Question

Please help me with these two problems. I will rate the answers.

1. The HIV-1 protease, an important enzyme in the life cycle of the human immunodeficiency virus-1 (HIV-1), is a good drug target for the treatment of HIV and AIDS. A protein produced by the virus, p6*, is an HIV-1 protease inhibitor. The activity of the HIV-1 protease was measured in the presence and absence of p6* using an assay involving an artificial substrate, as shown below. NH Lys Ala Nle 0 Ne-NH-CH-Otmucho CH Lys NO2 NH Ala Nle HO HIV-1 protease Arg Ala Val-Nle --NH-CH-C00 NH-Glu Absorbs light at 295 nm + CH2 NO, vo without p6* vo with p6* (S) (UM) (nmol. min) (nmol. min) 10 4.63 2.70 15 5.88 3.46 20 6.94 4.74 25 9.26 6.06 30 10.78 6.49 40 12.14 8.06 50 14.93 9.71 (a) Construct a Lineweaver-Burk plot and determine the Km and Vmax in the presence and in the absence of inhibitor (use Excel and including all the data manipulation) (6) What type of inhibitor is p6*? Explain.

Answer 1

The most important equation for determining the kinetic parameters of an enzyme: Km and Vmax is the Michaelis-Menten equation, which is mentioned as the following:

vo= Vmax. [S]/ Km +[S]

where, vo or is the initial velocity, Vmax is the maximum velocity, [S] is the substrate concentration, and Km is the Michaelis constant or the enzyme affinity for the substrate.

A graph can be plotted using the values of [S] and vo, through which Km and Vmax can be determined but approximately. To get the accurate values of Km and Vmax, the inverse of the Michaelis-Menten equation is determined which is called Lineweaver Burk equation, which is mentioned below:

1/vo= (Km/Vmax).1/[S]+ 1/Vmax

On comparing the equation with a straight line equation, we can get:

1/vo= (Km/Vmax).1/[S]+ 1/Vmax

y = m  . x + c

From the comparison, we can interpret that:

y=1/vo; m= Km/Vmax (m= slope of the straight line); x= 1/[S]; c=1/Vmax (c= constant)

Therefore, when we plot the Lineweaver Burk Plot using inverse values of [S] and vo, y-axis is taken as 1/vo, and x-axis is taken as 1/[S].

Also, y-intercept= 1/Vmax, and x-intercept= -1/Km

When the Michaelis-Menten graph is plotted, it is seen that the velocity of the reaction increases with the increases in the substrate concentration, and then after a limit becomes constant, even if the substrate concentration is increased. This maximum velocity attained is the Vmax of the enzyme. At this point, the enzyme is completely saturated with substrate. The substrate concentration at which the velocity is equal to half of Vmax is called the Km of the enzyme or Michaelis constant which determines the affinity of the enzyme towards the substrate. An enzyme with low Km has higher affinity towards the substrate and is considered an efficient enzyme. The type of inhibition can easily be determined by determining the values of Km and Vmax in presence and absence of inhibitor as the inhibitor affects these kinetic parameters and different types of inhibitions have different characteristic effect on Km and Vmax.

1. a. Table obtained from the data provided in the question:

[S] (uM)	vo without p6 (nM/min)	vo with p6 (nM/min)	1/[S] (1/uM)	1/vo (1/(nM/min))	1/vo with p6 (1/(nM/min))
10	4.63	2.7	0.1	0.215982721	0.37037037
15	5.88	3.46	0.066666667	0.170068027	0.289017341
20	6.94	4.74	0.05	0.144092219	0.210970464
25	9.26	6.06	0.04	0.107991361	0.165016502
30	10.78	6.49	0.033333333	0.092764378	0.154083205
40	12.14	8.06	0.025	0.082372323	0.124069479
50	14.93	9.71	0.02	0.066979236	0.102986612

