Question

10. The table below is the results of the study of enzymatic reaction with and without inhibitor. What type of inhibitor is it? Explain! Enzyme concentration in all reactions is 1 uM [Inh], μΜ S], mM 0 0.2 0.4 0.7 rate, umol/min 0 0 0 0 0 0.5 23.5 18.8 14.1 7.0 1 32.2 25.8 19.4 9.7 1.5 36.9 29.5 22.1 11.1 2.5 41.8 33.3 25.0 12.5 3.5 44 35.2 26.4 13.2

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question:...in...typed & image format...

$\Rightarrow$Answer:

  The type: of Inhibitor:   Non-competitive Inhibitor  ...

$\Rightarrow$Explanation:

Following is the complete Explanation, for the above Answer.

• Given:

Following is the given: data table, prepared by using MS Excel, presented here in...image format....

[Inh), uM ( inhibitor concentration, in micro-mol/L) 0.4 0.71 Information Reqd. for Lineweaver-Burk plot Substrate concentration/S), MM (milli-mol/L) Rxn. Rate [No Rxn. Inh) Rate.at.[Inh.1], (umol/min), Vo(umol/min), vi Rxn. Rxn. Rate:[Inh.2], (umol/min),V2 Rate[Inh.3), (um ol/min), V3 1/[S] 1/11 1/V2 0.5 23.5 32.2 36.9 41.8 44 1.5 2.5 3.5 18.8 25.8 29.5 33.3 35.2 14.1 19.4 22.1 20.04255319 0.05319149 0.07092199 9.7 0.0310559| 0.038759691 0.05154639 11.1 0.6667 0.027100271 0.03389831] 0.04524887 12.5 0.4 0.02392344 0.03003003 0.04 13.21 0.28571 0.02272727 0.02840909| 0.03787879|| 25 3003 26.4

$\Rightarrow$Where:

1. [S] = Substrate concentration, in  mM ( i.e. in milli-mol/L )
2. V_o = Reaction rate, for No Inhibitor , in [=] $\mu$M/ min.   ( micro. mol/ min. )
3. V₁ = Reaction rate, for inhibitor concentration at: [ Inh .1 ]
4. V₂ =   Reaction rate, for inhibitor concentration at: [ Inh .2 ]
5. V₃ =    Reaction rate, for inhibitor concentration at: [ Inh .3 ]

• ​​​​​​​Step - 1:

​​​​​​​We know, the following as the : Michaelis - Menten Equation:

$\Rightarrow$1/V = K_m / V_max x ( 1/[S] ) +   1 / V_max   --------------------------------Equation - 1

$\Rightarrow$ Where:

1. K_m = Michaelis - Menten constant
2. V_max = Maximum reaction velocity attained by the reaction.
3. V =   Reaction velocity [=]   $\mu$M/ min.

​​​​​​​$\Rightarrow$ Therefore: if we plot: 1/V vs. 1/[S] , and from the Liner TrendLine: we will get to see, the following:

1. slope =   K_m / V_max
2. y -intercept =    1 / V_max
3. x - intercept = - 1/K_m   

• ​​​​​​​Step - 2:

​​​​​​​Following is the graph : of   1/V vs. 1/[S], based on the the given data table above, prepared by using MS Excel, presented here...in....image format...

[Inh), uM (inhibitor concentration, in micro-moi/L) Information Reqd. for Lineweaver-Burk plot Substrate concentration(s), mM (milli- mol/L) Rxn Rate [No Rxn Inh] Rate.at.[Inh.1), (umol/min), Vo (umol/min), V1 Rxn Rate:[Inh.2], (umol/min),V2 Rxn Rate[Inh.3) (umo 1/[S] 1/min), V3 1/VO 11/01 11/12 1/03 0 23.5 18.8 25.8 29.5 33.3 35.2 19.4 22.1 25 26.4 2 0.042553191 0.053191489 0.070921986 0.142857143 1 0.031055901 0.03875969 0.051546392 0.103092784 11.11 0.66671 0.027100271 0.033898305 0.0452488691 0.09009009 12.51 0.4 0.0239234451 0.03003003 0.04 13.2 0.28571 0.022727273 0.028409091 0.0378787881 0.075757576 2.5 41.8 0.08

• Step - 3:

​​​​​​​Following is the complete graph: of 1/V  vs. 1/[S], prepared by using MS Excel, presented here in...image format...

HOS y = 0.039x + 0.0647 1/Vo vs. 1/(S) graph +1/V1 vs. 1/(S) graph +1/V2 vs. 1/[S] graph *1/V3 vs. 1/[s] graph Linear (1/Vovs. 1/[S] graph) Linear (1/V1 vs. 1/(S) graph) — Linear (1/V2 vs. 1/[S] graph) — Linear (1/V3 vs. 1/(S) graph) y = 0.019x + 0.032 & y=0.01 4x + 0.034 V = 0.011x +0.010 HH H 1/[S]

: Fig. 1 :

• Step - 4:

​​​​​​​As we can see, from the above graph: obtained under : Step - 3: i.e. the Lineweaver - Burk plot... for all the inhibitors, the , linear TrendLines, are converging at a certain point..., i.e. the value of K_m, remains unaffected, while the value of V_max , gradually reduces, from: V_oto ...V₃...

Thus, we can say: the type of the given inhibitor : is :

$\Rightarrow$    Non - Competitive Inhibitor ....