For a receptor-mediated signal transduction pathway to be sensitive to changes i

Question

For a receptor-mediated signal transduction pathway to be sensitive to changes in signal concentration, it is important that the magnitude of the cellular response to the signal be proportional to the extracellular concentration of signal present. In other words, an increase in the concentration of signal should stimulate a proportional increase in the cellular response. An important factor in the sensitivity of a receptor to its signal (or ligand) is the dissociation constant (Kd). For a population of receptors on a cell’s surface to respond proportionally to increases in signal concentration, the resting concentration of signal must be below the Kd of the receptor for the ligand.

Please use the following information to answer questions A through E (show your work):

Receptor A in the plasma membrane binds to ligand B with a Kd of 2.5 X 10-7 M. The resting concentration of ligand B in the extracellular space is 4.0 X 10-9 M.

Receptor X in the plasma membrane binds to ligand Y with a Kd of 3.0 X 10-8 M. The resting concentration of ligand Y in the extracellular space is the same as the Kd, 3.0 X 10-8.

A. What fraction of receptor A is bound to ligand B ([RL]/[RT]) in the resting state?

B. What fraction of receptor A is bound if the concentration of ligand B in the extracellular space increases 5-fold?

C. What fraction of receptor X is bound to ligand Y in the resting state?

D. What fraction of receptor X is bound if the concentration of ligand Y in the extracellular space increases 5-fold?

E. Please write a paragraph that uses the results of your calculations to illustrate that surface receptors will respond proportionally to increases in signal concentration only when the resting concentration of signal is below the Kd of the receptor.

Explanation / Answer

a. [L]/Kd + [L]

= 4.0 X 10-9 M/ 2.5 X 10-7 M + 4.0 X 10-9 M = 4.0 X 10-9 M/2.54 x 10-7

1.54 x 10-2

b. increased concentration 5(4.0 X 10-9 M) = 20 X 10-9 M

20 X 10-9 M/ 2.5 X 10-7 M + 20 X 10-9 M = 0.88 x 10-2

c. 3.0 X 10-8/3.0 X 10-8 +3.0 X 10-8 = 0.5

d.increased concentration 5(3.0 X 10-8)

15X 10-8/ 15.0 X 10-8 + 3.0 X 10-8

0.83

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