danielsgirl0403

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Jun 1, 2017
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Can someone please explain how to find this answer to me?

Researchers collected the following data for an enzyme-catalyzed reaction.

[Substrate] (muM)
3
6
10
20
50
100
1000
10,000

Vo (muM/min)
22
53
100
129
162
180
199
200

What is the approximate numerical value of the slope of the Lineweaver-Burk plot of these kinetic data?
Choices:
A) 0.01
B)0.05
C)10
D) 50

The answer is B- 0.05, and this was done by doing 10/200 using the slope equation m=Km/Vmax.
I understand how the Vmax is 200, because once the concentration of substrate is saturated, the max velocity shows that it would be 200. I don't understand why 10 is the Km and how to know that it is 10?

Thanks in advance.
 

quackwhacker

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Mar 4, 2008
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  1. Dental Student
Can someone please explain how to find this answer to me?

Researchers collected the following data for an enzyme-catalyzed reaction.

[Substrate] (muM)
3
6
10
20
50
100
1000
10,000

Vo (muM/min)
22
53
100
129
162
180
199
200

What is the approximate numerical value of the slope of the Lineweaver-Burk plot of these kinetic data?
Choices:
A) 0.01
B)0.05
C)10
D) 50

The answer is B- 0.05, and this was done by doing 10/200 using the slope equation m=Km/Vmax.
I understand how the Vmax is 200, because once the concentration of substrate is saturated, the max velocity shows that it would be 200. I don't understand why 10 is the Km and how to know that it is 10?

Thanks in advance.





The Km is the concentration at 1/2 vmax. Since the Vmax is 200, go to the substrate conc when it is 100 and youll see its 10. Hope this helps.
 

danielsgirl0403

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Jun 1, 2017
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The Km is the concentration at 1/2 vmax. Since the Vmax is 200, go to the substrate conc when it is 100 and youll see its 10. Hope this helps.

But if Km=Vmax/2, then Km would equal 100, how do you know to go look at the substrate concentration then? I would have just done m=100/200 = 0.5, which isn't an answer choice, but if it was thats probably the one I'd pick. I'm confused on why you have to pick the substrate concentration for the slope equation when the slope equation is m=Km/Vmas, not [ S] / Vmax
 

danielsgirl0403

2+ Year Member
Jun 1, 2017
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Km does not equal Vmax/2.

Km is the Michaelis-Menton constant that is defined as the substrate concentration that allows for half of the Vmax. So Km equals [ S] [/S]

I got it now!! wow thank you, that cleared up so much. Km is inversely proportional to the enzyme-substrate affinity though, correct? i just want to make sure that is correct since i have it written in my notes.
 
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Doogie.Howser

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Jun 6, 2017
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But if Km=Vmax/2, then Km would equal 100, how do you know to go look at the substrate concentration then? I would have just done m=100/200 = 0.5, which isn't an answer choice, but if it was thats probably the one I'd pick. I'm confused on why you have to pick the substrate concentration for the slope equation when the slope equation is m=Km/Vmas, not [ S] / Vmax

Km does not equal Vmax/2.

Km is the Michaelis-Menton constant that is defined as the substrate that allows for half of the Vmax. So Km equals [ S] at Vmax/2.

This is significant because when the substrate concentration equals Km, you know that exactly half of your enzymes are bound by substrate. This Km value therefore allows you to determine the binding affinity of your enzyme.
 

Doogie.Howser

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Jun 6, 2017
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I got it now!! wow thank you, that cleared up so much. Km is inversely proportional to the enzyme-substrate affinity though, correct? i just want to make sure that is correct since i have it written in my notes.

Sorry I had formatting issues so I reposted my answer haha.

Correct. Km is inversely proportional to the enzyme-substrate affinity. An enzyme with a high Km has low affinity, and an enzyme with a low Km has a high affinity.

This makes sense because if the Km is high, which you know just means that it takes a lot of substrate to bind half of your enzyme, then its affinity is low. And an enzyme with a high affinity for substrate will have a relatively small Km because only a tiny amount of substrate is needed to reach half the Vmax.
 
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