Hi,
I'm working with human genomic DNA and try to analyse SNPs after PCR and digestion.
If i get now 3 bands after PCR and restiction enzyme incubation for partly digested it could mean 2 different alleles or and incomplete digestion, couldn't it`?
Is there a smart way to get a positiv control?
Thanks alot in advance
Hi Max
if you can add 3 controls for the 3 possible genotypes, it should be enough. the 3 controls should be genotyped by direct sequencing.
you also need to adjust the quantity of amplicon to the right quantity of enzyme to ensure complete digestion.
all the best
fred
Hi Fred,
thanks alot. Yes I'm going to sequence the outcoming fragments after PCR and digestion.
And as you say I'm calculating the amount of DNA I expect after PCR with equasion N = n(1 + E)c N = amount of DNA, n = amount of template DNA, E = efficiency (1 = 100 %) und c = number of cycles. But I'm not quite sure how to set the E.
Dear Max Klasmeier,
the use of a competitive internal positive control (IPC) spiked into each experimental sample allows to differentiate in PCR-RLFP analysis between incomplete digestion and a heterozygous genotype. Note that a competitive IPC exhibits the same primer-binding sites as the sequence of interest. We described the procedure in: “Allelic variation of the COMTgene in a despotic primate society: A haplotype is related to cortisol excretion in Macaca fuscata. Hormones and Behavior 78: 220–30”.
Dear Ralf Steinborn ,
thank you very much. An internal positiv control seem to be the most accurate way to me.
One other effect that you may see with pcr rflp is a slightly larger band than expected that will not cut. For instance your product is 300bp cutting to 100 + 200. This works but in some heterozygous samples you see a 300 and a 310 band and the 310 does not cut. This can happen with heterozygous samples where the last pcr cycle of a strong amplification allows one mutant and 1 normal single strand of product to anneal and form a heteroduplex with a single stranded bubble at the mutation point and this cannot cut because it is single stranded. This can only happen with heterozygous samples Ralf Steinborn
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