Based on the size you determined for your PCR product in the previous quesion (2

Question

Based on the size you determined for your PCR product in the previous quesion (244 bp), theoretically how long of a polypeptide could be encoded in the PCR product (assume there are no introns). What would be the approximate molecular weight (in kilodaltons) of the polypepide (average size of an amino acid = 111 Daltons)? SHow all work please.

WHat do you predict would happen to the results of the PCR if you lowered the annealing temperature by 15 degrees C? (First annealing temperature was 39 degrees C)

Explanation / Answer

Answer:

1. Size of the PCR product = 244 bp

In absence of the sequence and assuming that the whole sequence encodes for the protein, the size of the polypeptide would be

= 244/3 amino acids = 81.33 amino acids = 81 amino acids

From the number of base pairs given in the question (244bp), the whole sequence cannot encode for the protein since 244 is not divisible by 3. The value so obtained (81 amino acids) is thus theoretical.

2. Approximate MW of the encoded polypeptide = 81 * 111 = 8991 daltons = 8.991 kDa

(9.027 kDa if we assume the theoretical number of amino acids = 81.33)

3. If the annealing temperature is lowered by 15 degrees C, no annealing of the primers to the template DNA are expected to take place, assuming that the previous annealing temperature of 39 degree C is near to the Tm.

(Annealing temperature is generally taken to be (Tm-5). Since the sequence and the Tm of the PCR product is unknown, it would be difficult to predict the exact outcome. The prediction can be validated by running a gradient PCR in the given temperature range)

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