Question #44959

Consider the enzymatic hydrolysis of cellobiose ($\ce{C12H22O11}$) by the enzyme cellulase. The reaction proceeds rapidly at $298\text{ K}$ and $\text{pH} = 5.0$, but ceases completely at $360\text{ K}$. Evaluate the following six statements regarding the substrate, the enzyme, and the reaction thermodynamics:

I. Cellobiose contains a $\beta(1 \rightarrow 4)$ glycosidic linkage, which is a hemiacetal linkage that undergoes rapid mutarotation at $360\text{ K}$, preventing enzyme recognition. II. The cessation of hydrolysis at $360\text{ K}$ is primarily due to the endothermic disruption of non-covalent interactions (e.g., hydrogen bonding, hydrophobic effects) within cellulase, altering its tertiary structure. III. The 'Lock and Key' specificity is lost at $360\text{ K}$ because the thermal energy exceeds the activation energy for the cleavage of the enzyme's primary peptide bonds. IV. The hydrolysis of cellobiose into two molecules of D-glucose is thermodynamically spontaneous ($\Delta G < 0$) at both $298\text{ K}$ and $360\text{ K}$. V. In the most stable chair conformation of the major product formed upon hydrolysis (in aqueous equilibrium), all hydroxyl groups and the hydroxymethyl group occupy equatorial positions. VI. Cellulase acts by lowering the activation energy of the hydrolysis, which simultaneously shifts the equilibrium constant ($K_{eq}$) of the reaction towards the formation of D-glucose at $298\text{ K}$.

How many of the above statements are CORRECT?