Sprint LIE 6:57 PM gastate.view.usg.edu Lisa is running an experiment to determi

Question

Sprint LIE 6:57 PM gastate.view.usg.edu Lisa is running an experiment to determine the kinetic parameters (Km and Vx) of the enzyme that she is studying, She has run a series of experiments where she changed the amount of substrate added and measured the accumulation of product over a 5 second time interval. She used this data to calculate the dependence of the rate of the reaction on substrate concentration. Her results listed below Substrate Concentration (mM)Velocity (mM/sec) 2.5 2 5 10 20 6.3 7.6 1. Using this information construct a Michaelis-Menten curve 2. Using this information construct a Lineweaver-Burk plot. 3. Using the Lineweaver-Burk plot determine the values of K. and Vm To learn more about how this enzyme functions Lisa decides to perform an experiment using two different types of inhibitors. She has collected the data below. Substrate Concentration Velocity Va mM 2.5 1.17 2.1 0.77 .25 6.3 7.6 20 7.2 2.86 1. Using this information construct a Michaelis-Menten curve that contains the data for all three data sets (velocity, va and va).All three data sets should be on the same graph. 2. Using this information construct a Lineweaver-Burk plot for all three data sets (velocity, va and va). All three data sets should be on the same graph 3. Using the Lineweaver-Burk plot determine the values of K and Vmax for the new data sets for all three data sets (Va and va). 4. What type of inhibitor (competitive, noncompetitive or neither) is Inhibitor (S,Vi)and Inhibitor(Sa). Explain your answer.

Explanation / Answer

Michaelis-Menten kinetics is V= VmaxS/(KM+S)

S =subsrate concentration, V= rate, Vamx= maximum rate and KM= Michelis- menten constant.

So a plot of V vs S need to be drawn. The data points along with the graph are shown below.

Linerweaver- burk plot is 1/V= (KM/Vmax)*1/S+ 1/Vmax

so a plot of KM/Vmax vs 1/S is drawn whose slope is KM/Vmax and intercept is 1/Vmax

when inhibitor is there, KM becomes Kmapp and Vmax becomes Vmax app

these plots are also generated ans shown below along with data points.

from the plot and equations of best fit, when there is no inhibitor

1/Vmax= intercept =0.095, Vmax= 1/0.1= 10 mM/sec, KM/Vmax= slope = 0.312, Km= 0.312*10 mM=3.12 mM

2. when there is inhibitor-1, 1/Vmaxapp =0.1. Vmaxapp=1/0.1= 10 mM/sec, Kmapp/Vmaxapp= 0.754, KMapp= 10*0.754= 7.54 mM, since Vmax remain the same for no inhibitor and inhibitor-1, this is the case of competitive inhibition where the inhibitor competes with the enzyme for the active sites.

3. for the inhibitor-2, 1/Vmaxapp=0.3, Vmaxapp=1/0.3=3.3 mM/sec and Kmapp/Vmaxapp = 0.998, KMapp=3.3*0.998=3.32 mM, since KM app remains almost equal to Km , this is non- competitive inhibition.Non-competitive inhibitor reacts with the enzyme-substrate complex, and slows the rate of reaction to form the enzyme-product complex.

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