A protein has two strcutural forms R and T These structural forms are in equilib

Question

A protein has two strcutural forms R and T These structural forms are in equilibrium with the equilibrium constant: K=[T]/[R]. Also, both T and R each have a single binding site for ligand L and have equilibrium constants: kT=[TL]/[T][L] and kR=[RL]/[R][L].

a) For this model the binding polynomial is defined as qb=[R]+[T]+[TL]+[RL]. Assume K=10, kT=0.100 and kR=1.00. Calculate the value of qb/[R]. Assume also [L]=0.1M. Assume qb/[R] is dimensionless.

b)For the model in part A calculate the binding fraction fB.

c)For the model in part A the equilibrium constant between RL and TL is defined as L=[TL]/[RL]. Calculate the value of L assuming the model in part A.

Explanation / Answer

This the question of Hill equation

Given: K=[T]/[R] ; kT=[TL]/[T][L]; kR= [RL]/[R][L]

(a) Given: qb= [R]+[T]+[TL]+[RL]; eq 1

K=10; kT=0.100; kR= 1.00; [L]=0.1

To Find qb/[R]=?; Dividing eq1 by [R] we get qb/[R]= 1+ [T]/[R]+[TL]/[R]+[RL]/[R], eq2

Therefore qb/[R]= 1+K +[TL]/[R] + kR [L] (given in question)

now we have value of K, kR & [L] as 10, 1 and 0.1 respectively. We have to find the value of [TL]/[R]

since we know kT=[TL]/[T][L] Dividing this equation by [R] we get

kT/[R]= ([TL]/[R])*(1/[T][L])

kT/([T]/K)= ([TL]/[R])*(1/[T][L])

In above equation cancelling [T] from both sides and putting the value of K, kT and [L], we get

[TL]/[R]=[L]*K *kT= 0.1*10*0.1= 0.1

Putting value of [TL]/[R] in eq 2, we get

qb/[R]= 1+10+0.1+0.1= 11.2

(b) using hill equation

Q(binding Fraction)= [L]n/ K+[L]n we know value of K and [L] from above data, value of n=2.5 as this is the case of positive binding which has value of n=2.5 so,

Q= [0.1]2.5/10+[0.1]2.5 =0.0031/10+.0031= 3.099*10-4

(c) Given [L]= [TL]/[RL]

Now we know kT/kR= ([TL]*[R]*[L])/([T]*[L]*[RL])

kT/kR= ([TL]/[RL])*([R]/[T])

[TL]/[RL]= (kT/kR) * ([T]/[R])

From above data [TL]/[RL]= [L]= 0.1*10= 1M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.