C16H34 + 16.28 O2 +1.42 NH3 → 1.65 C4.4H7.301.2No.86 + 8.74 CO2 + 13.11 H2O For a particular bacterial strain, the molecular formula was determined to be C4.4H7.301.2No.86. These bacterial cells are grown under aerobic conditions with hexadecane (C16H34) as substrate. Assume that 60% of the hexadecane is used for producing cells (also called biomass) and the remaining 40% of the hexadecane is used for other cell functions. You have been put in charge of a small batch fermenter for growing the bacteria and aim to produce 10 kg of cells for inoculation of a pilot-scale reactor. a) Assuming this 60% conversion, what is the yield of cells from hexadecane in kg/kg?
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- The photosynthetic process used by the green sulfur bacteria, under anaerobic conditions, is best represented by which of the following balanced equations? 6CO2 + 12H2O C6H12O6 + 6O2 + 6H2O 6CO2 + 12H2S C6H12O6 + 6S2 + 6H2O C12H22O11 + H2O C6H12O6 + C6H12O6 C6H12O6 + C6H12O6 C12H22O11 + H2O C6H12O6 + 6O2 + 6H2O 6CO2 + 12H2OThe photosynthetic process used by the green sulfur bacteria, under anaerobic conditions, is best represented by which of the following balanced equations? cell bio 6CO2 + 12H2S C6H12O6 + 6S2 + 6H2O C6H12O6 + C6H12O6 C12H22O11 + H2O 6CO2 + 12H2O C6H12O6 + 6O2 + 6H2O C12H22O11 + H2O C6H12O6 + C6H12O6 C6H12O6 + 6O2 + 6H2O 6CO2 + 12H2OC6H12O6 + a*O2 + b*NH3 --> c*CH1.8O0.6N0.12 + d*CO2 + e*H2O + f*C6H8O7 Citric acid can be produced according to the above formula by aerobically growing Aspergillus niger mold in a solution containing glucose as a substrate. The chemical composition of this mold is CH1.8O0.6N0.12. The yield of biomass from glucose is 0.34 g g-1 and the yield of citric acid from glucose is 0.535 g g-1. Using a reduction balance, calculate how many moles of oxygen are needed per mole of glucose. In this calculation, use the following rounded even values as atomic weights for all calculations: C=12 g/mol, H=1 g/mol, O=16 g/mol ja N=14 g/mol. So how many moles of oxygen are needed per mole of glucose ? Calculate the answer to three decimal places.
- Aerobic respiration, used by the mitochondria of eukaryotic cells, is best represented by which of the following balanced equations? 6CO2 + 12H2S g C6H12O6 + 6S2 + 6H2O C6H12O6 + 6O2 + 6H2O g 6CO2 + 12H2O C12H22O11 + H2O g C6H12O6 + C6H12O6 6CO2 + 12H2O g C6H12O6 + 6O2 + 6H2O C6H12O6 + C6H12O6 g C12H22O11 + H2OA genetically modified bacterium was cultivated to produce 1,3-propanediol (1,3-PDO), and the following yields were obtained: YX/S = 0.106 gX/gS and YP/S = 0.28 gP/gS. Knowing that the culture medium was formulated with glucose, ammonia and some mineral salts, and the cultivation was carried out under aerobic conditions, answer:a) What is the oxygen demand in this process?b) What is the maximum theoretical yield of 1,3-PDO in this process?c) Projecting the scale-up, what cell concentration must be reached to obtain 10 kg of 1,3-PDO in a batch reactor with 1 m3 of culture?Define the following: 1. Cofactor 2. Coenzyme 3. "-ase" ending
- 1. Know the structures of DNP-glycine and of the chromophore associated with cytochrome C. Low Salt Volume of blue band = 5.2 ml (Void Volume) Volume of yellow band = 2.6ml Volume of red band = 11ml High Salt Volume of blue band = 4.2 ml -> Void Volume Volume of yellow band = 5.2 Volume of red band = 1.2 2. What is the exclusion limit and the fractionation range for CM-Sephadex resin used? 3. What are the ion exchange reactions which occur when the coloured mixture is first applied to the column under low salt conditions? What exchange reactions occur upon the change of buffer from low salt to high salt concentrations? 4. What were the values for Vo and Vi for this column? Where the values for Vo and Vi the same under both conditions? 5. Explain the order in which the various components of the mixture elute under each condition in terms of their physical properties and the nature of the CMSephadex G-50 column. Would the same order be observed with a G-50 Sephadex column (i.e. no…Potassium nitrate is widely used in industries. The reaction for the industrial production of KNO3 is summarized in the equation below; 4KCl + 4 HNO3 + O2 → 4 KNO3 + 2 Cl2 + 2 H2O Which of the following statements below is correct about the production of KNO3? a. A redox reaction; KCl is a reducing agent and O2 is an oxidizing agent b. A redox reaction; HNO3 is a reducing agent and Cl is an oxidizing agent c. Not a redox reaction d. A redox reaction; KCl is a reducing agent and KNO3 is an oxidizing agent e. A redox reaction; HNO3 is a reducing agent and KNO3 is an oxidizing agentList four (4) individual alloantibodies which react preferentially at room temp or below.
- What are the donor atoms involved in Aspartame-Cu(II) binding? A) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of alpha carboxyilic acid group of Aspartic acid B) Nitrogen atom of alpha amine group of Phenyl alanine methyl ester and an oxygen atom of alpha carboxyilic acid group of Aspartic acid C) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of alpha carboxyilic acid group of Phenyl alanine methyl ester D) Oxygen atoms of the dipeptide Aspartame E) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of carboxyilic acid group of the R group of Aspartic acid33. Which one of the following is NOT a biochemical property associated with exotoxins Group of answer choices binding to cholesterol N glycosidase activity peptidase activity ADP ribosyl transferase activity cAMP activity 34. Temperature requirement is important for bacterial growth becos, Group of answer choices due to all of the above reasons promotes synchronized metabolic reactions temperature influences both lipids and proteins of cells the 3D structure of proteins is affected / altered if the temperature is too high if bacteria grow below the minimal growth temperature, the cells may dieBased on the kinetic constants below, which enzyme will most efficiently catalyze conversion of the substrate into product? A) Vmax = 10 uM s-1, KM = 10 µM B) Vmax = 10 uM s-1, KM = 0.01 µM C) Vmax = 1000 uM s-1, KM = 500 µM D) Vmax = 1 uM s-1, KM = 1 µM E) Vmax = 200 uM s-1, KM = 10 µM