Team:LCG-UNAM-Mexico/Notebook/2008-August

From 2008.igem.org

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<td class="bodyText"><p><p><strong>Hill's cooperativity</strong><br />
+
<td class="bodyText"><p><strong>Hill's cooperativity</strong><br />
     <strong>5th Reaction </strong> <br />
     <strong>5th Reaction </strong> <br />
     <strong>Reminder:</strong> </p>
     <strong>Reminder:</strong> </p>
-
<p>A  + B &lt;--&gt; AB&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
<p>A  + B &lt;--&gt; AB            <br />
-
     <strong>Ka=Keq=[AB]/[A][B]=1/Kd</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
     <strong>Ka=Keq=[AB]/[A][B]=1/Kd</strong>      <br />
   θ=[AB]/([AB]+[A])=[B]/([B]+Kd) </p>
   θ=[AB]/([AB]+[A])=[B]/([B]+Kd) </p>
-
<p><strong><u>MWC  Model</u></strong> (Cooperativity)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />
+
<p><strong><u>MWC  Model</u></strong> (Cooperativity)           <br />
-
A + nB &lt;--&gt; ABn&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
  A + nB &lt;--&gt; ABn        <br />
-
<strong>Ka=Keq=[ABn]/[A][B]n=1/Kd</strong>&nbsp;&nbsp; <br />
+
  <strong>Ka=Keq=[ABn]/[A][B]n=1/Kd</strong>   <br />
-
θ=[B]n/([B]n+Kd)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
  θ=[B]n/([B]n+Kd)        <br />
-
log(θ/(1- θ))=nlog(B)-log(kd)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; …Hill's equation</p>
+
  log(θ/(1- θ))=nlog(B)-log(kd)                        …Hill's equation</p>
-
<p>&nbsp;</p>
+
<p> </p>
<p><strong>Suppression mediated by cI:</strong> <br />
<p><strong>Suppression mediated by cI:</strong> <br />
-
   ρ  + nCI &lt;--&gt; ρ:CIn&nbsp;&nbsp;&nbsp;&nbsp; (k+,  k-)&nbsp; <br />
+
   ρ  + nCI &lt;--&gt; ρ:CIn     (k+,  k-)  <br />
-
   <strong>Keq=Ka=[ρ:CIn]/[ρ][CI]n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
   <strong>Keq=Ka=[ρ:CIn]/[ρ][CI]n       <br />
-
   </strong>Si ρ0=[ρ]+[ρ:CIn]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
   </strong>Si ρ0=[ρ]+[ρ:CIn]       <br />
-
   &nbsp;… ρ0=[ρ]+Keq[ρ][CI]n <br />
+
   … ρ0=[ρ]+Keq[ρ][CI]n <br />
-
   &nbsp; =&gt;  ρ= (ρ0/Keq)/((1/keq)+[CI]n) </p>
+
   =&gt;  ρ= (ρ0/Keq)/((1/keq)+[CI]n) </p>
<p>Flow=  k+[ρ][CI]n = K+((ρ0/Keq)/((1/Keq)+[CI]n))[CI]n </p>
<p>Flow=  k+[ρ][CI]n = K+((ρ0/Keq)/((1/Keq)+[CI]n))[CI]n </p>
<p><strong>Flow= k+</strong><strong>([ρ</strong><strong>0</strong><strong>]/K</strong><strong>eq</strong><strong>)</strong> <strong>[CI]n / ((1/Keq)+[CI]n)</strong></p>
<p><strong>Flow= k+</strong><strong>([ρ</strong><strong>0</strong><strong>]/K</strong><strong>eq</strong><strong>)</strong> <strong>[CI]n / ((1/Keq)+[CI]n)</strong></p>
-
<p><strong>=&gt; </strong><strong>Vm=  k</strong><strong>+</strong>([ρ0]/Keq)<strong>&nbsp;&nbsp; &amp;&nbsp;&nbsp; Kp=1/Keq=K</strong>d </p>
+
<p><strong>=&gt; </strong><strong>Vm=  k</strong><strong>+</strong>([ρ0]/Keq)<strong>   &amp;   Kp=1/Keq=K</strong>d </p>
<p><strong>So:</strong> <br />
<p><strong>So:</strong> <br />
-
   Keq = exp(  -ΔG / R T )&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />
+
   Keq = exp(  -ΔG / R T )           <br />
   k+ = (KB/h) T exp( -ΔG / R T ) = (KB/h) T Keq </p>
   k+ = (KB/h) T exp( -ΔG / R T ) = (KB/h) T Keq </p>
-
