Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose that the change in V\(_m\) was caused by the entry of Ca\(^{2+}\) instead of Na\(^+\). How many Ca\(^{2+}\) ions would have to enter the cell per unit membrane to produce the change? (a) Half as many as for Na\(^+\); (b) the same as for Na\(^+\); (c) twice as many as for Na\(^+\); (d) cannot say without knowing the inside and outside concentrations of Ca\(^{2+}\).

Short Answer

Expert verified
(a) Half as many as for Na\(^+\).

Step by step solution

01

Understand the question

We need to determine how the entry of Ca\(^{2+}\) ions affects the change in membrane potential \(V_m\) in comparison to Na\(^+\) ions. Different options are given based on how many Ca\(^{2+}\) ions are needed relative to Na\(^+\).
02

Recall the charge difference

Ca\(^{2+}\) ions have a charge of +2, compared to Na\(^+\) ions which have a charge of +1. This means each Ca\(^{2+}\) ion carries double the charge compared to each Na\(^+\) ion.
03

Determine required ions

Since Ca\(^{2+}\) is doubly charged, the entry of one Ca\(^{2+}\) ion would produce the same charge difference as two Na\(^+\) ions. Therefore, half as many Ca\(^{2+}\) ions are needed to effect the same change in \(V_m\) as Na\(^+\) ions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calcium Ions
Calcium ions, denoted as Ca\(^{2+}\), are positively charged ions that play a crucial role in various cellular processes. These ions carry a positive charge of +2, which means they are doubly charged compared to other common cations like sodium ions. This double charge is significant in the context of membrane potential, as it impacts how these ions influence charge distribution and changes in potential.

In cellular environments, calcium ions are often found in low concentrations inside cells but higher concentrations outside. This gradient is essential for signal transduction and muscle contraction. When calcium ions enter a cell, they can trigger various responses, such as the release of neurotransmitters or the initiation of muscle contraction.
Sodium Ions
Sodium ions, or Na\(^+\), are fundamental to maintaining the cell's resting membrane potential and facilitating action potentials. With a single positive charge, sodium ions are involved in transporting other ions or molecules across cell membranes.

In neurons, the rapid influx of sodium ions during an action potential leads to depolarization, a critical step in the transmission of nerve impulses. Sodium-potassium pumps work to maintain a high concentration of sodium ions outside cells and potassium ions inside, which is vital for various cellular functions.
  • Sodium ions have a +1 charge.
  • Key players in nerve impulse transmission.
  • Integral to maintaining balance in ion concentrations and membrane potential.
Ion Concentration
Ion concentration refers to the amount of specific ions present in and around a cell. This concentration plays a pivotal role in determining the membrane potential, which is essentially the voltage difference across a cell's membrane.

The membrane potential is influenced by the differential distribution of ions like calcium and sodium across the membrane, as quantified by the Nernst equation. This equation considers the intracellular and extracellular concentrations of ions to calculate the equilibrium potential.

Changes in ion concentration can lead to alterations in membrane potential, thereby impacting cellular activities such as muscle contraction, nerve firing, and the opening or closing of ion channels.
Electrochemical Gradient
The electrochemical gradient is a combination of two forces: the chemical gradient and the electrical gradient. It drives the movement of ions across membranes.

The chemical gradient refers to the difference in ion concentration between the inside and outside of the cell, whereas the electrical gradient is the difference in charge across the membrane.
  • Driving force for ion movement.
  • Maintains crucial cellular activities.
  • Essential for establishing membrane potential.

These gradients are at the heart of many physiological processes, including the transmission of nerve impulses and the operation of cardiac muscles. By maintaining a balance between these gradients, cells can ensure proper function and responsiveness to external stimuli.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free