Interpretation:
The plot of [I]0 against (δν)−1 is a straight line with the slope [E]0Δν and y-intercept −K1.
Concept introduction:
Many nuclei and electrons have spin, due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an
Answer to Problem 12.4IA
The plot of [I]0 against (δν)−1 is a straight line with the slope [E]0Δν and y-intercept −K1 has been rightfully shown.
Explanation of Solution
If the plot of [I]0 against (δν)−1 is a straight line with the slope [E]0Δν and y-intercept −K1, then the relation given below will hold true.
[I]0=[E]0Δνδν−K1
The dissociation reaction of the enzyme-inhibitor complex is shown below.
EI⇌E+I
The expression for the resonance frequency (ν) of the single peak in the NMR spectrum is shown below.
ν=fIνI+fEIνEI (1)
Where,
- νI is the frequency of free inhibitor, I.
- νEI is the frequency of bound inhibitor, EI.
- fI is the frequency of free inhibitor, I.
- fEI is the frequency of bound inhibitor, I.
The expression for fI is given below.
fI=[I][I]+[EI]
The expression for fEI is given below.
fI=[EI][I]+[EI]
Where,
- [I] is the concentration of free inhibitor, I.
- [EI] is the concentration of bound inhibitor, EI.
Substitute the value of fI and fEI in equation (1).
ν=[I][I]+[EI]νI+[EI][I]+[EI]νEI (2)
The expression for [I] is given below.
[I]0=[I]+[EI][I]=[I]0−[EI]
Where,
- [I]0 is the initial concentration of the inhibitor, I.
Substitute the value of [I]+[EI] and [I] in equation (2).
ν=[I]0−[EI][I]0νI+[EI][I]0νEI (3)
Equation (3) was solved further as shown below.
ν=[I]0νI−[EI]νI+[EI]νEI[I]0ν[I]0=[I]0νI+[EI](νEI−νI)ν[I]0−[I]0νIνEI−νI=[EI][EI]=[I]0(ν−νI)νEI−νI (4)
The dissociation constant (K1) is given by the expression shown below.
K1=[E][I][EI] (5)
Where,
- [E] is the concentration of the enzyme.
The expression for [E] is given below.
[E]=[E]0−[EI]
Substitute the value of [E] and [I] in equation (5).
K1=([E]0−[EI])([I]0−[EI])[EI] (6)
As the initial concentration of the inhibitor is greater than the concentration of the bound inhibitor, the equation (6) can be written as shown below.
K1=([E]0−[EI])[I]0[EI]K1[EI]=[E]0[I]0−[EI][I]0[E]0[I]0=K1[EI]+[EI][I]0[E]0=[EI](K1+[I]0)[I]0 (7)
Substitute the value of [EI] from equation (4) in equation (7).
[E]0=[I]0(ν−νI)νEI−νI×(K1+[I]0)[I]0=(K1+[I]0)ν−νIνEI−νI (8)
The expression for ν−νI is given below.
ν−νI=δν
The expression for νEI−νI is given below.
νEI−νI=Δν
Substitute the value of ν−νI and νEI−νI in equation (8).
[E]0=(K1+[I]0)δνΔν[E]0Δν=K1δν+[I]0δν[I]0δν=[E]0Δν−K1δν[I]0=[E]0Δνδν−K1 (9)
The equation (9) is in the form of an equation for straight line. Therefore, the plot of [I]0 against (δν)−1 is a straight line with the slope [E]0Δν and y-intercept of −K1.
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