# data[].attributes.headline
About Inhibitor value (I) to calculate Ki

# data[].attributes.subject
About Inhibitor value (I) to calculate Ki

# data[].attributes.body
Assalamualaikum dr, saya ingin bertanya sekiranya soalan suruh cari Ki untuk inhibitor.

Dalam formula:

Km app= Km ( 1 + (I) / Ki)

Untuk dapatkan nilai I tersebut kena guna dari salah satu inhibitor sama ada 3mM atau 5mM.

Katalah saya nak guna 3mM

Jadinya, nilai I yang saya perlu guna adalah : 3

atau nilai I ni saya perlu bahagikan dia jadikan 1/3 = 0.33 ?

Sebab saya agak keliru sama ada nilai I tersebut perlu guna dalam bentuk 3 atau 0.33.

Ini kerana jika nilai I yang guna adalah 0.33, maka nilai Ki yang diperoleh akan jadi lebih kecil berbanding menggunakan nilai 3.

# data[].attributes.body
Hi dr , to measure kcat do we need to include the volume of enzyme assay in the formula?

# data[].attributes.body
Good day Dr, Noncompetitive inhibitors bind to an allosteric site on the enzyme. this binding induces a conformational change in the enzyme that affects its catalytic activity, As a result, the enzyme's ability to convert substrate to product is reduced, leading to a decrease in the turnover number. Is it correct Dr?

# data[].attributes.body
Hello dr, Good morning. Can you please explain why uncompetitive inhibition reduces the km and vmax again? Thank you dr

# data[].attributes.body
Is there any possibilities that we have to draw mm plot to determine the type of inhibitors being used?

# data[].attributes.body
Hi Dr., sorry for disturbing. I have a question regarding the considerations in enzyme inhibition. For all competitive , non-competitive and uncompetitive inhibition, which category do they fall in terms of reversible and irreversible. In the notes, it is written that competitive inhibition usually reversible but not for the other two inhibitions. So for the other two, non-competitive and uncompetitive, which category do they fall? or both of these type of inhibitions can have both reversible and irreversible?

# data[].attributes.body
hello dr, i have a question regarding the Km of the competitive inhibitor. It is said that the Km will be higher if the the inhibitor binds to the enzyme (therefore has a lower affinity towards substrate). Yes. But it is also said that when we increase the substrate concentration, the substrate can replace the inhibitor from the enzyme right, so does this mean that when we increase substrate concentration the affinity of enzyme towards substrate increases ? No, adding more substrate will NOT increase the affinity of the enzyme towards the substrate. Adding the inhibitior, will reduce the enzyme activity(that is it is called inhibition), so when we add more substrate, more will form ES complex, and the activity will increase back to the value of without inhibitor. For e.g, the enzyme activity without any inhibitor will be 1.0, with inhibitor, may be reduced to 0.5, but when we add more substrate (or dilute the inhibitor conc), we will get enzyme activity 0.8. I thought the Km is unchanged throughout the reaction? yes, the Km of the enzyme (without any presence of inhibitor) remains the same, and at any conc of the substrate. But when we add an inhibitor, the Km of the enzyme towards the substrate will change, increasing it to Km(1 = [I]/Ki). Thank u in advance Dr! You're welcome :)

# data[].attributes.body
Good evening Dr, I have a question regarding the Malonate-enzyme complex. You said that it has ionic bonds since the malonate is not in enzyme analogue right ? ionic binds or weak covalent bonds, such as van der Waals interactions. But isnt Malonate the competitve inhibitor to succinate dehydrogenase so when it tries to bind to enzyme (lock and key), wont the enzyme change its active site shape so that the inhibitor can bind ? Wont it form covalent bond after that ? it can fit in but it will not form any 'extra bonds' or cause any significant conformational changes cos it is not the substrate. Conformational changes only occur when the true substrate enters the active site. If not, any molecules that is substrate analog, can be changed to product, which is wrong , right? Similar if you use a wrong key to open a lock. It can fit into the keyhole, but can you open the lock? Or is the lock and key theory is only applicable to substrates and enzymes? No, lock and key theory also applies to competitive inhibition.

# data[].attributes.body
Hello Dr, when the question asked us to find Vmax, we just find the  Vmax of enzymes with no inhibitor? unless it mentioned us to find alike - Find Vmax of enzyme with inhibitor?
Hi Emily. Yes, or in other words "find the Vmax(app) for the enzyme with inhibitor"

And when the question asked us to find Km, we just have to find Km for enzyme without inhibitors right? Yes. Because if for Km of inhibitors it should be known as Ki? No, cause there is also Km of the enzyme with inhibitor, which we called Km (app). And depending on the type of inhibition, say competitive inhbition, Km(app) = Km (1 + [I]/Ki)

# data[].attributes.body
Hello Dr, since both non competitive and uncompetitive bind to allosteric sites then why Km for non competitive remain unchanged but Km for uncompetitive reduced?

# data[].attributes.body
Hello Dr, is there any inhibitors from what we have learnt which will cause permanent damage to enzyme that will cause irreversible reaction?

# data[].attributes.body
Hello Dr, for non competitive inhibitor, inhibitor binds to either E or ES and will only cause conformational changes lead to difficulty substrate binds to active site or ES form product.
For uncompetitive inhibitor, when ES form, allosteric conformational change will occur which lead to inhibitor able to bind and stop the reaction.

Dr, is this correct?

# data[].attributes.body
salam dr, do we need to change the unit for enzyme concentration in order to find k3 value? Does it also apply on Ki too?

# data[].attributes.body
Dr, may i know, if LB plot consists of more than 2 lines (1 without enzyme, 2 inhibitor with different concentration),  the KI have to divide by 2 since there is two conditions?
If a graph consists of competitive inhibitor and non competitive, the Ki for both will be different?

# data[].attributes.body
Salam dr, for competitive inhibition, it can overcome by increase the concentration of the substrate right? So for noncompetitive and uncompetitive, is it necessary to increase the concentration [S]? if not why? and for both noncompetitive and uncompetitive inhibition, is there any way to overcome the inhibitor? Since both are not binding to the active site.

Thank you so much, Dr.

# data[].attributes.body
Dr, is it possible for Ki value to be in negative ?

# data[].attributes.body
Dear all, below is a good video that explains about dialysis.

# data[].attributes.attachment_link.title
Continuous dialysis in the laboratory (IQOG-CSIC)

# data[].attributes.attachment_link.description
Continuous dialysis in the laboratory (IQOG-CSIC) The dialysis technique is used to purify large molecules (proteins, DNA, polysaccharides...) Instituto de Química Orgánica General (IQOG-CSIC) Music: Carlos Estella - Inspiring Epic Glory CanalDivulgación http://www.youtube.com/user/CanalDivulgacion.