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- 2B. S. aureus hemolysin B attacks the RBC cell membrane by hydrolyzing the sphingomyelin headgroup: ОН HN .R hemolysin B cuts this bond i) Draw a plausible mechanism of hydrolysis for this lipid headgroup. Let B- and BH be general base and general acid. 00-P-O LOR2 OR, ii) Why is this damaging to the overall membrane architecture of the RBC?Draw the products formed when -D-idose is treated with C6H5NHNH2 (excess), H+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-…
- 3. Identify the oxidizing and reducing agent from the following a) MnQ, + 4AF 3Mn+ b) 2CUSO, + S0,+ 2KB1 + 2H02CUBR +2H SO, + K,SO, 2A1 Q: derlined atoms (1x3=3)5) In the below two step transformation, the first step AG is positive i.e. 1.7 kJ/mol. Yet this two step coupled reaction takes place to form Fructose-1.6-bisphosphate. Explain why?" C-H CH-OH CH,-0 H-C-OH AG"- 1.7 KJ/mol (+0.4 kcalmol) C=0 AG"=-14.2 k/mol(-3.4 kcalmol) HO-C-H HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH H-C-OH H-C-OH ATP ADP H-C-OH CH-0-P. CH,-0-P-o- CH,-o- Glucose-6-phosphate Fructose-6-phosphate Fructose-1,6-bisphosphateIII. Protein: QQICIMFELTQISS Predict the products of the following reactions with the protein given, if there is none, write NO RXN. Also indicate, if the reaction is fast or slow. 9.) КОНРЫ(СH-CО)» 10.) Glyoxilic Acid/Conc. H2SO4
- 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).Which of the following is an anomer of B-D-gulopyranose? O O го ОН H ОН H ОН H H ОН CH₂OH -II- H H ОН ОН CH2OH H ОН ОН CH2OH О 0. H CH2OH то Н H H ОН H H - о H ОН ОН H ОН H ОН ОН H О ОН ОН H N1. Consider the following parameters related to an enzyme that follows Michaelis-Menten kinetics for the reaction: k(1) k(2) S ----> ES ----> P k(-1)
- 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 %3DBalance each of the following redox reactions occurring in acidic solution. |-(aq)+NO-2(aq)→12(s)+NO(g)l-(aq)+NO2- (aq)→12(s)+NO(g) Express your answer as a chemical equation. Identify all of the phases in your answer. ClO-4(aq)+Cl-(aq)→CIO-3(aq)+CI2(g)CIO4- (aq)+Cl-(aq)→Cİ03-(aq)+Cl2(g) Express your answer as a chemical equation. Identify all of the phases in your answer. NO-3(aq)+Sn2+(aq)→Sn4+(aq)+NO(g)NO3- (aq)+Sn2+(aq)→Sn4+(aq)+NO(g) Express your answer as a chemical equation. Identify all of the phases in your answer.5. For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s-1, k-1 = 3.1 ⅹ 104 s-1, and k2 = 3.4 ⅹ 105 s-1. a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a