Prostaglandins are a class of eicosanoids, fatty acid derivatives with a variety

Question

Prostaglandins are a class of eicosanoids, fatty acid derivatives with a variety of extremely potent actions on vertebrate tissues. They are responsible for producing fever and inflammation and its associated pain Prostaglandins are derived from the 20-carbon fatty acid arachidonic acid in a reaction catalyzed by the enzyme prostaglandin endoperoxide synthase. This enzyme, a cyclooxygenase, uses oxygen to convert arachidonic acid to PGG, the intermediate precursor of many different prostaglandins A. Use the kinetic data below to make a Lineweaver- Burke for the reaction catalyzed by prostaglandin endoperosice synthase. Determine V and K for the uninhibited prostaglandin endoperoxide synthase. (4 pts) B. The concentration of the enzyme HM. Calculate k. (2 pts) used in the assays was 2.25 C.Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the synthesis of prostaglandins, Ibuprofen reduces inflammation and pain. Using the data measured in the presence of ibuprofen, create a Lineweaver-Burke plot to calculate the V and K. What type of inhibitor is Ibuprofen? Give your rationale. To which form of the enzyme does Ibuprofen bind? Free enzyme, Enzyme bound to substrate or either form? (6 pts) Arachidonic acid V. V,with 10 mg/mL (mM) 0.50 (mM/min) Ibuprofen (mM/min) 23.5 32.2 36.9 41.8 44.0 16.67 25.25 30.49 37.04 38.91 2.5 3.5

Explanation / Answer

Lineweaver-burke plot is 1/V= KM/Vmax)*1/S + 1/Vmax (1), so a plot of 1/V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax. Kcat= Vmax/ET, ET= Concentration of Enzyme

for the case of inhibition, KM in Eq.1 will be replaced with Kmapp and Vmax will be replaced with Vmaxapp.

The plot of 1/V vs 1/S is generated and shown below.

from the plot, when there is no inhibitor, 1/Vmax= intercept =0.019, Vmax=1/0.019=52.63 mM/min and

slope is KM/Vmax= 0.011, KM= 0.011*52.63 mM=0.58 mM

Kcat= 52.63*10-3 M/min/(2.25*10-6 M)=23391/min

for the case of inhibition, 1/Vmaxapp =0.019, Vmaxapp =52.63 mM/min =Vmax and hence this in competitive inhibiton where the enzyme competes with the substrate for the active sites of an enzyme.

Slope for inhibitopn, KMapp/Vmaxapp= 0.020, KMapp =0.020*52.63mM=1.0526 mM

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