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1a) During one round of PCR, the DNA double helix is separated into
its constituent strands, and new strands are built up against each
of these. The final products are two helices, and so the DNA
content of the cell
doubles.
b) If there is originally 1mg of DNA in the sample, 30 rounds of
PCR, will lead to doubling thirty times:
$Mass_{final} = 0.001 \times 2^{15} = 32.8$g of DNA
c) We must first calculate the number of moles of DNA in the
sample:
$Moles = \text{concentration} \times \text{volume} = 0.1 \times
10^{-12} \times 0.5 \times 10^{-3} = 5 \times 10^{-17}$ moles
DNA
The number of molecules of DNA is calculated by multiplying moles
by Avogadro's number:
$Molecules = 5 \times 10^{-17} \times 6.02 \times 10^{23} = 3.01
\times 10^7$
Thus to work out the number of rounds of PCR necessary:
$3.01 \times 10^7 \times 2^n = 10^{12}$
$2^n = \frac{10^{12}}{3.01 \times 10^7}$
$n = \frac{ln(\frac{10^{12}}{3.01 \times 10^7})}{ln(2)} =
15.02$
Therefore, we require 16 rounds of PCR to exceed $10^{12}$
molecules of DNA.
2a) Number of minutes = $\frac{24000}{6000} = 4$
b) In the first round of PCR, we will incorporate $2 \times 24000 =
48000$ bases into DNA. In the next round there will be four
template strands, and so $4 \times 24000 = 96000$ bases will be
incorporated into DNA. Thus over a total of ten rounds:
Number of incorporated bases = $ 24000 \times ( 2^1 + 2^2 +2^3
+...+2^{10})$
The part in brackets is a geometric series, which can be summed as
follows:
Let $ x = 2^1 + 2^2 +2^3 +...+2^{10}$
Then $2x = 2^2 +2^3 +2^4 +...+2^{11}$
$\therefore 2x -x = 2^{11} - 2$
$\rightarrow x = 2^{11} -2= 2(2^{10} - 1)$
Thus, the number of incorporated bases = $24000 \times 2(2^{10}
-1)= 48000(2^{10} -1) = 4.91 \times 10^7$
3) Solving the restriction map is done by partly by trial and
error, and partly by building the map up with more and more
information until the solution is found: