BIOCHEM - Lineweaver Burk enzyme kinetics question

[free99]

Please help!

In order to solve the problem below, you need the following information:

Question 6 and 7 provided this equation "y = 2 x + 0.5" and asked us to solve for Vmax and Km, giving Vmax = 2 "s" and Km=4 mM (s were some arbitrary units given). Easy enough...

Now to the actual question:

Following questions 6 and 7, in the presence of 3 mM inhibitor Y, Lineweaver-Burk plot of enzyme X in question 6 can be fit to equation y = 6 x + 0.5. This inhibitor is:

a. competitive with KI of 0.6 mM
b. competitive with KI of 1.5 mM
c. competitive with KI of 3 mM
d. uncompetitive with KI' of 3 mM
e. uncompetitive with KI' of 6 mM

According to my Biochem professor, the answer is "B".

Here's my sort-of solution:

We know off the bat it's a competitive inhibitor because the y-intercept stays constant, so D and E are out immediately.

Now, I use the equation for competitive inhibition:

[(1/Vo) = (αKm)/Vmax][1/] + (1/Vmax)

We want to solve for Ki, we're given = 3 mM, and we know α = 1 + /Ki

We know that when y = 0, the x intercept is equal to (-1/km). So, setting "y" (1/Vo) equal to 0 and solving for "x" (1/), we get -0.08333. Solving for km, we get km = 12

So now that we have km, I set [(αKm)/Vmax] equal to 6 (from the equation given in the question), and solved for α, giving α = 1

Finally, using the equation α = 1 + /Ki, I tried to solve for Ki using the values (α = 1, = 3 mM) and Ki ends up being undefined (1 = 1 + 3/x)

Does anyone see where I'm going wrong?

Thanks in advance for trying to solve this problem... Maybe it'll help you understand enzyme kinetics more in the process for your biochem final!

---

Members don't see this ad.

 

[free99]

Anyone got an idea?

 

[NDPitch]

I took medical biochem about a year ago, and we never did any problems this hard. Is d school biochem this hard? Haha

 

[free99]

It looks crazy, but it's actually not that bad - just a matter of substituting eqn's and variables, using some algebra, and knowing what the intercepts on the lineweaver-burk plot represent. I think I did everything correctly and he gave us wrong numbers for the problem... certainly not the first time it's happened.

I didn't know until a few days ago, but Chad's biochem videos are actually pretty solid, and free! Check them out if you need some clarification on some select topics.

 

[lutopie0730]

Please help!

In order to solve the problem below, you need the following information:

Question 6 and 7 provided this equation "y = 2 x + 0.5" and asked us to solve for Vmax and Km, giving Vmax = 2 "s" and Km=4 mM (s were some arbitrary units given). Easy enough...

Now to the actual question:

Following questions 6 and 7, in the presence of 3 mM inhibitor Y, Lineweaver-Burk plot of enzyme X in question 6 can be fit to equation y = 6 x + 0.5. This inhibitor is:

a. competitive with KI of 0.6 mM
b. competitive with KI of 1.5 mM
c. competitive with KI of 3 mM
d. uncompetitive with KI’ of 3 mM
e. uncompetitive with KI’ of 6 mM

According to my Biochem professor, the answer is "B".

Here's my sort-of solution:

We know off the bat it's a competitive inhibitor because the y-intercept stays constant, so D and E are out immediately.

Now, I use the equation for competitive inhibition:

[(1/Vo) = (αKm)/Vmax][1/] + (1/Vmax)

We want to solve for Ki, we're given = 3 mM, and we know α = 1 + /Ki

We know that when y = 0, the x intercept is equal to (-1/km). So, setting "y" (1/Vo) equal to 0 and solving for "x" (1/), we get -0.08333. Solving for km, we get km = 12

So now that we have km, I set [(αKm)/Vmax] equal to 6 (from the equation given in the question), and solved for α, giving α = 1

Finally, using the equation α = 1 + /Ki, I tried to solve for Ki using the values (α = 1, = 3 mM) and Ki ends up being undefined (1 = 1 + 3/x)

Does anyone see where I'm going wrong?

Thanks in advance for trying to solve this problem... Maybe it'll help you understand enzyme kinetics more in the process for your biochem final!

My biochem professor did not really cover KI calculation part, but based on your information, I've got some idea here

(α Km) / Vmax = 6

α = (6 Vmax) / Km where Km=4 and Vmax=2

This is the only part that I did differently; I used Km=4 instead of Km=12.


Therefore, α =3

Plugging into KI = / (α -1),

KI = 3 / 2 = 1.5


This could be more fun if I found this post before my final

 

[free99]

We know that when y = 0, the x intercept is equal to (-1/km). So, setting "y" (1/Vo) equal to 0 and solving for "x" (1/), we get -0.08333. Solving for km, we get km = 12

α = (6 Vmax) / Km where Km=4 and Vmax=2
I used Km=4 instead of Km=12.

I think you might be mixing up competitive vs. non-competitive inhibition. Under competitive inhibition, Vmax stays the same and Km changes; therefore Km cannot equal 4 mM (the Km from the first equation).

Line 1 --> y = 2x + 0.5 (Km = 4 mM, Vmax = 2)

Line 2 --> y = 6x + 0.5

I honestly think it's just an error on the professor's end - I believe the solution is sound. Thanks for weighing in on this, lutopie!

 

[KittySquared]

What language is that

 

[free99]

What language is that

HA! did you not have to go through this in biochem?

 

[pinpoint212]

I got
Km= 12
Vmax = 2
Vo = 1
= 4mM

does that look right to you?

 

[free99]

I got
Km= 12
Vmax = 2
Vo = 1
= 4mM

does that look right to you?

Km and Vmax are right on - those are the values I calculated as well.

The relationship between the rate (Vo) and the substrate concentration () is given by the Michaelis-Menten eqn - Vo and can be an infinite number of combinations. So, if the rate was "1", as you stated above, we would expect the substrate concentration to be 12 mM, not 4 mM (plug Vo = 1 into this equation and solve for : [(1/Vo) = (αKm)/Vmax][1/] + (1/Vmax)).

At any rate, the question was asking for the dissociation constant of the enzyme-inhibitor complex (Ki).

 

[KittySquared]

HA! did you not have to go through this in biochem?

Yes but I forgot everything as soon as I took my final. It was traumatizing. lol