Problem 1 In the chapter, we argued that t... [FREE SOLUTION]

Chapter 14: Problem 1

In the chapter, we argued that the mean spacing between molecules in an in vitro biochemical experiment is roughly \(100 \mathrm{nm}\) at \(\mu \mathrm{M}\) concentrations while in the cell the spacings are a factor of 10 smaller. Justify these statements with simple estimates. The biochemical "standard state" is often taken as \(1 \mathrm{M}\). Work out the mean spacing between molecules at this concentration.

Short Answer

Expert verified

The mean spacing between molecules is: about 118 nm at 碌M concentrations, roughly 10 times smaller within a cell (12 nm), and approximately 0.117 nm at 1M concentration.

Step by step solution

Convert micromolar (碌M) to molar (M)

First, convert the concentration from micromolar (碌M) to molar (M). Remember that 1 碌M = 1e-6 M.

Calculate number of molecules per volume

Next, we calculate the number of molecules per liter using Avogadro's number (NA), with \[NA = 6.022 * 10^{23} \text{ molec/mol}\]. There are NA molecules in 1 M of any substance. So, in a 1 碌M = 1e-6 M, there are \(6.022 * 10^{23} * 1e-6 = 6.022 * 10^{17}\) molecules per liter.

Calculate the Volume occupied by each molecule

Given that each micromole distributed evenly through 1 L of water, then each molecule must occupy a volume of \[1/ (6.022 * 10^{17}) = 1.66 * 10^{-21} L/molecule = 1.66 * 10^{-18} \, cm鲁/molecule\]. This is a cubic volume of space.

Calculate the distance between molecules

A key step is converting the volume per molecule to a distance, or the mean molecular spacing. We visualize each molecule as being located at the center of a cube. The cube root of the volume gives the average edge length, which is comparable to the mean spacing between molecules at a concentration of 1 碌M. The cube root of \(1.66 * 10^{-18} cm鲁\), or average edge length, gives an approximate molecular spacing of 118 nm.

Calculate the distance between molecules at a concentration of 1M

Lastly, the same calculation is done for a concentration of 1M, for a volume of \(1/(6.022 * 10^{23}) = 1.66 * 10^{-24} cm鲁/molecule\). The mean spacing in this circumstance, taking the cube root of this volume, gives a value of 0.117 nm

Compare the distances at different concentrations

By comparing these calculated distances, you can see that as the concentration increases from 碌M to M, the distance between molecules significantly decreases, justifying the given statements.

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Short Answer

Step by step solution

Convert micromolar (碌M) to molar (M)

Calculate number of molecules per volume

Calculate the Volume occupied by each molecule

Calculate the distance between molecules

Calculate the distance between molecules at a concentration of 1M

Compare the distances at different concentrations

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