Difference between revisions of "DNA Melting Thermodynamics"

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{{LecturePoint|Some of the strands combine to form double stranded DNA (dsDNA). The reaction is governed by the equation <math>1 A + 1 A' \Leftrightarrow 1 A \cdot A'</math>}}
 
{{LecturePoint|Some of the strands combine to form double stranded DNA (dsDNA). The reaction is governed by the equation <math>1 A + 1 A' \Leftrightarrow 1 A \cdot A'</math>}}
  
==Equilibrium concentrations of ssDNA and dsDNA depend on temperature==
+
==Equilibrium concentrations of ssDNA and dsDNA==
  
{{LecturePoint|The concentrations of the reaction products are related by the equilibrium constant: <math>K = \frac{\left [ A \cdot A' \right ]}{\left [ A \right ] \left [ A' \right ]}</math>}}
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{{LecturePoint|The concentrations of the reaction products are related by the equilibrium constant: <math>K_{eq} = \frac{\left [ A \cdot A' \right ]}{\left [ A \right ] \left [ A' \right ]}</math>}}
  
{{LecturePoint|The value of <math>\left . K \right .</math> is a function of temperature. According to the van't Hoff equation:}}
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{{LecturePoint|The value of <math>\left . K_{eq} \right .</math> is a function of temperature. According to the van't Hoff equation:}}
 
:<math>
 
:<math>
 
\begin{align}
 
\begin{align}
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{{LecturePoint|Solving for <math>\left . K \right .</math>:}}
 
{{LecturePoint|Solving for <math>\left . K \right .</math>:}}
 
:<math>
 
:<math>
K = \exp{\frac{\Delta S}{R} - \frac{\Delta H}{R T}}
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K_{eq} = e^\left [\frac{\Delta S}{R} - \frac{\Delta H}{R T} \right ] \quad (1)
 
</math>
 
</math>
  
{LecturePoint|At low temperatures, dsDNA is favored. As the temperature increases, more of the strands separate into their component ssDNA oligos.}}
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{{LecturePoint|At low temperatures, dsDNA is favored. As the temperature increases, more of the strands separate into their component ssDNA oligos.}}
  
{{LecturePoint|The transformation from dsDNA to dsDNA is called denaturation or melting.}}
+
{{LecturePoint|The transformation from dsDNA to ssDNA is called denaturation or melting.}}
  
 
{{LecturePoint|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.}}
 
{{LecturePoint|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.}}
  
{{LecturePoint|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 a lower temperature than G-C rich ones.}}
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{{LecturePoint|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.<ref>Breslauer et al., PNAS 83: 3746, 1986</ref>}}
  
 
==Fraction of dsDNA as a function of temperature==
 
==Fraction of dsDNA as a function of temperature==
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{{LecturePoint|Similarly, let <math>\left . C_{DS} \right .</math> be the concentration of double stranded DNA: <math>C_{DS} = {\left [ A \cdot A' \right ]}</math>}}
 
{{LecturePoint|Similarly, let <math>\left . C_{DS} \right .</math> be the concentration of double stranded DNA: <math>C_{DS} = {\left [ A \cdot A' \right ]}</math>}}
  
{{LecturePoint|<math>\left . C_T \right .</math> is the total concentration of DNA strands. <math>\left . C_T = 2 C_{SS} + 2 C_{DS}\right .</math>}}
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{{LecturePoint|<math>\left . C_T \right .</math> is the total concentration of DNA. <math>\left . C_T = 2 C_{SS} + 2 C_{DS}\right .</math>}}
  
 
{{LecturePoint|Let <math>\left . f \right .</math> be the fraction of total DNA that is double stranded}}
 
{{LecturePoint|Let <math>\left . f \right .</math> be the fraction of total DNA that is double stranded}}
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{{LecturePoint|Therefore, <math>C_{SS} = \frac{(1 - f)C_T}{2}</math>}}
 
{{LecturePoint|Therefore, <math>C_{SS} = \frac{(1 - f)C_T}{2}</math>}}
  
{{LecturePoint|Now we can solve for <math>\left . f \right .</math>:
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{{LecturePoint|Now we can solve for <math>\left . K \right .</math> in terms of <math>\left . f \right .</math> and <math>\left . C_T \right .</math>:}}
 
