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Ch. 20 - Recombinant DNA Technology
Klug - Concepts of Genetics 12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
All textbooksKlug 12th EditionCh. 20 - Recombinant DNA TechnologyProblem 31b
Chapter 20, Problem 31b

Most of the techniques described in this chapter (blotting, cloning, PCR, etc.) are dependent on hybridization (annealing) between different populations of nucleic acids. The 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 Na⁺ ions (M[Na⁺]) and formamide concentrations into consideration to compute a Tₘ (temperature of melting) is as follows:
Tₘ=81.5+16.6(log M[Na+])+0.41(%GC)−0.72(%formamide)
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 Tₘ varies as described in part (a).

Verified step by step guidance
1
Understand the concept of Tₘ (melting temperature): Tₘ is the temperature at which half of the DNA strands in a sample are denatured, meaning the hydrogen bonds between complementary bases are broken, and the strands separate.
Analyze the role of monovalent Na⁺ ions: Monovalent cations like Na⁺ stabilize the negative charges on the phosphate backbone of DNA, reducing repulsion between strands and increasing the Tₘ. This is reflected in the formula as the term 16.6(log M[Na⁺]).
Examine the impact of GC content: GC base pairs have three hydrogen bonds compared to two in AT base pairs, making GC-rich regions more thermally stable. The formula accounts for this with the term 0.41(%GC), which increases Tₘ as GC content rises.
Consider the effect of formamide: Formamide disrupts hydrogen bonding between nucleic acid bases, competing for bonding sites and lowering the stability of the DNA duplex. This decreases Tₘ, as shown by the term -0.72(%formamide).
Combine these factors to explain Tₘ variation: The formula integrates the stabilizing effects of Na⁺ ions and GC content with the destabilizing effect of formamide. Changes in these variables directly influence the Tₘ, demonstrating how hybridization conditions are affected by chemical and physical factors.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hybridization

Hybridization refers to the process where two complementary strands of nucleic acids (DNA or RNA) bind together through base pairing. This interaction is crucial for various molecular biology techniques, as it allows for the formation of stable double-stranded structures. Factors such as strand length, temperature, and GC content significantly influence the stability of these hybrids, affecting the overall efficiency of techniques like PCR and blotting.
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Melting Temperature (Tₘ)

The melting temperature (Tₘ) is the temperature at which half of the DNA strands are in the double-helix state and half are in the 'melted' single-strand state. Tₘ is influenced by the nucleotide composition, particularly the percentage of guanine (G) and cytosine (C) bases, as they form three hydrogen bonds compared to the two formed by adenine (A) and thymine (T). Additionally, the presence of ions and formamide can alter Tₘ by stabilizing or destabilizing the nucleic acid structure.
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Role of Monovalent Ions and Formamide

Monovalent ions, such as sodium (Na⁺), play a critical role in stabilizing nucleic acid structures by neutralizing the negative charges on the phosphate backbone, which enhances hybridization. Formamide, on the other hand, disrupts hydrogen bonding by competing for binding sites on nucleic acid bases, effectively lowering Tₘ. The balance between these components is essential for optimizing hybridization conditions in various molecular techniques.
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Related Practice
Textbook Question

The gel presented here shows the pattern of bands of fragments produced with several restriction enzymes. The enzymes used are identified above the lanes of the gel, and six possible restriction maps are shown in the column to the right.

One of the six restriction maps shown is consistent with the pattern of bands shown in the gel.

The highlighted bands (magenta) in the gel were hybridized with a probe for the gene pep during a Southern blot. Where in the gel is the pep gene located? 

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Textbook Question

A widely used method for calculating the annealing temperature for a primer used in PCR is 5 degrees below the melting temperature, Tₘ(°C), which is computed by the equation 81.5+0.41×(%GC)−(675/N), where %GC is the percentage of GC nucleotides in the oligonucleotide and N is the length of the oligonucleotide. Notice from the formula that both the GC content and the length of the oligonucleotide are variables. Assuming you have the following oligonucleotide as a primer,

5′-TTGAAAATATTTCCCATTGCC-3′

Compute the annealing temperature for PCR. What is the relationship between %GC and? Why? (Note: In reality, this computation provides only a starting point for empirical determination of the most useful annealing temperature.)

1016
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Textbook Question

Most of the techniques (blotting, cloning, PCR, etc.) are dependent on hybridization (annealing) between different populations of nucleic acids. The 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 Na⁺ ions (M[Na⁺]) and formamide concentrations into consideration to compute a Tₘ (temperature of melting) is as follows:

Tₘ=81.5+16.6(log M[Na+])+0.41(%GC)−0.72(%formamide)

For the following concentrations of Na⁺ and formamide, calculate the Tₘ. Assume 45% GC content.

  [Na⁺]  % Formamide

  0.825      20

  0.825      40

  0.165      20

  0.165      40

526
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Textbook Question

In humans, congenital heart disease is a common birth defect that affects approximately 1 out of 125 live births. Using reverse transcription PCR (RT-PCR), Samir Zaidi and colleagues [(2013) Nature 498:220.223] determined that approximately 10 percent of the cases resulted from point mutations, often involving histone function. To capture products of gene expression in developing hearts, they used oligo(dT) in their reverse transcription protocol.

How would such a high %T in a primer influence annealing temperature?

671
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Textbook Question

In humans, congenital heart disease is a common birth defect that affects approximately 1 out of 125 live births. Using reverse transcription PCR (RT-PCR), Samir Zaidi and colleagues [(2013) Nature 498:220.223] determined that approximately 10 percent of the cases resulted from point mutations, often involving histone function. To capture products of gene expression in developing hearts, they used oligo(dT) in their reverse transcription protocol.

Compared with oligo(dT) primers, a pool of random sequence primers requires a trickier assessment of annealing temperature. Why?

486
views
Textbook Question

In humans, congenital heart disease is a common birth defect that affects approximately 1 out of 125 live births. Using reverse transcription PCR (RT-PCR), Samir Zaidi and colleagues [(2013) Nature 498:220.223] determined that approximately 10 percent of the cases resulted from point mutations, often involving histone function. To capture products of gene expression in developing hearts, they used oligo(dT) in their reverse transcription protocol.

If one were interested in comparing the quantitative distribution of gene expression in, say, the right and left sides of a developing heart, how might one proceed using RT-PCR?

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