Chapter 10, Problem 4Q

A.

Summary Introduction

To explain: The number of DNA fragments labeled in gray, green, or red, or outlined in yellow produced if the PCR shown in Fig 10-12 runs additional two rounds of amplification and which will predominate over many cycles.

Introduction: Polymerase chain reaction (PCR) is a device used to amplify small fragments of DNA for various analyses. PCR is a cost-effective, dependable, and a simple way to repeatedly amplify, that is, to replicate a small fragment of DNA of interest. PCR is the most widely used molecular technique all over the world.

Explanation of Solution

The number of DNA produced in a generation is 2ⁿ, where ‘n’ is the number of generations. At the end of one additional round of amplification which is 2⁵=32 DNA, there will be gray-2; green-4; red-4; and yellow outlined-22 fragments. At the end of one more additional round of amplification which is 2⁶=64 DNA, there will be gray-2; green-5; red-5; and yellow outlined-52 fragments. 2ⁿ-2 is the actual formula, where 2 is the parent DNA being conserved. The yellow outlined DNA fragment will predominate over many cycles of reactions.

B.

Summary Introduction

To calculate: The number of cycles of PCR amplification required to produce 100 ng of DNA from a double-stranded DNA with 500 nucleotide paired sequence.

Introduction: PCR used to amplify small fragments of DNA for various analyses. PCR is done in simple steps: Denaturation of the DNA fragmentàAnnealing of the primersàElongation of the DNA fragment along the primer sideàDenaturationàCycle continues until the required DNA quantity is obtained. PCR is exclusively quantitative.

Explanation of Solution

$\begin{array}{c}\text{1}\text{ds}\text{DNA}\text{=}\text{500}\text{Nucleotides pairs}\\ \text{=1000 Nucleotides}\end{array}$

$\begin{array}{c}\text{Average}\text{molecular}\text{mass}\text{of}\text{1}\text{nucleotide}=\text{330}\text{g/mole}\\ \left(\begin{array}{l}\text{Required}\text{weight}\text{of}\text{DNA}\\ \text{from}\text{500}\text{nucleotide}\text{pairs},\text{W}\end{array}\right)=\text{100}\text{ng}\\ ={10}^{-7}\text{g}\end{array}$

$\begin{array}{c}\text{Mass}\text{of}\text{1}\text{ds}{\text{DNA, W}}_{\text{0}}\text{=Total No}\text{.}\text{of}\text{nucleotides×}\frac{\text{Average}\text{mass}\text{of}\text{1}\text{Nucleotide}}{\text{Avagadro}\text{constant}}\\ =1000\times \frac{\text{330}\text{g/mole}}{{\text{6×10}}^{\text{23}}\text{/mole}}\\ =\frac{33}{6}\times {10}^{3+1-23}\\ =5.5\times {10}^{-19}\text{g}\end{array}$

To find the number of PCR cycles:

$\begin{array}{c}\text{Final}\text{weight}\text{of}\text{DNA}=\text{Initial}\text{weight}\text{of}\text{DNA}\times 2^N\\ \text{where}N=\text{number}\text{of}\text{PCR}\text{cycles}\\ \text{Tharefore, W}={\text{W}}_{{}^{\text{0}}}{\text{×2}}^N\\ {10}^{-7}=5.5\times {10}^{-19}{\text{×2}}^N\end{array}$

  $\begin{array}{c}{10}^{-7+19}=5.5{\text{×2}}^N\\ {10}^{12}=5.5{\text{×2}}^N\end{array}$

$\begin{array}{c}\text{Taking}{\text{log}}_{\text{10}}\text{on}\text{both}\text{sides:}\\ \text{log}{10}^{12}=\log \left(5.5{\text{×2}}^N\right)\\ 12\log 10=\log 5.5+N\log 2\\ 12=0.7+N\times 0.3\\ N=\frac{12-0.7}{0.3}\end{array}$

     $\begin{array}{c}=\frac{11.3}{0.3}\\ =37.6cycles\\ \approx 40cycles\end{array}$

Conclusion

Thus, approximately 40 cycles of PCR are required to amplify 500 nucleotide pairs double-stranded DNA to obtain 100 ng of DNA. 100 ng of DNA is the minimum amount of DNA required for biochemical analysis. As PCR is automated, if the protocols are followed correctly, the entire procedure will take less than a day.