You wish to amplify a 1 Kbp single copy sequence from 1 µg of a single stranded
ID: 313330 • Letter: Y
Question
You wish to amplify a 1 Kbp single copy sequence from 1 µg of a single stranded genome which is 109 Kbp using a pair of primers “A” and “B”.
1. What is the minimum number of cycles required to obtain a double stranded amplification product delimited by primers “A” and “B”? 3
2. What mass of the double stranded PCR product (in µg) delimited by primers “A” and “B” would you have after the number of cycles indicated in the previous question? 2 x 10-9 µg
3. What mass of the double stranded PCR product (in µg) delimited by primers “A” and “B” would you have after a total of 30? Approx. 0.27 µg
4. How many additional PCR cycles would be required to attain the same yield of product which was obtained with 1 µg of genomic DNA if you had started with 1 ng of genomic DNA? Approx. 10
explain how to get these answers
Explanation / Answer
1. In cycle 1, there is one original (single stranded) template molecule, one anchored product (comprised of one single stranded DNA) and zero amplicons. in cycle 2, there is one anchored product(used as template), another anchored product (comprised of one single stranded DNA) and zero amplicons.Ifamplification efficiency is 100%, i.e., at the end of every cycle the number of molecules exactly doubles and that we start out with one original template molecule and primers in excess. so at the end of third cycle a complete amplicon is observed.
2. mass of PCR product obtained is
template used=1ug of 10^9kb(i suppose which is diff from question)=330x10^9x10^3=330x10^12 Da
330 is amu of single base, 1kilo=10^3
no of moles=wt/gm.molwt =1ug/330x10^12 =3.03x10^-15
no of particles=3.03x10^-15x6.023x10^23=18.24x10^8particles of template used
if these many particles used to make 1kb pcr product which is 660 x10^3, then
1N(6.023x10^23)=660*10^3. then 18.24x10^8 particles make 1998x10^-12 which is approximately 2x10-9ug
3. now calculate the no of copies from cycles30 which is 2^(n), 2^30. but as we observed amplicon after 3 cycles using the ss genome, it is 2^27 copies. so, 2^27X18.24x10^8x660/6.023x10^23 =.27ug
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