Question

Required calculation to be answered, with all of your work displayed in WORD: For proteins to be generated by the ribosome, tRNAs charged with their respective amino acids must continually be produced by aminoacyl RNA synthetases. a. Generate general reaction schemes (i.e. reactants → products) for: 1. The loading of tRNA with an amino acid by the aminoacyl tRNA synthetase II. Peptide bond formation by the ribosome b. The overall process of charging a tRNA with an amino acid (catalyzed by aminoacyl tRNA synthetases) is estimated to have a AG near zero. Let's consider the production of aminoacylated-tRNABY (assume the overall process has a AGⓇ-1 kJ/mol). When the process is coupled with the reaction of inorganic pyrophosphatase (hydrolysis of PP.), the overall process resulting in charged tRNABY for peptide synthesis can be driven forward (overall AG° <0 kJ/mol). Assuming the standard free energy change of PP, hydrolysis is -33.5 kJ/mol, calculate the ratio of free RNAsly to aminoacylated-tRNADY needed to drive the aminoacyl RNA synthetase reaction forward with and without coupling to the inorganic pyrophosphatase reaction. Assume approximately physiological concentrations given in the list below for your calculation. R*T=2.478 kJ/mol at 25°C. Molecule Concentration (mm) 15.7 Pi in erythrocytes PPi in rat liver -0.022 0.2 AMP АТР 3 Glycine Reminder: You must comert concentrations to units of molarity (M) In your Literature Review, use this example to help tie the importance of human inorganic pyrophosphatase to protein production by the ribosome.

Answer 1

The reaction catalyzed by aminoacyl tRNA synthetase is:

Free tRNA + Glycine + ATP $\rightleftharpoons$ Aminoacylated tRNA^​​​gly + AMP + PPi, $\Delta$ G⁰ = 1 kJ/mol

Equilibrium constant for this reaction, K_{​​​​​​eq} = [Aminoacylated tRNA] [AMP] [PPi] / [Free tRNA] [ATP] [Glycine]

The reaction catalyzed by inorganic pyrophosphatase is:

PPi$\rightleftharpoons$ 2Pi, $\Delta$ G⁰ = -33.5 kJ/mol

After coupling these two reactions, the reaction becomes:

Free tRNA + ATP + [Glycine] $\rightleftharpoons$ Aminoacylated tRNA^​​​gly + AMP + 2Pi

The standard free energy change becomes $\Delta$ G⁰ = 1 - 33.5 = -32.5 kJ/mol

After coupling, equilibrium constant K_{​​​​​​eq} = [Aminoacylated tRNA^​​​gly ] [AMP] [Pi]² / [Free tRNA] [ATP] [Glycine]

Concentrations after conversion: [Pi] = 15.7 mM = 15.7/1000 M = 0.0157M

[PPi] = 0.022 mM = 0.000022M

[AMP] = 0.2mM = 0.0002M

[ATP] = 3mM = 0.003M

[Glycine] = 1.5mM = 0.0015M

Now $\Delta$ G⁰ = -RT lnK_{​​​​eq}​​

RT = 2.478 kJ/mol

Without coupling, after putting the values in the formula,

1 = -2.478 x ln K_{​​​​​​eq} , or K_{​​​​​​eq} = e^{​​​​​​-(1/2.478)} = 0.668

Or, [Aminoacylated tRNA^​​​gly ] [AMP] [PPi] / [Free tRNA] [ATP] [Glycine] = 0.668

Or, [Aminoacylated tRNAgly] x 0.0002 x 0.000022 / [Free tRNA] 0.003 x 0.0015 = 0.0068

Or, [Free tRNA] / [Aminoacylated tRNAgly] = 0.0002 x 0.000022 / 0.0068 x 0.003 x 0.0015 = 0.143

So for uncoupled reaction, the required ratio of free tRNA to Aminoacylated tRNAgly is 0.143

Similarly, for coupled reaction, putting the values we get,

-32.5 = -2.478 x ln K^{​​​​​​}_{​​​​​​eq} , or K_​​​eq = e^{(-32.5/-2.478)} = 4.9 x 10⁵

Or, [Aminoacylated tRNAgly] [AMP] [Pi]² / [Free tRNA] [ATP] [Glycine] = 4.9 x 10⁵

Or, [Free tRNA] / [Aminoacylated tRNAgly] = 0.0002 x 0.0157² / 4.9 x 10⁵ x 0.003 x 0.0015 = 2.235 x 10⁸

For the coupled reaction, the required ratio of free tRNA and Aminoacylated tRNAgly is 2.235 x 10⁸