Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

One can use FRET to follow the hybridization of two complimentary strands of DNA. When the two strands bind to form a double helix, FRET can occur from the donor to the acceptor allowing one to monitor the kinetics of helix formation (for an example of this technique see Biochemistry 34 \((1995): 285) .\) You are designing an experiment using a FRET pair with \(\mathrm{R}_{0}=59.6\) A. For B-form DNA the length of the helix increases \(3.46 \AA\) per residue. If you can accurately measure FRET efficiency down to 0.10 , how many residues is the longest piece of DNA that you can study using this FRET pair?

Short Answer

Expert verified
Answer: Approximately 24 residues.

Step by step solution

01

Recall the relationship between FRET efficiency and distance

Before starting with the calculation, remember that FRET efficiency (E) is related to the distance (r) between the donor and acceptor pairs according to the following formula: \(E = 1 - \frac{1}{(1 + (\frac{r}{R_0})^6)}\) In this exercise, the FRET efficiency (E) is given to be 0.10, and the Förster radius (R0) is 59.6 A.
02

Calculate the distance between the donor and acceptor using FRET efficiency and Förster radius

Plug the given values of FRET efficiency and Förster radius into the formula to solve for the distance (r): \(0.10 = 1 - \frac{1}{(1 + (\frac{r}{59.6})^6)}\) Solve for r: \(r = 59.6 \cdot (1 - \frac{1}{(1 - 0.10)})^{\frac{1}{6}}\) Calculate r: \(r \approx 83.61\, Å\)
03

Determine the number of residues

It is given that the length of the helix increases by 3.46 Å per residue. To find the number of residues in the longest piece of DNA that can be studied using this FRET pair, divide the calculated distance by the increase in length per residue: \( \text{Number of residues} = \frac{r}{\text{increase in length per residue}}\) \( \text{Number of residues} = \frac{83.61\, Å}{3.46\, Å/\text{residue}}\) Calculate the number of residues: \( \text{Number of residues} \approx 24\) Thus, the longest piece of DNA that can be studied using this FRET pair has approximately 24 residues.

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!

Key Concepts

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

Förster Resonance Energy Transfer
Understanding the principle of Förster Resonance Energy Transfer (FRET) is crucial when examining interactions between biomolecules, such as the hybridization of DNA strands. FRET is a distance-dependent interaction between two light-sensitive molecules, a donor and an acceptor. When the donor molecule is excited by an external light source, it can transfer energy to the acceptor molecule without any physical contact or emission of photons, provided they are close enough (typically 1-10 nm apart).

The efficiency of this energy transfer is inversely proportional to the sixth power of the distance between the donor and acceptor molecules, a relationship described by the equation:
\[E = 1 - \frac{1}{{(1 + (\frac{r}{{R_0}})^6)}}\]
where \(E\) is the FRET efficiency, \(r\) is the distance between donor and acceptor, and \(R_0\) is the Förster radius, the distance at which the transfer efficiency is 50%. This principle is widely used in biochemistry for studying molecular interactions, such as DNA hybridization, as it provides a sensitive measure of molecular distances.
DNA Double Helix Formation
The DNA double helix is the fundamental structure of the genetic code. When two complementary strands of DNA come together, they form this iconic twisted ladder shape. Each strand is made up of a sugar-phosphate backbone and nucleotide bases which pair up according to specific base-pairing rules: adenine (A) with thymine (T) and cytosine (C) with guanine (G).

Double helix formation is a spontaneous process driven by hydrogen bonding between complementary bases and stacking interactions among the aromatic base pairs. These interactions result in the characteristic helical structure, which is stabilized further by the hydrophobic effect and has a specific geometric configuration. In B-form DNA, which is the most common form of DNA in vivo, the helix makes a complete turn every 10.5 base pairs, and each base pair adds approximately 3.46 Å to the length of the DNA duplex. This information is paramount when using FRET to study the length of DNA that can form under specific conditions.
Biochemistry of Nucleic Acids
Nucleic acids, including deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), are biomolecules essential for life. They are composed of monomers known as nucleotides, which include a phosphate group, a five-carbon sugar (deoxyribose in DNA and ribose in RNA), and a nitrogenous base.

