Based on your reading of this chapter, what would you expect to be the most immediate effect on glycolysis if the steady-state concentration of glucose- 6 -P were \(8.3 \mathrm{m}\) M instead of \(0.083 \mathrm{mM}\) ?

Short Answer

Expert verified
An immediate effect on glycolysis of an increase in steady-state concentration of glucose-6-P from 0.083 mM to 8.3 mM would generally be an increased rate of glycolysis due to increased substrate concentration. However, considering cellular complexity, potential feedback inhibition might occur, slowing down glycolysis.

Step by step solution

01

Understand the Role of Glucose-6-P in Glycolysis

The first step in this exercise involves understanding what the glucose-6-P does in glycolysis. Glucose-6-phosphate (glucose-6-P) is an intermediate in the glycolytic pathway and serves as a substrate for the enzyme phosphoglucose isomerase, leading to the formation of fructose-6-phosphate.
02

Consider the Effect of Increased Substrate Concentration on Enzyme Activity

Next, consider that, as a general rule, an increase in substrate concentration (up to a point) leads to an increase in the rate of an enzyme-catalyzed reaction. This is due to the higher probability of a collision between the enzyme and substrate molecules.
03

Apply Understanding to Glucose-6-P and Glycolysis

Given this understanding, the increase in glucose-6-P concentration from 0.083 mM to 8.3 mM would increase the rate of the reaction converting glucose-6-P to fructose-6-phosphate, due to increased substrate availability. This, in turn, would be expected to increase the overall rate of glycolysis, assuming no other limiting factors.
04

Other considerations

However, keep in mind that this is a simplification and in a biological system other factors such as the availability of other substrates and products, feedback inhibition, and cellular energy status can influence reaction rates. In this case, particularly relevant could be the phenomenon of feedback inhibition where accumulation of glucose-6-P could inhibit hexokinase, the enzyme responsible for the formation of glucose-6-P from glucose, slowing down glycolysis.

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Most popular questions from this chapter

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(Integrates with Chapter \(3 .)\) The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for hydrolysis of fructose- 1,6 -bisphosphate (FBP) to fructose6-phosphate (F-6-P) and \(\mathrm{P}_{\mathrm{i}}\) is \(-16.7 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{FBP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { fructose- } 6-\mathrm{P}+\mathrm{P}_{\mathrm{i}} \\] The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for ATP hydrolysis is \(-30.5 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \\] a. What is the standard free energy change for the phosphofructokinase reaction: \\[ \text { ATP + fructose- } 6 \text { -P } \longrightarrow \mathrm{ADP}+\mathrm{FBP} \\] b. What is the equilibrium constant for this reaction? c. Assuming the intracellular concentrations of [ATP] and [ADP] are maintained constant at \(4 \mathrm{m}\) Mand \(1.6 \mathrm{m} M\), respectively, in a rat liver cell, what will be the ratio of [FBP]/[fructose-6-P] when the phosphofructokinase reaction reaches equilibrium?

If \(^{32}\) P-labeled inorganic phosphate were introduced to erythrocytes undergoing glycolysis, would you expect to detect \(^{32} \mathrm{P}\) in glycolytic intermediates? If so, describe the relevant reactions and the \(^{32} \mathrm{P}\) incorporation you would observe.

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