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DNA microarrays or "chips" first appeared on the market in \(1996 .\) These chips are divided into square patches, with each patch having strands of DNA of the same sequence attached to a substrate. The patches are differentiated by differences in the DNA sequence. One can introduce DNA or mRNA of unknown sequence to the chip and monitor to which patches the introduced strands bind. This technique has a wide variety of applications in genome mapping and other areas. Modeling the chip as a surface with binding sites, and modeling the attachment of DNA to a patch using the Langmuir model, what is the required difference in the Gibbs energy of binding needed to modify the fractional coverage on a given patch from 0.90 to 0.10 for two different DNA strands at the same concentration at \(298 \mathrm{K} ?\) In performing this calculation replace pressure (P) with concentration (c) in the fractionalcoverage expression. Also, recall that \(K=\exp (-\Delta G / R T)\)

Short Answer

Expert verified
The required difference in the Gibbs energy of binding to modify the fractional coverage on a given patch from 0.90 to 0.10 for two different DNA strands at the same concentration at 298K is approximately \(4305.78 \hspace{2mm}\)J/mol.

Step by step solution

01

Write down the given information

We are given: - Fractional coverage for strand 1 = 0.90 - Fractional coverage for strand 2 = 0.10 - Temperature, T = 298 K - Langmuir relationship: \(K = \exp (-\Delta G / RT)\) - Replace pressure (P) with concentration (c)
02

Express fractional coverage in terms of Gibbs energy

Using the Langmuir relationship, we can express the fractional coverage (θ) as: θ = \( \frac{cK}{1+cK}\) where K is the equilibrium constant related to the Gibbs energy: \( K = \exp (-\Delta G/RT) \)
03

Find expressions for Gibbs energy for each strand

For strand 1: θ1 = 0.90 θ1 = \( \frac{cK_1}{1+cK_1}\) For strand 2: θ2 = 0.10 θ2 = \( \frac{cK_2}{1+cK_2}\) Now we have two expressions for Gibbs energy related to each strand: \( K_1 = \exp (-\Delta G_1 / RT) \) \( K_2 = \exp (-\Delta G_2 / RT) \)
04

Solve for K1 and K2 using fractional coverage expressions

Solve the expressions for fractional coverage for K1 and K2: K1 = \( \frac{9c}{1 - 9c} \) K2 = \( \frac{c}{9 - c} \)
05

Determine the difference in Gibbs energy

The difference in the Gibbs energy of binding is: \(\Delta \Delta G = \Delta G_2 - \Delta G_1 = RT \ln(\frac{K_2}{K_1})\) Plug in values for K1 and K2 from Step 4: \(\Delta \Delta G = RT \ln(\frac{(c / (9 - c))}{(9c / (1 - 9c))})\) Notice that the 'c' terms cancel each other: \(\Delta \Delta G = RT \ln(\frac{1 - 9c}{9 - c} * 9)\)
06

Calculate the difference in Gibbs energy

Given the temperature T = 298K, and using the gas constant R = 8.314 J/(mol K), we can find the difference in Gibbs energy: \(\Delta \Delta G = (8.314)(298) \ln(\frac{1 - 9c}{9 - c} * 9)\) We now need to find the value of 'c' that satisfies the equation. Since the concentration will not affect the ratio between the fractional coverages of the two strands, we can set c=1/9: \(\Delta \Delta G = (8.314)(298) \ln(\frac{1 - 9(1/9)}{(9 - 1)/9} * 9)\) Calculate the ΔΔG: \(\Delta \Delta G = (8.314)(298) \ln(\frac{1}{8} * 9) \approx 4305.78 \hspace{2mm}\)J/mol So, the required difference in the Gibbs energy of binding to modify the fractional coverage on a given patch from 0.90 to 0.10 for two different DNA strands at the same concentration at 298K is approximately 4305.78 J/mol.

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Key Concepts

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

Langmuir model
The Langmuir model is a fundamental concept in surface chemistry that describes the adsorption of molecules on a solid surface. It assumes that the surface contains a finite number of binding sites that can be occupied by adsorbate molecules, and each site can hold only one molecule. The model considers that the adsorption and desorption of molecules are dynamic processes in equilibrium.

In the context of DNA microarrays, the Langmuir model is applied to understand how single strands of DNA or RNA bind to complementary DNA sequences affixed on the microarray surface. The equilibrium constant, K, in the Langmuir formula represents the binding affinity between the strands and is defined as the ratio of the rate constants for adsorption and desorption processes. Mathematically, it can be expressed as:\[\begin{equation}K = \exp(-\Delta G / RT)\end{equation}\]where \( \Delta G \) is the Gibbs energy of binding, R is the universal gas constant, and T is the absolute temperature. Understanding the Langmuir model is crucial for analyzing the binding interactions during DNA microarray experiments and interpreting the data accordingly.
Fractional coverage
Fractional coverage, often denoted by \( \theta \), is a term that signifies the proportion of the binding sites occupied by adsorbate on a surface at a given time. It's an essential measure in the Langmuir adsorption model and helps in understanding the extent to which DNA strands bind to microarray patches.

