4 In aqueous solution, sucrose, 3, can react to form glucose and fructose. Thisprocess is acid catalysed, beginning with ether oxygen protonation.Table 2 gives the calculated proton affinities (PAS) for the ether oxygen atomsin sucrose. Protonation at position B leads to cleavage of the C-O bond to thesix membered ring, B6, or the C-O bond with the five membered ring, B5.HOH₂COHHO OHOH₂CBC-CH₂OHOH3H-Q CH₂OH;HAон нHOTable 2. Calculated protonaffinities for sucrose.a) Based only on the proton affinity values, which is the least likely position forinitial protonation of an ether oxygen in sucrose?PositionAB6B5Cb) Sketch two pathways that could follow protonation at ether oxygen B inaqueous solution and would lead to glucose and fructose.c) Figure 2 shows a simulation configuration used to study sucrose protonation atposition B in an aqueous solvent. Discuss the rationale for choosing the numberand location of water molecules in the model.BProton affinity / eV9.119.239.349.43HOH₂CCH HOOH HCH₂OH-HFigure 2. Configuration used to study sucrose protonation at ether oxygen B.

Proton affinity value is defined as the negative of enthalpy change (ΔH) of process. Higher the…

Answered: a) Based only on the proton affinity…

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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complete the following mechanism showing the cleavge of the peptide bond for the following cysteine protease reaction. you only need a catalytic dyad for cystein proteases because the pKa of cysteine is low enough (8) and close enough to the pH of 7.4 that is can easily be deprotonated by histidine.
Although the first two carbons of fructose and glucose are identical in structure to DHAP and GADP (from glycolysis), DHAP and GADP equilibriate on their in solution to favor the ketone over the aldehyde, while fructose and glucose do not. Why? a)The larger size of the molecule sterically hinders the isomerization b)The larger sugars have more OH groups which hydrogen bond and disrupt isomerization c)The larger sugars cyclize, and there is no carbonyl to isomerize in the cyclic form d)The larger sugars cyclize, and in the cyclic form the hydrogen bonding is very strong e)The larger sugars are less soluble in water than the smaller sugars
The dissociation of lactic acid to lactate is shown in the reaction. Lactic acid has a pKa of 3.86. OH Lactic acid OH OH Lactate + H+ A solution containing a mixture of lactic acid and lactate was found to have a pH of 5.10. Calculate the ratio of the lactic acid concentration to the lactate concentration in this solution. [lactic acid]/[lactate] =
The OH group of the serine side chain (R-OH) can undergo dissociation to form R-O- and H+. However, the ratio of the protonated to unprotonated forms of this group at pH 7 is approximately 1,000,000 to 1. Calculate the pKa of the serine OH group.
Draw the structures of L-Cysteine from highly protonated to depronated form. Note Structure A as the most protonated form. Provide the charge of each structure. (Answer structure A and B pls)
Based on the structure of sucrose below, is it classified as a reducing sugar? ОН 6CH OH 5 5 ОН 3 6CH OH a-Glucose 2 ОН 0. ОН HO 3 1 ОН ОН 2 CH₂OH 1 B-Fructose ОН CH₂OH ОН CH₂OH ОН OH НО a, B (1-2) linkage CH₂OH Sucrose + H2O
Write the names of the above molecules (A and B respectivley) ,If the 5’OH group was phosphorylated with triphosphate, what would the names then be (provide the full names of each)?
4- Salicylamides are inhibitors for an enzyme called scytalone dehydratase. SAR shows that there are three important hydrogen bonding interactions. Explain whether you think quinazolines could act as a bioisostere for salicylamides. HBA OH O 00 N NHR Quinazolines Salicylamides Ph HBA 5- Structure IX (X = NH) is an inhibitor of a metalloenzyme called thermolysin and forms interactions as shown. Explain why the analogue (X = O) has reduced binding affinity by a factor of 1000 and why the analogue (X = SH) has roughly the same binding affinity. Structure IX H HBD Ala-113 R
Dehalogenase enzymes catalyze the clevage of C-X bonds. One such dehalogenase catalyzes the following reaction. An active site aspartate is thought to carry out the initial nucleophilic attack that expels the chloride. Select the most likely intermediate in the reaction. (Note the stereochemistry. Refer to P11.19 in the textbook if needed.) O O A C D OE -CH₂ H H₂C™ CO₂ H B CO₂ CI N H₂O Cr CO₂ H CO₂ нотум CH3 D CO₂ five N E CH₂
Diagram the cyclization of monosaccharides which leads to the formationof pyran rings. Required: Please show me a correct and step by step process of the diagram and show me the correct drawing of the pyran ring.
The pKa for histidine is pKa = 6.1 while that for cysteine is pKa = 8.0 2. Assume that both histidine and cysteine are catalytic groups for a particular enzyme. Assume also that the side chain of cysteine must be in the deprotonated form. Estimate the pH at which the catalytic activity of this enzyme is the maximum and sketch a pH-activity graph.
The enzyme lysozyme hydrolyzes glycosidic bonds in peptidoglycan, an oligosaccharide found in bacterial cell walls. The active site of lysozyme contains two amino acid residues essential for catalysis: E35 and D52. Which of the following statements about lysozyme is true? More than one may apply The graph shows the pH-activity profile of lysozyme. 100 50 4. 6. 8. 10 pH Residue E35 exhibits general base catalysis O The pKa of E35 is approximately 4.3 O The pKa of D52 is approximately 4.5 O The optimal activity of lysozyme is approximately 5.2 Residue E35 exhibits covalent catalysis The pka of E35 is approximately 6.1 The pka of D52 is approximately 3.7 Activity (% of maximal)

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