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 =
The size of the cell = 10-15 m3
That is,
So, the volume of the cell is 10-12 liters.
Therefore, in the cell there are 60,000 numbers of Ca2+ ions present.
The concentration of Ca2+ ions should rise to
Therefore, the concentration rises up to 5µm, and the number of Ca2+ ions in the cell becomes
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.
We need to pass
In 1 millisecond, there are 106 ions that pass through the cell.
So, to pass
Therefore, to pass
At the cytosolic Ca2+ concentration 5µM, the channel would take 3 milliseconds to stay open.
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