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If you performed a PCR experiment starting with only one copy of double- stranded DNA, approximately how many DNA molecules would be present in the reaction tube after 15 cycles of amplification?

Short Answer

Expert verified
Answer: Approximately 32,768 DNA molecules will be present after 15 cycles of amplification.

Step by step solution

01

Identify the given values

The initial number of DNA molecules is 1. The amplification factor is 2, and the number of cycles is 15.
02

Use the exponential growth formula

The formula for exponential growth is: Final number of molecules = Initial number of molecules × (Amplification factor)^(Number of cycles) We can plug our values into this formula: Final number of molecules = 1 × (2)^(15)
03

Calculate the final number of DNA molecules

After plugging in the values, we can calculate the final number of DNA molecules: Final number of molecules = 1 × (2)^(15) = 32768 Approximately 32,768 DNA molecules will be present in the reaction tube after 15 cycles of amplification.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Growth in PCR
Understanding the concept of exponential growth is crucial when studying Polymerase Chain Reaction (PCR) amplification. Exponential growth refers to the process in which the quantity of DNA doubles with each cycle of the PCR process. To visualize this, imagine a single copy of a DNA molecule. After just one cycle, there will be two copies; after two cycles, four copies, and this doubling continues. This is because each newly synthesized strand serves as a template in the next cycle, compounding the increase in DNA quantity.

In mathematical terms, the exponential growth can be modeled as a function of the number of cycles. If you have a starting point, or an initial number of molecules, the amplification factor (which is 2 for PCR because the DNA doubles each cycle), and the number of cycles, you can predict the amount of DNA after a series of cycles using an exponential growth formula.
DNA Molecules Calculation
To calculate the number of DNA molecules present after a certain number of PCR cycles, a simple yet powerful formula can be used. This formula is expressed as:
Final number of molecules = Initial number of molecules \times (Amplification factor)^(Number of cycles).

For example, starting with a single double-stranded DNA molecule and assuming that every cycle doubles the number of DNA molecules, after 15 cycles, the calculation is straightforward. We can plug the values into the formula to find out that the final number of DNA molecules will be:\[ 1 \times (2)^{15} = 32,768 \].

This calculation allows students not just to predict the outcome of a PCR experiment but also to gain insight into the nature of exponential processes, which are a fundamental concept in biology, population dynamics, and even finance.
PCR Cycle Number
The number of PCR cycles directly influences the total amount of DNA produced in the amplification process. Each cycle theoretically doubles the amount of DNA from the previous cycle; hence, more cycles result in a greater yield of DNA. It's critical to choose an appropriate cycle number to avoid problems like nonspecific amplification or the depletion of reagents.

For instance, in the given exercise, 15 cycles are specified. This number is chosen based on the goal of the experiment, the sensitivity of the detection method, and the initial amount of DNA. By understanding the implications of the PCR cycle number, students can better plan their experiments and anticipate the amount of DNA yield they can achieve. It's important to note that after a certain number of cycles, the reaction may plateau, and thus the DNA amplification may no longer strictly follow an exponential trend due to limitations in the reaction components and conditions.

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Most popular questions from this chapter

In a typical PCR reaction, describe what is happening in stages occurring at temperature ranges (a) \(92-95^{\circ} \mathrm{C},\) (b) \(45-65^{\circ} \mathrm{C},\) and (c) \(65-75^{\circ} \mathrm{C}\)

You have recovered a cloned DNA segment from a vector and determine that the insert is 1300 bp in length. To characterize this cloned segment, you isolate the insert and decide to construct a restriction map. Using enzyme I and enzyme II, followed by gel electrophoresis, you determine the number and size of the fragments produced by enzymes I and II alone and in combination, as recorded in the following table. Construct a restriction map from these data, showing the positions of the restriction-enzyme cutting sites relative to one another and the distance between them in units of base pairs. \(\begin{array}{cc}\text { Enzyme } & \text { Restriction Fragment Sizes (bp) } \\\ \text { I } & 350,950 \\ \text { II } & 200,1100 \\ \text { I and II } & 150,200,950\end{array}\)

Most of the techniques described in this chapter (blotting, cloning. \(\mathrm{PCR},\) etc.) are dependent on hybridization (annealing) between different populations of nucleic acids. Length of the strands, temperature, and percentage of GC nucleotides weigh considerably on hybridization. Two other components commonly used in hybridization protocols are monovalent ions and formamide. A formula that takes monovalent \(\mathrm{Na}^{+}\) ions \(\left(\mathrm{M}\left|\mathrm{Na}^{+}\right|\right)\) and formamide concentrations into consideration to compute a \(T_{m}\) (temperature of melting is as follows: $$T_{m}=81.5+16.6\left(\log \mathrm{M}\left[\mathrm{Na}^{+}\right]\right)+0.41(96 \mathrm{GC})-0.72(\% \text { formamide })$$ (a) For the following concentrations of Na' and formamide, calculate the \(T_{m \cdot}\) Assume \(45 \%\) GC content. (b) Given that formamide competes for hydrogen bond locations on nucleic acid bases and monovalent cations are attracted to the negative charges on nucleic acids, explain why the \(T_{m}\) varies as described in part (a).

In the context of recombinant DNA technology, of what use is a probe?

As you will learn later in the text (Special Topics Chapter \(1-\) CRISPR-Cas and Genome Editing, the CRISPR-Cas system has great potential but also raises many ethical issues about its potential applications because theoretically it can be used to edit any gene in the genome. What do you think are some of the concerns about the use of CRISPR-Cas on humans? Should CRISPR-Cas applications be limited for use on only certain human genes but not others? Explain your answers.

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