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DNA of the Human Body If completely unraveled, all of a human's DNA would be able to reach a distance of nearly \(3.2 \times 10^{5} \mathrm{~km}\), the distance from Earth to the moon. Given that each base pair in a DNA helix extends a distance of \(3.4 \AA\), calculate the number of base pairs found within the entirety of a human's DNA.

Short Answer

Expert verified
The total number of base pairs is approximately \(9.41 \times 10^{17}\).

Step by step solution

01

Understanding the Units

First, we need to understand the units involved in the problem. The total length of the DNA is given in kilometers, while the extension of each base pair is given in angstroms (\(\text{Å}))\). We need to convert these numbers to a common unit for accurate computation.
02

Convert Length from Kilometers to Angstroms

1 kilometer is equal to \(10^{13}\) angstroms. Therefore, we can convert \(3.2 \times 10^{5} \text{ km}\) to angstroms:\[3.2 \times 10^{5} \text{ km} = 3.2 \times 10^{5} \times 10^{13} \, \text{Å} = 3.2 \times 10^{18} \, \text{Å}\]
03

Calculate the Number of Base Pairs

Now that we have converted the total length of the DNA to angstroms, we can determine the number of base pairs. Each base pair extends \(3.4\ \, \text{Å}\). Therefore, to find the number of base pairs:\[\frac{3.2 \times 10^{18} \, \text{Å}}{3.4 \, \text{Å/base pair}} = \frac{3.2}{3.4} \times 10^{18} \, \text{base pairs} \approx 0.941 \times 10^{18} \, \text{base pairs}\]
04

Simple Calculation

Calculate the value of \(0.941 \times 10^{18}\):\[0.941 \times 10^{18} \approx 9.41 \times 10^{17}\ \, \, \text{base pairs}\]

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Key Concepts

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

Base Pairs
When we talk about DNA, one crucial aspect to understand is the concept of base pairs. DNA, or deoxyribonucleic acid, is structured as a double helix. This double helix is composed of two strands that are held together by base pairs. Base pairs are formed by the bonding of specific nucleotides: adenine with thymine, and guanine with cytosine.
In this problem, base pairs play a key role because we're looking at how they repeat along the DNA to give it length. Each base pair provides a measure of length for the DNA helix. They are like building blocks that repeat over and over, ultimately defining the structural length of the DNA strand.
This exercise simplifies the complexity of DNA by representing its length through these repeating units. Understanding base pairs helps us appreciate how a seemingly small structure can be mapped out to cover an immense length.
Angstrom Conversion
Converting units is essential when working with DNA strand measurements, as in this example. An angstrom (\( \text{Å} \)) is a unit of length that is commonly used to measure things at an atomic scale, such as bonds and DNA base pairs. One angstrom equals \( 10^{-10} \) meters. In contrast, larger dimensions like the total length of all human DNA can be given in kilometers.
  • To solve the problem accurately, we need to convert kilometers to angstroms.
  • This requires understanding conversion factors – specifically \(1 \text{ km} = 10^{13} \text{ Å} \).
By converting, we change the total length of DNA from kilometers to angstroms. This gives us a consistent unit of measurement to calculate how many base pairs fit the given length. While conversions might seem tedious at first, they simplify complex multiscale problems significantly.
Step-by-Step Problem Solving
Breaking down problems into steps is a key approach to effective problem-solving. The original solution provides a perfect example of a clear, methodical process. Start by understanding the given units, which is fundamental, as mismatched units can lead to erroneous conclusions.
Next, convert the units so that they match across the problem context. This step narrows down discrepancies and sets the ground for accurate calculations. Once the units are harmonized, focus on solving the main question, which is calculating the number we seek – in this case, how many base pairs fit into a given length of DNA.
Finally, execute the calculation with attention to detail. Once you reach a conclusion like \(0.941 \times 10^{18}\), simplify if necessary to provide a clear and concise answer. This method is an excellent way to tackle a variety of scientific problems, not just those related to DNA.

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Most popular questions from this chapter

Genomic Sequencing In large-genome sequencing projects, the initial data usually reveal gaps between contigs where no sequence information has been obtained. To close the gaps, DNA primers complementary to the \(5^{\prime}\)-ending strand at the end of each contig are especially useful. Explain how researchers could use these primers to close the gaps between contigs.

Sanger Sequencing Logic In the Sanger (dideoxy) method for DNA sequencing, researchers add a small amount of a dideoxynucleoside triphosphate, such as ddCTP, to the sequencing reaction along with a larger amount of the corresponding deoxynucleoside, such as dCTP. What result would researchers observe if they omitted dCTP from the sequencing reaction?

The Structure of DNA Elucidation of the threedimensional structure of DNA helped researchers understand how this molecule conveys information that can be faithfully replicated from one generation to the next. To see the secondary structure of double-stranded DNA, go to the Protein Data Bank website (www.rcsb.org). Use the PDB identifiers provided in parts (a) and (b) below to retrieve the structure summary for a double-stranded DNA segment. View the 3D structure using JSmol. The viewer select menu is below the right corner of the image box. Once in JSmol, you will need to use both the display menus on the screen and the scripting controls in the JSmol menu. Access the JSmol menu by clicking on the JSmol logo in the lower right corner of the image screen. Refer to the JSmol help links as needed. a. Access PDB ID 141D, a highly conserved, repeated DNA sequence from the end of the genome of HIV-1 (the virus that causes AIDS). Set the Style to Ball and Stick. Then use the scripting controls to color by element (Color > Atoms > By Scheme > Element

Preserving DNA in Bacterial Endospores Bacterial endospores form when the environment is no longer conducive to active cell metabolism. The soil bacterium Bacillus subtilis, for example, begins the process of sporulation when one or more nutrients are depleted. The end product is a small, metabolically dormant structure that can survive almost indefinitely with no detectable metabolism. Spores have mechanisms to prevent accumulation of potentially lethal mutations in their DNA over periods of dormancy that can exceed 1,000 years. \(B\). subtilis spores are much more resistant than are the organism's growing cells to heat, UV radiation, and oxidizing agents, all of which promote mutations. a. One factor that prevents potential DNA damage in spores is their greatly decreased water content. How would this affect some types of mutations? b. Endospores have a category of proteins called small acid-soluble proteins (SASPs) that bind to their DNA, preventing formation of cyclobutane-type dimers. What causes cyclobutane dimers, and why do bacterial endospores need mechanisms to prevent their formation?

Nucleotide Structure Which positions in the purine ring of a purine nucleotide in DNA have the potential to form hydrogen bonds but are not involved in Watson-Crick base pairing?

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