and equal HE EFFECT OF INHIBITORS ON ENZYME KINETICS 6. Refer to the initial sho

Question

and equal HE EFFECT OF INHIBITORS ON ENZYME KINETICS 6. Refer to the initial shown above. Imagine that the experimen following conditions, fill in the table indicating the effect on Vmax a velocity vs. substrate concentration graph for a typical Michaelis Menten enzyme enzyme. Be as specific as possible. (0.4 points: 0.05 points per tal conditions are altered as described below. For each of the nd Km on the Michaelis-Menten Experimental Condition Twice as much enzyme is used Half as much enzyme is used A competitive inhibitor is present A non-competitive inhibitor is present Vmax CONTROLLED PARALYSIS Succinylcholine is a fast-acting, short-duration muscle relaxant that is used when a tube is insert patient's trachea or when a bronchoscope is used to examine the trachea and bronchi for signs Mithin serends of the administration of succinylcholine, the patient experiences muscle paralys 7.

Explanation / Answer

Michaelis-menten equation

Vo=Vmax[S]/km+[S]

where Vo=initial velocity of enzyme-substrate reaction

[S]=substrate concentration

km=(K2+k-1)/k1 =michaelis constant

Vmax=maximum velocity of th rxn

At very high [S], Vo=Vmax[S]/km+[S]

for [S]=km

or,Vo=1/2 *Vmax

Now, when enzyme is saturated with substrate, kcat=catalytic constant=Vmax/[E] [E]=enzyme concentration

1) twice as much as enzyme is used

[E]''=2[E]

V''max=kcat* [E]''=kcat*2[E]=2 (kcat*[E]=2*Vmax

Vmax is doubled

Vo=kcat [E] [S]/km+[S]

or,1/Vo=(km/Vmax)1/[S]+1/[Vmax]

if [E]''=2[E] then V'max=2Vmax

km/Vmax=slope=constant

or, k''m=constant*V''max

k''m=2(const*Vmax)=2km

So km is also doubled

2)half as much enzyme is used

[E]''=1/2[E]

V''max=kcat* [E]''=kcat*1/2[E]=1/2 (kcat*[E]=1/2*Vmax

Vmax is also half its value.

km/Vmax=constant

km''=Vmax''*constant=1/2(Vmax)*const=1/2 km

km also reduced to half

3)A competitive inhibitor is present

for the presence of competitive inhibitor,michaelis -menten equation becomes,

Vo=Vmax [S]/km(app)+[S]

Vmax =unchanged

km=increased by [I]/ki times

km(apparant)=km(1+[I]/ki] ki=inhibitor dissociation constant,[I]-inhibitor concentration

4)non-competitive inhibitor

Vmax reduced

Vmax(apparant)=Vmax/(1+[I]/ki)

km unaffected

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