Solution:
True. (Because I said so.)
Solution:
False. (You get a grade of "XF" meaning failure due to academic dishonesty.)
Solution:
True. (The web pages, as long as they are equally available
to everyone in the class, are for you to use as you see fit, but
you must give proper credit to the source, as is always the case.
On the other hand, it is not ok to receive a Mathcad file from
your friend, move things around a bit, and turn it in for credit,
because that is basically copying and is unfair to others who do
not have access to the same information. If you work together
with your classmate to produce one outcome, you need to credit
that person; not giving proper credit means you pass off work
as your own.
Solution:
True. (It is everyone's responsibility, not just the instructor's.)
Solution:
False. (Your are innocent until proven guilty, and your fellow
students in the Student Honor Council are the judges of your innocence.)
Solution:
Biochemical engineering is the part of chemical engineering that
deals with biological components (biomolecules, biopolymers,
enzymes, cells). Concepts in transport, thermodynamics,
reaction, separation, unit operation, etc. remain largely
unchanged.
Solution:
Life: Having the ability to reproduce on its own.
Organism: a living entity, i.e., an entity that has the ability
to reproduce on its own.
Solution:
Species: A group of organisms that can interbreed with other members of
the same species. A new organism (like a liger) that cannot propagate
is not a new species. A mutant that can propagate with the wild type from
which it evolved is not a new species.
Solution:
Fats/lipids, polysaccharides ... structural/energy
Protein, polynucleotides (DNA & RNA) ... informational/regulatory
Solution:
stacked plates
spiral wound (or pleated)
hollow fiber (shell-and-tube)
Solution:
Enzyme-substrate
Antibody-antigen
DNA-DNA, DNA-RNA
Receptor-ligand
Solution:
reuse expensive enzyme
keep product free of contamination from enzyme
Solution:
advantage disadvantage
---------------------------------------------------------------
polyacrylamide gel permanent toxicity & thermo inactivation
calcium alginate mild soft & disintegration of gel
Solution:
The pH at which a zwitterion (such as a protein) has no net charge (i.e., neutral), thus, will
not migrate in an electric field.
Solution:
Electrophoresis on a gel that has a pH gradient. The electric
field drives the charged protein molecules to a pH environment
until the protein molecules has no net charge and stops
migrating.
Lysine: H2N-(CH2)4-CH(NH2)-COOH pK1 (for -COOH) = 2.18 pK2 (for -NH2) = 8.95 pKR (for R) = 10.53Based on these dissociation constants, give equation(s) that lets you calculate the isoelectric point.
Solution:
Lysine can take on four forms: A++, A+, A, A-
at pH<pK1: +H3N-(CH2)4-CH(NH3+)-COOH abbreviated as A++
at pK1<pH<pK2: +H3N-(CH2)4-CH(NH3+)-COO- abbreviated as A+
at pK2<pH<pKR: +H3N-(CH2)4-CH(NH2)-COO- abbreviated as A
at pKR<pH: H2N-(CH2)4-CH(NH2)-COO- abbreviated as A-
pK1 (for -COOH) = 2.18 = [A+][H+]/[A++]
pK2 (for -NH2) = 8.95 = [A][H+]/[A+]
pKR (for R) = 10.53 = [A-][H+]/[A]
conservation of species : [A++]+[A+]+[A]+[A-]=[A]total
isoelectric point: 2*[A++]+[A+]=[A-]
Solve the above 5 equations for 5 unknowns: [A++], [A+], [A], [A-], [H+]
Solution:
Half way between pK2 and pKR.
pKI = (pK2+pKR)/2 = (8.95+10.53)/2 = 9.74
The reported value is pKI=9.59
Solution:
You competitor copied your creation. The complementary sequence of
5'-GATTACA-3' is 5'-TGTAATC-3'.
5'-GATTACA-3'
3'-CTAATGT-5'
Solution:
expected number of occurrences = (1000-7+1)/47 (roughly 1000 in 16K, or 6%)
More precisely...
probability of finding GATTACA starting at base #1 = 1/47 = 4-7
probabiliy of not finding GATTACA starting at base #1 = 1-4-7
probabiliy of not finding GATTACA starting at base #2 = 1-4-7
probabiliy of not finding GATTACA starting at base #1 through base #994 = (1-4-7)994
probabiliy of finding GATTACA somewhere starting at base #1 through base #994 = 1-(1-4-7)994 = 6%
Solution:
GATTACA
CTAATGT
base relative mole molecular relative wt
mole fraction wt wt fraction
------------------------------------------------------------
A 5 5/14 MWA 5*MWA 5*MWA/Σ
T 5 5/14 MWT 5*MWT 5*MWT/Σ
C 2 2/14 MWC 2*MWC 2*MWC/Σ
G 2 2/14 MWG 2*MWG 2*MWG/Σ
------------------------------------------------------------
total 14 1 Σ 1
Solution:
There are 20 possible amini acid at each position.
20x20x20x20x20x20=207=27*107 (over a billion)
Solution:
expected number of occurrences = (100-7+1)/207 (roughly 100 in a billion)
More precisely...
probabiliy of finding GATTACA starting at peptide #1 = 1/207 = 20-7
probabiliy of not finding GATTACA starting at peptide #1 = 1-20-7
probabiliy of not finding GATTACA starting at peptide #2 = 1-20-7
probabiliy of not finding GATTACA starting at peptide #1 through peptide #94 = (1-20-7)94
probabiliy of finding GATTACA somewhere starting at base #1 through base #94 = 1-(1-20-7)94 = 7x10-8 (roughly 100 in a billion)
S1 --> S2 --> PYou design an enzyme E by combining the active sites of both E1 and E2 into one single protein. E can complex with S1 and S2 to form ES1 and ES2, respectively. ES1 and ES2 can further complex with S2 and S1, respectively; however, only ES2 can lead to the final product P.
K1 K3
E + S1 <---> ES1 <---> E + S2
K2 k2
E + S2 <---> ES2 ---> E + P
K2
ES1 + S2 <---> ES1S2
K1
ES2 + S1 <---> ES1S2
Based the above mechanism, derive the reaction rate v2
(which is the rate of formation of product P. Likewise, derive
the reaction rate v1 (which is the rate of formation
of intermediate S2). List only applicable equations
and the variables to be solved for, but do not actually solve
them.
Solution:
definition of rate: v1=k3*ES1-k-3*E*S2-k2*E*S2+k-2*ES2-k-2*ES1*S2+k-2ES1S2
definition of rate: v2=k2*ES2
conservation of enzyme species: E0=E+ES1+ES2+ES1S2
equilibrium assumptions: K1=E*S1/ES1
K2=E*S2/ES2
K1=ES2*S1/ES1S2
(Or, in lieu of the above 3 equilibrium assumptions, 3 quasi-steady-state assumptions on intermediate species: d(ES1)/dt=0, d(ES2)/dt=0, d(ES1S2)/dt=0)
Solve the above 6 equations for 6 unknowns: v1, v2, E, ES1, ES2, ES1S2
Note that the following equilibrium relationship is redundant; it
is not an independent equation and should not be included along
with others that have already been included.
K2=ES1*S2/ES1S2
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