Chapter 12, Problem 18Q

Summary Introduction

To explain: How much time the channel would take to stay open in order for the cytosolic Ca²⁺ concentration to rise to 5µM.

Concept introduction:

In cytosol,Ca²⁺ is kept in a low concentration as compared to its extracellular fluid. The movement ofCa²⁺ ion across the cell membrane is nonetheless critical, as Ca²⁺ can bind tightly to a range of proteins in the cell by changing their activities. The low the backgroundconcentration of free Ca²⁺ ion in the cytosol, themore sensitive the cell is to an increase in cytosolic concentration of Ca²⁺ ions. Thus in eukaryotic cells, a very small concentration of free Ca²⁺is maintained in theircytosol, in the face of a very much higher extracellularCa²⁺ concentration. This huge concentration gradientis achieved primarily by help of ATP-driven Ca²⁺ pumps in boththe plasma membrane and the endoplasmic reticulum membrane that vigorously pump Ca²⁺ out of the cytosol.

Explanation of Solution

Given,

The number of Ca²⁺ channels in the plasma membrane = 1000 in numbers

The cytosolic concentration of Ca²⁺ = 100nM = $100\times {10}^{-9}\text{M}$=10^-7 M

The size of the cell = 10^-15 m³

That is,

$\begin{array}{l}\text{1}{\text{m}}^{\text{3}}\text{=1000 liters}\\ {\text{10}}^{\text{-15}}{\text{m}}^{\text{3}}\text{=}{\text{10}}^{\text{-15-3}}\text{liters}\text{=}{\text{10}}^{\text{-12}}\text{liters}\end{array}$

So, the volume of the cell is 10^-12 liters.

${\text{Ca}}^{\text{2+}}\text{ions in the cell = Avogadro constant }\times \text{cell volume }\times {\text{cytosolic Ca}}^{\text{2+}}\text{concentration}$

$\begin{array}{l}{\text{Ca}}^{\text{2+}}{\text{ions in the cell = 6×10}}^{\text{23}}{\text{×10}}^{\text{-12}}{\text{×10}}^{\text{-7}}\\ \text{=}{\text{6×10}}^{\text{23}}{\text{×10}}^{\text{-19}}\\ \text{=}\text{6}\text{×}{\text{10}}^{\text{4}}{\text{Ca}}^{\text{2+}}\text{ions}\\ \text{=}\text{60,000}{\text{Ca}}^{\text{2+}}\text{ions}\end{array}$

Therefore, in the cell there are 60,000 numbers of Ca²⁺ ions present.

The concentration of Ca²⁺ ions should rise to $\text{5µm = 5}\times {\text{10}}^{\text{-6}}\text{M}$. Then, theCa²⁺ ion present in the cell is about

$\begin{array}{l}{\text{Ca}}^{\text{2+}}{\text{ions in the cell = 6×10}}^{\text{23}}{\text{×10}}^{\text{-12}}\text{×5}\times {\text{10}}^{\text{-6}}\\ \text{=}{\text{30×10}}^{\text{23}}{\text{×10}}^{\text{-19}}\\ \text{=}\text{3}\text{×}{\text{10}}^{\text{6}}{\text{Ca}}^{\text{2+}}\text{ions}\end{array}$

Therefore, the concentration rises up to 5µm, and the number of Ca²⁺ ions in the cell becomes $\text{3}\text{×}{\text{10}}^{\text{6}}{\text{Ca}}^{\text{2+}}\text{ ions}$.

Already, we have 60,000 Ca²⁺ ions. Now, we need only 2,940,000Ca²⁺ ions.

Each of the channels of the cell allows 10^-6 ions/sec.

$\begin{array}{l}{\text{So, 1000 channels = 1000 10}}^{\text{6}}\text{ions/sec}\hfill \\ \begin{array}{l}\text{ = }{\text{10}}^{\text{9}}\text{ions/sec}\\ \text{=}{\text{10}}^{\text{6}}\text{ions/millisecond}\end{array}\hfill \end{array}$

We need to pass $\text{3}\text{×}{\text{10}}^{\text{6}}{\text{Ca}}^{\text{2+}}\text{ions}$ through the cell.

In 1 millisecond, there are 10⁶ ions that pass through the cell.

So, to pass $\text{3}\text{×}{\text{10}}^{\text{6}}{\text{Ca}}^{\text{2+}}\text{ions}$ we need $\frac{\text{3}\text{×}{\text{10}}^{\text{6}}\text{ions}}{{\text{10}}^{\text{6}}\text{ions/millisecond}}\text{=3 milliseconds}$.

Therefore, to pass $\text{3}\text{×}{\text{10}}^{\text{6}}{\text{Ca}}^{\text{2+}}\text{ ions}$ through the cell membrane, the channel needs 3 milliseconds.

Conclusion

At the cytosolic Ca²⁺ concentration 5µM, the channel would take 3 milliseconds to stay open.