Attachment: Genecode table
Solution:
Electron Acceptor: a chemical species that can be reduced, typically oxygen
for aerobic metabolism.
Electron Donor: a chemical species that can be oxidized (e.g., NADH).
Solution:
NADH (nicotinamide adenine dinucleotide): carrier of reducing power in cells;
carrier of H (aka fuel).
Solution:
ATP (adenosine triphosphate): carrier of energy in cells.
P/O ratio is the number of ATP generated per oxygen,
equivalently per NADH2 or FADH2; it is a
measure of energy capture efficiency from oxidizing H.
Solution:
Glycolysis is repressed in the presence of oxygen. Lower
oxygen partial pressure (dissolved oxygen) causes higher rate of
glycolysis.
Solution:
High glucose level causes ethanol production (even under
aerobic condition).
Solution:
doubling time = tdouble = 1hr
# cells after 24*7 divisions = 224*7 bacteria = 824*7/3 ≈ 1024*7*0.3 = 1050.?
mass of bacterium @ 103kg/m3 = (10-6m)3*(103kg/m3) = 10-15kg
EMU = mass of earth @ 6*103kg/m3 = (volume of earth)(density)
= (1021m3)*(6*103kg/m3) = 6*1024kg
cell mass after a week = (1050.? bacteria)*(10-15kg/bacteria)*(EMU/6*1024kg) ≈ 1011 Earth Mass Unit
Basically, it is an astronomical number by any measure!
Solution:
specific growth rate = ln(2)/tdouble = ln(2)/(1 hr) ≈ 0.7 hr-1
Solution:
Every time a bacterium doubles, it has to replicate one new
set of chromosome. Thus, the average rate of DNA synthesis is 1
chromosome per hour in each bacterium.
Solution:
Once replication starts, two replication forks travel
away from the origin of replication (ORI) and meet half way at
the opposite side of a circular chromosome. It takes 0.5 hour
for the replication fork to travel half way.
Solution:
It takes 1/3 as long to double as a given number of times
to reach a given level.
average number of replication forks = (1 chromosome/20min)*(fork-hr/chromosome) = 3 forks
------------------------------------------------------------- Bases Functions ------------------------------------------------------------- 142- 147 Promoter for beta-gal (-30 signal), induced by IPTG 176- 180 Promoter for beta-gal (-10 signal) 216- 539 beta-galactosidase codons 230- 289 mcs (polylinker) 885-1745 beta-lactamase codons -------------------------------------------------------------
Bases Number
-------------
A 659+666=1325 Note: A=T
T 659+666=1325
C 686+657=1343 Note: C=G
G 686+657=1343
Solution:
base # 180 216 539
DNA ...promoter-----rbs--codons------terminator...
mRNA <-------mRNA------>
The mRNA starts after the promoters (i.e., after 180) and ends at
the terminator (which is not given but should be after the
beta-galactosidase stop codon. (i.e., somewhere after 539).
mRNA is more than just the codons; it normally includes both
extra nucleotides before the start codon (including rbs) and
nucleaotides after the stop codon. A cell uses the strand that
is complementary to the given strand as a template to transcribe
mRNA; the given strand is NOT the template strand.
Solution:
A little after the promoters (i.e., after 180), but a little before
the beta-galactosidase codons (i.e., before 216).
Thus, somewhere between 180 and 216.
Solution:
216- 539 beta-galactosidase codons
pUC19 181 tgtgagcgga taacaatttc acacaggaaa cagctATG ACC ...
translation NH2-M- -T-
:
pUC19 481 GCGGTATTTC ACACCGCATA TGGTGCACTC TCAGTACAAT CTGCTCTGAT GCC GCA TAGt
translation -A- -A-COOH
Note that the stop codon "TAG" is NOT a peptide; translation
stops precisely because it does not code for any peptide.
Solution:
A cloning vector is a tool for carrying the heterologous gene
into a cell (host). A vector is generally a small
extrachromosomal piece of DNA (e.g., a plasmid, a phage, or a
virus) that contains, in addition to the target gene, the
regulatory elements needed for its replication (ORI),
transcription of the target gene (promoter/operator and
terminator), and translation (ribosome binding site). If just
the target gene (without all the necessary regulatory components
for replication, transcription, and translation) is inserted into
a host, nothing will happen.
Solution:
A plasmid is a parasite that cannot replicate on its own, nor
transcribe mRNA on its own, nor synthesize protein on its own.
The host helps the vector propagate and expresses the target gene
to synthesize the target protein. Nothing will happen without
the host.
DNA sequence: .....??????...??????.....
Solution:
We facilitate insertion by digesting with restriction
enzyme(s); the "..." before/after the codons are restriction
sites.
peptide M Y T Y Ter
DNA sequence: .....atg tat...act tat taa.....
Solution:
Insert either a) after the beta-galactosidase promoter/rbs but
before the beta-galactosidase start codon (i.e., 216), or b)
after the beta-galactosidase stop codon (i.e., 539) but before
the terminator. In short, either right before or right after an
intact beta-galactosidase gene so that the peptide
"MyDearestPatty" and beta-galactosidase (which generates blue
color) are both under the same promoter as an operon and are both
synthesized.
Solution:
Swap a few bases via PCR to allow recognition. Because we do
not want to chop up the vector into many small pieces or have
other unintended insertion sites, it is best to identify an
enzyme that does NOT make any other cut on the original pUC19.
(Hereafter is not tested in the interest of time.) For example,
by examining the restriction sites on pUC19, we see Xma III (with
recognition c/ggccg) is a candidate. Below, we examine the
location right after the beta-galactosidase condons.
pUC19 541 taagccagcc ccgacacccg ccaacacccg ctgacgcgcc ctgacgggct tgtctgctcc
line up cggccg
# mismatch 423345 3354255236 4454254246 2444332455 2355234654 36534
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The best is a number of mismatch of 0, meaning "bingo!" a perfect
match (i.e., restriction enzyme recognition). The second best is
a number of mismatch of 1 (which we do not have in this region).
The third best is a number of mismatch of 2 (which there are
several, as marked above). For example, we can create a
recognition by Xma III at 555 by swapping two bases with PCR to
change it from "cacccg" to "cggccg". (Describe how to swap
bases with PCR.)
Solution:
High productivity.
Solution:
dx/dt = 0 = -D*x + μ*x
ds/dt = 0 = D*(sf-s) - μ*x/Y
Solve the above 2 algebraic equations for steady-state values of x and s.
maxD,sf added_value = D*(γ*x-sf)
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