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Chapter 19: Problem 31

The melting temperature of a DNA molecule can be determined by differential scanning calorimetry (Impact 12.1) The following data were obtained in aqueous solutions containing the specified concentration \(\mathrm{c}_{\text {salt }}\) of an soluble ionic solid for a series of DNA molecules with varying base pair composition, with \(f\) the fraction of \(\mathrm{G}-\mathrm{C}\) base pairs: $$ C_{\text {salt }}=1.0 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} $$ $$ \begin{array}{llllll} f & 0.375 & 0.509 & 0.589 & 0.688 & 0.750 \\ T_{\mathrm{m}} / \mathrm{K} & 339 & 344 & 348 & 351 & 354 \end{array} $$ \(c_{\text {salt }}=0.15 \mathrm{~mol} \mathrm{dm}^{-3}\) $$ \begin{array}{llllll} f & 0.375 & 0.509 & 0.589 & 0.688 & 0.750 \\ T_{\mathrm{m}} / \mathrm{K} & 359 & 364 & 368 & 371 & 374 \end{array} $$ (a) Estimate the melting temperature of a DNA molecule containing \(40.0\) per cent \(\mathrm{G}-\mathrm{C}\) base pairs in both samples. Himt. Begin by plotting \(T_{\mathrm{m}}\) against fraction of \(G-C\) base pairs and examining the shape of the curve. (b) Do the data show an effect of concentration of ions in solution on the melting temperature of DNA? If so, provide a molecular interpretation for the effect you observe.

Short Answer

Expert verified
The melting temperatures can be estimated by linear interpolation from the given data points for both salt concentrations. The effect of salt concentration on melting temperatures can be observed, with higher salt concentrations resulting in higher melting temperatures, which can be explained by the stabilizing effect of ions on the DNA double helix, reducing repulsion between negatively charged phosphate groups.

Step by step solution

01

Plot the Data

Plot the provided data points, with fraction of G-C base pairs (\f$) on the horizontal axis and melting temperature (\(T_{\text{m}}\) in Kelvin) on the vertical axis. Create two separate plots—one for the data at the salt concentration of \(1.0 \times 10^{-2} \text{ mol dm}^{-3}\) and one for \(0.15 \text{ mol dm}^{-3}\).
02

Analyze the Plot

Examine the shape of the plotted curves to determine if they suggest a linear relationship between the fraction of G-C base pairs and the melting temperature. If the relationship appears linear, fit a straight line to the data points. This straight line will be used to interpolate the melting temperature for a DNA molecule containing 40.0% of G-C base pairs.
03

Linear Interpolation

Using the linear equation for both salt concentrations obtained from the plotting, perform a linear interpolation to estimate the melting temperature (\(T_{\text{m}}\) for a DNA molecule with 40.0% G-C base pairs, which is a fraction of 0.400 for both salt concentrations.
04

Compare Melting Temperatures at Different Salt Concentrations

Once both melting temperatures are estimated for a fraction of 0.400 at two different salt concentrations, compare the values to determine the effect of ion concentration in solution on the melting temperature of DNA.
05

Provide a Molecular Interpretation

Based on the comparison from Step 4, explain the molecular basis behind the effect of salt concentration on DNA melting temperature. Consider the role of ions in stabilizing the DNA double helix through shielding the negative charges on the phosphate backbone.

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Key Concepts

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

Differential Scanning Calorimetry
Differential Scanning Calorimetry (DSC) is a powerful analytical technique used to measure the thermal stability of various substances, including biological macromolecules like DNA. When applied to DNA, DSC measures how its melting temperature, or Tm, changes as it is heated up. This melting temperature is the point at which half of the DNA double helix has dissociated into single strands.

During a DSC experiment, a small DNA sample is heated at a controlled rate while the amount of energy absorbed or released is recorded. This gives us a detailed profile of the DNA's thermal behavior. The point at which the DNA absorbs the most energy corresponds to its melting temperature. This is a critical parameter in understanding the stability of the DNA molecule under various conditions. The information gained from DSC is invaluable in fields such as genetics, forensics, and biotechnology.
Base Pair Composition
The sequence of a DNA molecule is made up of four different nucleotide bases - adenine (A), thymine (T), guanine (G), and cytosine (C). These bases pair up with each other (A with T and G with C) to form what we refer to as base pairs. The proportion of G-C to A-T base pairs in a DNA molecule, known as its base pair composition, can significantly impact the molecule's properties, including its melting temperature.

G-C base pairs form three hydrogen bonds between them, unlike A-T pairs that only form two. Consequently, the more G-C pairs in a DNA molecule, the more stable and the higher its melting temperature. This implies that the melting temperature of DNA is a reflection of its base pair composition. Understanding the correlation between Tm and the fraction of G-C base pairs helps in predicting how different DNA molecules will respond to thermal changes.
Ion Concentration Effect
The surrounding ion concentration significantly affects the melting temperature of DNA. Ions in solution play a crucial role in the stability of the DNA helix. In our exercise, we see different melting temperatures for DNA in solutions with varying salt concentrations.

