Many antibiotics alter the rate of reaction for enzymes that produce and then cross-link the strands of this polymer together. The image below shows the structure of one antibiotic called penicillin as compared to the portion of the peptidoglycan that undergoes cross linkage. Condon A HS. CH CH CH N-CH cooo Cand cooo wDAla-D-Ala [S] 2. Using the graph above, which condition do you predict was the penicillin condition? 3. Explain how penicillin effects the enzyme. 4. Why it is important to take the full dosage of antibiotics. Briefly explain 5. Is peptidoglycan still being produced in the portion of the graph marked by the circle in condition B?
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- Identify the following by describing their functions: EF-G, EF-Tu, EF-Ts, EF-P, and peptidyl transferaseProtease enzymes cleave proteins by hydrolyzing peptide bonds. The strategy for each type of metalloprotease begins with generating a nucleophile that attacks the peptide bond that attacks the peptide carbonyl group. Macmillan Learning On the basis of the information provided in the figure, show the next step in the mechanism for peptide-bond cleavage by a cysteine protease. H Cysteine proteases R₂ HN R₁ Draw curved arrows on the pre-drawn structures to show the cysteine protease mechanism. If you need to reset the structures, click More followed by Reset Drawing. Select Draw Templates Groups More N R2 H \.. C O S R1 Erase NMatch the antibiotic to its target with bacteria. erythromycin [ Choose ] [Choose ] streptomycin, an aminoglycoside bacterial enzymes involved in folic acid synthesis Bacterial gyrase tetracycline bacterial transpeptidase 70S ribosomes Bacterial RNA polymerase ampicillin Bactrim= trimethoprim + sulfa drug [ Choose ] rifampin [Choose ] ciprofloxacin, a fluoroquinolone [ Choose ] >
- 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…A different Fab fragment binds to lysozyme with a dissociation constant of Kd=10-6 M. A 1 nM (10-9 M) solution of lysozyme is treated with increasing concentrations of this Fab fragment. At what concentration of added Fab will half of the lysozyme be bound to this Fab?A researcher creates random copolymers of three nucleotides each by mixing polynucleotide phosphorylase with guanine and adenine nucleotides in a ratio of 5 guanine nucleotides to 1 adenine. Give the different copolymers produced and their theoretical proportions.
- Explain why bacteria with the gene for the enzyme beta-lactamase are not susceptible to penicillin G. Support your explanation by creating a diagram of penicillin G. Label the beta-lactam ring.The alkaline hydrolysis of pAUGCAGC oligonucleotide produces: O A. Uridine 2'-monophosphate, uridine 3'-monophosphate, cytosine 2'-monophosphate O B. Adenosine 2'-monophosphate, adenosine 3'-monophosphate, adenosine 21,5'-bisphosphate OC. Guanosine 2'-monophosphate, guanosine 3'-monophosphate, cytosine 3'-monophosphate O D. Cytidine 3'-monophosphate, guanosine 2'-monophosphate, adenine 2'-monophosphate O E. Adenine 3,5'-bisphosphate, guanine 2,5'-bisphosphate, uridine 2'-monophosphate O F. Uridine 2'-monophosphate, uridine 3'-monophosphate, guanine 3'-monophosphate= A different Fab fragment binds to lysozyme with a dissociation constant of Ka 10-6 M. A 1 nM (10-9 M) solution of lysozyme is treated with increasing concentrations of this Fab fragment. At what concentration of added Fab will half of the lysozyme be bound to this Fab? [F] = ab M
- A toxin is glycopeptide, X that is composed of an octapeptide and a saccharide Y. Glycopeptide X is composed of a glycan Y and an octapeptide. The octapeptide was cleaved by trypsin, giving 2 tetrapeptides of exactly the same composition except for its N-terminals. When the octapeptide was cleaved by chymotrypsin, it gave a pentapeptide and a tripeptide. DNFB treatment gave DNP-ser. The composition of the octapeptide is lys, ser, trp.and leu. The leu:ser ratio is 4:1. 1. What is the sequence of this octapeptide? Saccharide Y is a tetrasaccharide. To determine its linkages, a person methylated glycopeptide X effectively methylating saccharide Y. Cleavage gave the peptide and the following products: 2,3,4-tri-O-methyl-a-D-glucopyranoside acid, 2,3-di-O-methyl-ß-D- glucopyranoside, 1,3,4-tri-O- methyl-ß-D-glucuronic acid treatments to the isolated tetrasaccharide gave the following results: and 1,4,6-tri-O-methyl-ß-D-N- acetylgalactosamine. Further 1. A. Treatment of saccharide Y with an…A nonapeptide was treated with dithiothreitol to reduce any disulfide bridges, then partially hydrolyzed, giving rise to the following fragments. What was the sequence of the original polypeptide? Phe-Gln-Asn Pro-Arg-Gly-NH2 Cys-Tyr-Phe Asn-Cys-Pro-Arg Tyr-Phe-Gln-Asn O Cys-Tyr-Phe-GIn-Asn-Cys-Pro-Arg-Gly-NH2 O H2N-Gly-Arg-Pro-Cys-Asn-Gln-Phe-Tyr-Cys O Gly-Arg-Pro-Cys-Tyr-Phe-Gln-Asn-Cys-NH2 Cys-Pro-Arg-Gly-NH2-Cys-Tyr-Phe-Gln-Asn O None of these are correct.A spheroidal bacterium with a diameter of 1.0 μm (micrometer, 1 μm = 10-6) contains 25,000 molecules of the protein hexokinase. What is the molar concentration of the protein inside the cell?