6. If inhibitor X changes lactase activity to a Vo of 0.333333333333 mM per minute when [S] = 2.0 mM, and a Vo of 0.50 mM per minute when [S] = 3.33333333333 mM, calculate the new Vmax. 1.0 mM per minute 2.0 mM per minute 2.5 mM per minute 5.0 mM per minute 10.0 mM per minute 7. Calculate the new Km of the lactase reaction with inhibitor X. 1.0 mM 2.0 mM 2.5 mM 5.0 mM 10.0 mM 8. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. 1.0 per minute 2.0 per minute 2.5 per minute 5.0 per minute 10.0 per minute

6. If inhibitor X changes lactase activity to a Vo of 0.333333333333 mM per minute when [S] = 2.0 mM, and a Vo of 0.50 mM per minute when [S] = 3.33333333333 mM, calculate the new Vmax. 1.0 mM per minute 2.0 mM per minute 2.5 mM per minute 5.0 mM per minute 10.0 mM per minute 7. Calculate the new Km of the lactase reaction with inhibitor X. 1.0 mM 2.0 mM 2.5 mM 5.0 mM 10.0 mM 8. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. 1.0 per minute 2.0 per minute 2.5 per minute 5.0 per minute 10.0 per minute

Curren'S Math For Meds: Dosages & Sol

11th Edition

ISBN:9781305143531

Author:CURREN

Publisher:CURREN

Chapter9: Parenteral Medication Labels And Dosage Calculation

Section: Chapter Questions

Problem 7.7P

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-1/aka 1/v. Increasing a=1 (no inhibitor) Slope = a KµV…... Slope ak 7-15 satino inhibitori B D MELIN H W MIN 061 d'Nat 1/V a = 2 a=1.5 Increasing (1) o1 (no inhibitor) Slope = KNALLE
If inhibitor X changes lactase activity to a Vo of 0.111111111111 mM per minute when [S] = 0.40mM, and a Vo of 0.166666666667 mM per minute when [S] = 1.0 mM, calculate the new Vmax.A. 0.10 mM per minuteB. 0.25 mM per minuteC. 0.50 mM per minuteD. 1.0 mM per minuteE. 2.0 mM per minute
Calculate the new Km of the above lactase reaction, with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 mM 0.50 mM 1.0 mM 2.0 mM 5.0 mM
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?
Determine the Ki for the inhibitor at 30 °Cand decide what type of inhibitor is being used. Eo T I S V (g/I) (°C) (mmol/ml) (mmol/ml) (mmol/ml-min) 1.6 30 0.1 2.63 1.6 30 0.033 1.92 1.6 30 0.02 1.47 1.6 30 0.01 0.96 1.6 30 0.005 0.56 1.6 49.6 0.1 5.13 1.6 49.6 0.033 3.70 1.6 49.6 0.01 1.89 1.6 49.6 0.0067 1.43 1.6 49.6 0.005 1.11 0.92 30 0.1 1.64 0.92 30 0.02 0.90 0.92 30 0.01 0.58 0.92 30 0.6 0.1 1.33 0.92 30 0.6 0.033 0.80 0.92 30 0.6 0.02 0.57
PTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG 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 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minute
2. The two diagrams to the right il- lustrate plots of steady-state ki- netic studies to characterize the in- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is smaller than that for the natural 5 0.8- NH = 3.5 0.6- 0.4- 0.2- 10 μΜ 20 μΜ 48 µM substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 2 0.2- 0.4- concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8F 4 12 20 28 36 44 52 60 68 76 84 92 .2 2.0 2.4 ΑMP () , 10 μΜ AMP (+ ) 20 μΜ AMP (-), and 48 µM AMP (-). [AΤP] (μΜ) log[ATP] (µM) (a) . Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe how ATP interacts with the enzyme in the case of no AMP (•). (b) ! with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement…
6. The Vmax and KM values for an unusual hexokinase found in Try- pansoma cruzi (the causative agent of Chagas disease) are shown in the presence and absence of a bisphonate inhibitor (structure shown). Without inhibitor With inhibitor Км (mM) 90 125 Vn max (umol min-1. mL-1) 0.30 0.12 ОН O=P-O- A bisphonate compound a. 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 good candidate for a drug to treat the disease?
2. The two diagrams to the right il- lustrate plots of steady-state ki- 5FA 0.8 netic studies to characterize the in- nH = 3.5 0.6- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is 0.4- 0.2- smaller than that for the natural 10 μΜ 20 μΜ 48 μΜ substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 0.2 0.4 concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8 4 12 20 28 36 44 52 60 68 76 84 92 1.2 2.0 2.4 AMP (•), 10 µM AMP (•), 20 µM AMP (-), and 48 µM AMP (). [ΑΤPΙ (μM) log[ATP] (µM) (а) how ATP interacts with the enzyme in the case of no AMP (•). Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe (b) with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement of the…
9. For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in both reactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitive or uncompetitive. 110 100 1/Va 00 80 70 60 1/V 50- 40 30- 20- 10 F0008
1. The diagram below plots the free energy of glycolytic intermediates in mature red blood cells. +10 AG°' = - 130.9 kJ AGU -10 AGO 130 9 kJ (-31.3 kcal) or -20 - 31.3 kcal -30 -40 AG = AG 103.8 KJ (-24.6kcal) -50 AG 103.8 kJ -60 or - 24.6 kcal -70 GLU G6P F6P FBP GAP GBP PG3 PG2 PEP PYR LAC Note that Gor free energy in kcal is indicated along the vertical axis. The abbreviations in upper case letters along the horizontal axis at the bottom represent the various intermediates in the glycolytic conversion of glucose to pyruvate and to lactate. Therefore, the difference in any two points on the graph represents the difference in free energy between the two points. For instance, the AG'and the AG indicated with arrows on the left-hand end of the graph represent the two types of free energy changes associated with the conversion of glucose to glucose-6-phosphate catalyzed by hexokinase. bB S dD, write the equation (a) showing the relationship of AG and AG°' with respect to the mass-action…