You have a treatment process in which a microbial population consumes a substrate (chemical A) according to Monod kinetics, with Hmax=2 d', Ks = 350 mg/L, and Y 0.15 g cells/g A. What biomass concentration (in mg/L) must be maintained to result in a substrate utilization rate of -1000 mg/L-hr for a concentration of A equal to 5000 mg/L?
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- A certain drug has a half-Iife in the body of 2.5 h. What should the interval between doses be, if the concentration of drug in the body should not fall belc 40.% of its initial concentration? Round your answer to 2 significant digits.What is the Ksp of Zn3(PO4)2 (MW = 320.8 g/mol) if its solubility in water at 25 °C is 4.9 x 10-5 g/L? format for the scientifc notation 1.0 x 10^-1An enzyme catalyzed the reaction: S -> P has a AG" value of +4.8 kJ/mol. If you start with a solution containing 500 mM substrate [S] prior to the addition of enzyme and initiating the catalyzed reaction, then what is the substrate concentration [S] after reaching equilibrium? [R = 8.315 J/ mol* K; T = 25 C] NOTE - Provide your numeric answer in mM units as a TWO digit significant figure, and do NOT include units in your answer (report the numeric answer ONLY).
- 6. A 190 lb worker in a factory received an exposure to AB which has a molecular weight (MW) of 278 g/mole. AB is completely metabolized in the body to A (MW 194) and B. A is completely excreted in the urine and B is further metabolized to five different compounds. A 24-hr total urine sample was 2060 ml, Analysis showed the concentration of A in the urine was 1.28 ng/mL What was the absorbed dose of AB in ug/kg body wt?Drug Concentration A pharmaceutical companyclaims that the concentration of a drug in a patient’sbloodstream will be at least 10% for 8 hours. Supposeclinical tests show that the concentration of a drug (aspercent) t hours after injection is given byC(t) = 200t/2t2 + 32During what time period is the concentration at least10%? Is the company’s claim supported by the evidence?With the enzyme activity results in Table 1, without plotting a graph, determine: a) b) c) Vmax (explain) Why is the velocity v constant at [S] greater than 2 x 10° M? What is the free [E] at [S] = 2 x 102 M? Table 1 [S] (mol/L) 2 x 10-1 2 x 10-2 2 x 10-3 2 x 10-4 1,5 x 10-4 1.3 x 10-5 v (umol/min) 60.00 60.00 60.00 48.00 45.00 12.00
- If p50 = 3.1 torr for elephant Mb (myoglobin) and p50 = 5.1 torr for zebra Mb, please pick ALL of the the statements that are correct: Elephant Mb binds oxygen more tightly than zebra Mb, since the p50 value is lower, which may help to explain some of the differences in physiology between elephants and zebras When pO2 = 10 torr, YO2 or the fractional saturation of zebra Mb with oxygen is LOWER than for elephant Mb. When pO2 = 40 torr, zebra Mb is more than 90% saturated with oxygen. When pO2 = 40 torr, elephant Mb is more than 90% saturated with oxygen. When pO2 = 5.1 torr, zebra Mb is 50% saturated with oxygen.6. Five reaction mixtures containing equal concentrations of an enzyme are made up to the substrate concentrations [S] indicated in the table below and the initial rates of the reaction (V) are measured. The experiment is repeated with an enzyme inhibitor I present at 0.22 mM in each reaction mixture. (a) Determine Km for the substrate and K, for the inhibitor. (b) Can you say anything about the type of the inhibitor? [S], mM V (µmol/min) (no I) V (µmol/min) (+ 0.22 mM I) 0.1 28 17 0.15 36 23 0.2 43 29 0.5 65 50 0.75 74 61The catalytic activity of an insect aminopeptidase was investigatedusing an artificial peptide substrate. The Vmax was 4.0 × 10-7 M/s and the KM was 1.4 × 10-4 M. The enzyme concentration used in the assay was 1.0 × 10-7 M. What is the value of kcat? 2.86 x 10-3 M 0.25 M/s 4.0 s-1 2.86 x 104 s/M 0.25 s-1
- Given the Ksp for Ag2CrO4 is equal to 1.20 x 10-12, calculate the percent error in your experimentally determined value for average Ksp (this will be an extremely large percentage – think about why).For an enzyme -catalyzed reaction, KM=10.0 mmol dm-3, vmax=0.250 mmol dm²³ s-'; [E]o= 2.3 x 10-6 mmol dm-3. Calculate the catalytic efficiency (=kb/KM) of the enzyme. 0.920 dm³ mol-l s-1 3.1 x 107 dm³ mol-1 s-1 1.1 x107 dm³ mol-l s-1 0.250 dm³ mol-l s-1Draw the nullclines and find equilibria of the following extensions of the basic disease model. Find the nullclines and equilibria of this model when α = 2.0, µ = 1.0, and k = 4.0.