Considering the general structure of penicillins, how are they rendered ineffective?Select one:O a B-Lactamase performs nucleophilic attack of the amide in the penicillin side chainOb. B-Lactamase performs nucleophilic attack of the amide in the B-lactam ring and subsequently hydrolyzes this intermediateOC. B-Lactamase performs nucleophilic attack of the thiazolidone ring on the carbon next to the electronegative sulfur leading to acovalently inhibited B-Lactamaseo d. All of theseO e. None of these

•Penicillin nucleus consists of fused Thiaolidine and B- Lactam linkage to which side chains are…

Answered: Considering the general structure of…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter18: Glycolysis

Section: Chapter Questions

Problem 25P: Using the ActiveModel for phosphofructokinase (Trypanosoma), describe the difference between the...

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Considering the general structure of penicillins, how are they rendered ineffective? Select one: a. β-Lactamase performs nucleophilic attack of the amide in the penicillin side chain b. β-Lactamase performs nucleophilic attack of the amide in the β-lactam ring and subsequently hydrolyzes this intermediate c. β-Lactamase performs nucleophilic attack of the thiazolidone ring on the carbon next to the electronegative sulfur leading to a covalently inhibited β-Lactamase d. All of these e. None of these
Draw the amino acid–Schiff base that forms in the breakdown of 3-hydroxykynurenine to yield 3-hydroxyanthranilate in the tryptophan degradation pathway (Fig.) and indicate which bond is to be cleaved.
The enzyme cytidine deaminase catalyzes the conversion of cytidine to uridine. Cytidine deaminase catalyzes the reaction through an addition of water across the cytidine 3,4-bond, forming a tetrahedral intermediate followed by the elimination of NH3 to form the product uridine. This is like the addition-elimination mechanism that we studied for adenosine deaminase. Cytidine deaminase НО. NH₂ N пOH OH cytosine R + H₂O cytidine The Km value for the substrate cytidine is 2.5 × 10-4 M, and the K; for competitive inhibition by the product uridine is 2.5 × 10-³ M. N R A reduced derivative of the product, 3,4,5,6-tetrahydrouridine was shown to be a fully reversible competitive inhibitor with a Ki of 2.4 x 10-7 M, a value approximately 10,000 times lower than that of the product uridine. NH₂ NH uridine HO. N R R = D-ribose HOH OH uridine 3NH uracil ring numbering H NH H H + NH3 H OH H 3,4,5,6-tetra- hydrouridine a.) Draw a structure of the intermediate that we predict to form during the…
6-Mercaptopurine , after its conversion to the corresponding nucleotide through salvage reactions, is a potent competitive inhibitor of IMP in the pathways for AMP and GMP biosynthesis. It is therefore a clinically useful anticancer agent. The chemotherapeutic effectiveness of 6 mercaptopurine is enhanced when it is administered with allopurinol. Explain the mechanism of this enhancement.
Please answer the question below:
Lysosyme catalyzes the hydrolysis of the carbohydrate linkage in part of the bacterial cell wall (peptidoglycan layer). 1. Propose a mechanism using general acid/base catalysis 2. Propose a mechanism using covalent catalysis. (An intermediate has been observed that indicates a glycosidic bond to the aspartate).
Draw a stepwise mechanism for the following reaction. The fourmemberedring in the starting material and product is called a β-lactam.This functional group confers biological activity on penicillin and many related antibiotics. (Hint: The mechanism begins with β elimination and involves only two steps.)
Tazobactam is a BETA-lactam that inhibits the enzyme that bacteria secrete to resist penicillin. Draw a mechanism for the formation of the tazobactam-trapped BETA-lactamase conjugate that proceeds through the following steps.
In serine protease catalytic triad mechanism, the first and second products that leaves the active site are Carboxylic acid and amide respectively Amide and carboxylic acid respectively Ester and amide respectively Carboxylic acid and ester respectively
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
Please answer the following questions 1) based on the structures of adenine, cytosine, guanine, and thyme. Why does A prefer to bind to T and C prefer to bind to G 2) were the binding energies the same for AT and CG? Explain the difference 3) which of the four know bases is the unknown synthetic base most like in terms of its binding characteristics? What does this indicate about its structure? 4) if the unknown synthetic base was used to study DNA replication, which natural bases would it interact with? 5) is the picture question
A schematic representation of the enzyme IspD complexed to inhibitor 3, and a series of inhibitors 3-5 are shown below. Ala202 lle240 mwww NH NH Val263 ОН www HN N- lle177 HN 'N' CI 3 X = N 4 X = C-CN 5 X = C-COO IC50 274 µM IC50 140 nM IC50 35 nM NH2 HN Val266 N -N O-H---- N HN %3D Arg157 HN wwww lle265 Explain why structure 4 is a more potent inhibitor (lower IC50 value) than inhibitor 3 and why structure 5 is a much weaker inhibitor (higher IC50 value) than 3 and 4.

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