Question

One way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. Enzyme Substrate Turnover number (per second) Ribonuclease RNA 100 Fumarase fumarate 800 Lactate dehydrogenase lactate Urease urea 10,000 1000 How many molecules of fumarate can one molecule of fumarase act on in 28.7 min? fumarate molecules: 1008000

help onthese 2

Answer 1

turnover number of fumerase = 800

It means in one second one fumerase molecule can act on 800 fumerate molecules.

Then in 28.7 min. = 28.7*60 = 1722 seconds, one molecule of fumerase can act on 800 *1722 = 1377600 fumerate molecules.

thus fumerate molecules = 1377600

given [X] = 0.10M

rate =0.0040M/s

unit of k = M^(-1) s^(-1)

it means reaction is of second order.

then rate = k [X]^(2)

put values

0.0040 = k (0.1)^2

k = 0.40M^(-1) s^(-1)