAGAAGA GAAGGAAGAG GAGAGGGGGC CGCAGTGGCG
5821 ACTCGGCGCT CGGAAGCCGG GCTCATGGAC GGGTGAGGCG GCGGTGTGCG CAGACAGTGC
5881 TCCAGCCGCG CGCGCTCCCC AGGCCCTGGC CCGGGCCTCG GGCCGGGGAG GAAGAGTAGC
5941 TCGCCGAGGC GCCGAGGAGA GCGGGCCGCC CCACAGCCCG AGCCGGAGAG GGAGCGCGAG
6001 CCGCGCCGGC CCCGGTCGGG CCTCCGAAAC CATGAACTTT CTGCTGTCTT GGGTGCATTG
6061 GAGCCTTGCC TTGCTGCTCT ACCTCCACCA TGCCAAG
EXON #2 9126.. 9177 52 BASES
9121 TGGTC CCAGGCTGCA CCCATGGCAG AAGGAGGAGG GCAGAATCAT CACGAAG
EXON #3 12254..12450 197 BASES
12241 TGGTGAA GTTCATGGAT GTCTATCAGC GCAGCTACTG CCATCCAATC
12301 GAGACCCTGG TGGACATCTT CCAGGAGTAC CCTGATGAGA TCGAGTACAT CTTCAAGCCA
12361 TCCTGTGTGC CCCTGATGCG ATGCGGGGGC TGCTGCAATG ACGAGGGCCT GGAGTGTGTG
12421 CCCACTGAGG AGTCCAACAT CACCATGCAG
EXON #4 13245..13321 77 BASES
13201 ATTATG CGGATCAAAC
13261 CTCACCAAGG CCAGCACATA GGAGAGATGA GCTTCCTACA GCACAACAAA TGTGAATGCA
13321 G
EXON #5 13674..13703 30 BASES
13621 ACCAAAG
13681 AAAGATAGAG CAAGACAAGA AAA
EXON #6 15517..15639 123 BASES
15481 AAAA TCAGTTCGAG GAAAGGGAAA
15541 GGGGCAAAAA CGAAAGCGCA AGAAATCCCG GTATAAGTCC TGGAGCGTGT ACGTTGGTGC
15601 CCGCTGCTGT CTAATGCCCT GGAGCCTCCC TGGCCCCCA
EXON #7 16741..16872 132 BASES
16741 TCCCTGTGGG CCTTGCTCAG AGCGGAGAAA GCATTTGTTT GTACAAGATC CGCAGACGTG
16801 TAAATGTTCC TGCAAAAACA CAGACTCGCG TTGCAAGGCG AGGCAGCTTG AGTTAAACGA
16861 ACGTACTTGC AG
EXON #8 19326..21272 1947 BASES
19321 ATGTG ACAAGCCGAG GCGGtgagcc gggcaggagg aaggagcctc cctcagggtt
19381 tcgggaacca gatctctcac caggaaagac tgatacagaa cgatcgatac agaaaccacg
19441 ctgccgccac cacaccatca ccatcgacag aacagtcctt aatccagaaa cctgaaatga
19501 aggaagagga gactctgcgc agagcacttt gggtccggag ggcgagactc cggcggaagc
19561 attcccgggc gggtgaccca gcacggtccc tcttggaatt ggattcgcca ttttattttt
19621 cttgctgcta aatcaccgag cccggaagat tagagagttt tatttctggg attcctgtag
19681 acacacccac ccacatacat acatttatat atatatatat tatatatata taaaaataaa
19741 tatctctatt ttatatatat aaaatatata tattcttttt ttaaattaac agtgctaatg
19801 ttattggtgt cttcactgga tgtatttgac tgctgtggac ttgagttggg aggggaatgt
19861 tcccactcag atcctgacag ggaagaggag gagatgagag actctggcat gatctttttt
19921 ttgtcccact tggtggggcc agggtcctct cccctgccca ggaatgtgca aggccagggc
19981 atgggggcaa atatgaccca gttttgggaa caccgacaaa cccagccctg gcgctgagcc
20041 tctctacccc aggtcagacg gacagaaaga cagatcacag gtacagggat gaggacaccg
20101 gctctgacca ggagtttggg gagcttcagg acattgctgt gctttgggga ttccctccac
20161 atgctgcacg cgcatctcgc ccccaggggc actgcctgga agattcagga gcctgggcgg
20221 ccttcgctta ctctcacctg cttctgagtt gcccaggaga ccactggcag atgtcccggc
20281 gaagagaaga gacacattgt tggaagaagc agcccatgac agctcccctt cctgggactc
20341 gccctcatcc tcttcctgct ccccttcctg gggtgcagcc taaaaggacc tatgtcctca
20401 caccattgaa accactagtt ctgtcccccc aggagacctg gttgtgtgtg tgtgagtggt
20461 tgaccttcct ccatcccctg gtccttccct tcccttcccg aggcacagag agacagggca