<table width="423" border="1" cellpadding="1" cellspacing="1" bordercolor="#000000" bgcolor="#CC99FF">
+
<table bgcolor="#cc99ff" border="1" bordercolor="#000000" cellpadding="1" cellspacing="1" width="423">
-
   <tr>
+
   <tbody>
-
     <td width="53" nowrap="nowrap"><p>Keq= </p></td>
+
     <tr>
-
    <td width="90" nowrap="nowrap"><p>2.89517E+17 </p></td>
+
      <td width="53" nowrap="nowrap"><p>Keq= </p></td>
-
    <td width="52" nowrap="nowrap"><p>&nbsp; </p></td>
+
      <td width="90" nowrap="nowrap"><p>2.89517E+17 </p></td>
-
    <td width="39" nowrap="nowrap"><p><strong>K</strong><strong>B</strong><strong>=</strong> </p></td>
+
      <td width="52" nowrap="nowrap"><p>  </p></td>
-
    <td width="69" nowrap="nowrap"><p>1.38E-23 </p></td>
+
      <td width="39" nowrap="nowrap"><p><strong>K</strong><strong>B</strong><strong>=</strong> </p></td>
-
    <td width="87" nowrap="nowrap"><p>J/K </p></td>
+
      <td width="69" nowrap="nowrap"><p>1.38E-23 </p></td>
-
  </tr>
+
      <td width="87" nowrap="nowrap"><p>J/K </p></td>
-
  <tr>
+
    </tr>
-
    <td width="53" nowrap="nowrap"><p>k+= </p></td>
+
    <tr>
-
    <td width="90" nowrap="nowrap"><p>1.79764E+30 </p></td>
+
      <td width="53" nowrap="nowrap"><p>k+= </p></td>
-
    <td width="52" nowrap="nowrap"><p>/s </p></td>
+
      <td width="90" nowrap="nowrap"><p>1.79764E+30 </p></td>
-
    <td width="39" nowrap="nowrap"><p><strong>h=</strong> </p></td>
+
      <td width="52" nowrap="nowrap"><p>/s </p></td>
-
    <td width="69" nowrap="nowrap"><p>6.63E-34 </p></td>
+
      <td width="39" nowrap="nowrap"><p><strong>h=</strong> </p></td>
-
    <td width="87" nowrap="nowrap"><p>J s </p></td>
+
      <td width="69" nowrap="nowrap"><p>6.63E-34 </p></td>
-
  </tr>
+
      <td width="87" nowrap="nowrap"><p>J s </p></td>
-
  <tr>
+
    </tr>
-
    <td colspan="3" nowrap="nowrap"></td>
+
    <tr>
-
    <td width="39" nowrap="nowrap"><p><strong>R=</strong> </p></td>
+
      <td colspan="3" nowrap="nowrap"></td>
-
    <td width="69" nowrap="nowrap"><p>1.9872 </p></td>
+
      <td width="39" nowrap="nowrap"><p><strong>R=</strong> </p></td>
-
    <td width="87" nowrap="nowrap"><p>cal/(K mol) </p></td>
+
      <td width="69" nowrap="nowrap"><p>1.9872 </p></td>
-
  </tr>
+
      <td width="87" nowrap="nowrap"><p>cal/(K mol) </p></td>
-
  <tr>
+
    </tr>
-
    <td width="53" nowrap="nowrap"><p><strong>ΔG=</strong> </p></td>
+
    <tr>
-
    <td width="90" nowrap="nowrap"><p><strong>-23810</strong> </p></td>
+
      <td width="53" nowrap="nowrap"><p><strong>ΔG=</strong> </p></td>
-
    <td width="52" nowrap="nowrap"><p><strong>cal/mol</strong> </p></td>
+
      <td width="90" nowrap="nowrap"><p><strong>-23810</strong> </p></td>
-
    <td width="39" nowrap="nowrap"><p><strong>T=</strong> </p></td>
+
      <td width="52" nowrap="nowrap"><p><strong>cal/mol</strong> </p></td>
-
    <td width="69" nowrap="nowrap"><p>298 </p></td>
+
      <td width="39" nowrap="nowrap"><p><strong>T=</strong> </p></td>
-
    <td width="87" nowrap="nowrap"><p>K </p></td>
+
      <td width="69" nowrap="nowrap"><p>298 </p></td>
-
  </tr>
+
      <td width="87" nowrap="nowrap"><p>K </p></td>
-
</table> </p>
+
    </tr>
 +
  </tbody>
 +
</table>
 +
<p>&nbsp;</p>
         </td>
         </td>
       </tr>  
       </tr>  
