19 FumarateSuccinateMalateCitrate<0.110.0 ± 1.0<0.15.5 1.57.1 ± 0.50.5 ± 0.24.2 ± 0.4150 ± 25Fumarase; because the citrate cycle is blocked.Fumarase; because pyruvate is used to make oxaloacetate.Lactate dehydrogenase; because the defect is in this enzyme.Succinate dehydrogenase; because pyruvate is converted to lactate.Malate dehydrogenase; because the citrate cycle is blocked. Defects in the Citrate Cycle are rare but have been described. Based on thelevel of metabolites shown in a patient's blood in the table below, whichhuman metabolic enzyme is most likely defective; and why is the lactate levso high in this patient?CompoundLactateFumarateSuccinateMalateCitratePatient (mmol/L)25.5 ± 1.2<0.110.01.0<0.15.5 ± 1.5Normal control(mmol/L)1.5 ± 0.37.1 ± 0.50.5 ± 0.24.2 ± 0.4150 ± 25Fumarase: because the trato cycle is blocked

Glycolysis consists of 10 enzymatically catalysed reactions that convert one 6-carbon molecule of…

Answered: Defects in the Citrate Cycle are rare…

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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Because of the position of arsenic in the periodic table, arsenate (AsO43- ) is chemically similar to inorganic phosphate and is used by phosphaterequiring enzymes as an alternative substrate. Organic arsenates are quite unstable, however, and spontaneously hydrolyze. Arsenate is known to inhibit ATP production in glycolysis. Identify the target enzyme, and explain the mechanism of inhibition.
A new drug, Proinebrium, that reduces Kcat (Ki = 2.0 uM) has been developed to treat ethylene glycol poisoning. (1) What concentration of Proinebrium is required to achieve 50% inhibition of ethylene glycol metabolism by alcohol dehydrogenase when the concentraion of ethlyene glycol in the blood is 50 uM?
What concentration of Proinebrium is required to achieve 50% inhibition of ethylene glycol metabolism by alcohol dehydrogenase when the concentraion of ethlyene glycol in the blood is 50 uM?
Table 1 shows the kinetic data that have been obtained for glucoamylase from Aspergillus niger at different temperatures in the production of glucose from maltodextrin. The rates of reaction measured are using an enzymes concentration of 0.003 μmolml-1. Outline and describe the reaction scheme of the production of glucose from maltodextrin by using glucoamylase.
For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.142
Draw the products of the reaction of xylulose-5-phosphate and erythrose-4-phosphate catalyzed by transketolase in the pentose phosphate pathway. Provide the structure in the protonation state found in physiological conditions. H H H OH FO HO-H H-OH H OPO3²- Q transketolase Draw glyceraldehyde-3- phosphate H H- H H H O OH OH OPO3²- Draw fructose-6- phosphate Q I I
Years ago there was interest in using uncouplers such as dinitrophenol asweight control agents. Presumably, fat could be oxidized without concomitant ATP synthesis for re-formation of fat or carbohydrate. Why was this a bad (i.e., fatal) idea?
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The glucose/glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconeogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP]; [P;] = 1 mM. Consider the glycolytic reaction catalyzed by hexokinase: ATP + glucose ADP + glucose-6-phosphate AG = - 16.7 kJ/mol (a) Calculate the equilibrium constant (K) for this reaction at 298°K, and from that, calculate the maximum [glucose-6-phosphate]/ Iglucose] ratio that would exist under conditions where the reaction is still thermodynamically favorable. (b) Reversal of this interconversion in gluconeogenesis is catalyzed by glucose-6-phosphatase: glucose-6-phosphate + H20 = glucose + P AG" = -13.8 kJ/mol
The standard free energy change for this reaction in the direction written is +23.8 kuimol. The tabie shows the concentrations of the three intermediates in the hepatocyte of a mammal. Intermediate Concentration (M) Fructose 1.0-bisphosphate 0.000028 Gyoeraldehyde 3phosphate 0.0000068 Ditydroxyacetone phosphate 0.000032 At body temperature (37 "C). what is the actual free energy change for the reaction (in kimol) ?
You make reaction progress curve by plotting absorbance vs time (seconds) and find the equation of the line to be y = -0.00235x + 0.7129. Calculate the U/µL and U/mL of lactate dehydrogenase activity in this fraction. The LDH activity is done identical to what is indicated in the lab manual. Show each step of the calculation from AU/time to M/min, to mol/min, to µmol/min to µmol/min/µL (=U/µL).

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