A receptor-ligand complex has a dissociation constant of Kd = 20 nM. The rate of receptor-ligand complex formation with an added ligand concentration of 10 µM is 5 × 10³ s-¹. What is the value of the reverse rate constant, k_₁? S
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You have not correctly calculated the
value of the reverse rate constant.
To begin, calculate the value of k₁ by
dividing the rate of receptor-ligand
complex formation (the rate of the
forward reaction) by the
ligand concentration.
k₁ =
rate forward
[L]
X
Then, calculate the value of the reverse
rate constant, k_1, by multiplying the
dissociation constant, Kd, and the
forward rate constant, k₁.
k_1 = Kakı](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85c5dbd4-804e-42e1-b4a2-adf26b9b73d6%2F9687a795-9751-402e-9927-ed464518d754%2F02ri2v_processed.png&w=3840&q=75)
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- the number of low affinity binding sites occupied by a ligand at 100% saturation is greater than the number of high affinity binding sites occupied by a ligand at 100% saturation true or false?Protein A binds to Ligand X with a Kd of 1 μM. Protein B binds to Ligand Y with a Kd of 100 nM. Which of the following statements is TRUE? A. Protein A binds Ligand X ten times tighter than Protein B binds Ligand Y B. At a sufficiently high concentration of ligand, the binding becomes irreversible C. When the concentration of Ligand X is 1 μM, 50% of Protein A is bound to ligand D. Protein B binds to Ligand Y with a Kd of 1 × 10-8 mol/L. E. When both binding reactions are at equilibrium, Protein A has more Ligand X bound than Protein B has ligand Y bound The answer is C, could you show how to get the answer? Thanks12 mM of protein A is combined with 6 mM of ligand X in water. After the protein-ligand complex binding reaches equilibrium, you measure that the free ligand concentration is 3 mM and the concentration of protein-ligand complex is 3 mM. What is the Kd for protein A? Although they would be in mM, do not include units in your answer, only the number as a whole integer.
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