6. The Vmax and KM values for an unusual hexokinase found in Try-pansoma cruzi (the causative agent of Chagas disease) are shown inthe presence and absence of a bisphonate inhibitor (structure shown).Without inhibitorWith inhibitorКм (mM)90125Vnmax (umol min-1. mL-1)0.300.12ОНO=P-O-A bisphonate compounda. What type of inhibitor is bisphonate? b. The parasite hexokinase,unlike the mammalian enzyme, is not inhibited by glucose-6-phos-phate but is inhibited by pyrophosphate (PP;). Is this observation con-sistent with your answer to part a? c. Might bisphonate be a goodcandidate for a drug to treat the disease?

Mixed inhibition is a type of inhibition in which the inhibitor may bind to the enzyme whether or…

Answered: 6. The V, and KM values for an unusual…

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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explain why A liposome of 1;1 molar ratio of cholesterol to phospholipids will be the optimum combination for the highest concentration of drug entrapment’
PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
3
4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. (8) mg V₂ μmol s1 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00 0 100 200 300 [s], mM 400 500 WT (a) Estimate the KM and Vmax for the wild-type and mutant enzyme from the graph. V105A (b) Calculate the keat and keat/KM for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
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Available cloxacillin 2g match the following reconstitution option with the order ordered: 1500 mg of cloxacillin IV q8h ordered: 800 mg of cloaxacillin IM q6h ordered: 500 mg of cloxacillin IM q8h ordered: 1800 mg of cloxacillin IV q8h 1. 2. add 4.8 mL sterile water for concentration of 400mg/mL add 9.8 mL of sterile water for concentration of 200 mg/mL
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