C16H34 + 16.28 O2 +1.42 NH3 → 1.65 C4.4H7.301.2No.86 + 8.74 CO2 + 13.11 H20For a particular bacterial strain, the molecular formula was determined to be C4.4H7.301.2No.86. Thesebacterial cells are grown under aerobic conditions with hexadecane (C16H34) as substrate.IAssume that 60% of the hexadecane is used for producing cells (also called biomass) and the remaining40% of the hexadecane is used for other cell functions. You have been put in charge of a small batchfermenter for growing the bacteria and aim to produce 10 kg of cells for inoculation of a pilot-scalereactor.a) Assuming this 60% conversion, what is the yield of cells from hexadecane in kg/kg?b) What minimum amount of hexadecane substrate (in kg) must be contained in your culturemedium? Remember that only 60% of the hexadecane is converted to cells.c) What must be the minimum concentration of hexadecane (in kg/m³) in the medium if thefermenter working volume is 4.3 cubic meters? Remember that only 60% of the hexadecane isconverted to cells.

Answered: Image /qna-images/answer/cb3785a3-e895-4452-8a4e-30da58d03f1f.jpg

Answered: C16H34 + 16.28 O2 +1.42 NH3 → 1.65…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Which step is catalyzed by the enzyme responsible for the Met 2 mutant? (picture attached) A) homoserine → cysteine B) cysteine → cystathionine C) cystathionine → homocysteine D) homocysteine → methionine E) methionine → homoserine
Hemoglobin is considered to be a tetrameric complex with a 64 kDa (α β)2. When attempting to purify hemoglobin, we must first purify the α and β monomers (about 16 kDa each) to prepare the tetramer. This is formed from the dimer intermediate: 2 α + 2 β -> 2 αβ -> (α β)2. The graph given represents a size-exclusion chromatogram after the refolding of the hemoglobin tetramer Using the size-exclusion chromatogram given, 1. Draw an SDS-Page Gel with a reducing agent such as BME using the three peaks listed on the graph.
You have discovered a new hemoglobin variant you are calling Hb-21. You have noticed the following from your studies of Hb-21: - The binding constant of Hb-21 is the same at pH 7.1 and 7.4.- The binding constant curves generated by Hb-21 and O2 are sigmoidal.- The Hill coefficient for Hb-21 is 3.-The binding constant for Hb-21 with O2 in the presence of 5mM 2,3-bisphosphglycerate is about 6-times lower instead of the 8-times lower that is observed under these conditions for the wild type enzyme. Use this information to answer questions: Based on this information which of these features of hemoglobin would you conclude are altered in Hb-21 as compared to the wild type hemoglobin? ____ the cooperativity.____ the amount of T versus R state in the tissues. ____ the Bohr effect. What is the major issue for an individual with the Hb-21 variant? Explain exactly what aspect of hemoglobin’s function is altered.
Consider the reaction catalyzed by PFK. In the presence of AMP, which of the following will be expected? Check all that apply: a)The Km for substrate would be decreased b) the initial velocity plot would show the curve shifted to the right c) the R state is stabilized d)the rate of the reaction is diminished
PTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 ± 0.22 0.53 + 0.02 34.7 + 0.25 52.7 ± 0.7 0.066 + 0.001 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
Imagine that a mutation has been discovered in the HBA1 gene, where amino acid residue # 1 in the alpha subunit is changed from valine to lysine. The structure of these two amino acids is shown below. How would this mutation affect the mobility of HbA in the gel?a) HbA with this mutation would have the same mobility as normal HbAb) HbA with this mutation would have the same mobility as HbSc) HbA with this mutation would have a faster mobility than normal HbA O d) HbA with this mutation would have a mobility somewhere in between HbS and normal HbA
Consider the positively charged amino acid lysine Lys2+ 21 COOH I H&N-C-H I pH 14 12 10 8 6 4 2 0 CH₂ I CH₂ I CH₂ I CH₂ T NH₂+ 0 Nelson p85 2.18 = 2.18 PK₁ Lys+ COO™ I H₂N-C-H H₂N-C-H ī I -----) 8.95 Lysº 8.95 pK₂ pka carboxyl = 2.19 pkaamino = 9.67 pka sidechain = 4.25 COO™ I CH₂ I CH₂ I CH₂ I CH₂ I NH₂¹ 1.0 2.0 Equivalents of OH added- COO™ I H₂N-C-H I 10.79 1 10.79 pk Isoelectric point Lys CH₂2 I CH₂ I CH₂ I CH₂ T NH₂ 3.0 +H3N NH3+ T CH₂ T CH₂ CH₂ CH₂ -COO™ H Lysine (Lys, K) Physiological pH = 7.4 < pl → Amino acid is positively charged at physiological pH 1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. 3. Draw the Aa structure at:…
Using the molecular formula C6H12O6, show the following:a. Where does this molecule originate.b. Show its movement from its origin to where it is consumed (be sure to add any proteins that are needed for this process and also define the process of its movement.c. Show the FATE of every carbon in the C6H12O6 molecule being sure to label the important intermediate molecules that are formed during its consumption and make sure they are placed in the proper cellular location.

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