What is the optimal amount of DNA template that should be used for PCR?

The optimal amount of template required depends on the complexity of the template and the copy number of the target sequence. Approximately 104 copies of the target DNA sequence are required to detect the amplification product in 25–30 PCR cycles.

-Typically, 1 µg of human genomic DNA contains 3.04 x 105 molecules of DNA. For most PCR applications, 30–100 ng of human genomic DNA is sufficient. High-copy targets, such as housekeeping genes, require only 10 ng of template. Template amounts for higher-complexity templates range between 10 ng and 500 ng.
-Typically, 1 µg of E. coli genomic DNA contains 2 x 108 molecules of DNA; therefore, the recommended amount of template is between 100 pg and 1 ng.
-Typically, 1 µg of lambda DNA contains 1.9 x 1010 molecules of DNA; therefore, the template input can be as little as 100 pg.
-The amount of cDNA template depends on the copy number of the target. cDNA input is typically described in terms of equivalent RNA input. The amount of cDNA in a PCR reaction can be as little as 10 pg (RNA equivalent).

It is important to note that not all polymerases can tolerate excessive amounts of template. For samples containing excess template (up to 1 µg), we recommend PrimeSTAR GXL DNA Polymerase.

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