Question

a.) You have performed a series of experiments determining the K_i values for three competitive inhibitors. The following table lists the results:

Inhibitor	Ki (uM)
A	6.5
B	1.5
C	0.25

i) Which inhibitor binds with higher affinity to the free enzyme? Explain.

ii) If the same concentration of inhibitor were used in each experiment, which inhibitor would give the smallest value of K_M? Explain your answer.

b.) You want to load 10ug of protein in 15uL into one of the 10% polyacrylamide gel wells. The protein needs to be in 1X buffer and in a total volume of 0.250ml. You are given a 5.58mg/ml protein solution, a 20X sample buffer, and distilled water. How much of each would you mix together to make the required volume? Show all steps please.

Answer 1

a) i) Enzyme inhibition rxn can be represented as:

EI<--->E +I

Kidissociation constant[E][I]/[EI]

where E:enzyme

I:inhibitor

Thus, higher the Ki value less EI or enzyme-inhibitor complex is formed,lesser is the inhibition.

Lowest Ki gives highest affinity of inhibitor for enzyme.

Answer:C

ii) According to michaelis-menten equation, considering enzyme inhibition,(for competitive inhibition)

rate of rxn VVmax[S]/Km +[S]

where (1+[I]/Ki)

[S]:substrate concentration

Vmax:maximum rate

Km:michaelis -menten constant

So, Km(apparant) or KappKm(1+[I]/Ki)

Higher Ki means lower Km (michaelis -menten constant) and vice versa

so , A with highest Ki gives smallest value for Km

Also, theoretically, Km is the concentration of substrate that allows the enzyme to reach half Vmax.So, if Ki is higher , less inhibition will take place ,lesser substrate is required to quench the enzyme to reach (Vmax /2)

for VVmax/2

Km[S]

b)Amount of protein to be loaded10ug/15uL well

total volume of the well 0.250ml250uL

So, amount of protein loaded(10ug/15uL)*250uL166.7 ug

So protein concentration in 1X buffer 166.7ug/250uL0.6668 ug/uL

Now ,protein stock concentration 5.58mg/ml5.58ug/ul

(concentration of protein stock)*(volume of protein stock)(concentration of protein required)*(volume of protein required)

or, (5.58ug/ul)*(volume of protein stock)(0.6668 ug/uL)*250uL

(volume of protein stock) to be taken (0.6668 ug/uL)*250uL/(5.58ug/ul)29.9 uL30uL(approx)

So, u need to add 30uL of stock protein in 220 ml of 1X buffer

To prepare 10ml 1000uL buffer:

Volume to be prepared1mL

dilution factor1/20stock volume/Volume to be prepared

stock volume1/20*1000 uL50uL

take 50uL stock buffer and dilute it to 1000 uL,and use 220 uL0.22ml of it to prepare protein solution

[check:

30 uL of stock protein contains protein (5.58ug/ul)*30uL167.4 ug

Added protein concentration 167.4 ug/250uL0.6696 ug/uL

As well is15 uL ,so added protein amount 0.6696 ug/uL *15uL10 ug(as expected)]