The oxygen requirement for cell growth in glucose can be represented by the following equation (Mateles, 1971) ro2 32Nc+8(Nh2)+16(No2) Yx/sM +yo2 – 2.67yc + 1.714yN2 – 8y H2 In which ro2 is the oxygen required for each gram of cells produced, N stands for no. of atoms present in each molecule substrate, y stands for mass fractions and M is the MW of the substrate. The yeast cell may be considered to be CH1.800.5NO.2. Calculate rO2 if the yield factor (Yx/s) is 0.46g of cells produced for each gram of substrate consumed.
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- On the right the Hill plot com- (b) pares the O2 binding properties of Hb Ya- kima with those of HbA in 0.1 M NaCl buff- Hb-Yakima ered to pH 7 with 0.01 M bis-Tris. Focus first on the line for "stripped Hb". This is the term for hemoglobin isolated from erythro- cytes with removal of all organic phosphate molecules that might bind to the protein in RBCS. You can see that 2,3-bisphospho- glycerate (BPG; labeled DPG according to old terminology) does not alter the O2 bind- ing affinity of Hb Yakima in contrast to HbA (although it was shown that BPG did bind to the deoxyHb Yakima molecule). Also, Hb Yakima is associated with markedly decreased allostery in the absence and presence of BPG, in comparison to HbA. IHP = inositol hexaphosphate, an artificial allosteric modifying ligand that binds more tightly than BPG. stripped Hb Hb-A •DPG +DPG n= 1.0 n = 2.3 n=2,5 +THP +IHP 0.5 0.5 1 5 10 50 po, ( mm Hg ) %3D On the right is a diagram copied from the lecture handout "Hemoglobin and Allo-…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:…) To better understand how the lacR works, let's start let's approximate the volume of a bacterial cell by treating it as a cylinder with radius that is 0.5 um and length that is 2 um. What is the volume of the cylinder in liters, L)? Next, let's calculate the concentrations of the lacR (ligand) and the operator site (Op) (primary receptor) in this volume. There are 10 lacR molecules in the cell and 1 operator site. Using this information calculate [lacR] and [Op] in molar units in one cell? (see eq. 13.17 and 13.18).
- We humans do not express an alpha-Galactosidase enzyme and therefore cannot break down stachyose but the bacteria that live in our large intestine can. These bacteria contain an alpha-Galactosidase enzyme that hydrolyzes the galactopyranosyl-alpha (1-6) glycosidic bonds. The bacteria then ferment the products to generate the gas that is famously associated with eating beans. Draw the Haworth projections of the products of alpha-Galactosidase catalyzed reaction. Name the products of this enzyme catalyzed reaction.We want to measure the activity of alanine aminotransferase (ALAT) present in a serum. The reaction catalyzed by the enzyme is: Reaction 1: I- *H₂N- glutamate H - C-COO CH₂ CH₂ COO 0.1 M phosphate buffer pH 7.4 : 550 μL 1.2 M alanine: 100 μL CH3 pyruvate CH3 C time (min) A340 CIO O COO The enzyme reaction is alized in the following conditions: In a 1 cm-cuvette are added: COO™ pyruvate lactate dehydrogenase* (LDH, 300 µμg.mL-¹): 50 μL 1.5 mM NADH : 200 μL 0.04 M a-ketoglutarate: 500 µL serum containing ALAT: 600 μµL ALAT NADH + H+ 0 0.915 a-cétoglutarate COO LDH * Lactate dehydrogenase (LDH) reduces pyruvate into lactate, with the concomitant oxydation of NADH. This allows to indirectly measure the amount of product formed. с=0 CH₂ 1 0.741 Reaction 2: NAD+ CH₂ COO™ H-C CH3 OH COO™ lactate alanine H + *H3N-C The reaction is performed at 25 °C and the absorbance at 340 nm is monitored every minute, for 5 min. The absorbance values are given in the table below: Data: ENADH at 340 nm =…The diagram below shows the substrate binding cleft for a protease, providing the substrate structure, and indicating the residues (using one-letter code) that line the four specificity pockets. 1 M F H₂N K R IZ 2 3 P F S W оо E 4 The protease is known to cleave the amide linkage between W and E residues for substrates containing the WEFD sequence. Using 3-letter code with amino acids linked by a "dash" (ex. GLY-ALA), the N-terminal product is A and the C-terminal product is A
- 16. The overall reaction for the glycolysis reaction is C6H₁2O6(aq) + 2NAD+ (aq) + 2ADP³(aq) + 2HPO(aq) + 2H₂O(1) 2CH3COCO₂ (aq) + 2NADH(aq) + 2ATP4 (aq) + 2H3O+ (aq). What is A,G at chemical equilibrium?Shown below is a substrate for a Trypsin. Draw the mechanism for this serine protease using the artificial substrate. Be sure to draw the catalytic triad, and show the role of the oxyanion hole. Draw the complete structure of every intermediate and product and PUSH ARROWS!!!!! Do not abbreviate structures using R and R' H₂N _N_CH. сно CH₂ CH₂ CH₂ NH d=19H₂ NH₂ O CH- H₂C HN O CHThe compound below, doxorubicin, is a substrate for p-glycoprotein. Cancer cells with a higher expression of P-gp (more P-gp protein per cell) than normal cells were treated with doxorubicin. What would be the expected observation? o= 0= OH O OH HCI OH -OH NH₂ & OH more doxorubicin inside the cell doxorubicin would be oxidized doxorubicin would have more cytotoxic effect on these cancer cells less doxorubicin inside the cell
- 10. Chymotrypsin is a serine protease enzyme. The Km for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8*102, and the Km for the reaction of chymotrypsin with N- acetyltyrosine ethyl ester is 6.6*10“ M. catalytic triad Ser 195 His 57 Gly 193 N-H OH R-N- Ca N-H O-C- Asp 102 N-acetyl valine N-acetyl tyrosine Chymotrypsin Active Site a. What is the nucleophile here and how is it activated? b. Which substrate has an apparent higher affinity for the enzyme. c. Propose a reason for the difference in affinity based on the shape of each of the substrates (see active site figure, cleaves on the C-side of aromatic residues).Lysozyme catalyzes a "bi-bi" reaction, which means there are (how many) reactants and (how many) products. List, in order, the reactants that bind and the products that are released during a lysozyme-catalyzed reaction cycle -- be succinct but be specific. 1. First reactant = 2. First product = 3. Second reactant = 4. Second product %3D1. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins ofbacterial wall) are 37 C- temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. 2 Consider the matic reaction schee: Asnaragine + H20 Aspartate+ NH3: