/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 BIO Electrophoresis. Electrophor... [FREE SOLUTION] | 91Ó°ÊÓ

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BIO Electrophoresis. Electrophoresis is a process used by biologists to separate different biological molecules (such as proteins) from each other according to their ratio of charge to size. The materials to be separated are in a viscous solution that produces a drag force \(F_{\mathrm{D}}\) proportional to the size and speed of the molecule. We can express this relation- ship as \(F_{\mathrm{D}}=K R v,\) where \(R\) is the radius of the molecule (modeled as being spherical), \(v\) is its its speed, and \(K\) is a constant that depends on the viscosity of the solution. The solution is placed in an external electric field \(E\) so that the electric force on a particle of charge \(q\) is \(F=q E\) . (a) Show that when the electric field is adjusted so that the two forces (electric and viscous drag) just balance, the ratio of \(q\) to \(R\) is \(K v / E\) . (b) Show that if we leave the electric field on for a time \(T,\) the distance \(x\) that the molecule moves during that time is \(x=(E T / k)(q / R)\) . (c) Suppose you have a sample containing three different biological molecules for which the molecular ratio \(q / R\) for material 2 is twice that of material 1 and the ratio for material 3 is three times that of material 1. Show that the distances migrated by these molecules after the same amount of time are \(x_{2}=2 x_{1}\) and \(x_{3}=3 x_{1}\) . In other words, material 2 travels twice as far as material \(1,\) and material 3 travels three times as far as material \(1 .\) Therefore, we have separated these molecules according to their ratio of charge to size. In practice, this process can be carried out in a special gel or paper, along which the biological molecules migrate. (Fig. P21.94). The process can be rather slow, requiring several hours for separations of just a centimeter or so.

Short Answer

Expert verified
The solution shows how electrophoresis separates molecules by charge-to-radius ratio based on migration distance.

Step by step solution

01

Equate Electric Force and Drag Force

First, we set the electric force \( F = qE \) equal to the drag force \( F_D = KRv \). Since they balance, we have:\[ qE = KRv \]
02

Solve for the Ratio \( \frac{q}{R} \)

From the equation \( qE = KRv \), solve for \( \frac{q}{R} \):\[ \frac{q}{R} = \frac{Kv}{E} \]This shows the relationship between the charge-to-radius ratio and the constants involved.
03

Calculate Distance Travelled \( x \)

When the electric field is left on for time \( T \), the velocity \( v \) of the molecule becomes constant. The distance \( x \) traveled is:\[ x = vT \]Using \( qE = KRv \) and substituting \( v = \frac{qE}{KR} \), we have:\[ x = \left(\frac{qE}{KR}\right)T \]Rewriting, we get:\[ x = \left(\frac{ET}{K}\right)\left(\frac{q}{R}\right) \]
04

Analyze the Ratios for Different Materials

Given that material 2 has a \( \frac{q}{R} \) ratio twice that of material 1 and material 3 is three times that of material 1, we examine their distances:- For material 1: \( x_1 = \left(\frac{ET}{K}\right)\left(\frac{q_1}{R_1}\right) \)- For material 2: \( x_2 = \left(\frac{ET}{K}\right)\left(2 \times \frac{q_1}{R_1}\right) = 2x_1 \)- For material 3: \( x_3 = \left(\frac{ET}{K}\right)\left(3 \times \frac{q_1}{R_1}\right) = 3x_1 \)
05

Conclude the Separation Based on Migration Distances

The distances calculated show that material 2 travels twice as far and material 3 travels three times as far as material 1. Thus, they are separated according to their charge-to-size ratios, achieving the purpose of electrophoresis.

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Key Concepts

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

Charge-to-size ratio
In electrophoresis, the charge-to-size ratio is an essential concept, as it determines how molecules move relative to one another in an electric field. Each molecule, depending on its charge and size, will experience varying influences from the electric field. A higher charge combined with a smaller size results in a higher charge-to-size ratio, allowing the molecule to move faster and further.

To better understand this, consider a charged sphere moving through a gel under an electric field. The sphere's charge provides the force necessary to propel it, while its size affects the resistance it encounters, known as viscous drag. Therefore, the overall movement or separation during electrophoresis primarily depends on this ratio.

Practical applications include separating proteins or nucleic acids in molecular biology, where differences in the charge-to-size ratio allow researchers to identify and analyze distinct biomolecules based on their migration rates.
Viscous drag force
Viscous drag force is a resistance experienced by a particle moving through a viscous medium, such as a gel used in electrophoresis. This force is a product of the viscosity of the medium and the particle's velocity and size. This relationship can be mathematically expressed as:

\[ F_D = K R v \]
Where:
  • \( F_D \) is the viscous drag force
  • \( K \) is a constant that describes the medium's viscosity
  • \( R \) is the radius of the particle
  • \( v \) is the speed of the particle
As much as the electric field tries to push the particle forward, the viscous drag opposes this motion based on the size and speed of the particle. Larger molecules usually experience a higher drag force because they have a larger surface area interacting with the medium.

This resistance must be overcome by the electric force for successful migration of molecules, making viscous drag a critical factor in molecular separation techniques like electrophoresis.
Electric field in biology
An electric field in biology, especially in electrophoresis, serves as the driving force that moves charged particles through a medium. The electric field is defined by the relationship:

\[ F = qE \]
Where:
  • \( F \) is the force
  • \( q \) is the charge of the particle
  • \( E \) is the electric field strength
This field acts on charged molecules, causing them to accelerate in the direction of the field's applied force. By adjusting the strength of the electric field, different charged particles can be manipulated to travel specific distances, helping to separate and analyze them.

For biological molecules like DNA or proteins, which naturally have varying charges and sizes, the electric field allows for their separation based on the charge-to-size ratio. The effectiveness of separation techniques, such as those in electrophoresis, often relies heavily on how well this field is controlled and maintained during experiments.
Molecular separation techniques
Molecular separation techniques are pivotal in biology for distinguishing and analyzing different molecules. Electrophoresis is one such technique that separates molecules based on their charge-to-size ratios using an electric field. In a typical setup, molecules are placed in a gel matrix that provides resistance while an electric current is applied.

These techniques are crucial for:
  • Identifying proteins and nucleic acids
  • Purifying compounds for further analysis
  • Diagnosing genetic or infectious diseases
In practice, variations in electrophoresis, like agarose gel electrophoresis or SDS-PAGE, cater to specific types of molecules and purposes. The clarity of molecular separation achieved through techniques like this allows scientists to perform quantitative and qualitative analyses of complex biological mixtures.

Thus, mastering molecular separation techniques opens up vast possibilities in scientific research, medical diagnostics, and biotechnology applications.

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Most popular questions from this chapter

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is \(-3.20 \times 10^{-9} \mathrm{C}\) (a) Find the M number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is \(82,\) and its atomic mass is 207 \(\mathrm{g} / \mathrm{mol}\) .

CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively,

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

BI0 Signal Propagation in Neurons. Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e\) , enter the axon. How many coulombs of charge enter a \(1.5-\mathrm{cm}\) length of the axon during this process?

LA A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{N} / \mathrm{C}\) . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00\(\mu\) in the field, assuming it starts from rest.

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