0.5 0.4 1/Vmax app 1/vo (1/(nmol/min)) 0.3 0.2 1/vo (1/(nmol/min)) 0.1 -1/Km 1/Vmax 1/vo with p6 (1/(nmol/min)) O 0.05 0.1 0.15 -0.05 -0.1 -1/Km app -0.2 1/[S] (1/UM)

From the Lineweaver Burk plot, the following values have been calculated:

1/Km (1/uM)	1/Km app (1/uM)	1/Vmax (1/(nmol/min))	1/Vmax app (1/(nmol/min))	Km (uM)	Km app (uM)	Vmax (nmol/min)	Vmax app (nmol/min)
0.019	0.01	0.35	0.4	52.63158	100	2.857143	2.5

Km app means the Km of the enzyme in presence of the inhibitor.

Vmax app means the Vmax of the enzyme in presence of the inhibitor.

b. The values obtained shows that the Km of the enzyme in absence of p⁶ inhibitor was 52.63 uM which increased to 100 uM in presence of the inhibitor.

Also, Vmax of the enzyme in absence of p⁶ inhibitor was 2.86 nmol. min^-1, which decreased to 2.5 nmol. min^-1.

This change in Km and Vmax both, indicates that the inhibition is of mixed type. In mixed type of inhibition, the inhibitor binds to both, free enzyme as well as the enzyme-substrate complex, with different rate constants which eventually affects both the reaction rate velocity and the enzyme affinity for the substrate. Hence, both Vmax and Km are affected, Km increases with decrease in affinity of the enzyme towards the substrate while Vmax decreases as the inhibitor slows down the reaction rate.

2. a. Table obtained from the data provided in the question:

Phenyl phosphate [S](mM)	vo no inhibitor (nM/min)	vo with inhibitor (nM/min)	1/[S] (1/mM)	1/vo no inhibitor (1/(nM/min))	1/vo with inhibitor (1/(nM/min))
4	1.176	0.476	0.25	0.850340136	2.100840336
2	0.909	0.436	0.5	1.100110011	2.293577982
1	0.667	0.385	1	1.499250375	2.597402597
0.67	0.556	0.333	1.492537313	1.798561151	3.003003003
0.5	0.455	0.286	2	2.197802198	3.496503497
0.33	0.345	0.244	3.03030303	2.898550725	4.098360656

From the Lineweaver Burk plot, the following values have been calculated:

5 4 1/Vmax app 1/vo (1(/nM/min)) 3 2 -1/Km app -1/vo no inhibitor (1/(nM/min)) 1/vo with inhibitor (1/(nM/min)) 1 1 1/Vmax o -4 -2 ( 2 4 1 1/[S] (1/mM) -1/Km -2

1/Km (1/mM)	1/Km app (1/mM)	1/Vmax (1/(nM/min))	1/Vmax app (1/(nM/min))	Km (mM)	Km app (mM)	Vmax (nM/min)	Vmax app (nM/min)
1	2.6	0.7	1.9	1	0.384615	1.428571	0.526316

Km app means the Km of the enzyme in presence of the inhibitor.

Vmax app means the Vmax of the enzyme in presence of the inhibitor.

b. The values obtained shows that the Km of the enzyme in absence of homoarginine inhibitor was 1mM which decreased to 0.38 mM in presence of the inhibitor.

Also, Vmax of the enzyme in absence of homoarginine inhibitor was 1.43 nM. min^-1, which decreased to 0.53 nM. min^-1.

This change in Km and Vmax both, indicates that the inhibition is of uncompetitive type. In uncompetitive type of inhibition, the inhibitor binds to the enzyme-substrate complex, and affects both the reaction rate velocity and the enzyme affinity for the substrate. It increases the binding of the substrate to the enzyme, thus increasing the affinity of the enzyme towards the substrate. Hence, Km decreases. But, do to the enzyme-substrate -inhibitor complex formation, the reaction rate or the velocity decreases, due to which the Vmax decreases.