Assignment 2

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University of Ottawa *

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Biology

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Dec 6, 2023

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Assignment 2: September 25, 2023 1. Tm & Ta a. Melting Temperatures of DNA Hybrids 𝐓? = 𝟔? . ? ° ? + ?? ° ? × [( # 𝐨? ? 𝐚𝐧? ? – ?𝟔.4)/ N] Fwd primer: 5’-GTG TAT AAG CTT ATG CTG CCG ATC AGC GAG GAG CA-3’ Rev primer: 5’-CTG TAT GGT ACC TTA GCA TCG CTC CAG TTG CTC GCA-3’ i) First Cycle : Tm fwd = 64.9 + 41 × [ ( G + C − 16.4 ) n ] Tm fwd = 64.9 + 41 × [ ( 8 + 6 − 16.4 ) 23 ] Tm fwd = 64.9 + 41 × − 12 115 Tm fwd = 64.9 − 4.27826 Tm fwd = 60.62 C Tm 1 st cycle = Tm fwd + Tm rev 2 = 60.62 + 59.09 2 = 119.71 2 = 59.855 C ii) Last Cycle : Tm fwd = 64.9 + 41 × [ ( G + C − 16.4 ) n ] Tm fwd = 64.9 + 41 × [ ( 11 + 7 − 16.4 ) 35 ] Tm fwd = 64.9 + 41 × 8 175 Tm fwd = 64.9 + 1.8743 Tm fwd = 66.77 C Tm la st cycle = Tm fwd + Tm rev 2 = 66.77 + 67.86 2 = 134.63 2 = 67.315 C b. Annealing Temperature i) Ta 1 st Cycle = 59.09 − 5 = 54.09 C Ta la st Cycle = 66.77 − 5 = 61.77 C ii) As discussed on page 2 of the lab manual, the value of the annealing temperature ( Ta ) should be “about 5 C below the lowest Tm of the two primers”. During the first PCR cycle, the reverse primer has the lower of the two melting temperatures, while the forward primer has the lower Tm during last PCR cycle. Therefore, the Ta for Tm rev = 64.9 + 41 × [ ( G + C − 16.4 ) n ] Tm rev = 64.9 + 41 × [ ( 5 + 8 − 16.4 ) 24 ] Tm rev = 64.9 + 41 × − 17 120 Tm rev = 64.9 − 5.80833 Tm rev = 59.09 C Tm rev = 64.9 + 41 × [ ( G + C − 16.4 ) n ] Tm rev = 64.9 + 41 × [ ( 8 + 11 − 16.4 ) 36 ] Tm rev = 64.9 + 41 × 13 180 Tm rev = 64.9 + 2.96111 Tm rev = 67.86 C

the first cycle is 5 degrees below the Tm of the reverse primer at the first cycle (59.09 C ) and the Ta of the last cycle is 5 degrees below the Tm of the forward primer during the last cycle (66.77 C ). 2. Recovery Yield Well 1 shows that a band formed from the purified amplicon sample. The intensity is proportional to the amount of SYBR Safe molecules. Because wells 2 and 4 were loaded with unpurified amplicon, we would expect the purified band (in well 1) to be twice as intense. Because no band appeared in well 2, it can be concluded that the PCR failed, and so the amount of amplicon in well 1 is actually proportional to the amount in well 4. Because the sample in well 1 is approximately twice the intensity of the DNA marker (which represents 3000bp or 30ng/10µL), one can conclude that the purified sample contains 60ng of DNA. The sample was taken from the 2µL of PCR amplicon, therefore the elution solution contains 60ng/2µL, or 30ng/µL, of DNA. Because the solution contained 30µL of elution buffer, it can be concluded that there are m= CV= 30ng/µL(30µL)= 900ng of DNA. Because the band in well 4 has an intensity factor that is approximately 3 times stronger than that of the marker in well 7, we can predict that there is ~90ng/10µL. With 90ng of DNA in the well, taken from a 2µL sample, the concentration of the PCR product is 90ng/2µL= 45ng/µL. Because the PCR sample had a volume of 50µL, we can conclude that there are m= CV= 45ng/µL(50µL)= 2250ng of total DNA content. % yeild = experimental yeild theoretical yeild × 100 % yeild = 900 ng 2250 ng × 100 % yeild = 40% 3. Digest a) Estimate the concentration of purified digest insert, we can look observe the band in well 2 (corresponding to the purified amplicon insert). The DNA marker displays that the sample is roughly 2500bp in length= 25ng of DNA. The intensity of the band is roughly twice as intense as the marker, therefore the concentration of DNA in well 2 is equal to 50ng/10µL= 5ng/µL. To deduce the concentration of purified amplicon insert: 10 000 bp Well 7 Figure 1: 1% Agarose Gel Electrophoresis of Purified & Unpurified T7 RNA Polymerase PCR Amplicon. Well 1 is loaded with purified amplicon. Wells 2 and 4 are loaded with unpurified amplicon, while wells 3 and 5 are loaded with negative controls. The lack of a band in well 2 suggests that the PCR was incomplete for that sample. The MassRuler Express Forward DNA Ladder Marker was loaded into well 7. Since 10µL of Marker were loaded into the gel, and the band in well 1 indicates 3000bp, there are 30ng of the DNA fragment in well 1. 1 000 bp 1 500 bp 2 000 bp 3 000 bp 5 000 bp 7 000 bp Well 5 Well 3 Well 4 Well 2 Well 1 100 bp 200 bp 300 bp 500 bp 700 bp 700 bp

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