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| | = \frac{2 f}{(1 - f)^2 C_T} | | = \frac{2 f}{(1 - f)^2 C_T} |
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| − | {{LecturePoint|At the melting point, <math>f = \frac{1}{2}</math> and <math>K_{eq} = \frac {4}{C_T}.}} | + | {{LecturePoint|At the melting point, <math>f = \frac{1}{2}</math> and <math>K_{eq} = \frac {4}{C_T}</math>.}} |
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| | </math> | | </math> |
Revision as of 18:04, 9 April 2008
DNA solution
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Consider a solution containing equal quantities of complementary single stranded DNA (ssDNA) oligonucleotides $ \left . A \right . $ and $ \left . A' \right . $.
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Some of the strands combine to form double stranded DNA (dsDNA). The reaction is governed by the equation $ 1 A + 1 A' \Leftrightarrow 1 A \cdot A' $
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Equilibrium concentrations of ssDNA and dsDNA
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The concentrations of the reaction products are related by the equilibrium constant: $ K_{eq} = \frac{\left [ A \cdot A' \right ]}{\left [ A \right ] \left [ A' \right ]} $
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The value of $ \left . K_{eq} \right . $ is a function of temperature. According to the van't Hoff equation:
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- $ \begin{align} \Delta G & = \Delta H - T \Delta S\\ & = -R T \ln K\\ \end{align} $
- where
- $ \Delta G $ is the change in free energy
- $ \Delta H $ is the enthalpy change
- T is the absolute temperature
- $ \Delta S $ is the entropy change
- R is the gas constant
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Solving for $ \left . K \right . $:
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- $ K_{eq} = e^\left [\frac{\Delta S}{R} - \frac{\Delta H}{R T} \right ] \quad (1) $
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At low temperatures, dsDNA is favored. As the temperature increases, more of the strands separate into their component ssDNA oligos.
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The transformation from dsDNA to ssDNA is called denaturation or melting.
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Short sequences of about 10-40 base pairs (such as those used in the DNA Melting lab) tend to denature all at once, while longer sequences may melt in segments.
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Less energy is required to split the double hydrogen bond of A-T pairs than the triple bond of G-C pairs. Thus, A-T rich sequences tend to melt at lower temperatures than G-C rich ones.[1]
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Fraction of dsDNA as a function of temperature
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Let $ \left . C_{SS} \right . $ represent the concentration of either single stranded oligonucleotide: $ C_{SS} = {\left [ A \right ] = \left [ A' \right ]} $.
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Similarly, let $ \left . C_{DS} \right . $ be the concentration of double stranded DNA: $ C_{DS} = {\left [ A \cdot A' \right ]} $
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$ \left . C_T \right . $ is the total concentration of DNA. $ \left . C_T = 2 C_{SS} + 2 C_{DS}\right . $
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Let $ \left . f \right . $ be the fraction of total DNA that is double stranded
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- $ f = \frac{2 C_{DS}}{C_T} = \frac{C_T - 2 C_{SS}}{C_T} = 1 - 2 \frac{C_{SS}}{C_T} $
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Therefore, $ C_{SS} = \frac{(1 - f)C_T}{2} $
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Now we can solve for $ \left . K \right . $ in terms of $ \left . f \right . $ and $ \left . C_T \right . $:
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- $ K_{eq} = \frac{C_{DS}}{C_{SS}^2} = \frac{f C_T / 2}{ [(1 - f) C_T / 2] ^ 2} = \frac{2 f}{(1 - f)^2 C_T} {{LecturePoint|At the melting point, <math>f = \frac{1}{2} $ and $ K_{eq} = \frac {4}{C_T} $.}}
</math>
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Substituting from equation 1,
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- $ e^\left [\frac{\Delta S}{R} - \frac{\Delta H}{R T} \right ] = \frac{2 f}{(1 - f)^2 C_T} $
{{LecturePoint|Taking the log of both sides and applying the quadratic formula gives $ \left . f \right . $ as a function of
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