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Ch. 11 - Chromosome Structure and DNA Sequence Organization
Klug - Essentials of Genetics 10th Edition
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook
Chapter 11, Problem 17

How many base pairs are in a molecule of phage T2 DNA 52-µm long?

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1
Understand that the problem asks for the number of base pairs in a DNA molecule given its length in micrometers (µm).
Recall that the length of DNA can be related to the number of base pairs by knowing the length per base pair. For double-stranded DNA, the average distance between adjacent base pairs is approximately 0.34 nanometers (nm).
Convert the length of the DNA molecule from micrometers to nanometers to match the units of base pair spacing. Since 1 µm = 1000 nm, multiply the length by 1000: \$52 \, \mu m \times 1000 = 52000 \text{ nm}$.
Use the formula to calculate the number of base pairs: \(\text{Number of base pairs} = \frac{\text{Total length in nm}}{\text{Length per base pair in nm}} = \frac{52000}{0.34}\).
Perform the division to find the number of base pairs in the DNA molecule.

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Key Concepts

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

DNA Structure and Base Pairing

DNA is composed of two strands forming a double helix, with base pairs (adenine-thymine and guanine-cytosine) connecting the strands. The length of DNA can be measured in base pairs, which represent the number of paired nucleotides along the molecule.
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DNA Length Measurement and Conversion

The physical length of DNA molecules is often measured in micrometers (µm), but to find the number of base pairs, one must convert this length using the known distance between base pairs, approximately 0.34 nanometers (nm) per base pair.
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Phage T2 DNA Characteristics

Phage T2 is a bacteriophage with a well-studied DNA genome. Knowing its DNA length in micrometers allows calculation of the total base pairs by applying the standard base pair spacing, linking physical measurements to genetic information content.
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Related Practice
Textbook Question

Contrast the structure of SINE and LINE DNA sequences. Why are LINEs referred to as retrotransposons?

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Textbook Question

Mammals contain a diploid genome consisting of at least 10⁹ bp. If this amount of DNA is present as chromatin fibers, where each group of 200 bp of DNA is combined with 9 histones into a nucleosome and each group of 6 nucleosomes is combined into a solenoid, achieving a final packing ratio of 50, determine:

(a) the total number of nucleosomes in all fibers,

(b) the total number of histone molecules combined with DNA in the diploid genome, and

(c) the combined length of all fibers.

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Textbook Question

Assume that a viral DNA molecule is a 50-µm-long circular strand with a uniform 20-Å diameter. If this molecule is contained in a viral head that is a 0.08-µm-diameter sphere, will the DNA molecule fit into the viral head, assuming complete flexibility of the molecule? Justify your answer mathematically.

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Textbook Question

The human genome contains approximately 106 copies of an Alu sequence, one of the best-studied classes of short interspersed elements (SINEs), per haploid genome. Individual Alu units share a 282-nucleotide consensus sequence followed by a 3'-adenine-rich tail region [Schmid (1998)]. Given that there are approximately 3 x 109 base pairs per human haploid genome, about how many base pairs are spaced between each Alu sequence?

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Textbook Question

The following is a diagram of the general structure of the bacteriophage chromosome. Speculate on the mechanism by which it forms a closed ring upon infection of the host cell.

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