20521 ggatccacgt gcccattgtg gaggcagaga aaagagaaag tgttttatat acggtactta
20581 tttaatatcc ctttttaatt agaaattaaa acagttaatt taattaaaga gtagggtttt
20641 ttttcagtat tcttggttaa tatttaattt caactattta tgagatgtat cttttgctct
20701 ctcttgctct cttatttgta ccggtttttg tatataaaat tcatgtttcc aatctctctc
20761 tccctgatcg gtgacagtca ctagcttatc ttgaacagat atttaatttt gctaacactc
20821 agctctgccc tccccgatcc cctggctccc cagcacacat tcctttgaaa taaggtttca
20881 atatacatct acatactata tatatatttg gcaacttgta tttgtgtgta tatatatata
20941 tatatgttta tgtatatatg tgattctgat aaaatagaca ttgctattct gttttttata
21001 tgtaaaaaca aaacaagaaa aaatagagaa ttctacatac taaatctctc tcctttttta
21061 attttaatat ttgttatcat ttatttattg gtgctactgt ttatccgtaa taattgtggg
21121 gaaaagatat taacatcacg tctttgtctc tagtgcagtt tttcgagata ttccgtagta
21181 catatttatt tttaaacaac gacaaagaaa tacagatata tcttaaaaaa aaaaaagcat
21241 tttgtattaa agaatttaat tctgatctca aa
Solution:
CDS join(5492..6097,9126..9177,12254..12450,13245..13321,
13674..13703,15517..15639,16741..16872,19326..19347)
5461 ccccttggga tcccgcagct gaccagtcgc gCTGACGGAC AGACAGACAG ACACCGCCCC
:
:
19321 ATGTG ACAAGCCGAG GCGGtgagcc gggcaggagg aaggagcctc cctcagggtt
(Note that we have an exception in the start codon; insted of ATG, we have CTG.)
From the above information, we have:
Nucleotides: CTG ACG ....... AGG CGG tga
N-terminus: M T ....... R R (stop) C-terminus
The full sequence for the precursor protein is (from
Transcript Variant 1):
MTDRQTDTAPSPSYHLLPGRRRTVDAAASRGQGPEPAPGGGVEG
VGARGVALKLFVQLLGCSRFGGAVVRAGEAEPSGAARSASSGREEPQPEEGEEEEEKE
EERGPQWRLGARKPGSWTGEAAVCADSAPAARAPQALARASGRGGRVARRGAEESGPP
HSPSRRGSASRAGPGRASETMNFLLSWVHWSLALLLYLHHAKWSQAAPMAEGGGQNHH
EVVKFMDVYQRSYCHPIETLVDIFQEYPDEIEYIFKPSCVPLMRCGGCCNDEGLECVP
TEESNITMQIMRIKPHQGQHIGEMSFLQHNKCECRPKKDRARQEKKSVRGKGKGQKRK
RKKSRYKSWSVYVGARCCLMPWSLPGPHPCGPCSERRKHLFVQDPQTCKCSCKNTDSR
CKARQLELNERTCRCDKPRR"
The full sequence for the signal+mature VEGFA is (the above sequence minus the "pre"-junk):
sig_peptide 1032..1109
/gene="VEGFA"
/gene_synonym="MGC70609; MVCD1; VEGF; VEGF-A; VPF"
mat_peptide 1110..1727
/gene="VEGFA"
/gene_synonym="MGC70609; MVCD1; VEGF; VEGF-A; VPF"
/product="vascular endothelial growth factor isoform a"
1021 cctccgaaac cATGAACTTT CTGCTGTCTT GGGTGCATTG GAGCCTTGCC TTGCTGCTCT
1081 ACCTCCACCA TGCCAAGTGG TCCCAGGCTG CACCCATGGC AGAAGGAGGA GGGCAGAATC
:
1681 AGCTTGAGTT AAACGAACGT ACTTGCAGAT GTGACAAGCC GAGGCGGtga gccgggcagg
From the above information, we have:
Nucleotides: ATG AAC ....... AGG CGG tga
N-terminus: M N ....... R R (stop) C-terminus
The full sequence for the signal+mature VEGFA protein is (from
Transcript Variant 1):
MNFLLSWVHWSLALLLYLHHAKWSQAAPMAEGGGQNHH
EVVKFMDVYQRSYCHPIETLVDIFQEYPDEIEYIFKPSCVPLMRCGGCCNDEGLECVP
TEESNITMQIMRIKPHQGQHIGEMSFLQHNKCECRPKKDRARQEKKSVRGKGKGQKRK
RKKSRYKSWSVYVGARCCLMPWSLPGPHPCGPCSERRKHLFVQDPQTCKCSCKNTDSR
