bio q

[RDH]

which statement about enzymes is Not true?

a. Many enzymes have a distinctpH optimum

b. Many enzymes show linear increase in Rx rate with increasing T

c. Competitive inhiibitors compete with substrate for binding at the enzyme active site

d. The binding of noncompetitive inhibitors outside the active site distorts the active site

I thought it was d but its not according to high yield bio !


ansewer is b if any one knows the why could you please explain?

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[Mrbubbles]

which statement about enzymes is Not true?

a. Many enzymes have a distinctpH optimum

b. Many enzymes show linear increase in Rx rate with increasing T

c. Competitive inhiibitors compete with substrate for binding at the enzyme active site

d. The binding of noncompetitive inhibitors outside the active site distorts the active site

I thought it was d but its not according to high yield bio !


ansewer is b if any one knows the why could you please explain?

b. with increasing t, enzymes start to denature , therefore cannot increase rx rate in a linear fashion.
d. competitive inhibitors bind to the enzyme binding site, therefore enzyme cannot go forward with rxn.
noncompetitive inhibitors bind to a different region on the enzyme, causing a conformational change, therefore the substrate that's suppose to bind to the enzyme to cause a rxn can no longer do so, cuz the enzyme is no longer in original shape =)

it's noncompetitive cuz the inhibitor is not competiting for the same binding site, cuz there are 2 binding sites, as opposed to one in the competitors

 

[doc3232]

b. with increasing t, enzymes start to denature , therefore cannot increase rx rate in a linear fashion.
d. competitive inhibitors bind to the enzyme binding site, therefore enzyme cannot go forward with rxn.
noncompetitive inhibitors bind to a different region on the enzyme, causing a conformational change, therefore the substrate that's suppose to bind to the enzyme to cause a rxn can no longer do so, cuz the enzyme is no longer in original shape =)

it's noncompetitive cuz the inhibitor is not competiting for the same binding site, cuz there are 2 binding sites, as opposed to one in the competitors

It is true that enzymes denature at a very high temperature, but for some time their enzyme activity does increase with increased temperature.
A, C and D are always true though. I have seen questions that say B is true though so this is a tricky question.

 

[RDH]

I know that a noncompetitive inhibitors binds to an area other than the active site, but the reason I thought d was the wrong ansewer is because I did not think it destortes the enzyme active site, and I know that at some point with increasing temp the enzyme denatures but the rate increases for some point so it would be linear for some point and then drops due to enzyme denature.

So what your saying is that w/ a noncompetitive inhibitors the enzyme' active site becomes destorted.

and increasing T would we have more of a sigmoidal than a linear increase.

 

[td4azklz]

I know that a noncompetitive inhibitors binds to an area other than the active site, but the reason I thought d was the wrong ansewer is because I did not think it denatures the enzyme and I know that at some point with increasing temp the enzyme denatures but the rate increases for some point and the drops due to enzyme denature.

So what your saying is that w/ a noncompetitive inhibitors the enzyme becomes denatured.

and increasing T we probably would have more of a sigmoidal than a linear increase.

With a noncompetitive inhibitor, the enzyme doesnt get denatured, it just changes its conformation to the inactive form.

Does anybody know if this is reversible or not?

 

[futuredent22]

With a noncompetitive inhibitor, the enzyme doesnt get denatured, it just changes its conformation to the inactive form.

Does anybody know if this is reversible or not?

i'm pretty sure both of these are reversible

 

[RDH]

Irreversible inhibition occurs when permanent damage is done to the active site either by tight covalentbonding of the inhibitor to the enzyme or the DENATURATION of the active site which result in the prevention of an enzyme substrate complex... so I would think that noncompetitive inhibitor would be irriversible since it denatures the active site.

 

[acupofchai]

NONCOMPETITIVE inhibition is IRREVERSIBLE. It lowers the affinity of the enzyme for substrate by binding to the allosteric site, which distorts the active site.
(Vmax same, Kb increases)

COMPETITIVE inhibition is REVERSIBLE. Affinity of the enzyme for the substrate is the same, and the competitor or substrate can be interchanged. However, the overall rate decreases.
(Vmax lower, Kb same)

 

[Dencology]

which statement about enzymes is Not true?

a. Many enzymes have a distinctpH optimum

b. Many enzymes show linear increase in Rx rate with increasing T

c. Competitive inhiibitors compete with substrate for binding at the enzyme active site

d. The binding of noncompetitive inhibitors outside the active site distorts the active site

I thought it was d but its not according to high yield bio !


ansewer is b if any one knows the why could you please explain?

Ok, so if you look at choice A, you will see that that it say distint pH optimum. which is correct. most Enzymes work best at 7.2 pH but you also have enzymes that work at pH 2 like in stomach. so they have their own optimum enzymes. so A is correct.
For choice b, as temperature increase enzymes (with are proteins) denature, what i mean is that they loose they 3D structure and they don't work as they should in decreasing the activation energy. so they also have their own temp which is around 37 celcious( it is written in Kaplan BB)

For choice C, for conpetetive inhibitor, you have an active site and not an active site. what i mean is that think of active site as an primary binding site for substrate, now if this site is filled, then the enzyme can not bind to the specific substrate and and not act. think of it and there is one chair around a table and you want to eat your food. if that chair is taken by someone else then you can not eat your food at that table. so it is the same concept. now for noncompetetive, think of the chair and table analogy and now instead of one chair your have two chair and one is take, but there is lots of food on this table by the person who is occuping the first chair that you can not put y our food on the table. in relation to what we talk abou there, in noncompetetive a substrate bind a site which is not the active site and change the conformation of the enzyme active site. so this enzyem is not active as well.
Hope this helped.

 

[futuredent22]

NONCOMPETITIVE inhibition is IRREVERSIBLE. It lowers the affinity of the enzyme for substrate by binding to the allosteric site, which distorts the active site.
(Vmax same, Kb increases)

COMPETITIVE inhibition is REVERSIBLE. Affinity of the enzyme for the substrate is the same, and the competitor or substrate can be interchanged. However, the overall rate decreases.
(Vmax lower, Kb same)

you've got some of that backwards. In competitive, Vmax remains the same. Thats because theoretically the substrate can bind to every single available enzyme and proceed through the course of the rxn. This would happen if there was a whole lot more substrate than inhibitor.

In noncompetitive, Vmax decreases because it doesn't matter how much substrate there is. enzymes that become bound to the inhibitor will not work, regardless of whether the substrate managed to bind as well. The V of the reaction will lower with increasing inhibitor.

 

[hdu]

Non competitive inhibition can be reversible. Allosteric inhibition is reversible. It is a normal way of regulating metabolism. Ex. ATP inhibits some enzymes of krebs cycle. When ATP is high, the enzymes are inhibited and the Krebs Cycle decreases its activity. While the cell consumes ATP, the concentration of ATP decreases, so the allosteric inhibition in those enzymes dissapears, and the Krebs Cycle increase the activity again.

My blog: www.biochemistryquestions.wordpress.com

 

[sang86]

the answer is clearly B. even if you do not know the trend of graph, you can simply get the answer by elminating all the correct choices.

 

[ksksks]

no bio graph is EVER linear (i think )

everything = LOGARITHMIC

 

[yakuza]

nvm. read the question wrong