August Library (8/24/09): Difference between revisions

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  Assume the concentration of probe Jan09-1 to be 10ng/ul. Knowing the length of the probe is ~100 bases and 1 base = 330 Daltons (1bp = 2 bases = 660 Da).
  Assume the concentration of probe Jan09-1 to be 10ng/ul. Knowing the length of the probe is ~100 bases and 1 base = 330 Daltons (1bp = 2 bases = 660 Da).
  Solve: Concentration of Jan09-1 probe: 10ng/ul / 33000 = 303 nM
  Solve: Concentration of Jan09-1 probe: 10ng/ul / 33000 = 303 nM
     Volume of probe needed=  
     Volume of probe needed= (200ng/3pg * 100 * 46417)/(6.0 * 10^23 * 303 nM) = 1.73ul
(200ng/3pg * 100 * 46417)/(6.0 * 10^23 * 303 nM) = 1.73ul
 
Volumes needed of each sample:
                Sample 1 – 200ng / 102.7ng/ul = 1.947ul
Sample 2 – 200ng / 183.4ng/ul = 1.091ul
Sample 3 – 200ng/ 100.0ng/ul =  2.000ul
Sample 4 – 200ng/ 200.0ng/ul =  1.000ul
Sample 5 – 200ng/ 58.9ng/ul =    3.396ul
Sample 6 – 200ng/ 55.9ng/ul =    3.578ul
Sample 7 – 200ng/ 89.4ng/ul =    2.237ul
Sample 8 – 200ng/ 99.7ng/ul =    2.006ul

Revision as of 03:42, 14 September 2009

Probes, Samples, and Calculations

Probes:

Jan09-1:  12.4 ng/ul,  46417 probes;  376 nM
Jan09-2:  16.3 ng/ul,  53493 probes;  494 nM
Jan09-3:  7.8 ng/ul,   53828 probes;  236 nM

Samples:

Sample #	        Sample name	       Amount	     Concentration
  1	            MH001 1+2 xHeaIII	       10 ul	      102.7 ng/ul
  2	                 MH010B xMboI	        6 ul	      183.4 ng/ul
  3	                 MH010B xDpnI	       10 ul  	      100 ng/ul
  4	                   TCF #3	        6 ul	      200 ng/ul
  5                        DK4 #2	        17 ul	      58.9 ng/ul
  6	                   TK4 #2	        18 ul	      55.9 ng/ul
  7	                   TK9	                12 ul	      89.4 ng/ul 
  8	                   Pol II	        10 ul	      99.7 ng/ul

Calculations

Probe: Target ratio is 100:1
The mass of one genome is 3pg: 200ng of DNA is equal to 200ng/3pg.
The amount of probe needed is 200ng/3pg * 250 for each target, and the total amount of probe is 
(200ng/3pg * 100 * 46417)/(6.0 * 10^23)

Example
Assume the concentration of probe Jan09-1 to be 10ng/ul. Knowing the length of the probe is ~100 bases and 1 base = 330 Daltons (1bp = 2 bases = 660 Da).
Solve: Concentration of Jan09-1 probe: 10ng/ul / 33000 = 303 nM
    	Volume of probe needed= (200ng/3pg * 100 * 46417)/(6.0 * 10^23 * 303 nM) = 1.73ul
Volumes needed of each sample:
               Sample 1 – 200ng / 102.7ng/ul = 1.947ul

Sample 2 – 200ng / 183.4ng/ul = 1.091ul Sample 3 – 200ng/ 100.0ng/ul = 2.000ul Sample 4 – 200ng/ 200.0ng/ul = 1.000ul Sample 5 – 200ng/ 58.9ng/ul = 3.396ul Sample 6 – 200ng/ 55.9ng/ul = 3.578ul Sample 7 – 200ng/ 89.4ng/ul = 2.237ul Sample 8 – 200ng/ 99.7ng/ul = 2.006ul