CKARQLELNERTCRCDKPRR
The full sequence for the mature VEGFA is (the above sequence minus the signal peptide):
mat_peptide 1110..1727
/gene="VEGFA"
/gene_synonym="MGC70609; MVCD1; VEGF; VEGF-A; VPF"
/product="vascular endothelial growth factor isoform a"
1081 acctccacca tgccaagtgg tcccaggctG CACCCATGGC AGAAGGAGGA GGGCAGAATC
:
1681 AGCTTGAGTT AAACGAACGT ACTTGCAGAT GTGACAAGCC GAGGCGGtga gccgggcagg
From the above information, we have:
Nucleotides: GCA CCC ....... AGG CGG tga
N-terminus: A P ....... R R (stop) C-terminus
The full sequence for the mature VEGFA protein is (from
Transcript Variant 1):
APMAEGGGQNHH
EVVKFMDVYQRSYCHPIETLVDIFQEYPDEIEYIFKPSCVPLMRCGGCCNDEGLECVP
TEESNITMQIMRIKPHQGQHIGEMSFLQHNKCECRPKKDRARQEKKSVRGKGKGQKRK
RKKSRYKSWSVYVGARCCLMPWSLPGPHPCGPCSERRKHLFVQDPQTCKCSCKNTDSR
CKARQLELNERTCRCDKPRR
Solution:
The following is for precursor (including the signal sequence)
as it is synthesized at the ribosome.
The following is for the mature peptide (excluding the signal
peptide). Copied from Transcript Variant 1):
CDS join(5492..6097,9126..9177,12254..12450,13245..13321,
13674..13703,15517..15639,16741..16872,19326..19347)
Based on the above information, we have the following range for codons:
exon # range in gene size
------------------------------
exon #1 5492.. 6097 606
exon #2 9126.. 9177 52
exon #3 12254..12450 197
exon #4 13245..13321 77
exon #5 13674..13703 30
exon #6 15517..15639 123
exon #7 16741..16872 132
exon #8 19326..19347 22
------------------------------
total number of bases 1239 (including 3 for the stop codon)
total number of bases 1236 (excluding 3 for the stop codon)
# of codons = # of peptieds = (1236 bases)/(3 bases/codon) = 412 codons = 412 peptides
VEGFA MW = (412 peptides)*(~120 Dalton/peptide) = ~49.4K Dalton
mat_peptide 1110..1727
/gene="VEGFA"
/gene_synonym="MGC70609; MVCD1; VEGF; VEGF-A; VPF"
/product="vascular endothelial growth factor isoform a"
Based on the above information, we have the following range for codons:
total number of bases = (1110..1727) = 618 bases
# of codons = # of peptieds = (618 bases)/(3 bases/codon) = 206 codons = 206 peptides
Solution:
Precursor protein as it is synthesized at the ribosome
----------------------------------------------------
# Amino Acid Abbrev. Count No. % Wt. %
----------------------------------------------------
1 Alanine (Ala, A) 37 8.98 5.79
2 Arginine (Arg, R) 42 10.19 14.43
3 Asparagine (Asn, N) 7 1.70 1.76
4 Aspartic Acid (Asp, D) 12 2.91 3.04
5 Cysteine (Cys, C) 20 4.85 4.54
6 Glutamic Acid (Glu, E) 36 8.74 10.23
7 Glutamine (Gln, Q) 19 4.61 5.36
8 Glycine (Gly, G) 42 10.19 5.27
9 Histidine (His, H) 13 3.16 3.92
10 Isoleucine (Ile, I) 8 1.94 1.99
11 Leucine (Leu, L) 24 5.83 5.98
12 Lysine (Lys, K) 24 5.83 6.77
13 Methionine (Met, M) 9 2.18 2.60
14 Phenylalanine (Phe, F) 8 1.94 2.59