:<math>
 
:<math>
\begin{align}
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K_{eq} = \frac{C_{DS}}{C_{SS}^2}
K & = \frac{\left [ AA' \right ]}{\left ( \frac{1}{2} C_T - \left [ AA' \right ] \right ) ^ 2}
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      = \frac{f C_T / 2}{ [(1 - f) C_T / 2] ^ 2}
  = \frac{\left [ AA' \right ]}{C_T^2 \left ( \frac{1}{2} - \frac{\left [ AA' \right ]}{C_T} \right ) ^ 2}
+
      = \frac{2 f}{(1 - f)^2 C_T}
  = \frac{\frac{2 \left [ AA' \right ]}{C_T}}{2 C_T \left ( \frac{1}{2} - \frac{1}{2}\frac{2 \left [ AA' \right ]}{C_T} \right ) ^ 2} \\
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  & = \frac{f}{2 C_T \left ( \frac{1}{2} - \frac{1}{2} f \right ) ^2}
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\end{align}
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</math>
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==Free energy==
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{{LecturePoint|At the melting point, <math>f = \frac{1}{2}</math> and <math>K_{eq} = \frac {4}{C_T}.}}
  
 +
</math>
  
Let <math>C_T \quad</math> be the total concentration of ssDNA.
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{{LecturePoint|Substituting from equation 1,}}
 
:<math>
 
:<math>
\begin{align}
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e^\left [\frac{\Delta S}{R} - \frac{\Delta H}{R T} \right ] = \frac{2 f}{(1 - f)^2 C_T}
C_{ss} & = \left [ A \right ] = \left [ A' \right ] \quad (3) \\
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C_{ds} & = \left [ AA' \right ] \quad (4) \\
+
\end{align}
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</math>
 
</math>
 +
 +
{{LecturePoint|Taking the log of both sides and applying the quadratic formula gives <math>\left . f \right .</math> as a function of

Revision as of 18:03, 9 April 2008

DNA solution

$ \bullet $ Consider a solution containing equal quantities of complementary single stranded DNA (ssDNA) oligonucleotides $ \left . A \right . $ and $ \left . A' \right . $.
$ \bullet $ 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' $

Equilibrium concentrations of ssDNA and dsDNA

$ \bullet $ 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 ]} $
$ \bullet $ The value of $ \left . K_{eq} \right . $ is a function of temperature. According to the van't Hoff equation:
$ \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
$ \bullet $ Solving for $ \left . K \right . $:
$ K_{eq} = e^\left [\frac{\Delta S}{R} - \frac{\Delta H}{R T} \right ] \quad (1) $
$ \bullet $ At low temperatures, dsDNA is favored. As the temperature increases, more of the strands separate into their component ssDNA oligos.
$ \bullet $ The transformation from dsDNA to ssDNA is called denaturation or melting.
$ \bullet $ 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.
$ \bullet $ 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]

Fraction of dsDNA as a function of temperature

$ \bullet $ Let $ \left . C_{SS} \right . $ represent the concentration of either single stranded oligonucleotide: $ C_{SS} = {\left [ A \right ] = \left [ A' \right ]} $.
$ \bullet $ Similarly, let $ \left . C_{DS} \right . $ be the concentration of double stranded DNA: $ C_{DS} = {\left [ A \cdot A' \right ]} $
$ \bullet $ $ \left . C_T \right . $ is the total concentration of DNA. $ \left . C_T = 2 C_{SS} + 2 C_{DS}\right . $
$ \bullet $ Let $ \left . f \right . $ be the fraction of total DNA that is double stranded
$ f = \frac{2 C_{DS}}{C_T} = \frac{C_T - 2 C_{SS}}{C_T} = 1 - 2 \frac{C_{SS}}{C_T} $
$ \bullet $ Therefore, $ C_{SS} = \frac{(1 - f)C_T}{2} $
$ \bullet $ Now we can solve for $ \left . K \right . $ in terms of $ \left . f \right . $ and $ \left . C_T \right . $:
$ 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}.}} $
$ \bullet $ Substituting from equation 1,
$ 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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