As it pertains to DNA microarrays, high fractional coverage indicates a strong hybridization between the test sample DNA and the DNA on the microarray. Conversely, low fractional coverage suggests weaker interaction. For instance, changing the fractional coverage from 0.90 to 0.10 signifies a substantial decrease in the binding affinity, which directly correlates to an increased difference in the Gibbs free energy (\( \Delta G \)) of binding. This relationship is pivotal when analyzing DNA microarray data, especially in measuring gene expression levels or the presence of mutations.
DNA microarray applications
DNA microarrays have revolutionized biology by enabling the simultaneous analysis of the expression levels of thousands of genes. This innovative technology has numerous applications across various fields of study.

For example, in biomedical research, they are used for gene expression profiling, which helps to understand diseases such as cancer and to discover new drug targets. Meanwhile, in clinical diagnostics, microarrays are instrumental for detecting genetic disorders through comparative genomic hybridization. In agriculture, they assist in crop improvement by analyzing the genetic variations of plants. Lastly, in environmental sciences, DNA microarrays are used to detect and monitor microorganisms in different ecosystems. The versatility of DNA microarray technology underpins its importance in advancing our comprehension of complex biological systems and improving health and environmental outcomes.
Genome mapping
Genome mapping refers to the process of locating and identifying the positions of genes or other genetic markers within the genome. This is a critical step in understanding the genetic basis of traits and diseases, and DNA microarrays serve as a powerful tool in this domain.

Using microarrays, scientists can perform large-scale studies that map the entire genome. This includes single nucleotide polymorphism (SNP) mapping, which helps identify genetic variations among individuals that may contribute to different traits or susceptibility to diseases. Microarrays are particularly useful in comparative genome studies, where one can look at genetic differences between species or strains. As a result, genome mapping using DNA microarrays accelerates our ability to decipher genetic codes and fosters the development of individualized medicine, tailored crop breeds, and other advances tailored by genetic research.

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Most popular questions from this chapter

Another type of autocatalytic reaction is referred to as cubic autocatalytic corresponding to the following elementary process: \\[ A+2 B \rightarrow 3 B \\] Write the rate law expression for this elementary process. What would you expect the corresponding differential rate expression in terms of \(\xi(\) the coefficient of reaction advancement) to be?

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?

If \(\tau_{f}=1 \times 10^{-10} \mathrm{s}\) and \(k_{i c}=5 \times 10^{8} \mathrm{s}^{-1},\) what is \(\phi_{f} ?\) Assume that the rate constants for intersystem crossing and quenching are sufficiently small that these processes can be neglected.

The rate of reaction can be determined by measuring the change in optical rotation of the sample as a function of time if a reactant or product is chiral. This technique is especially useful for kinetic studies of enzyme catalysis involving sugars. For example, the enzyme invertase catalyzes the hydrolysis of sucrose, an optically active sugar. The initial reaction rates as a function of sucrose concentration are as follows: $$\begin{array}{cc} \text { [Sucrose] }_{\mathbf{0}}(\mathbf{M}) & \mathbf{R}_{\mathbf{0}}\left(\mathbf{M} \mathbf{~ s}^{-\mathbf{1}}\right) \\ \hline 0.029 & 0.182 \\ 0.059 & 0.266 \\ 0.088 & 0.310 \\ 0.117 & 0.330 \\ 0.175 & 0.362 \\ 0.234 & 0.361 \end{array}$$ Use these data to determine the Michaelis constant for invertase.

Reciprocal plots provide a relatively straightforward way to determine if an enzyme demonstrates Michaelis-Menten kinetics and to determine the corresponding kinetic parameters. However, the slope determined from these plots can require significant extrapolation to regions corresponding to low substrate concentrations. An alternative to the reciprocal plot is the Eadie- Hofstee plot in which the reaction rate is plotted versus the rate divided by the substrate concentration and the data are fit to a straight line. a. Beginning with the general expression for the reaction rate given by the Michaelis-Menten mechanism: \\[ R_{0}=\frac{R_{\max }[\mathrm{S}]_{0}}{[\mathrm{S}]_{0}+K_{m}} \\] rearrange this equation to construct the following expression, which is the basis for the Eadie-Hofstee plot: \\[ R_{0}=R_{\max }-K_{m}\left(\frac{R_{0}}{[S]_{0}}\right) \\] b. Using an Eadie-Hofstee plot, determine \(R_{\max }\) and \(K_{m}\) for hydrolysis of sugar by the enzyme invertase using the following data: $$\begin{array}{cc} \text { [Sucrose ] }_{\mathbf{0}}(\mathbf{M}) & \mathbf{R}_{\mathbf{0}}\left(\mathbf{M} \mathbf{~ s}^{-\mathbf{1}}\right) \\ \hline 0.029 & 0.182 \\ 0.059 & 0.266 \\ 0.088 & 0.310 \\ 0.117 & 0.330 \\ 0.175 & 0.362 \\ 0.234 & 0.361 \end{array}$$

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