Question

Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L

Answer 1

Answer:

The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.

Explanation:

Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.

Nernst equation:

E = 58 millivolts/z. [Log₁₀ (C-out/C- in)

Where,    

• E = Equilibrium potential

• 58 millivolts/z = Constant

• z = Ion charge + positive or negative symbol

• C-out = Ion concentration out of the cell

• C-In = Ion concentration inside the cell

By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.

Now let us see the provided values,

• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+

• C-out = Ion concentration out of the cell ⇒ 5 mEq/L

• C-In = Ion concentration inside the cell ⇒ 150 mEq/L

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)

E = 58 millivolts (Log₁₀ 30)

E = 58 millivolts (1.477)

E = 85.67 millivolts

85.7 mV is the absolute value of equilibrium potential.

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)

E = 58 millivolts (Log₁₀ 42.85)

E = 58 millivolts (1.63)

E = 94.65 millivolts

94.7 mV is the absolute value of equilibrium potential.

If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.  

Answer 2

The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.

Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.

Nernst equation:

$E = 58 millivolts/z. [Log₁0 (C-out/ -in)$

Where,

• E = Equilibrium ability.
• 58 millivolts/z =Constant.
• z=lon charge + advantageous or terrible symbol.
• C - out = Ion awareness out of the cell.
• C-ln= ion awareness in the cell.

For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,

$z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)$$E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.$

What happens if the extracellular attention of potassium is modified from 5.0 to 3.5?

If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.

Thus it is clear from this that the potassium equilibrium potential is affected.

To know more about the potassium equilibrium refer to the link :

https://brainly.com/question/2742049