I’m so confused guys if someone can explain this in detail with the steps so I k

Question

I’m so confused guys if someone can explain this in detail with the steps so I know what to do I would appreciate it a lot, and if you could include the Units because I’m getting them mixed up :/
5.- (1 point) Using Lineweaver-Burk, you determined an inhibitor to cause a mixed inhibition when assayed at a concentration of 5 mM. If the equation that characterized the reaction without inhibitor is y-0.603 + 0.1978 (units: /(mM/min) and x=mM), the reaction in the presence of the inhibitor y=2.4502 + 0.3514. Calculate the values ki and ki'.

Explanation / Answer

when there is no inhibitior, 1/V= (KM/Vmax)*1/S + 1/Vmax, also known as Lineweaver-burk plot, so a plot of 1/V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax.

for inhibition, KM need to be replaced with Kmapp and Vmax has to be replaced with Vmaxapp

from the equation of best fit, intercept is 1/Vmax=0.1978, Vmax= 1/0.1978=5.05 mM/min, KMapp/Vmaxapp= 0.603, KMapp=0.603*5.05mM=3.045mM

Mixed type inhibition is an inhibition in which binding of the substrate or the inhibitor affect the enzyme’s binding affinity for the other.

for the case of mixed inhibition, 1/Vmax app = 0.3514, Vmax app = 2.845 mM/min, KMapp/Vmaxapp=slope = 2.4502, KMapp= 2.845*2.4502 mM=6.97mM

Kmapp= KM*(1+I/Ki) and Vmaxapp= Vmax*(1+[I]/Ki*alpha)

6.97= 3.045*(1+I/KI)

I/KI=1.288

KI= I/1.28= 5/1.288 mM=3.88mM

1/Vmaxapp= (1/Vmax)*(1+I/Kialpha)

1/0.1978= (1/2.4502)*Ki', KI= (1+I/KIalpha)

KI' = 12.38mM

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