Question

3.46 The cell membrane is covered with protein complexes called Na-K+ pumps. Each pump moves 3 Nations from the intracellular space to the extracellular milieu for every 2 K ions it moves into the cell. During normal function, the pumps work constantly. Under normal conditions, there is a gradient of Nat and K ions between the cell and the extracellular space; the concentrations of these ions in these spaces can be found in Table 3.13 9. Since the ions are pumped against their respective gradients, energy in the form of ATP is required. TABLE 3.13 Normal Extracellular and Intracellular lon Concentrations Extracellular concentration (mm) Intracellular concentration (mm) Na+ 145 15 K+ 5.0 140 An experiment is conducted using ouabain, which blocks Na -K pumps in cells. During this time, the gradient collapses; the intracellular Na concentration becomes 80 mM and the intracellular K concentration 72.5 mM. After the experiment, ouabain is removed from the cells by rinsing with PBS (phosphate buffered saline). The pumps begin to operate again to reestablish the gradients. The recovery phase of the experiment lasts 4.0 hr, during which time the cells work to regain the previous ion balance. Model the ion pumps on a cell membrane during the recovery phase. Assume that the volume of the cell is 65.4 um. Assume that there are 1.0 x 10' ion pumps per cell. Assume that the pumping rate is constant (i.e., it is not dependent on the gradient of the ions). Assume that there is no diffusion of Na* or K ions across the cell membrane and that no other ion pumps or channels are active. a. Calculate the Na pumping rate for one cell (molecules/ (pump.s)) needed to reestablish the intracellular Na concentration in 4.0 hr without consideration of the K pumping rate. b. Calculate the K pumping rate for one cell (molecules/ (pump.s)) needed to reestablish the intracellular K+ concentration in 4.0 hr without consideration of the Na+ pumping rate. c. Will the cell be able to reestablish the ion balance for both Na+ and K+ to the equilibrium intracellular concentrations listed in Table 3.13 with just the described Na- K pump? Why or why not? d. In a different experiment, you calculate a Na pumping rate of 1.6 molecules/ (pump.s). What intracellular K concentration (mm) can be established in 3.0 hr? Assume the collapsed intracellular conditions above as the starting point for the recovery phase.

please explain steps and show work

no pictures were provided. table 3.13 is given above.

Answer 1

Cell volume = 65.4µm³ = 65.4µl

No. of ion pumps = 1 x 10⁵

1. Calculate Na+⁺ pumping rate (molecules/ pump/s)

Initial intracellular conc of Na+⁺ = 80mM

Final intracellular conc of Na+⁺ = 15mM

Total conc of Na+⁺ pumped out = 80 – 15 = 65mM

1M of Na+ = 22.989gm in 1L

1000mM = 22.989gm in 1L

65mM = 1.494gm in 1000000µl

65mM = 97.7µg in 65.4µl

97.7 µg of Na+ pumped out by 1 x 10⁵ pumps in 4 hours

97.7 µg of Na+ pumped out by 1 x 10⁵ pumps in 14400 seconds

0.00678 µg of Na+ pumped out by 1 x 10⁵ pumps in 1second

0.00000000678 µg of Na+ pumped out by 1 pump in 1 second

Number of moles = weight of Na+ / molecular weight of Na+

= 0.00000000678 µg/ 22989000ug

= 2.949 x 10 ^-16 molecules/pump

2. Calculate K⁺ pumping rate (molecules/ pump/s)

Initial intracellular conc of K⁺ = 72.5mM

Final intracellular conc of K⁺ = 140mM

Total conc of K⁺ pumped in = 140-72.5 = 67.5mM

1M of K+ =39.098gm in 1L

1000mM = 39.098gm in 1L

67.5mM = 2.639gm in 1000000µl

67.5mM = 172.59µg in 65.4µl

172.59 µg of K+ pumped out by 1 x 10⁵ pumps in 4 hours

172.59 µg of K+ pumped out by 1 x 10⁵ pumps in 14400 seconds

0.01198 µg of K+ pumped out by 1 x 10⁵ pumps in 1second

0.0000001198µg of Na+ pumped out by 1 pump in 1 second

Number of moles = weight of K+ / molecular weight of K+

= 0.0000001198µg / 39098000ug

= 3.064 x 10 ^-15 molecules/pump