15 Proline (Pro, P) 32 7.77 6.84
16 Serine (Ser, S) 32 7.77 6.13
17 Threonine (Thr, T) 12 2.91 2.67
18 Tryptophan (Trp, W) 7 1.70 2.87
19 Tyrosine (Tyr, Y) 8 1.94 2.87
20 Valine (Val, V) 20 4.85 4.36
----------------------------------------------------
Total: 412 100.00 100.00
The precursor protein's molecular weight is 45,474 Dalton
Mature protein
----------------------------------------------------
# Amino Acid Abbrev. Count No. % Wt. %
----------------------------------------------------
1 Alanine (Ala, A) 5 2.43 1.49
2 Arginine (Arg, R) 19 9.22 12.43
3 Asparagine (Asn, N) 6 2.91 2.87
4 Aspartic Acid (Asp, D) 8 3.88 3.86
5 Cysteine (Cys, C) 18 8.74 7.78
6 Glutamic Acid (Glu, E) 16 7.77 8.65
7 Glutamine (Gln, Q) 12 5.83 6.44
8 Glycine (Gly, G) 14 6.80 3.35
9 Histidine (His, H) 8 3.88 4.59
10 Isoleucine (Ile, I) 8 3.88 3.79
11 Leucine (Leu, L) 9 4.37 4.27
12 Lysine (Lys, K) 20 9.71 10.74
13 Methionine (Met, M) 7 3.40 3.85
14 Phenylalanine (Phe, F) 5 2.43 3.08
15 Proline (Pro, P) 15 7.28 6.10
16 Serine (Ser, S) 12 5.83 4.38
17 Threonine (Thr, T) 6 2.91 2.54
18 Tryptophan (Trp, W) 2 .97 1.56
19 Tyrosine (Tyr, Y) 6 2.91 4.10
20 Valine (Val, V) 10 4.85 4.15
----------------------------------------------------
Total: 206 100.00 100.00
The mature protein's molecular weight is 23,898 Dalton
(Electrophoresis will not show this wt, because, based on the
"COMMENT" section, an active VEGF is further glycosylated and
appears as a difulfide linked homodimer.
Solution:
The starting plasmid vector should already possess all
regulatory elements: ORI for replication, promoter/operator
terminator for transcription of the penicillin-resistant gene,
and start codon and stop codon within the penicillin-resistant
gene for translation. Steps: identify a restriction
endonuclease(s) that cleaves the target gene to expose sticky
ends. Ideally, this endonuclease should also cleave at either
before the start codon (but after the promoter sequence) or after
the stop codon (but before the termination sequence) of the
penicillin-resistant gene. After cleaving a mixture of the
target gene and the given plasmid and allowing the fragments to
recombine, treat with ligase. Transform E. coli with the
recombinant library. Spread out E. coli on a plate containing
penicillin and the colorless substrate. Those lacking the
plasmid will not grow to form colonies; while those harboring the
plasmid with an intact penicillin gene can grow to form colonies.
Since the target converts the colorless substrate into a blue
product, pick the blue colony; these are the ones that harbor the
desired gene.
Solution:
We follow the process outlined in one of the class handouts
"Cloning Example: Cloning Luciferase into pBR322".
except that we substitute luciferase with VEGFA and pBR322 with pUC19.
