You then treat enzyme a with a phosphatase. you measure km in the presence of fructose 2,6 bis phosphate, and find that km = 15 um and without fructose 2,6 bis phosphate km = 16 um. what can you conclude from this treatment? what is the effect on glycolysis?

Competitive inhibition of enzymes is accompanied by an increase in the km value. this can be understood qualitatively as well as derived in a manner similar to michealis-menten kinetics :
km value for an enzyme is roughly used as a measure of the affinity between the enzyme molecule and its substrate and is obtained by plotting reaction rates at different substrate concentrations and then determining the substrate concentration at which the reaction rate reaches half its maximum (saturation) value (1/2 vmax). a higher affinity is indicated by a lower km value, as the enzyme would require lower substrate concentration to reach 1/2 vmax.
competitive inhibitors are usually structurally analogous to the substrate and work by occupying the active site of the enzyme, reducing the sites available for binding of the substrate. as a result, a greater concentration of substrate is required to reach the same previous reaction rate of 1/2 vmax, explaining why there is a reduction in the value of km.
a treatment of competitive inhibition in a manner similar to michealis-menten kinetics leads us to the following equation :
max s vo = akms where п а 3 1+ кі
here vmax is the rate of the reaction at saturation, [s] is the substrate concentration, [i] is the inhibitor concentration, k1 is the equilibrium constant for the enzyme-inhibitor reaction.
by definition, \alpha is greater than 1, in presence of inhibitor, thus having increasing the effective value of km.
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