Assume the energy of hydrogen bonds per base pair to be \(10 \mathrm{~kJ} / \mathrm{mol}\). Given two complementary strands of DNA containing 10 base pairs each, estimate the ratio of two separate strands to hydrogen-bonded double helix in solution at \(300 \mathrm{~K}\). The ratio for a single base pair is given by the formula \(\exp (-\Delta E / R T)\) where \(\Delta E\) is the energy of hydrogen bond per base pair, \(R\) is the gas constant, and \(T\) is the temperature in kelvins.

Short Answer

Expert verified
The ratio of two separate strands to hydrogen-bonded double helix in solution at 300 K is approximately 1.15E-54

Step by step solution

01

Substitution of values

First, substitute the given values into the formula for the ratio of a single base pair. The energy of hydrogen bond per base pair, \( \Delta E \) is given as 10 kJ/mol. So, convert it to joules (1 kJ = 1000 J) to be consistent with the SI unit of R. It becomes 10000 J/mol. The gas constant, \( R \), is 8.3145 J/(mol.K) and the temperature, \( T \), is 300 K. The formula becomes \(\exp (-10000 / (8.3145 * 300))\).
02

Calculate the ratio for a single base pair

Calculate the expression in the formula to get the ratio for a single base pair. The resulting ratio is approximately 2.08E-06.
03

Calculate the overall ratio for all base pairs

Since there are 10 base pairs, the overall ratio is given by the ratio for a single base pair raised to the power of the number of base pairs. So, it's \((2.08E-06)^{10}\). Calculate this expression to get the overall ratio.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free