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-
<td class="bodyText"><p>August 5th(summary)- -- -- -- -- -- -- --   - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ---- --- --- -- - ------ -- ---- --- --- --- -- -- ----- ---, --- - - --- -- -- -- -- -- -- --  - ------ -----.
+
<td class="bodyText"> <p>Hill's Cooperativity<br />
-
  --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ----. </p>
+
</p>
 +
<p>5th Reaction, conflict ...<br />
 +
</p>
 +
<p>If we consider that: <br />
 +
</p>
 +
<p>Keq = exp (-ΔG / R T) <br />
 +
</p>
 +
<p>k + = (KB / h) T exp (-ΔG / R T) = (KB / h) T Keq</p>
 +
<p> and given that the flow is (k + / Keq) [ρ0] [CI] n / ((1/Keq) + [CI]  n), the value of the maximum speed of the flow loses its meaning. </p>
 +
<p>   The speed limit is being determined by (k + / Keq) [ρ0], but k + / Keq  = (KB / h) * T, and we know that [ρ0] is arbitrary, i.e., Vmax is no longer based on the reaction as such, which does not make sense. </p>
 +
  <p>  For  example: Take the same reaction that we are considering, the maximum speed of the flow of the reaction would be the same with the promoter  that has the operators of CI, that if you used one with a random sequence,  so, whether we repeated the experiment, with the same temperature and  the same concentration of DNA and an equal number of copies of the sequence, the  maximum speed reached by the flow would be the same for the real  promoter as for for any sequence, without taking any consideration with their affinity for their substrates... That does not makes sense! </p>
 +
  <p>  The proposed explanation  is that the equation used to determine k + does not fit our model, we  should explore other possibilities. </p>
 +
<p>&nbsp;</p>
         </td>
         </td>
       </tr>  
       </tr>  
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         </tr>
         </tr>
         <tr>
         <tr>
-
<td class="bodyText"><p>August 7th(summary)- -- -- -- -- -- -- --   - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ---- --- --- -- - ------ -- ---- --- --- --- -- -- ----- ---, --- - - --- -- -- -- -- -- -- --   - ------ -----.
+
<td class="bodyText"><p>Hill's Cooperativity: <br />
-
--- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ----. </p>
+
  5th Reaction, resolving the conflict... <br />
 +
  <br />
 +
  The error in the previous approach is that we were considering  ΔG to be the same for both equations (for Keq &amp;  k+).<br />
 +
   <br />
 +
The explanation of why these two values are different is very clear  when we look at the graph below. Recalling what the two  constants represent:</p>
 +
<div id="urdd">
 +
  <div align="center"><img src="http://docs.google.com/File?id=dntmktb_99dz485zf8_b" alt="" name="sm1w6" id="sm1w6" /></div>
 +
</div>
 +
<p>We know that the balance depends solely on the difference between Gibbs  free energy of the substrate and the product (ΔG 'th), The one with less energy will be favored in the balance, while the rate of reaction  depends on the activation energy needed for the conversion (ΔG ‡). A  reaction reaches equilibrium faster or slower depending on the rate of reaction (depending on how big is ΔG ‡), but the balance of it as such does not change. <br />
 +
   <br />
 +
Thus: <br />
 +
       
