The reaction Glc + ATP rightarrow Glc-6-PO_4 + ADP can be catalyzed by two isozy

Question

The reaction Glc + ATP rightarrow Glc-6-PO_4 + ADP can be catalyzed by two isozymes: Hexokinase (H) and Glucokinase (G). H has a KM of 3 mu M for Glc, while G has a KM of 370 mu M for Glc. After successfully purifying H and G (separately), you mistakenly forget to label one of your samples of pure enzyme. You know it must be either H or G, but you're not sure which one it is. How can you experimentally determine which enzyme (H or G) is in your sample? Explain how your experiment will allow you to identify one enzyme over the other.

Explanation / Answer

A small Km indicates that the enzyme requires only a small amount of substrate to become saturated. Hence, the maximum velocity is reached at relatively low substrate concentrations. A large Km indicates the need for high substrate concentrations to achieve maximum reaction velocity.

Here hexokinase has low Km than glucokinase.This suggests that hexokinase will work at low concentrations of glucose in comparison to glucokinase.Thus in an experiment if an enzyme gets fully saturated by adding small amounts of substrate i.e glucose (by saturation it means that at fairly low levels of substrate concentration hexokinase will convert all the glucose into its product i.e glucose 6 phosphate).

In other experiment it will be clear that relatively high concentration of glucose will be required for its conversion to the product i.e glucose 6 phosphate.

Thus if a concentration of glucose required is low and in that low conc. it is able to change into its product, the enzyme would be hexokinase and if the glucose conc. is high then the enzyme present would be glucokinase.

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