In humans, the biosynthesis of L-glutamine is catalyzed by the L-glutamine synthetase. The global reaction can be written as follows: ATP + glutamate + NH3 ADP + phosphate + glutamine The A,Gº' value for this reaction is -16.3 kJ.mol¯¹ (from left to right). 1- Explain the significance this information. 2- This reaction can be considered as the sum of two component reactions which are exergonic and endergonic, respectively, under standard conditions. Write these two component reactions and evaluate A,Gº of the endergonic reaction. Data: A,Gº of ATP hydrolysis = -30.4 kJ mol¯¹
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- The inilial reactions in the biosynthesis of the amino acid aspartate at 298 K are: Carbamoyphoephate Cartamoyi + phosphate ADP + phoaphate - ATP AO=-12,300 calmol AG-T800 calimol Which of the following statements is gorrect? A) The energy which is released fron the conversion of 1 mole of carbamoyphosphate to carbamoyl+ phosphate (P) in reaction I is sufficient to drive the synthesis of 1 mole of ATP in reaction II. B) AG for the reaction Carbamoyphosphate + Carbamoyl + phosphate is +123 kcalimol. c) AHP for reaction I cannot be detemined trom the information given. D) Al of the above. E) None of the above.The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1In the hydrolysis of ATP to ADP and Pi, the equilibrium concentration of ATP is too small to be measured accurately. A better way of determining K’eq, and hence ΔG◦’ of this reaction, is to break it up into two steps whose values of ΔG◦’ can be accurately determined. This has been done using the following pair of reactions (the first being catalyzed by glutamine synthetase): (1) ATP + glutamate + NH3 ⇌ ADP + Pi + glutamine + H+ ΔG1◦’= -16.3 kJ/mol (2) glutamate + NH3 ⇌ glutamine + H2O + H+ ΔG2◦’= 14.2 kJ/mol What is the ΔG◦’ of ATP hydrolysis according to these data and is the overall reaction spontaneous? What is the value of the equilibrium constant for the overall reaction at 25.0 °C. If the concentration of ATP at equilibrium is 20.0 mM and the concentration of ADP at equilibrium is 50 nM. What is the concentration the phosphate group (in mM) at equilibrium?
- 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) ?5) In an experiment to investigate the inhibition of the enzyme-glucosidase the following data for the rates of reaction with glucopyranoside for various substrate concentrations was obtained. By constructing a Leaver-Burk plot, determine the value of the Michaelis constant. [S]/ (10-6 mol dm-3) v/ (10-3 mol dm-3 s-1) 1.00 2.00 3.00 4.00 16.7 33.3 41.1 49.8The ATP hydrolysis reaction (shown below) has a AG of -31 kJ mol-1. NH2 NH2 Do you predict that AS is positive or negative for this reaction. Explain in one sentence. а. .N H20 -OCH2 OCH2 Energy Phosphate ion ÓH ÓH Adenosine triphosphate (ATP) Adenosine diphosphate (ADP) b. How does the hydrophobic effect contribute to the AS you predicted in part a. Make sure to discuss the ordering of water molecules in your answer. Include a minimum of 3 sentances in your explaination.
- During glycogen synthesis, glucose-1P is converted into a molecule called UDPG. This reaction also cleaves uridine triphosphate (UTP) forming uridine monophosphate and pyrophosphate (PPi). Provide four reasons why UTP can be used to power this reaction (no diagrams necessary).The degradation of glycogen is catalyzed by the enzyme phosphorylase and has AGO" equal to +3.1 kJ · mol1. The equation for this reaction is shown below. glycogen (n residues) + P;→ glycogen (n-1 residues) + G1P What is the ratio of [P;]/[G1P] under standard conditions? Use 2 significant figures. [P;] : [G1P] = i :1 What is the value of AG under cellular conditions when the [P;/[G1P] ratio is 50/1? Use 2 significant figures. AG = i kJ. mol-1For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1
- The KMof the enzyme for the substrate adenosine is 3 × 10ꟷ5M. The product inosine acts as an inhibitor of the reaction, with an inhibition constant (KI, the dissociation constant for enzyme-inhibitor binding) of 3 × 10ꟷ4M. However, a transition state analog,Inhibits the reaction with KIof 1.5 × 10ꟷ13M. Explain why 1,6-dihydroinosine serves as a better inhibitor of adenosine deaminase than inosine. Elaborate on your answeA particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.Triosephosphate isomerase (TIM) catalyzes the conversion of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate. The enzyme’s catalytic groups are Glu 165 and His 95. In the first step of the reaction, these catalytic groups function as a general-base and a general-acid catalyst, respectively. Propose a mechanism for the reaction.