The turnover numbers for the enzymes, E1 and E2 are 150

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The turnover numbers for the enzymes, E1 and E2 are 150 s^–1 and 15 s^–1 respectively. This means

1. E1 binds to its substrate with higher affinity than E2
2. The velocity of reactions catalyzed by E1 and E2 at their respective saturating substrate concentrations could be equal, if concentration of E2 used is 10 times that of E1
3. The velocity of E1 catalyzed reaction is always greater than that of E2
4. The velocity of E1 catalyzed reaction at a particular enzyme concentration and saturating substrate concentration is lower than that of E2 catalyzed reaction under the same conditions

Correct Option: B

Turnover number is expressed in molecules of product/active site/second. It can be defined as the number of molecules of a substrate acted upon in a period of 1 minute by a single enzyme molecule; with the enzyme working at a maximum rate.here the statement shows the dP/dt of the two enzymes at unit time. Again dP/dt is the velocity of the reaction i.e. the E1 has greater velocity than of E2.