Ions, especially positively charged ones like sodium (Na+) or magnesium (Mg2+), can shield the negatively charged phosphate groups in the DNA backbone. This ionic shielding helps to stabilize the DNA double helix by reducing repulsion between phosphate groups. Therefore, at higher ion concentrations, the melting temperature of DNA increases as it becomes more stable. This stabilization effect is a critical consideration in experimental settings where DNA is subject to temperature variations such as in PCR amplifications, and it also has implications in understanding natural biological systems where ion concentrations can vary widely.

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Most popular questions from this chapter

The fluidity of a lipid bilayer dispersed in aqueous solution depends on temperature and there are two important melting transitions. One transition is from a 'solid crystalline' state in which the hydrophobic chains are packed together tightly (hence move very little) to a 'liquid crystalline state', in which there is increased but still limited movement of the of the chains. The second transition, which occurs at a higher temperature than the first, is from the liquid crystalline state to a liquid state, in which the hydrophobic interactions holding the aggregate together are largely disrupted. (a) It is observed that the transition temperatures increase with the hydrophobic chain length and decrease with the number of \(\mathrm{C}=\mathrm{C}\) bonds in the chain. Explain these observations. (b) What effect is the inclusion of cholesterol likely to have on the transition temperatures of a lipid bilayer? Justify your answer.

The rate of sedimentation of a recently isolated protein was monitored at \(20^{\circ} \mathrm{C}\) and with a rotor speed of 50000 r.p.m. The boundary receded as follows: $$ \begin{array}{llllllll} t / \mathrm{s} & 0 & 300 & 600 & 900 & 1200 & 1500 & 1800 \\ r / \mathrm{cm} & 6.127 & 6.153 & 6.179 & 6.206 & 6.232 & 6.258 & 6.284 \end{array} $$ Calculate the sedimentation constant and the molar mass of the protein on the basis that its partial specific volume is \(0.728 \mathrm{~cm}^{3} \mathrm{~g}^{-1}\) and its diffusion coefficient is \(7.62 \times 10^{-11} \mathrm{~m}^{2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\), the density of the solution then being \(0.9981 \mathrm{~g} \mathrm{~cm}^{-3}\). suggest a shape for the protein given that the viscosity of the solution is \(1.00 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\).

Consider the thermodynamic description of stretching rubber. The observables are the tension, \(\mathrm{t}\), and length, \(/\) (the analogues of \(p\) and \(\mathrm{V}\) for gases). Because \(d w=t d l\), the basic equation is \(d U=\mathrm{TdS}+\mathrm{tdl}\). (The term \(\mathrm{pdV}\) is supposed negligible throughout.) If \(G=U-T S-t l\), find expressions for \(\mathrm{d} G\) and dA, and deduce the Maxwell relations $$ \left(\frac{\partial S}{\partial l}\right)_{T}=-\left(\frac{\partial t}{\partial T}\right)_{l} \quad\left(\frac{\partial S}{\partial t}\right)_{T}=-\left(\frac{\partial l}{\partial T}\right)_{t} $$ Go on to deduce the equation of state for rubber,$$ \left(\frac{\partial U}{\partial l}\right)_{T}=t-\left(\frac{\partial t}{\partial T}\right)_{l} $$

Sedimentation studies on haemoglobin in water gave a sedimentation constant \(S=4.5 \mathrm{~Sv}\) at \(20^{\circ} \mathrm{C}\). The diffusion coefficient is \(6.3 \times 10^{-11} \mathrm{~m}^{2} \mathrm{~s}^{-1}\) at the same temperature. Calculate the molar mass of haemoglobin using \(\mathrm{v}_{s}=0.75 \mathrm{~cm}^{3} \mathrm{~g}^{-1}\) for its partial specific volume and \(\rho=0.998 \mathrm{~g} \mathrm{~cm}^{-3}\) for the density of the solution. Estimate the effective radius of the haemoglobin molecule given that the viscosity of the solution is \(1.00 \times 10-{ }^{3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\).

The following table lists the glass transition temperatures, \(T g\), of several polymers. Discuss the reasons why the structure of the monomer unit has an effect on the value of \(T g\). $$ \begin{aligned} &\text { Polymer Poly(oxymethylenc) Polyethylene Poly(vinyl chloride) Polystyrene }\\\ &\begin{array}{llll} \text { Structure }-\left(\mathrm{OCH}_{2}\right)_{n}- & -\left(\mathrm{CH}_{2} \mathrm{CH}_{2}\right)_{*}- & -\left(\mathrm{CH}_{2}-\mathrm{CHCl}\right)_{n}- & -\left(\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{C}_{\mathrm{v}} \mathrm{H}_{3}\right)\right)_{2}- \\ T_{s} / \mathrm{K} & 198 & 253 & 354 & 381 \end{array} \end{aligned} $$
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