Step 1. Download the VEGFA gene (vegf1.seq) and cut out just the gene
sequence (vegf1.sq1).
Note the following range of bases:
Step 2. Map out the restriction sites.
The restriction enzyme recognition sites (after feeding the above
gene sequences into my gene4 program) are
below:
Step 3. Choose a restriction enzyme.
Since it is desirable to avoid cutting up the VEGFA
codons, we narrow down to only those that do not cleave in the
middle of the VEGFA codons.
largest precursor (492..1730)
signal peptide + mature peptide (1032..1730)
mature peptide (1110..1730)
Also download the pUC19gene (puc19.seq) and cut out just the gene
sequence (puc19.sq1).
vegf1.out
puc19.out
where the sites preceeding and following mature VEGFA are marked
with "*", the sites preceeding signal peptide of VEGFA are marked
with "**" (which are good cleavage locations), and the sites
preceeding the largest possible precursor are marked with "***".
puc-vegf.out
Further selecting only those that cleave both before and
after the VEGFA codons, we narrow down to the following:
VEGF1 pUC19
---------------------------------
c/ccggg Ava I*, Nsp III*, Xcy I, Xma I
910** 268 (mcs within beta-galactosidase)
1947*
a/gatct Xho II*
1773*
g/gatcc Xho II*, BamH I, Bst I
468*** 263 (mcs within beta-galactosidase)
2904*
Step 4. Check for codon alignment for Xma I.
Insertion of the segment (910, 1947) into pUC19 at 268 results
in the following recombinant sequence.
pUC19 1 gcgcccaata cgcaaaccgc ctctccccgc gcgttggccg attcattaat gcagctggca
pUC19 61 cgacaggttt cccgactgga aagcgggcag tgagcgcaac gcaattaatg tgagttagct
pUC19 121 cactcattag gcaccccagg ctttacactt tatgcttccg gctcgtatgt tgtgtggaat
pUC19 181 tgtgagcgga taacaatttc acacaggaaa cagctATGAC CATGATTACG CCAAGCTTGC
pUC19 241 ATGCCTGCAG GTCGACTCTA GAGGATCC
VEGF 901 ccgggcctcg ggccggggag gaagagtagc tcgccgaggc gccgaggaga
VEGF 961 gcgggccgcc ccacagcccg agccggagag ggagcgcgag ccgcgccggc cccggtcggg
VEGF 1021 cctccgaaac cATGAACTTT CTGCTGTCTT GGGTGCATTG GAGCCTTGCC TTGCTGCTCT
VEGF 1081 ACCTCCACCA TGCCAAGTGG TCCCAGGCTG CACCCATGGC AGAAGGAGGA GGGCAGAATC
VEGF 1141 ATCACGAAGT GGTGAAGTTC ATGGATGTCT ATCAGCGCAG CTACTGCCAT CCAATCGAGA
VEGF 1201 CCCTGGTGGA CATCTTCCAG GAGTACCCTG ATGAGATCGA GTACATCTTC AAGCCATCCT
VEGF 1261 GTGTGCCCCT GATGCGATGC GGGGGCTGCT GCAATGACGA GGGCCTGGAG TGTGTGCCCA
VEGF 1321 CTGAGGAGTC CAACATCACC ATGCAGATTA TGCGGATCAA ACCTCACCAA GGCCAGCACA
VEGF 1381 TAGGAGAGAT GAGCTTCCTA CAGCACAACA AATGTGAATG CAGACCAAAG AAAGATAGAG
VEGF 1441 CAAGACAAGA AAAAAAATCA GTTCGAGGAA AGGGAAAGGG GCAAAAACGA AAGCGCAAGA
VEGF 1501 AATCCCGGTA TAAGTCCTGG AGCGTGTACG TTGGTGCCCG CTGCTGTCTA ATGCCCTGGA
VEGF 1561 GCCTCCCTGG CCCCCATCCC TGTGGGCCTT GCTCAGAGCG GAGAAAGCAT TTGTTTGTAC
VEGF 1621 AAGATCCGCA GACGTGTAAA TGTTCCTGCA AAAACACAGA CTCGCGTTGC AAGGCGAGGC
VEGF 1681 AGCTTGAGTT AAACGAACGT ACTTGCAGAT GTGACAAGCC GAGGCGGtga gccgggcagg
:
There is no frame shift problem. Below is the translation
product, where the blue colored portion is the residual
beta-galactosidase and the red colored portion corrsponds to
signal+mature VEGFA. Thus, post translational processing is
needed to cleave off the residual beta-galactosidase and the
sigal peptide.
