Your goal is to amplify by PCR the following...

Your goal is to amplify by PCR the following sequence: 5'…GAATGTAAACTTCGGCGAATCAG………….GATGAAGTTCAAATCATCGACGGC…3' 1) The first and last nucleotides (bold, red) correspond to nucleotides 747 and 1134 and define the first (747) and last (1134) nucleotide of the amplified sequence. a) What are the sequences of the 18-nucleotides forward (or sense) and reverse (or antisense) primers that you plan to use to amplify the DNA sequence located between nucleotides 747 and 1134? Indicate 5’ and 3’ 6 pts b) What is the length of the PCR product (expressed in base pairs or bp)? Show your calculation. 6 pts 2) The PCR mixture contains the template DNA, the dNTP mixture, the reaction buffer, both primers (forward and reverse) and the Taq polymerase. a) You have stock solutions of your two primers at both a concentration of 20 ?mol/L. Your supervisor asked you to add in the PCR reaction 300ng of each primer. Determine the volume (expressed in ?l) of primers you need to add in your 100?l reaction, knowing that the two primers have the same molecular weight (7500 g/mol). Show your calculation. 8 pts Reminder: 1g =10³ mg = 10??g =10? ng. The same applies for mol, mmol, ?mol and nmol b) You know that 3 ?l of the Taq polymerase catalyze the incorporation of 0.03 ?mol of dNTP in 15 min at 70°C. Your advisor asked you to use 2 units of the Taq polymerase. Calculate the volume of Taq polymerase you need to add in your 100 ?l PCR reaction? To help you, your advisor gives you the information that came with the Taq polymerase solution and states that 1 unit of this polymerase catalyzes the incorporation of 10 nmoles of dNTP in 30 min at 70°C. Show your calculation. 8 pts

Transcript

00:01 So the first and last nucleotides of the nucleotide sequence correspond to nucleotide 4747 and 1134 of the sequence of a dna sequence.

00:17 So very first question, what are the sequence of the 18 nucleotide forward or reverse primer that you plan to amplify the dna sequence located in between these two? so first of all, you want to write down a complementary sequence.

00:45 So we can see it follows base pair rule a paired with t, c paired with g.

01:40 All right, so now let's say we have the bottom strand opposite orientation g paired with c, a paired with t.

01:49 The next step, you want to decide where to put your forward reverse primer.

01:55 The forward primer is going to be complementary to the bottom strand, the three priming.

02:00 So the forward primer is going to be the same as the top strand.

02:06 One, two, three, four, five, six, seven, eight, nine, 11, 12, 13, 14, 15, 16, 17, and 18.

02:28 18 nucleotide.

02:30 And so you can see that the new strand is going to extend this way.

02:34 The reverse primer is going to be put where the three priming of the top strand.

03:08 So then again, 18 complementary to the three priming of the top strand, three.

03:13 So it's going to go this way.

03:15 So you can see these are the two primers, green, reverse, red, forward, and each 18 nucleotide.

03:24 And the question says, what is the length of pcr product? so we know that the beginning of the sequence is going to be 747, which is where the blue g is.

03:41 And the end of the sequence is 1134 base pair.

03:49 So we put the primer lineup right on the spot with g and c, with the blue g and blue c.

03:58 So the distance between these two nucleotide is obviously 1134 base pair minus 747 base pair.

04:12 So you have 387 base pair in between these two nucleotide.

04:18 So this is the size of the pcr product.

04:23 So we amplify the dna in between the two nucleotide g and c.

04:31 Now the next part of the question, you want to do some calculation.

04:33 The you have stock solution of your two primers, both at 20 micromolar per liter.

04:44 And your supervisor asks you to add into the pcr reaction 300 nanogram of each primer.

04:55 So determine the volume of the primer you need to add to your 100 microliter, knowing that the two primers have similar molecular weight, which is 7 ,500 gram per mole.

05:08 So let's take a look.

05:10 So first of all, you know that the concentration at 20 molar per liter and the amount that you want to, amount of primer that you want to add it to the reaction is in the mass form, which is nanogram.

05:27 So the very first step, you want to convert the micromolar into a nanogram.

05:33 So then we can do our calculation.

05:36 So, uh, our first step, you want to know, this is question number two, the total primer that you need.

05:46 So you know, know that the primer is concentration multiplied by volume.

05:55 You need 20 micromolar per liter of primer multiplied by the reaction, which is a total of, of x microliter and with a total, which equals 300 nanogram.

06:22 So as you can see that this is micro molar and you have a nanogram.

06:28 So let's convert 20 micromolar to gram.

06:32 So according to molecular weight, let's put it right down there.

06:38 Um, a one micromolar.

06:43 So one micromole equals 10 to negative six mole.

06:55 So 20 micromole is going to be two, two times 10 to the negative five mole.

07:10 And we also know the molecular weight is 7 ,500 gram per mole.

07:17 So we know the mass of 20 micromole is going to be two times 10 to negative five mole multiplied by molecular weight, 7 ,500 gram per mole.

07:43 You end up having 0 .15 gram.

07:47 And we also know that one gram, uh, equals, uh, 10 to nine power nanogram.

07:55 So then 0 .15 gram times 10, uh, two to ninth equals 1 .5 times 10 to the eighth nanogram.

08:12 So from there, we know that the concentration can be converted into, um, 1 .5 times 10 to the eighth nanogram per liter...

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