From your Lineweaver-Burk plot, derive the values for the following table. Indicate the appropriate units. (b) By describing the potential changes in the kinetic parameters, identify and justify the type of inhibitor that was inhibiting the acid phosphatase activity.   Km Vmax Uninhibited     Inhibited

Intramitochondrial ATP concentrations are about 5 mM, and phosphate 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 condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]

The hexapeptide Ala-Met-Leu-Lys-Phe-Asp is digested in the same tube with both Cyanogen Bromide and Trypsin at the same time at pH=7. Draw the structure of the product(s) that would bind to a cation exchange column. (Relevant pKa values are 2.2, 3.9, 9.5 and 10.5. Assume pKa values for any newly generated a-amino and a-carboxyl groups are 9.5 and 2.2 respectively)

Calculate the actual, physiological AG for the reaction Phosphocreatine + ADP = creatine + ATP at 37 °C, as it occurs in the cytosol of neurons, where [phosphocreatine] = 4.7 mM, [creatine] = 1.0 mM, [ADP] = 0.73 mM, and [ATP] = 2.6 mM. Standard Free Energies of Hydrolysis of Some Phosphorylated Compounds Phosphorylated compound AGʻ° (kJ/mol) phosphoenolpyruvate -61.9 phosphocreatine -43.0 ADP (→ AMP + P;) -32.8 ATP (→ ADP + P;) -30.5

The ΔG°′ for hydrolytically removing a phosphoryl group from ATP is about twice as large as the ΔG°′ for hydrolytically removing a phosphoryl group from AMP (−14 kJ · mol−1). Explain the discrepancy.

Human blood serum contains a class of enzymes known as acid phosphatases, which hydrolyze biological phosphate esters under slightly acidic conditions (pH 5.0): R-O-P-O3-2 + H2O --> R-OH + HO-P-O3-2. Acid phosphatases are produced by erythrocytes, the liver, kidney, spleen, and prostate gland. The enzyme from the prostate gland is clinically important because an increased activity in the blood is frequently an indication of cancer of the prostate gland. The phosphatase from the prostate gland is strongly inhibited by tartrate ion, but acid phosphatases from other tissues are not. How can this information be used to develop a commercial specific procedure for measuring the activity of the acid phosphatase of the prostate gland in human blood serum? *         1. Prostate cancer cannot be diagnosed biochemically.       2. Use tartrate to inhibit phosphatase from prostate gland and then subtract the results from the total serum enzyme activities to get an…

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 49.8

Intracellular concentrations in resting muscle are as follows: fructose- 6-phosphate, 1.0 mM; fructose-1,6-bisphosphate, 10 mM; AMP, 0.1 mM; ADP, 0.5 mM; ATP, 5 mM; and P, 10 mM. Is the phosphofructokinase reac- tion in muscle more or less exergonic than under standard conditions? By how much?

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]

The isomerization of dihydroxyacetone phosphate (DHAP) to glyceraldehyde 3-phosphate (GAP) is catalyzed by triose phosphate isomerase. In the cell, the concentration ratio of DHAP/GAP = 5.5. Calculate [DHAP] (in M) when [GAP] = 0.00002