4. The kinetics of an enzyme are measured as a function of substrate concentrati

Question

4. The kinetics of an enzyme are measured as a function of substrate concentration in the presence and absence of 100 uM inhibitor. Include any necessary spreadsheets, graphs and show full calculations explanations. Remember, in a college class we use spreadsheet and chart programs such as Excet. If you absolutely must draw a spreadsheet or graph by hand, it must be on graph paper and drawn very carefully to scale. A sketch on a piece of notebook or white paper will not be accepted! Always always include units throughout your steps and make sure your answers are fully labeled with correct units. and provide full No inhibitor 10.4 14.5 22.5 33.8 40.5 2.1 2.9 4.5 6.8 8.1 10 30 i. finclude spreadsheet necessary to answer the following questions here] (1 pts) the following questions, parts ili and v, herej Fully (2 pts) ginclude the graphs/figures necessary labeled scatter plots correctly constructed, labeled, trendlines and equations

Explanation / Answer

Michaeles-Menten kinetic equation is

V= VmaxS/(KM+S)

or 1/V= (KM+S)/VmaxS

1/V= (KM/Vmax)*1/S + 1/Vmax

so a plot of 1/V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax

in case of inhibition, the slope is KMapp/Vmaxapp and intercept is 1/Vmaxapp

the data on 1/V vs 1/S is generated and shown along with the plot.

From the plot, when there is no inhibitor, intercept 1/Vmax= 0.022 and Vmax=1/0.022=45.45 uM/min

KM/Vmax= 0.224, KM= 0.224*Vmax=0.224*45.45= 10.2 uM

turn over number= Vmax/ET, ET= Enzyme concentration = 45.45/(1.25*10-10)/min=3.63*1011/min

this is the number per active site. When there are two active site, the turn over number = 2*3.63*1011/min=7.26*1011/min

in the presence of inhibitor, 1/Vmaxapp= 0.112, Vmaxapp= 1/0.112=8.93 uM/min and

KMapp/Vmaxapp= 1.108, KMapp= 1.108*8.93 uM=9.89 uM

So both KM and Vmax compared to no inhibitor have decreased. This is the case of uncompetitive inhibition where the inhibitor bind to the enzyme at the same time as the enzyme's substrate. However, the binding of the inhibitor affects the binding of the substrate, and vice-versa

for this type, KMapp= KM/(1+I/KI)

1+I/KI = KM/KMapp= 10.2/9.89=1.03

I/KI= 0.03, KI= I/0.03= 100/0.03 =3333.3uM, I = inhibitor concentration

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.