You two have identified mutants of an enzyme that catalyzes

You two have identified mutants of an enzyme that catalyzes a commercially useful reaction. Both mutants have the same Vmax as the original or wild type (WT) enzyme. However one mutant has a lower Km for substrate than the WT, the other has a higher Km for substrate. Which mutant is likely to be a \"better\" (i.e., faster) enzyme for catalyzing the reaction: a. Neither. Since they have the same Vmax they will have the same reaction velocities at all substrate concentrations b. The lower Km enzyme since it will show greater reaction velocities especially at low concentrations of substrate c. The higher Km enzyme since it will show greater reaction velocities especially at higher concentrations of substrate.

Solution

Ans. Given,

            Mutant 1 = V_max same,            Lower K_m

            Mutant 2 = V_max same,            Higher K_m

K_m is the substrate concentration at which half- V_max is achieved.

An enzyme with lower K_m can, thus, attain V_max at lower [S], whereas the enzyme with higher K_m requires greater [S] to attain same V_max. For example, (a theoretical scenario), if enzyme E1 (from mutant 1, has lower K_m) requires 0.1 M [S] to reach V_max; E2 (from mutant 2, has higher K_m) will require 1.0 M [S] to attain same V_max.

Both enzymes work with same efficiency at [S] >> K_m , i.e. in excess of substrate. However, as soon as [S] falls below 1.0 M, E2 gets lowered it velocity. Thus, plenty of substrate molecules are still present in the medium before E2’s velocity is lowered, or even no significant reaction occurs at [S] below 1.0 M. So, the whole media (say, thousands of liters) with [S] below 1.0 M (say, at 0.5 M) goes waste. It’s commercially disadvantageous as resources are being wasted.

However, E1 works well at [S] as low as 0.1M. SO, if E1 is added to the spent media (wasted using E2), it converts the remaining substrate into product with velocity V_max till [S] falls below 0.1 M. So, use of E1 minimizes wastage of substrate.

Therefore, using E1 reduces wastage of media (thousands liters or so on) because it can catalyze substrate at relatively low [S] that would be left un-catalyzed by E@ having higher K_m.

So, use of E1 is preferred for commercial purposes.      

Correct option: B.