fructose-6-phosphate + ATP <-> fructose-1,6- biphosphate + ADP AG = 30.5 and 16.3 respectively Standard Free Energy change: -14.2 kj/mol ATP: 3mM, ADP: 0.4mM, 7mM fructose-1,6-biphosphate: At what concentration of fructose-1,6-biphosphate would the reaction proceed backwards at 35°C?

Intramitochondrial ATP concentrations are about 5 mM, and phos- phate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these conditions. The energy charge is the concentra- tion of ATP plus half the concentration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]

1. a. Calculate the physiological DG of the reaction shown below at 37°C, as it occurs in the cytosol ofneurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. The standard free energy change for the overall reaction is –12.5 kJ/mol. Phosphocreatine + ADP ® creatine + ATP b. The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose6-phosphate. Calculate the standard free energy change of this reaction if incubation of 20 mMglucose 1-phosphate (no glucose-6 phosphate initially present) yields a final equilibrium mixtureof 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate at 25°C and pH 7.0.   c. If the rate of a nonenzymatic reaction is 1.2 x 10–2 μM s–1, what is the rate of the reaction at 37℃ inthe presence of an enzyme that reduces the activation energy by 30.5 kJ/mol?

Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex?   - Mixed inhibition, Ki= -21.2 uM - Mixed inhibition, Ki= 1/16.5 uM - Competitive inhibition, Ki= 16.5 uM - Competitive inhibition , Ki= -1/16.5 uM - None of the above

If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24

Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places

When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

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During glycolysis, glucose is converted into fructose-6- phosphate in two successive reactions: glucose + ATP glucose 6-phosphate + ADP glucose 6-phosphate fructose 6-phosphate The AG' for the overall reaction is: +15.0 kJ/mol. -18.4 kJ/mol. -16.7 kJ/mol. → -15.0 kJ/mol. +18.4 kJ/mol. AG¹⁰ = -16.7 kJ/mol AG¹⁰ = +1.7 kJ/mol

The free energy change of each step of glycolysis is given in the table below. ∆G°’ is the free energy under standard conditions (25°C, 1M each reactant, pH 7), while ∆G is the free energy change at presumed physiological conditions. Why must no step have a positive ∆G under physiological conditions?