A DNA molecule, with its double- helix structure, can in some situations behave like a spring. Measuring the force required to stretch single DNA molecules under various conditions can provide information about the biophysical properties of DNA. A technique for measuring the stretching force makes use of a very small cantilever, which consists of a beam that is supported at one end and is free to move at the other end, like a tiny diving board. The cantilever is constructed so that it obeys Hooke’s law—that is, the displacement of its free end is proportional to the force applied to it. Because different cantilevers have different force constants, the cantilever’s response must first be calibrated by applying a known force and determining the resulting deflection of the cantilever. Then one end of a DNA molecule is attached to the free end of the cantilever, and the other end of the DNA molecule is attached to a small stage that can be moved away from the cantilever, stretching the DNA. The stretched DNA pulls on the cantilever, deflecting the end of the cantilever very slightly. The measured deflection is then used to determine the force on the DNA molecule

Based on given figure below, how much elastic potential energy is stored in the DNA when it stretched\(0.1{\rm{ pN/nm}}\)?

1. \({\rm{2}}{\rm{.5}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\)
2. \(1.2 \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\)
3. \(5.0 \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\)
4. \({\rm{2}}{\rm{.5}} \times {\rm{1}}{{\rm{0}}^{ - 12}}{\rm{ J}}\)

Step by step solution

01

Elastic potential energy

Potential energy that is stored when a force is used to deform an elastic item is known as elastic potential energy.

Formula for finding elastic potential energy is

\({U_{el}} = \frac{1}{2}k{x^2}\)

02

Identification of given data

Here we have given that the DNA when it stretched \(0.1{\rm{ pN/nm}}\)

\( \Rightarrow k = 0.1{\rm{ pN/nm}}\)

Also, we have the graph

03

Finding elastic potential energy

So, let\(k = 0.1{\rm{ pN/nm}}\)and\(x = 50{\rm{ nm}}\)

\(\begin{aligned}{} \Rightarrow {U_{el}} &= \frac{1}{2}k{x^2}\\ \Rightarrow {U_{el}} &= \frac{1}{2} \times \left( {0.1 \times {{10}^{ - 12}} \times {\rm{1}}{{\rm{0}}^9}{\rm{ N/m}}} \right) \times {\left( {50 \times {{10}^{ - 9}}{\rm{ m}}} \right)^2}\\ \Rightarrow {U_{el}} &= 125 \times {10^{ - 21}}{\rm{ J}}\\ \Rightarrow {U_{el}}& = 1.2 \times {10^{ - 19}}{\rm{ J}}\end{aligned}\)

Hence elastic potential energy is \(1.2 \times {10^{ - 19}}{\rm{ J}}\)

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