Quiz #2

Biochemical Engineering

Fall 2007

Instruction

You may use one 8.5x11 sheet of notes, but nothing else (no calculators, laptop computers, etc). Any act of academic dishonesty will not be tolerated.

(25 pts.) Briefly describe/contrast the following concepts/terms.

Central Dogma of Biology.

Solution:

  Replication:   Synthesis of DNA molecule based on matching the nucleic acid
                 sequence of the complementary strand (A=T, C-=G).
  Transcription: Synthesis of mRNA from a DNA template
  Translation:   Synthesis of peptides/protein molecules based on the codon
                 sequence of an mRNA molecule.

Exon & Intron

Solution:

  Exon:  coding segment in DNA that remains in mature mRNA.
  Intron: non-coding segment in DNA that is excised out during post-transcriptional
         modification and is absent in mature mRNA.

Inhibition & Repression

Solution:

  Inhibition: The presence of an inhibitor decreases the enzyme activity.
              This modulation happens after translation (after protein is synthesized).
  Repression: A corepressor, which is usually a gene product, combines with a
              repressor, which is normally inactive, to form an active complex
              that binds to the operator of a gene to block a normally open gene.
              This modulation happens before transcription (before mRNA is synthesized).

Oxidation & Reduction

Solution:

   Oxidation: give off electrons
   Reduction: receive electrons

ATP & NADH

Solution:

   ATP (adenosine triphosphate): carrier of energy in cells
   NADH (nicotinamide adenine dinucleotide): carrier of reducing power in cells

(35 pts.) Immobilized Enzyme. Consider an enzyme immobilized on the surface of glass beads. The enzyme follows Michaelis-Menten kinetics
```
  v = v_m*s/(K_m+s)
```
where s is the substrate concentration.
1. Set up the equation(s) that leads to a solution of the substrate concentration at the reaction surface.
  Solution:
```
   Rate of Mass Transfer = Rate of Reaction
                       J = v
                k_l*(s_b-s)= v_m*s/(K_m+s) --> Solve for s
```
2. What are the physical units of the following parameters as they appear in the above equation?
```
  v
  s
  v_m
  K_m
  k_l
```
  Solution:
```
  v     g*sec^-1cm^-2
  s     g*cm^-3
  v_m    g*sec^-1cm^-2
  K_m    g*cm^-3
  k_l    cm*sec^-1
```
3. Plot the measured (apparent) reaction rate as a function of the substrate concentration in the bulk. Compare it to the intrinsic reaction rate. Does the apparent value of v_m increase or decrease as a result of immobilization? (Circle one.)
  |Increase|Decrease|Both|Neither|
  
  Solution:
  Neither. Immobilization does not affect the apparent value of v_m as the rate of reaction approaches v_m for enzymes in both free and immobilized forms.
4. (10 pts.) Define effectiveness factor in English and give the formula when the intrinsic rate expression follows the Michaelis-Menten kinetics.
  Solution:
```
                         rate /w mass transfer resistance
  Effectiveness factor = ---------------------------------
                         rate /wo mass transfer resistance

      v(s)    s/(K_m+s)
  h = ----- = ---------
      v(s_b)   s_b/(K_m+s_b)

  where s is the solution from Part a).
```
5. Plot effectiveness factor versus v_m while holding everything else constant. With twice as many immobilized enzyme beads in the same reactor, does the effectiveness factor increase or decrease? (Circle one.)
  |Increase|Decrease|Both|Neither|
  Solution:
  Plot: the effectiveness factor starts at 1 when v_m=0 and decreases with increasing v_m. The number of beads governs the surface area, but does not affect the effectiveness factor, provided all other parameters (k_l and s_b) remain unchanged.
6. Consider a CSTR of volume V containing immobilized enzyme beads with reaction surface area of A. An operator pumps in the feed and withdraws the reactor content at a constant flow rate F. The substrate concentrations in the feed is s_f, and there is no enzyme in the feed. Set up the equation(s) needed to find the steady-state level of conversion in the exit stream.
  Solution:
```
   Material balance on the entire CSTR:
     Accumulation = In - Out + Generation
          --> At steady-state, Accumulation = 0.
     V*(ds/dt) = 0 = F*(s_f-s_b)-A*v(s)   ... eqn(1)
   The value of s comes from the equation in Part a):
      k_l*(s_b-s)= v_m*s/(K_m+s)   ... eqn (2)
```
  Solve the above two equations for s and s_b (i.e., a system of 2 coupled algebraic eqns with 2 unknowns). We can solve these two equations sequentially. First, solve for s in terms of s_b with Eqn (2), then substitute s(s_b) into Eqn (1) to find s_b.
(40 pts.) Biology. Attached is the gene sequence file of a cloning vector pGREEN. You may directly mark these pages to answer the following questions.
1. A string of 3498 alphabet letters "a", "t", "c", and "g" spell out the gene sequence. If you were to count the number of occurrences of "a" in this string of 3498 letters, you will find that number to be (circle all the correct answer(s)):
```
  3498/2
  3498/4
  same as the number of "t"
  same as the number of "c"
  same as the number of "g"
  none of the above
```
  Solution:
  In a double stranded DNA, A & T are always present in equal numbers; so are C and G. However, only one single strand is given here and the number of "A" can be anything.
2. What is a "cloning vector"?
  Solution:
  A cloning vector is a vehicle to carry the target gene.
3. What regulatory elements do you expect to find in this cloning vector, which may or may not be identified in the "FEATURES" section? (Hint: Follow the central dogma of biology.)
  Solution:
```
   replication: ORI
   transcription: promoter/operator, terminator sequence
   translation: start codon, stop codon (as part of the marker proteins)
```
4. Is the given strand a sense strand (i.e., the template strand from which mRNA is transcribed) for green fluorescent protein (gfp)? For beta-lactamase (bla)?
```
  gfp:  |Sense|Antisense|Both|Neither|
  bla:  |Sense|Antisense|Both|Neither|
```
  Solution: Antisense strand for both green fluorescent protein and beta-lactamase.
5. How many amino acids are in green fluorescent protein?
  Solution:
```
Number of DNA bases: (33..749) = (749-33+1) = 717
Minus 3 stop codons: 717-3 = 714
Number of peptides in green fluorescent protein = 714/3 = 238 peptides
```
6. (10 pts.) Circle the three codons that correspond to the first three peptides at the amine terminus of green fluorescent protein. Repeat for the first two peptides at the acid terminus. Mark the amine and acid ends. Give the two peptides at the amine terminus and the two peptides at the acid terminus.
  Solution:
  Abbreviated green fluorescent protein: (33..749)
```
  base number               216                      749
                             |                        |
  DNA                  5'-...atg agt aaa....tac aaa tag...-3'
  gfp                    NH₂--M---S---K .... Y---K--COOH
                         NH₂-Met-Ser-Lys....Tyr-Lys-COOH
```
7. If you insert a target gene (say, insulin) into this pGREEN vector at the gfp site, and then insert this vector into a host cell (say, E. coli), briefly describe how you would subsequently isolate and identify those cells that may produce your target protein (insulin).
  Solution:
  Grow the transformed host cells (E. coli) on a dish containing the antibiotic ampicillin. Only cells contaning plasmids will grow. Of these proliferating colonies, cells containing plasmids with the target gene inserted at the intended gfp site will not be fluorescent. Pick out these non-fluorescent colonies and test them for expression of the target gene.

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Biochemical Engineering -- Quiz #2
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Nam Sun Wang

Department of Chemical & Biomolecular Engineering

University of Maryland

College Park, MD 20742-2111

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