HW4_2011_sol(1)

# HW4_2011_sol(1) - Problem 1 (25 points) Known: Chemostat So...

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Unformatted text preview: Problem 1 (25 points) Known: Chemostat So = 2 g/L YX/S = 0.45 g X/g S µm = 0.5 hr‐1 S = 0.1 g/L / = 0.25 g X /g O2 D = 0.28 hr‐1 C* = 8 mg/L Solve: a) For chemostat and steady state, µ = D, X = YX/S (So – S) = 0.45(2‐0.1) = 0.86 g X /L OUR = qo2 X = µX/YX/O2, qo2 = D/YX/O2 = 0.28 / 0.25 = 1.12 g O2/g X hr = 1120 mg O2/g X hr b) In oxygen being the rate‐limiting step, OTR = OUR qo2 X = kLa (C* ‐ CL) kLa = qo2 X / (C* ‐ CL) = 1120(0.86)/(8‐2) = 161 hr‐1 Problem 2 (15 points) Genome Size: 2.8 × 106 %G or C: 65 => %G = 65/2, %C = 65/2 %T or A: 35 => %T = 35/2, %A = 35/2 Sequence: CCATGG Probability = P(C) × P(C) × P(A) × P(T) × P(G) × P(G) = 0.325 × 0.325 × 0.175 × 0.325 × 0.325 = 0.000342 # of sites = Size × probability = 2.8 × 106 × 0.000342 = 957.6 = ~ 958 sites Problem 3 (60 points) 1. Gene has probably more than one BamH1 sites. But since only partial digestion was done so there would be all permutations of gene length possible and there may still be some BamH1 sites present in the fragments. 2. His gene of interest is at the most around 10 kb long. It may be significantly smaller than this. He cuts plasmid so he can ligate free BamHI sites. Also disrupts the tetracycline gene so that he can eliminate those plasmids which were not cloned. Inactivation of enzyme is necessary as his fragments are only partially digested and if he doesn't inactivate it then it will further fragment his fragments. Also inactivation is essential as he need to do ligation reaction and it will again fragment if BamH1 enzyme is not inactivated. 3. AmpR means plasmid is inside. TetS means that the DNA was cloned into the plasmid (in between Tet gene) and hence it is disrupted. 4. Finds clone that have sequences specific to elixir to find positive clones. Replica plate is made so as to have a ready stock (in original plate) if in case he find the required clones and also to perform probing on clones. 15 kb total, 4.5 for original, so 10.5 for DNA that was cloned. Assume complete digestion and each fragment inside the cloned gene is 5.5, 3, 1.5, 0.5 = 10.5. Now he wants a segment of this gene/actual elixir gene. So he completely digests the plasmid and repeats step from above to find subcloned plasmids Since 5 and 7.5 kb were found to be positive with his radioactive mRNA so it means that both fragments 0.5 and 3 kb are part of Elixir gene and hence there is a minimum of one BamH1 site in the gene Elixir 5. 6. ...
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## This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.

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