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Fundamental Organic chemistry

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Fundamental Organic chemistry Powered By Docstoc
					BIOCHEMISTRY OF MOLECULES
3.1 To plot the absorption spectra of NAD+ and NADH

Results:

             Wavelength                   NAD+                       NADH
                400                       0.071                      0.036
                380                       0.072                      0.038
                360                       0.097                      0.046
                340                       0.140                      0.047
                320                       0.139                      0.048
                300                       0.117                      0.052
                280                       0.134                      0.054
                260                       0.168                      0.076
                240                       0.138                      0.052

3.1.3 Reports and Questions

1.   Refer to graph on page 7.

2.   The main difference in the absorption spectra is that NADH has two peaks at 260nm and
     340nm while NAD+ has only one peak at 260nm.Wavelength of 340nm is used to
     distinguish NADH and NAD+ .

3.   The wavelength of choice for measuring NADH in a mixture containing both NAD+ and
     NADH would be 340nm . This is because when the absorption spectra for the mixture is
     plotted, NADH will show a peak at 340nm as NADH has specificity and sensitivity at
     340nm. NAD+ wouldn’t show the peak at wavelength of 340nm.




3.2 To study the relationship between absorbance and concentration of NADH

Results:

           Concentration of NADH ( mM )                       A340
                          0                                  0.000
                        0.01                                 0.071
                        0.05                                 0.305
                        0.10                                 0.608
                        0.15                                 0.938
                          0.20                                   1.254


3.2.3 Report

1. Refer to the table above.

2. Refer to graph on page 8.

3.                  A = ξcl,          whereby l = 1cm
                So, A = ξc
                    ξ= A
                         c
                    ξ = gradient of graph

                Hence, ξ = m = 1.254-0.000
                                 0.20-0.00
                             = 6.27 A/mM




3.3 To determine the Km and Vmax of lactose dehyrogenase

Results:

     Tube No.    10 mM NAD+          200 mM       0.05M buffer      LDH (ml)    mM lactate
                     (ml)          lactate (ml)    pH 9.0 (ml)                  (final conc)
        1            0.10              0.03           2.67               0.20         2
        2            0.10              0.06           2.64               0.20         4
        3            0.10              0.12           2.58               0.20         8
        4            0.10              0.24           2.46               0.20        16
     5                 0.10              0.30        2.40     0.20     20


 Time (min)                                          A340
                   Tube 1               Tube 2      Tube 3   Tube 4   Tube 5
      0            0.000                0.000       0.000    0.000    0.000
     0.5           0.015                0.018       0.057    0.088    0.104
     1.0           0.022                0.030       0.076    0.101    0.128
     1.5           0.027                0.041       0.093    0.128    0.146
     2.0           0.033                0.051       0.108    0.148    0.170
     2.5           0.040                0.060       0.123    0.168    0.191
     3.0           0.046                0.069       0.135    0.186    0.212
     3.5           0.051                0.078       0.147    0.202    0.230
     4.0           0.057                0.086       0.158    0.218    0.258


3.3.3 Report

1.    Refer to graph on page 9.

                                Absorbance
      Initial velocity, V0 =
                                Time (min)
                              = gradient of curve

      Tube 1: V0 = 0.057-0.015A
                     4.0-0.5 min
                = 0.012 A/min

      Tube 2: V0 = 0.086-0.018 A
                      4.0-0.5 min
                = 0.019 A/min

      Tube 3: V0 = 0.147-0.076 A
                     3.5-1.0 min
                 = 0.028 A/min

      Tube 4: V0= 0.186-0.088 A
                   3.0-0.5 min
                = 0.039 A/min

      Tube 5: V0= 0.258-0.104 A
                   4.0-0.5min
                = 0.044 A/min

2.    Using the Beer-Lambert Law :

               ξ= A
                   c
        = Absorbance
               mM
      V0 = A
           t
         = Absorbance
             Minutes
      V0 in mM/min = A/min
                        A/mM
                         mM
                       =
                         min
      V0 in moles/min = mM x 0.003
                          min


Tube 1: V0 in mM/min = 0.012 A/min
                       6.27 A/mM
                     = 3.21x10-3 mM/min

        V0 = ( 1.91 x 10-3 ) x 0.003
           = 5.730 x 10-6 mmoles/min
           = 5.730 x 10-9 moles/min

Tube 2: V0 in mM/min = 0.019 A/min
                        6.27 A/mM
                     = 3.03x10-3 mM/min

        V0 = ( 3.03 x 10-3 ) x 0.003
           = 9.909 x 10-6 mmoles/min
           = 9.909 x 10-9 moles/min

Tube 3: V0 in mM/min = 0.028 A/min
                       6.27 A/mM
                     = 4.47 x10-3 mM/min

        V0 = ( 4.47 x 10-3 ) x 0.003
           = 1.340 x 10-5 mmoles/min
           = 1.340 x 10-8 moles/min




Tube 4: V0 in mM/min = 0.039 A/min
                       6.27 A/mM
                     = 6.22 x 10-3 mM/min

        V0 in moles/min = ( 6.22 x 10-3 ) x 0.003
                        = 1.866 x 10-5 mmoles/min
                        = 1.866 x 10-8 moles/min
        Tube 5: V0 in mM/min = 0.044 A/min
                                6.27 A/mM
                             = 7.02 x10-3 mM/min

                  V0 in moles/min = ( 7.02 x 10-3 ) x 0.003
                                  = 2.105 x 10-5 mmoles/min
                                  = 2.105 x 10-8 moles/min

3.      Please refer to graphs on page 10 for Michaelis-Menten curve and page 11 for
        Lineweaver-Burk plot.

                                              Initial            1                  1
                         Concentration
                                           Velocity, V0
         Tube No.        of lactate, [S]                        [S]                 V
                                             ( x10-8                              8
                             (mM)
                                           moles/min)         (mM-1 )        ( x10 min/moles)
             1                  2             0.573           0.5000               1.74
             2                  4             0.991           0.2500               1.01
             3                  8             1.340           0.1250               0.75
             4                 16             1.866           0.0625               0.54
             5                 20             2.105           0.0500               0.48

        Hence,
                 Vmax obtained from the Michaelis-Menten curve= 2.1 x 10-8 moles/min

                 Vmax obtained from the Lineweaver-Burk curve = ____1___
                                                                  0.36 x108
                                                              = 2.77 x10-8 moles/min


                 Km obtained from the Michaelis-Menten curve = 5.2 mM

                 Km obtained from the Lineweaver-Burk curve= _1_
                                                             0.13
                                                           = 7.69 mM




         The Vmax values obtained using both plots are rather similar with a slight difference in
     terms of the results obtained. In the Michaelis-Menten curve, as substrate concentration
     increases, the initial velocity will increase asymptotically with it until it approaches the
     limiting value(Vmax). However the Km values obtained using both plots are rather far with a
     difference of 2.49mM. Km is then derived from Vmax/2. In the Lineweaver-Burk curve, the y-
     intercept is 1/Vmax whereas the x-intercept is -1/Km. The Lineweaver-Burk plot is derived
     from the Michaelis-Menten curve by rearrangement of the Michaelis-Menten equation into
     the equation below:
1    Km     1     1
  =(     )(   )+
V    Vmax [S]    Vmax

				
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