Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

2. Consider the following PCR amplification (not the one carried out in lab) whi

ID: 91118 • Letter: 2

Question

2. Consider the following PCR amplification (not the one carried out in lab) which produces a 1,084 bp product. At the start of the amplification, the concentrations of both For and Rev primers were 0.8 M, each dNTP was 250 M, Mg2+ was 7.5 mM and 0.10 ng of a 5,236 bp template was present in a reaction volume of 100 L. If the amplification were carried out for 30 cycles (sufficient to completely consume the limiting reagent), what component would limit the PCR product yield? (2 points). What is the theoretical yield (in moles) of the 1,084 bp PCR product under these conditions (assuming equal quantities of A, C, G, and T were incorporated into the PCR product)? (2 points). If a base pair has an average molecular weight of 650 g/mole, how many micrograms could be produced? (2 points). Be sure to show your calculations. Note that comparing this number to your yield provides an estimate of the efficiency of your PCR amplification.

3. Determining the size and concentration of DNA fragments is a common task in molecular biology laboratories. The gel below was used by Dr. Stewart to determine both the sizes and concentrations of two DNA fragments needed for his research. For each fragment, 1 L of the indicated dilution was run on the gel. Use the methods described in the lab manual to determine both the sizes and concentrations of each of the two fragments in the original (undiluted) stock. Data for HindIII lambda DNA are shown in the table below. Show your calculations clearly. (10 points).

using the image below

Report 1 instructions pc X O file///cyusers/rosaj/AppData/Local Temp/Report%201%20instructions.pdf 4 of 5 Type here to search Lane Sample 5x-diluted fragment 1 10x-diluted fragment 1 50x-diluted fragment 1 4 HindIII lambda DNA (300 ng) 5x-diluted fragment 2 10x-diluted fragment 2 50x-diluted fragment 2 a Cal 5.49 PM 6/12/2017 R2

Explanation / Answer

1. The limiting component of PCR product yield is that component of the PCR reaction mixture which runs out first.

While calculating the amount of each reactant needed in a PCR reaction, one should keep in mind that the DNA template doubles in each cycle.

Each strand doubles at the end of each cycle. At the end of amplification, it generates (2n - (n+1)) copies where n is the number of cycles.

Hence the reagent utilization also doubles up at each cycle, through out the 30 cycles.

The reagents in PCR mixture should be added accordingly.

The number of moles of product synthesized for each mole of the reagent utilized should be calulated.

2. The molecular weight of DNA template can be calcukated based on its size (length in base pairs).

The average weight of each base pair of DNA is 650 Daltons, that is 650 g/Mol.

Thus the molecular weight of a double stranded DNA can be calculated by multiplying the length of DNA in base pairs with 650.

Here the length of DNA is 1,084 bp.

Therefore the molecular weight of the given DNA = 1084X650 Daltons or g/Mol. (This is the molar mass of DNA)

To calculate the amount of DNA produced using Avagadro's number.

The PCR product can be calculated in copy number.

Number of copies= (amount of DNA template in nanogramsX 6.022X1023 /Length of DNA template in base pairs X 1 X 10 9 X 650 ; where 6.022X1023 is the Avagadro's constant number

= 0.1 X 6.022X1023 / 1084 X 1 X10 9 X 650

3. To determine the size of the DNA fragment using a gel,

Run the DNA of interest on agarose gel (Agarose gel electrophoresis).

Along with the sample, load a DNA ladder in another well, which serves as a marker to know the molecular weight.

The DNA migrates in the gel according to its size. Smaller the size of DNA, farther will be the migration.

Compare the loaded DNA with the marker DNA and determine the molecular weight of DNA.

Concentration of the PCR product can be obtained by spectrophotometric analysis.

Concentration in micrograms/ml is A260-A320X Dilution factor X 50micrograms/ml

where A is the absorbance.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Chat Now And Get Quote