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10.2_ For any particular ratio of an enzyme inhibitor to the inhibition constant, Kl, competitive inhibitor will decrease the reaction rate more that substrate-depe...

Question

10.2_ For any particular ratio of an enzyme inhibitor to the inhibition constant, Kl, competitive inhibitor will decrease the reaction rate more that substrate-dependent non-competitive inhibitor when [S] < Km but the reverse occurs when [S] > Km: Show this is true algebraically by Using the equations for inhibition in Michaelis-Menten kinetics Explain this result conceptually (i.e-, without the algebra)

10.2_ For any particular ratio of an enzyme inhibitor to the inhibition constant, Kl, competitive inhibitor will decrease the reaction rate more that substrate-dependent non-competitive inhibitor when [S] < Km but the reverse occurs when [S] > Km: Show this is true algebraically by Using the equations for inhibition in Michaelis-Menten kinetics Explain this result conceptually (i.e-, without the algebra)



Answers

A different mode. The kinetics of an enzyme is measured as a function of substrate concentration in the presence and absence of $100 \mu \mathrm{M}$ inhibitor. (a) What are the values of $V_{\max }$ and $K_{\mathrm{M}}$ in the presence of this inhibitor? (b) What type of inhibition is it? (c) What is the dissociation constant of this inhibitor? (GRAPH CANNOT COPY) (d) If $[S]=30 \mu \mathrm{M},$ what fraction of the enzyme molecules have a bound substrate in the presence and in the absence of $100 \mu \mathrm{M}$ inhibitor?

None competitive inhibited noncompetitive inhibitor. The next is lure is lower than for the normal enzyme. Mhm. Mhm. For the normal ensign. But GM is destined. Yeah. Is the same. Mhm. Yeah. Yeah. Mhm. Okay. Noncompetitive in hebrew effect the next or not. Such noncompetitive inhibitor reduce the max without affecting camp. Yeah. Without Yeah. Okay. Affecting P. M. You would not see the temporary increasing camp described above. For the irreversible Mhm. A reversible Mhm competitive. Mhm. In hebrew. Yeah. And then we're running the reaction. Mhm. We brought inhibitor and Mhm substrate present. Yeah.

Hello there students today for those questions they brought saying how a noncompetitive inhibitor does not prevent the enzyme from buying its substance. I'm substrate. What will be the effect of increasing the substrate concentration in the presence of non occupied not competitive and inhibitors do you expect a non competitive inhibitor to affect the enzyme maximum? Philosophy are okay I am and explain briefly. So in this case there would be okay no effect place of increasing substrate. Yeah, concentration. Yeah. Yeah. Mhm. The fee max what they reduced. Yeah. Bye. Uh Nine come had negative and hipper. Yeah. Yeah. The K the K. M. Would. Yeah. B unaffected. Yeah. Bye. Uh Non huh. I had to test and hit her. Oops. Okay. Yeah and, yep that is it. I found this helpful and please take care and stay safe. Okay bye.

So the rate equation for an end time subject to competitive inhibition is given by be not as equal to the max was quite by s divided by ALF. Okay, um God s so we can use another definition of the total enzyme. The right equation is determined as follows you have e t is equal to e concentration off ad and I'm substrate ad because I'm inhibitor this equation one So fast Unit's itself e I So we know that. Okay, Juan is equal to e multiplied by I divided by e I. We re arrange for the guy that's able to he wants quite by I started by K one So I substitute in Equation one on on the next page. What we get is e t is equal to e at the ass on e long her concentration of I divided by K one so we can simplify that. What we got is e t equals and I'm substrate ad and this time multiplied by Juan Odd inhibitor. Okay, want so as awful describes the effect of the competitive inhibitor we have e t is equal to eat us ad e multiplied by Alfa. This is equation too. Thank you. So for E, that's equal. Thio, eat us. Okay, Um, by by us. Substitute an equation too. What we get is B t is equal to e s and E s most. By K am, we divide that by concentration of Earth's multiplied by Arthur we re arrange for S e s is equal to eat tea. Good bye Bye. Earth divided by Alfa K Um, Ad s now substituting in vino equals K two multiplied by concentration of the s on defining with K two multiplied by the concentration of e t equals B max, we get the following equation which will right out on the next page we have seen on equals k two and I'm so straight we got being on is equal to K to multiply by E t Uh huh Divided by Alfa K um, concentration of s. But because K two most part by E thio e t concentrations be Max, what we get is be not is equal to feed Max provided by e t. No supplied by e t multiplied by us all divided by a k M add concentration of Earth's weaken and have some consolations here. So finally we get vino is equal to the max s divided by a K um God s

To determine the factor of the rate increase When for a first order reaction being catalyzed with an enzyme, we increased the enzyme concentration from 1.5 times 10 to the negative, 7 to 4.5 times 10 to the negative six. We can assume then that the concentration of the enzyme following first order kinetics can be expressed as the rate equals. Some rate constant, multiplied by the concentration of the enzyme raised to the first power. To calculate the factor, or the ratio will simply take a ratio of these two rate constants, each having the enzyme concentration raised to the first power. A ratio of these two rate laws, not rate constants allow for the rate constants to cancel, And the ratio of the rates that will then be our factor of increase, which is 30.7 or 31 times faster.


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