MTMITPSLHACRSTLEDPRASGRGGRVARRGAEESGPPHSPSRRGSASRAGPGRASETMN
FLLSWVHWSLALLLYLHHAKWSQAAPMAEGGGQNHHEVVKFMDVYQRSYCHPIETLVDIF
QEYPDEIEYIFKPSCVPLMRCGGCCNDEGLECVPTEESNITMQIMRIKPHQGQHIGEMSF
LQHNKCECRPKKDRARQEKKSVRGKGKGQKRKRKKSRYKSWSVYVGARCCLMPWSLPGPH
PCGPCSERRKHLFVQDPQTCKCSCKNTDSRCKARQLELNERTCRCDKPRR
Step 5. Selection. Transform into E. coli. Spread on an
agar plate containing both ampicillin and X-Gal (which is the
substrate for beta-galactosidase). Incubate. Select white
colonies (which indicates the absense of beta-galactosidase
because that gene is now destroyed) as candidates for further
analysis.
Solution (Prob 5.4):
Pasteur Effect: Lower oxygen partial pressure (dissolved
oxygen) causes higher rate of glycolysis.
Solution (Prob 5.7):
Photosynthesis occurs in stacked membranes in procaryotes, but
in chloroplast (an organelle) in eucaryotes.
Solution (Prob 5.9):
Crabtree Effect: High glucose level causes ethanol production
(even under aerobic condition).
Solution:
Solution:
We do not necessarily operate at the dilution rate
corresponding to maxium productivity (max Dx or max Dp). This
theoretical optimum dilution rate may be close to washout,
there may be fluctuations/upsets in the operating environment,
the steady-state assumption may be invalid, ...
Solution:
Upstream (cell culture/fermentation)
take out a vial of N2 frozen host culture from cell bank; thaw
grow starter culture and expand to form inoculum
prepare bioreactor (media preparation, sterilization)
inoculate fed-batch bioreactor
feed glucose & glutamine at rate to avoid their accumulation and minimize lactate & ammonia production
induction (perhaps with Zn)
harvest
Downstream (separation & purification)
collect & discard cells (via filtration or centrifugation)
concentrate broth (via cross-flow ultrafiltration)
collect antibody (via affninity chromatography with protein A column; loading and elution)
wash out salt (de-salting or salt-exchange via cross-flow ultrafiltration, or permeation chromatography)
concentrte antibody product
freeze dry (lyophilization)
vial filling & packaging
dx/dt = μ*x -qp*x/Yp
ds/dt = -μ*x/Yx
dp/dt = qp*x
where qp is the specific rate of product formation given by eqn 6.18 of Shuler & Kargi
qp=α*μ+β
(Note that my notation Yp above is the same as
the biomass-to-product yield Yp/x in Shuler & Kargi's
problem statement.) In a CSTR, we add the flow-transport terms to
the above batch equations. Again, plot cell productivity D*x and
product productivity D*p versus D and find the optimum dilution
rate for each case. Caution: negative concentration makes no
physical sense; be sure to stay away from such nonsense. Note
how drastically/insignificantly the addition of this
Yp term affect cell productivity and product
productivity.
Solution:
Variables Unit Start Run #2 Run #3 ... Run #n ----------------------------------------------------------- Glucose g/L 180 ... ... ... Soybean Meal g/L 25 ... ... ... Phosphate g/L 8 ... ... ... Salt g/L 10 ... ... ... Rpm rpm 70 ... ... ... Aeration vvm 0.9 ... ... ... Volume liter 12000 ... ... ... Inoculum % 7 ... ... ... Running Time hour 90 ... ... ... ----------------------------------------------------------- Daily Profit -$914 ... ... ...
Solution:
See problem statement.
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