Molecular Motors [VISUAL] Molecular motors are ubiquitously used by cells for many purposes, includi — Analytical Chemistry Chemistry Question
Molecular Motors
Molecular Motors
[VISUAL]
Molecular motors are ubiquitously used by cells for many purposes, including transporting various cargos from one part of the cell to another. One important motor protein is kinesin, which walks on filamentous tubes called microtubules made of another protein tubulin. In fact, kinesin is an enzyme, an ATPase, powered by hydrolysis of adenosinetriphosphate, ATP.
Consider placing a macroscopically long microtubule into a solution of free kinesin, Pfree, with the concentration [Pfree] and assume that there is equilibrium between tubule-bound kinesin (Pbound), free kinesin and binding sites (Site) available on the surface of the microtubule:
Pbound <=> Pfree + Site
The occupancy of single binding sites by kinesin molecules is governed by the law of mass action:
[Pfree][Site] / [Pbound] = Kd
where [Site] is the total concentration of binding sites on the microtubule, [Pbound] is the concentration of the kinesin molecules bound to the microtubule, and Kd is the equilibrium constant.
When the kinesin molecule is bound to a microtubule, it moves unidirectionally along its surface with a speed, v = 640 nm/s.
Imagine a geometric plane, which is oriented perpendicular to the microtubule and intersects the microtubule at some specific position along the tube. This plane is called a cross section.
Estimate the rate of passage of kinesin molecules through an arbitrary cross section of the microtubule in units of kinesin molecules per second. This rate of passage of kinesin molecules is related to the rate at which the microtubule–derived nanomotor moves in one or another direction. Use the following information:
* There are n = 16 kinesin binding sites per each l = 5 nm length of the microtubule.
* Kinesin molecules move independently of each other.
* Assume that kinesin molecules bound on the microtubule sites and free kinesin molecules in solution are in a dynamic equilibrium.
* Use the following parameters: Kd = 0.5 · 10–6, [PFree] = 100 . 10–9 mol dm–3, and [Site] = 10 . 10–6 mol dm–3.
Model Answer
Consider an imaginary cross section plane cutting the microtubule, and ask how many motors will pass through it during some time, τ . Then, it follows that any motor which is found within a distance of vτ from the cross section plane at the given instant of time will cross the plane within the subsequent time duration, τ . Hence, the total number of motors crossing the wall during time τ is (line-density of motors) · (length of vτ ), leading to ρM vτ , where ρM is the linear density of motors, in units of number of motors per nm. To obtain the rate of passage of motors per unit time, we divide the latter expression by τ, resulting in the rate of passage of motors through a cross section per unit time as, J_M = ρ_M · v.
Next, we have to find ρM. From the mass action law, the concentration of bound motors is,
[Pbound] = [Pfree][Site] / Kd = (0.1 [µM] × 10 [µM]) / 0.5 [µM] = 2 [µM].
(where µM = mol dm-3)
This means that the fraction of occupied sites on the microtubule is,
f = [Pbound] / [Site] = 2 [µM] / 10 [µM] = 0.2.
Per l = 5 nm, there are 16 kinesin binding sites, but only 20 % of them are occupied by kinesins at equilibrium. Hence, the line density of motors is:
ρM = (16 / 5 nm) × 0.2 = 0.64 nm^-1.
Substituting into J_M = ρ_M · v we obtain:
J_M = 0.64 nm^-1 × 640 nm s^-1 = 410 s^-1.
Consequently, 410 motors pass through the cross section of the microtubule per second.