Problem 88 - College Physics

Question

BIO The electric egg. The eggs of many species undergo a rapid change in the electrical potential difference across the outer membrane when they are fertilized. This change in potential difference affects the physiological development of the eggs. The poterntial difference across the membrane is called the membrane potential, \(V_{m},\) defined as the inside potential minus the outside potential. The membrane potential \(V_{m}\) arises when protein enzymes use the energy available in ATP to actively expel sodium ions (Na') and accumulate potassium ions \(\left(\mathrm{K}^{+}\right) .\) Because the membrane of the unfertilized egg is selectively permeable to \(\mathrm{K}^{+},\) the \(V_{m}\) of the resting sea urchin egg is about \(-70 \mathrm{mV}\) ; that is, the inside has a potential of 70 \(\mathrm{mV}\) less than that of the outside. The egg membrane behaves as a capacitor with a specific capacitance of about 1\(\mu \mathrm{F} / \mathrm{cm}^{2} .\) When a sea urchin egg is fertilized, Na' channels in the membrane are opened, \(\mathrm{Na}^{+}\) enters the egg, and \(V_{m}\) rapidly changes to \(+30 \mathrm{mV},\) where it remains for several minutes. The concentration of \(\mathrm{Na}^{+}\) in the egg's interior is about 30 mmoles/liter (30 \(\mathrm{mM} )\) and 450 \(\mathrm{mM}\) in the surrounding sea water. The inside \(\mathrm{K}^{*}\) concentration is about 200 \(\mathrm{mM}\) and the outside \(\mathrm{K}^{+}\) is 10 \(\mathrm{mM} .\) A useful constant that connects electrical and chemical units is the Faraday number, which has a value of approximately \(10^{5}\) coulomb/mole. That is, an Avogadro number (a mole) of monovalent ions such as Na^ + or \(\mathrm{K}^{+}\) carries a charge of \(10^{5} \mathrm{C}\) . How many moles of \(\mathrm{Na}^{+}\) must move per unit area of membrane to change \(V_{m}\) from \(-70 \mathrm{mV}\) to \(+30 \mathrm{mV},\) making the assumption that the membrane behaves purely as a capacitor? A. \(10^{-4}\) mole \(/ \mathrm{cm}^{2}\) B. \(10^{-9}\) mole/cm \(^{2}\) C. \(10^{-12} \mathrm{mole} / \mathrm{cm}^{2}\) D. \(10^{-14} \mathrm{mole} / \mathrm{cm}^{2}\)

Step-by-Step Solution

Answer

1 × 10^{-12} mole/cm² of Na+ ions must move, so the answer is C.

Step 1: Understanding the Problem

The problem asks us to find out how many moles of \( \mathrm{Na}^{+} \) ions must move across 1 \( \mathrm{cm}^2 \) of membrane to change the membrane potential \( V_m \) from \(-70 \mathrm{mV}\) to \(+30 \mathrm{mV}\) assuming the membrane behaves as a capacitor.

Step 2: Calculate Change in Voltage

The change in potential difference (voltage) across the membrane is calculated by subtracting the initial membrane potential from the final membrane potential: \[ \Delta V = 30 \mathrm{mV} - (-70 \mathrm{mV}) = 100 \mathrm{mV} = 0.1 \mathrm{V}. \]

Step 3: Understanding Capacitance and Charge

A capacitor's charge \( Q \) is related to capacitance \( C \) and voltage \( V \) by the formula: \[ Q = C \times \Delta V. \] The specific capacitance provided is \( 1 \mu F/cm^2 = 1 \times 10^{-6} F/cm^2 \).

Step 4: Calculate Change in Charge

Substitute the values into the formula for charge to find the change in charge per unit area: \[ Q = 1 \times 10^{-6} \mathrm{F/cm}^2 \times 0.1 \mathrm{V} = 1 \times 10^{-7} \mathrm{C/cm}^2. \]

Step 5: Convert Charge to Moles of Na+ Ions

Using the Faraday constant (\( 10^5 \mathrm{C/mole} \)), convert the charge obtained to moles of \( \mathrm{Na}^{+} \) ions: \[ \text{Moles of } \mathrm{Na}^{+} = \frac{1 \times 10^{-7} \mathrm{C/cm}^2}{10^5 \mathrm{C/mole}} = 1 \times 10^{-12} \mathrm{mole/cm}^2. \]

Step 6: Determine the Correct Answer

Based on the calculations, the required moles of \( \mathrm{Na}^{+} \) per square centimeter of membrane is \( 1 \times 10^{-12} \mathrm{mole/cm}^2 \), which corresponds to option C.

Key Concepts

Capacitance

Capacitance is an essential concept to grasp when studying membrane potential. It represents a system's ability to store an electric charge. In biological membranes, capacitance relates to how charges build up on either side of a membrane. In simpler terms, a capacitor stores and releases electrical energy. Consider the egg membrane discussed in the exercise. It acts as a natural capacitor. Capacitance is measured as farads per square centimeter, represented as \(\mu F/cm^2\).

Understanding the capacitance of a membrane is crucial. The specific capacitance given in the exercise is 1 \(\mu F/cm^2\), indicating how much charge is stored per voltage change per area. A change in voltage across the membrane leads to a shift in stored charge, typically related to the movement of ions such as sodium ions (Na^+).

This concept helps us understand how the egg can change its membrane potential during events like fertilization, where ionic movements and charge storage dynamics come into play.

Ion Channels

Ion channels are gateways in cell membranes that regulate the flow of ions in and out of a cell. These proteins are integral in establishing the membrane potential discussed in the original exercise. They permit selective permeability, enabling specific ions to move across the membrane when triggered.

For example, in the sea urchin egg problem, the controlled opening of sodium ion channels allows Na^+, in higher concentration outside the egg, to flow inside. This ion movement impacts the membrane potential. When Na^+ enters the egg, a positive charge accumulates within the cell, altering the potential from \(-70 \mathrm{mV}\) to \(30 \mathrm{mV}\).

Ion channels not only control potential but also influence vital cell functions, including signal transduction and energy production. Disturbances in these channels can lead to physiological disorders due to improper salt and energy balance in cells.

Faraday Constant

The Faraday constant is a fundamental constant in electrochemistry. It connects electrical units to chemical units, crucial for problems involving electrical charge changes. It tells us how much electric charge is carried by a mole of ions—specifically, monovalent ions like Na^+.

The constant is approximately \(10^5\mathrm{C/mole}\), meaning one mole of singly-charged ions will carry \(10^5\mathrm{C}\) of charge. In our exercise, the change in membrane potential translates into a specific amount of charge being required to move across the membrane.

• The Faraday constant helps in converting this charge (given in coulombs) to the number of moles of ions needed.
• By dividing the charge by the Faraday constant, we determine how many moles of ions, specifically Na^+ in the example, must move to achieve the desired voltage change.

Understanding the Faraday constant allows us to interpret the charge transfers in cellular processes, highly relevant in physiological and biochemical contexts.