 +
Keq = exp (- ΔG 'º / R T) <br />
 +
       
 +
k + = (KB / h) T exp (- ΔG ‡ / RT) ≠ (KB / h) T Keq</p>
 +
<p>&nbsp; </p>
         </td>
         </td>
       </tr>   
       </tr>   

Revision as of 19:34, 29 September 2008

LCG-UNAM-MexicoTeam

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August

2008-08-04

Hill's cooperativity
5th Reaction
Reminder:

A + B <--> AB           
Ka=Keq=[AB]/[A][B]=1/Kd     
θ=[AB]/([AB]+[A])=[B]/([B]+Kd)

MWC Model (Cooperativity)           
A + nB <--> ABn       
Ka=Keq=[ABn]/[A][B]n=1/Kd  
θ=[B]n/([B]n+Kd)       
log(θ/(1- θ))=nlog(B)-log(kd)                        …Hill's equation

 

Suppression mediated by cI:
ρ + nCI <--> ρ:CIn     (k+, k-) 
Keq=Ka=[ρ:CIn]/[ρ][CI]n      
Si ρ0=[ρ]+[ρ:CIn]      
… ρ0=[ρ]+Keq[ρ][CI]n
=> ρ= (ρ0/Keq)/((1/keq)+[CI]n)

Flow= k+[ρ][CI]n = K+((ρ0/Keq)/((1/Keq)+[CI]n))[CI]n

Flow= k+([ρ0]/Keq) [CI]n / ((1/Keq)+[CI]n)

=> Vm= k+([ρ0]/Keq)   &   Kp=1/Keq=Kd

So:
Keq = exp( -ΔG / R T )          
k+ = (KB/h) T exp( -ΔG / R T ) = (KB/h) T Keq

Keq=

2.89517E+17

 

KB=

1.38E-23

J/K

k+=

1.79764E+30

/s

h=

6.63E-34

J s

R=

1.9872

cal/(K mol)

ΔG=

-23810

cal/mol

T=

298

K

 
2008-08-05

Hill's Cooperativity

5th Reaction, conflict ...

If we consider that:

Keq = exp (-ΔG / R T)

k + = (KB / h) T exp (-ΔG / R T) = (KB / h) T Keq

and given that the flow is (k + / Keq) [ρ0] [CI] n / ((1/Keq) + [CI] n), the value of the maximum speed of the flow loses its meaning.

The speed limit is being determined by (k + / Keq) [ρ0], but k + / Keq = (KB / h) * T, and we know that [ρ0] is arbitrary, i.e., Vmax is no longer based on the reaction as such, which does not make sense.

For example: Take the same reaction that we are considering, the maximum speed of the flow of the reaction would be the same with the promoter that has the operators of CI, that if you used one with a random sequence, so, whether we repeated the experiment, with the same temperature and the same concentration of DNA and an equal number of copies of the sequence, the maximum speed reached by the flow would be the same for the real promoter as for for any sequence, without taking any consideration with their affinity for their substrates... That does not makes sense!

The proposed explanation is that the equation used to determine k + does not fit our model, we should explore other possibilities.

 
2008-08-07

Hill's Cooperativity:
5th Reaction, resolving the conflict...

The error in the previous approach is that we were considering ΔG to be the same for both equations (for Keq & k+).

The explanation of why these two values are different is very clear when we look at the graph below. Recalling what the two constants represent:

We know that the balance depends solely on the difference between Gibbs free energy of the substrate and the product (ΔG 'th), The one with less energy will be favored in the balance, while the rate of reaction depends on the activation energy needed for the conversion (ΔG ‡). A reaction reaches equilibrium faster or slower depending on the rate of reaction (depending on how big is ΔG ‡), but the balance of it as such does not change.

Thus:
        Keq = exp (- ΔG 'º / R T)
        k + = (KB / h) T exp (- ΔG ‡ / RT) ≠ (KB / h) T Keq

 

2008-08-11

August 11th(summary)- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ---- --- --- -- - ------ -- ---- --- --- --- -- -- ----- ---, --- - - --- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ----.

2008-08-20

August 20th(summary)- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ---- --- --- -- - ------ -- ---- --- --- --- -- -- ----- ---, --- - - --- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ----.

2008-08-21

August 21st(summary)- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ---- --- --- -- - ------ -- ---- --- --- --- -- -- ----- ---, --- - - --- -- -- -- -- -- -- -- - ------ -----. --- ------ --- ---- -- --- ----- -- -- -- ------ ---- ---- - -- - ---- - ----- ---- ---, - ----- --- ----- ---- ---- ---- -- --- --- --- --- -- ---- ---------- - ------ ---- --- ---- ----- --- --- ----- ---- ----- ------ ---- -- -- --- ------- - ----- --- ----.


 
 
             

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