Which of the following rate-enhancing features does lysozyme use to increase the rate of the hydrolysis reaction it catalyzes?

Arrange the events for the hydrolysis of amide bonds by chymotrypsin in their correct order.

1. attack by Ser 196 to give a tetrahedral intermediate
2. enzyme returns to initial state
3. protonation by His 57 and then cleavage of the C-N bond release the C-terminal fragment
4. protonation by His 57 and then cleavage of the C-O bond release the N-terminal fragment
5. attack by water, leading to formation of a tetrahedral intermediate
6. binding of substrate to properly position the scissile bond for cleavage

a. 6,2,4, 3, 1, 5
b. 6, 1, 3, 5,4,2
c. 6, 1,3,5,2,4
d. 6, 1, 3,4, 5, 2

Part B:

Which of the following rate-enhancing features does lysozyme use to increase the rate of the hydrolysis reaction it catalyzes?
1. general acid-base catalysis
2. electrostatic catalysis
3. distortion of substrate
4. covalent catalysis

a. Only choices I, 2, and 4 are correct-
b. Only choices 1 and 2 are correct.
c. Only choices 1, 3, and 4 are correct.
d. Al of the listed choices are correct.

b).6,1,3,5,4,2

d. All of the listed choices are correct.

Explanation:

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

58750.08068 is the answer i think