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If the \(\mathrm{GC}\) content of a DNA molecule is 48 percent, what are the percentages of the four bases \((\mathrm{A}, \mathrm{T}, \mathrm{G}, \text { and } \mathrm{C})\) in this mole cule?

Short Answer

Expert verified
A and T are each 26%, G and C are each 24%.

Step by step solution

01

Understand the Problem

In a DNA molecule, the four bases are adenine (A), thymine (T), guanine (G), and cytosine (C). The GC content of a DNA molecule is the percentage of bases that are either guanine (G) or cytosine (C). If the GC content is given as 48%, we need to find the individual percentages of each of the four bases.
02

Understanding the Base Pairing

In DNA, adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). Therefore, the percentage of A will be equal to the percentage of T, and the percentage of G will be equal to the percentage of C.
03

Calculate G and C Percentages

Given that the GC content is 48%, this means that the combined percentage of G and C is 48%. Let the percentage of G be x and C be x as well, because G and C are equal. This means G + C = 48%, or 2x = 48%. Solving for x gives x = 24%. Therefore, the percentage of G is 24%, and the percentage of C is also 24%.
04

Calculate A and T Percentages

Since the total percentage of all bases must equal 100%, and the percentage of G and C already account for 48% of the molecule, the remaining percentage must account for A and T. Therefore, A + T = 100% - 48% = 52%. Since A and T percentages are equal, let the percentage of A be y and the percentage of T be y, so A + T = 2y = 52%. Solving for y gives y = 26%. Therefore, the percentage of A is 26%, and the percentage of T is also 26%.

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Key Concepts

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

DNA Base Pairing
DNA consists of long chains of nucleotides, each containing one of four nitrogenous bases: adenine (A), thymine (T), guanine (G), and cytosine (C). These bases pair up in a specific manner to form the distinctive double-helix structure of DNA. It is vital to understand the base pairing rules because they ensure that DNA replicates correctly during cell division. This process allows genetic information to be accurately passed from one generation to the next.

Base pairing is based on hydrogen bonds forming between the bases, which ensures that the DNA strands stay bonded securely. The specificity of these pairings underpins the molecule's information storage capacity, helping it maintain the integrity of genetic codes.
Nucleotide Percentages
In the context of DNA, nucleotide percentages refer to the proportion of each of the four bases (A, T, G, C) within a DNA molecule. Since DNA is composed of two complementary strands, the percentages are key to understanding its composition. When given the GC content of a DNA molecule, you can easily deduce the percentages of the G and C nucleotides.

Simply put, if the GC content is provided as, say, 48%, this implies that guanine and cytosine together make up 48% of the DNA. Their individual contributions will each be half of this, i.e., 24% each. Through the knowledge of nucleotide pairings, we can also calculate the percentages of adenine and thymine, by subtracting the GC content from 100% and dividing the result evenly between A and T.
Adenine-Thymine Pairing
Adenine always pairs with thymine, forming one of the crucial complementary pairs in the DNA structure. This pairing is stabilized by two hydrogen bonds, which, although weaker than the three hydrogen bonds found in GC pairing, ensures proper DNA replication and function.

In any given strand of DNA, the amount of adenine will exactly match the amount of thymine. This equimolecular ratio allows one to infer their percentage if the other percentages (such as GC content) are known. For example, if the GC content accounts for a certain percentage, the AT content can be determined by subtracting the GC percentage from the total 100%.
Guanine-Cytosine Pairing
Guanine pairs with cytosine, forming a pairing that is essential to the double-helix structure of DNA. This pairing is bonded by three hydrogen bonds, making them more stable than AT pairs. This increased stability is significant because regions of DNA with a high GC content are often more thermally stable and partake in critical roles in genome stability.

Just like AT pairing, the percentage of guanine in a DNA molecule is equal to the percentage of cytosine. In a scenario where the GC content is given, this percentage accounts for the total combined presence of guanine and cytosine. Understanding these pairings not only aids in determining nucleotide percentages but also gives insights into the physical characteristics of DNA and its functional aspects during cellular processes.

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Most popular questions from this chapter

Explain what is meant by the terms conservative and semiconservative replication.

If thymine makes up 15 percent of the bases in a specific DNA molecule, what percentage of the bases is cytosine?

The DNA polymerases are positioned over the following DNA segment (which is part of a much larger molecule) and moving from right to left. If we assume that an Okazaki fragment is made from this segment, what will be the fragment's sequence? Label its \(5^{\prime}\) and \(3^{\prime}\) ends. \(5^{\prime} \ldots.\) CCTTAAGACTAACTACTTACTGGGATC....3' \(3^{\prime} \ldots .\) GGAATTCTGATTGATGAATGACCCTAG....5'

Which of the following would happen if DNA synthesis were discontinuous on both strands? a. The DNA fragments from the two new strands could become mixed, producing possible mutations. b. DNA synthesis would not take place, because the appropriate enzymes to carry out discontinuous replication on both strands would not be present. c. DNA synthesis might take longer, but otherwise there would be no noticeable difference. d. DNA synthesis would not take place, because the entire length of the chromosome would have to be unwound before both strands could be replicated in a discontinuous fashion.

E. coli chromosomes in which every nitrogen atom is labeled (that is, every nitrogen atom is the heavy isotope \(^{15} \mathrm{N}\) instead of the normal isotope \(^{14} \mathrm{N}\) ) are allowed to replicate in an environment in which all the nitrogen is \(^{14} \mathrm{N}\). Using a solid line to represent a heavy polynucleotide chain and a dashed line for a light chain, sketch each of the following descriptions: a. The heavy parental chromosome and the products of the first replication after transfer to a \(^{14} \mathrm{N}\) medium, assuming that the chromosome is one DNA double helix and that replication is semiconservative. b. Repeat part \(a\), but now assume that replication is conservative. c. Repeat part \(a,\) but assume that the chromosome is in fact two side-by-side double helices, each of which replicates semiconservatively. d. Repeat part \(c,\) but assume that each side-by-side double helix replicates conservatively and that the overall chromosome replication is semiconservative. e. If the daughter chromosomes from the first division in \(^{14} \mathrm{N}\) are spun in a cesium chloride density gradient and a single band is obtained, which of the possibilities in parts \(a\) through \(d\) can be ruled out? Reconsider the Meselson and Stahl experiment: What does it prove?

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