In a subset of voltage-gated K^+ channels, the N-terminus of each subunit acts like a tethered b

Name: Problem 9, Chapter 11: Molecular Biology of the Cell
Uploaded: 2025-06-26T02:05:17-08:00
Duration: 2 min 1 s
Channel: Hailey Tomashek

step 1 For this question, where we want to calculate the concentration for one ball in the hemisphere,

Question

In a subset of voltage-gated $\mathrm{K}^{+}$ channels, the N-terminus of each subunit acts like a tethered ball that occludes the cytoplasmic end of the pore soon after it opens, thereby inactivating the channel. This "ball-andchain" model for the rapid inactivation of voltage-gated $\mathrm{K}^{}$ channels has been elegantly supported for the shaker $\mathrm{K}^{+}$ channel from Drosophila melanogaster. (The shaker $\mathrm{K}^{}$ channel in Drosophila is named after a mutant form that causes excitable behavior-even anesthetized flies keep twitching.) Deletion of the N-terminal amino acids from the normal shaker channel gives rise to a channel that opens in response to membrane depolarization, but stays open instead of rapidly closing as the normal channel does. A peptide (MAAVAGLYGLGEDRQHRKKQ) that corresponds to the deleted N-terminus can inactivate the open channel at $100 \mu \mathrm{M}$ Is the concentration of free peptide $(100 \mu \mathrm{M})$ that is required to inactivate the defective $K^{+}$ channel anywhere near the local concentration of the tethered ball on a normal channel? Assume that the tethered ball can explore a hemisphere [volume $\left.=(2 / 3) \pi r^{3}\right]$ with a radius of $21.4 \mathrm{nm}$ which is the length of the polypeptide "chain" (Figure Q1 $1-2$ ). Calculate the concentration for one ball in this hemisphere. How does that value compare with the concentration of free peptide needed to inactivate the channel?

   In a subset of voltage-gated $\mathrm{K}^{+}$ channels, the N-terminus of each subunit acts like a tethered ball that occludes the cytoplasmic end of the pore soon after it opens, thereby inactivating the channel. This "ball-andchain" model for the rapid inactivation of voltage-gated $\mathrm{K}^{*}$ channels has been elegantly supported for the shaker $\mathrm{K}^{+}$ channel from Drosophila melanogaster. (The shaker $\mathrm{K}^{*}$ channel in Drosophila is named after a mutant form that causes excitable behavior-even anesthetized flies keep twitching.) Deletion of the N-terminal amino acids from the normal shaker channel gives rise to a channel that opens in response to membrane depolarization, but stays open instead of rapidly closing as the normal channel does. A peptide (MAAVAGLYGLGEDRQHRKKQ) that corresponds to the deleted N-terminus can inactivate the open channel at $100 \mu \mathrm{M}$
Is the concentration of free peptide $(100 \mu \mathrm{M})$ that is required to inactivate the defective $K^{+}$ channel anywhere near the local concentration of the tethered ball on a normal channel? Assume that the tethered ball can explore a hemisphere [volume $\left.=(2 / 3) \pi r^{3}\right]$ with a radius of $21.4 \mathrm{nm}$ which is the length of the polypeptide "chain" (Figure Q1 $1-2$ ). Calculate the concentration for one ball in this hemisphere. How does that value compare with the concentration of free peptide needed to inactivate the channel?

Molecular Biology of the Cell

Bruce Alberts,… 6th Edition

Chapter 11, Problem 9

Instant Answer

Solved by Expert Hailey Tomashek

06/26/2025

Step 1

First, we need to find the volume of the hemisphere that the tethered ball can explore. The volume of a hemisphere is given by the formula V = (2/3)πr^3, where r is the radius. In this case, the radius is 21.4 nm. Show more…

Show all steps

Thanks for your feedback!

In a subset of voltage-gated $\mathrm{K}^{+}$ channels, the N-terminus of each subunit acts like a tethered ball that occludes the cytoplasmic end of the pore soon after it opens, thereby inactivating the channel. This "ball-andchain" model for the rapid inactivation of voltage-gated $\mathrm{K}^{*}$ channels has been elegantly supported for the shaker $\mathrm{K}^{+}$ channel from Drosophila melanogaster. (The shaker $\mathrm{K}^{*}$ channel in Drosophila is named after a mutant form that causes excitable behavior-even anesthetized flies keep twitching.) Deletion of the N-terminal amino acids from the normal shaker channel gives rise to a channel that opens in response to membrane depolarization, but stays open instead of rapidly closing as the normal channel does. A peptide (MAAVAGLYGLGEDRQHRKKQ) that corresponds to the deleted N-terminus can inactivate the open channel at $100 \mu \mathrm{M}$ Is the concentration of free peptide $(100 \mu \mathrm{M})$ that is required to inactivate the defective $K^{+}$ channel anywhere near the local concentration of the tethered ball on a normal channel? Assume that the tethered ball can explore a hemisphere [volume $\left.=(2 / 3) \pi r^{3}\right]$ with a radius of $21.4 \mathrm{nm}$ which is the length of the polypeptide "chain" (Figure Q1 $1-2$ ). Calculate the concentration for one ball in this hemisphere. How does that value compare with the concentration of free peptide needed to inactivate the channel?

Play audio

Feedback

Hailey Tomashek and 53 other subject Biology educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Voltage-Gated Potassium Channels

These are ion channels that open or close in response to changes in membrane potential, allowing potassium ions to pass through the cell membrane. They play a critical role in generating and shaping action potentials in excitable cells by precisely controlling the timing and duration of the potassium currents, which, in turn, influence the electrical signaling of neurons and muscle cells.

Ball-and-Chain Inactivation Mechanism

This is a rapid inactivation process in certain voltage-gated channels where a tethered peptide segment (the 'ball') is attached to the channel by a flexible linker (the 'chain'). Upon channel activation, the ball quickly swings into and blocks the channel pore from the intracellular side, ensuring that the channel inactivates rapidly. This mechanism is essential for controlling the duration of the channel’s open state and for the proper repolarization of the membrane.

Effective Local Concentration in Confined Spaces

This concept involves determining the local concentration of a molecule that is tethered within a confined volume, such as a hemispherical region accessible to a tethered peptide segment. The effective local concentration is generally very high because the tethered molecule is restricted within a limited spatial volume relative to free molecules in bulk solution. This high local concentration can lead to efficient and rapid interactions with target sites, in contrast to the relatively lower concentration required when the molecule is free in solution.

Please give Ace some feedback