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A typical PCR experiment might involve the following thermal cycles: 25 cycles o

ID: 60871 • Letter: A

Question

A typical PCR experiment might involve the following thermal cycles: 25 cycles of (30 sec 95 degree C rightarrow 30 sec T_m - 5 degree C rightarrow 1 min 75 degree C) Identify the denaturation, annealing and elongation steps of the PCR in the above thermal cycle. Why does it make sense to use T_m - 5 degree C in the middle step? What would be the effect of using T_m - 20 degree C in the middle step? What would be the effect of using T_m + 10 degree C' in the middle step? In a perfect PCR amplification cycle, you would expect twice the amount of product after the cycle. Thus, you might expect 2^25 or 34 million fold amplification if you were able to achieve perfect amplification for a 25 cycle PCR. If you used T_m exactly for the middle step, what is the maximum fold amplification you might expect after 25 cycles. (To keep the calculation simple, assume that you only need to consider one primer with melting temperature T_m. The other primer magically anneals perfectly at all temperatures.)

Explanation / Answer

a) As per the PCR experiment the denaturation temp. is 95o C annealing step is -5oC and the elongation step is 75oC.

b) Tm (-5oC ) is actually melting temperature of primer and it is suggested to use primer annealing temperature as 5 or below degree celsius. But the Tm is used much below like -20oC or +10oC  then the primer may or may not bind with the template specifically and it would be unspecific binding which may result with unspecific bands or unwanted results.

c) at least 100, 000x amplification would be expected after 25 cycles of PCR reactions.

where, C25 = 10,000 = 105

so, c5 = 10 and also 5(log c) = log 10 = 1

so log c = 0.2 and c = 1.58 (approx.)

so it could be said as if each cycle 1 template would 1.58 templates for the first round of synthesis then there would 100,000x fold increase in DNA template at least at the end of 25 reactions.

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