Question

1. The following kinetic data was generated for the Hst enzyme. The overall reaction for this enzyme is: OH ОН + NH3 NH2 All reactions are carried out in a final reaction volume of 2 mL with 0.1ug of purified Hst protein present at pH 7.8. The Hst protein has a molecular weight of 90 kDa based on estimates from gel filtration column chromatography and a molecular weight of 45 kDa based on estimates from denaturing SDS PAGE gel electrophoresis. Here is the kinetics data generated with two graphs of this data: (Histidine] Velocity UM (MM/min) 0 10 30 90 250 600 0.215 0.494 0.869 1.15 1.28 y - 39.373x +0.7133

Answer 1

1. The enzyme Hst will come under the Lyases class of enzyme classification because it is involved in breaking of a bond, releasing ammonia and forming a double bond between the two carbons (one of them from which the C-N bond was broken to release ammonia).

1b. V_max is the maximum velocity or the rate of reaction that an enzyme can achieve. It is the rate at which the enzyme is completely saturated with substrate and further addition of substrate or increase in substrate concentration will not increase the rate of the reaction. K_m is the indicator of the enzymes's substrate affinity. It is the concentration of the substrate at which the enzyme will be at half of the maximum rate of reaction (Vmax). Higher the K_m, lower the affinity towards the substrate and lower the Km, higher the affinity. Enzymes with lower Km are considered efficient.

From the table given in the question, a Lineweaver Burk Plot was created and the following values were obtained.

1/Vmax	Vmax (μM/min)	1/Km	Km (μM)
0.7	1.428571	0.019	52.63158

5 ] Lineweaver Burk Plot 1/. M/min 1 1/Km 1 1/Vmax -0.06 -0.04 -0.02 0.02 0.04 0.06 0.08 0.1 0.12 1/[S]uM

1c. K_cat is the turn over number of an enzyme which is defined as the number of substrate molecules that a particular enzyme can convert into the product in a given period of time, preferably, 1 second. Enzymes with higher K_cat are more efficient.

K_cat = V_max (μM/min) / Total Enzyme concentration [E]_T (μM)

Since the [E]_T=0.1 μg we need to convert it into μM. The molar mass of the protein is not known but its molecular weight is 90 KDa (under reducing conditions 45 KDa which implies it has two subunits).

Upon concersion of 0.1 μg into moles we get [E]_T= 0.111 x 10^-5 μM or 1.11 pM

K_cat = 1.428571/ 0.111 x 10^-5= 12.87 x 10⁵ /min

Catalytic efficiency is a ratio of K_cat and K_m.

Catalytic efficiency= K_cat/ K_m= 12.87 x 10⁵ min^-/  52.63158 μM =0.244 x 10⁵= 244 x 10² min^- /μM

1d. For 0.2 μg, the molar mass will be 0.222 x 10^-5 μM of the enzyme [E]_T

K_cat is also equivalent to the rate constant k₂ of a simple enzyme catalyzed reaction

S+E forms ES complex which gives P+E. the formation of P from ES is governed by the rate constant k₂.

Hence, it can be considered as V_max= k₂.[E]_T (from Michaelis Menten Equation)

v_o= V_max. [S]/ K_m + [S]

v_o= k₂.[E]_T. [S]/ K_m + [S]

[E]_T=  0.222 x 10^-5 μM

k₂= 12.87 x 10⁵ /min

K_m = 52.63158 μM

[S]= 2.5 μM

v_o= (12.87 x 10⁵ x   0.222 x 10^-5 x  2.5)/ (52.631 + 2.5)= 7.14/ 55.131= 0.1295= 0.13 μM /min

v_o=0.13 μM /min