If k 1 values for the first order dissociation of bx and by...

Name: if k 1 values for the first order dissociation of bx and by are known to be equal which of the following is true and why circle the correct answer and explain your reasoning below a kdbx kdb 07677
Uploaded: 2019-05-31T17:44:46-08:00
Duration: 14 min 11 s
Channel: Md Mohibullah
Description: if k 1 values for the first order dissociation of bx and by are known to be equal which of the following is true and why circle the correct answer and explain your reasoning below a kdbx kdb 07677

If k-1 values for the first-order dissociation of BX and BY are known to be equal, which of the following is true and why? Circle the correct answer and explain your reasoning below. A. KD(BX) < KD(BY) ; k1X > k1Y B. KD(BX) = KD(BY) ; k1X = k1Y C. KD(BX) > KD(BY) ; k1X < k1Y D. No conclusion can be drawn from the data provided. Explanation:

Transcript

00:01 So in this problem we are given with two reactions of two drugs with a protein and we are given the initial concentration of the protein and both the drugs.

00:14 And we are also given the equilibrium concentration for the drug protein complexes.

00:22 And from all these values, we have to find out the value of kc for each of the drug protein reaction.

00:28 And from there we'll have to say which drug protein binding is more stronger than the other we'll start by writing the reaction equation for the protein with for the for the protein with drug a so since it's a one is to one reaction as it is mentioned in the question that means the reaction will like protein plus drug a will give us so if we write them and equilibrium will be the product will be a protein complex.

01:17 Now we are given the initial concentration of the protein and the drug initial concentration of the protein is given as 1 .5 times 10 to the power minus 6 molar and the initial concentration of the drug is given as 2 times 10 to the per minus 6 molar.

01:51 So now if we set up an eye stable for this reaction right away, we will have to write down the concentration of the drug protein complex initially zero since there is no drug protein complex formed at the initial stage.

02:16 And then if we consider the change as x then the change for both of the protein and drag will be minus x and for the drag protein complex it will be plus x since because it's a one -st -one reaction that means there we don't have to raise the value of the coefficient for any of the components here and now at the equilibrium we will have the protein concentration as 1 .5 times 10 to the power minus x and for the drug 8 will be 2 times 10 to the power minus xm and for the drug protein complex it will be only x now in the question we are given that the equilibrium concentration for this drug protein complex is one times 10 to the power minus six molar and that means the value of x is one times 10 to 1 minus 6 molar so you can write the value of x as one times 10 to the power minus 6 molar so now if we just plug in the value of x in these two in these two quantities we can find out the equilibrium concentration of both the protein and the drug a.

04:04 So we are asked to find out the value of kc here for each of the reaction.

04:11 And if we have to find out the value of kc, we know that the expression of kc will contain the equilibrium concentration of both the product and all the reactants.

04:22 So we already know the equilibrium concentration of the product, but we have to know the equilibrium concentration of the reactants...

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