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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.

During the calibration process, the cantilever is observed to deflect by\(0.10{\rm{ nm}}\)when a force of\(3.0{\rm{ pN}}\)is applied to it. What deflection of the cantilever would correspond to a force of\(6.0{\rm{ pN}}\)? (a)\(0.07{\rm{ nm}}\); (b)\(0.14{\rm{ nm}}\); (c)\(0.20{\rm{ nm}}\); (d)\(0.40{\rm{ nm}}\).

Step by step solution

01

Concept of Hooke’s law

Hooke’s law is given by,\(F = - kx\)

Where\(F{\rm{ is spring force}}\),\(k{\rm{ is constant}}\),\(x{\rm{ is deflection distance}}\).

02

Identification of given data

Here we have given that cantilever is deflected by\(0.10{\rm{ nm}}\)

\( \Rightarrow x = 0.10{\rm{ nm}}\)

Also, we have force apply on it is\({\rm{3}}{\rm{.0 pN}}\)

\( \Rightarrow F{\rm{ = 3}}{\rm{.0 pN}}\)

03

Finding deflection of the cantilever would correspond to a force of \(6.0{\rm{ pN}}\).

Now, by applying Hooke’s law we have,

\(\begin{aligned}{} \Rightarrow F{\rm{ = kx}}\\ \Rightarrow {\rm{k = }}\frac{F}{x}\\ \Rightarrow k = \frac{{3 \times {{10}^{ - 12}}{\rm{ N}}}}{{0.1 \times {{10}^{ - 9}}{\rm{ m}}}}\\ \Rightarrow k = 0.03{\rm{ N/m}}\end{aligned}\)

Now, when\(F{\rm{ = 6}}{\rm{.0 pN}}\)

\(\begin{aligned}{} \Rightarrow F{\rm{ = kx}}\\ \Rightarrow {\rm{x = }}\frac{F}{k}\\ \Rightarrow x = \frac{{6 \times {{10}^{ - 12}}{\rm{ N}}}}{{{\rm{0}}{\rm{.03 N/m}}}}\\ \Rightarrow x = 0.2 \times {10^{ - 9}}{\rm{ m}}\\ \Rightarrow x = 0.2{\rm{ nm}}\end{aligned}\)

Hence deflection of cantilever when \(F{\rm{ = 6}}{\rm{.0 pN}}\) is \(x = 0.2{\rm{ nm}}\).

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