We define a quantity called the turnover number to be the maximum number of subs

Question

We define a quantity called the turnover number to be the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by

turnover number = R_max/[E]_t = k_2

If the enzyme molecule has more than one active site, then we multiply [E]t by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that Rmax for carbonic anhydrase = 249 ?mol

Explanation / Answer

turnover number = k2 = Rmax/[E]t

Rmax =249 micro mol*L^-1*S^-1 = 249 *10^-6 mol*L^-1*S^-1    [ 1 micro = 10^-6]

[E]t= 2.34n mol*L^-1. = 2.34 *10^-9 mol*L^-1 [ 1 nano = 10^-9]

turn over = 249 *10^-6 mol*L^-1*S^-1 / 2.34 *10^-9 mol*L^-1

= 106.41026*10^3 s^-1= 106410.26 s^-1 (answer)

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