The sequence of these nucleotides in DNA encodes genetic information, which is read and translated into proteins during the process of gene expression. The biochemistry of nucleic acids is intricate, involving a variety of enzymes and proteins that assist in DNA replication, repair, and transcription. Understanding the chemical properties of nucleic acids is vital for many techniques in molecular biology, including the use of FRET in monitoring DNA interactions, such as hybridization and helix formation.
Kinetics of Helix Formation
The kinetics of helix formation refer to the rate at which DNA strands anneal to form a double helix. This process is temperature-dependent and follows the second order kinetics, as two separate strands must come together. Variables that influence the kinetics include the nucleotide sequence, salt concentration, and the presence of nucleation sites.

During the annealing process, strands search for complementary sequences, which, upon finding, will begin to zipper up, forming the double helix. Monitoring the kinetics is often done through methods such as FRET, which allows researchers to observe the rate of hybridization in real-time. Understanding the kinetics helps in elucidating the mechanisms behind genetic processes and can be crucial in applications such as DNA computing and the design of novel therapeutics.

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

Protein tyrosine phosphatases (PTPases) are a general class of enzymes that are involved in a variety of disease processes including diabetes and obesity. In a study by Z.-Y. Zhang and coworkers [J. Medicinal Chemistry 43 \((2000): 146]\) computational techniques were used to identify potential competitive inhibitors of a specific PTPase known as PTP1B. The structure of one of the identified potential competitive inhibitors is shown here: The reaction rate was determined in the presence and absence of inhibitor \(I\) and revealed the following initial reaction rates as a function of substrate concentration: $$\begin{array}{ccc} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~} \mathbf{s}^{-\mathbf{1}}\right) \\ {[\mathbf{S}](\boldsymbol{\mu} \mathbf{M})} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~} \mathbf{s}^{-1}\right),[\boldsymbol{I}]=\mathbf{0} & {\left[\begin{array}{cc} \boldsymbol{I} & =\mathbf{2 0 0} \boldsymbol{\mu} \mathbf{M} \end{array}\right]} \\ \hline 0.299 & 0.071 & 0.018 \\ 0.500 & 0.100 & 0.030 \\ 0.820 & 0.143 & 0.042 \\ 1.22 & 0.250 & 0.070 \\ 1.75 & 0.286 & 0.105 \\ 2.85 & 0.333 & 0.159 \\ 5.00 & 0.400 & 0.200 \\ 5.88 & 0.500 & 0.250 \end{array}$$ a. Determine \(K_{m}\) and \(R_{\max }\) for PTP1B. b. Demonstrate that the inhibition is competitive, and determine \(K_{i}\)

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

The overall reaction for the halogenation of a hydrocarbon (RH) using Br as the halogen is \(\mathrm{RH}(g)+\mathrm{Br}_{2}(g) \rightarrow\) \(\operatorname{RBr}(g)+\operatorname{HBr}(g) .\) The following mechanism has been proposed for this process: \\[ \begin{array}{l} \operatorname{Br}_{2}(g) \stackrel{k_{1}}{\longrightarrow} 2 \operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\operatorname{RH}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{R} \cdot(g)+\operatorname{HBr}(g) \\\ \mathrm{R} \cdot(g)+\operatorname{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{RBr}(g)+\operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\mathrm{R} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \operatorname{RBr}(g) \end{array} \\] Determine the rate law predicted by this mechanism.

A likely mechanism for the photolysis of acetaldehyde is \\[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g)+h \nu \rightarrow \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g)+\stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{3} \mathrm{CO} \cdot(g) \\ \mathrm{CH}_{3} \mathrm{CO} \cdot(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \\] Derive the rate law expression for the formation of \(\mathrm{CO}(g)\) based on this mechanism.

Consider the gas-phase isomerization of cyclopropane: Are the following data of the observed rate constant as a function of pressure consistent with the Lindemann mechanism? $$\begin{array}{cccc} \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) & \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) \\ \hline 84.1 & 2.98 & 1.36 & 1.30 \\ 34.0 & 2.82 & 0.569 & 0.857 \\ 11.0 & 2.23 & 0.170 & 0.486 \\ 6.07 & 2.00 & 0.120 & 0.392 \\ 2.89 & 1.54 & 0.067 & 0.303 \end{array}$$

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