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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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Ch. 11 - Chromosome Structure and DNA Sequence Organization
Klug10th EditionEssentials of GeneticsISBN: 9780135588789
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TOC.tableOfContents
- Ch. 1 - Introduction to Genetics16
- Ch. 2 - Mitosis and Meiosis28
- Ch. 3 - Mendelian Genetics37
- Ch. 4 - Modification of Mendelian Ratios53
- Ch. 5 - Sex Determination and Sex Chromosomes32
- Ch. 6 - Chromosome Mutations: Variation in Number and Arrangement37
- Ch. 7 - Linkage and Chromosome Mapping in Eukaryotes36
- Ch. 8 - Genetic Analysis and Mapping in Bacteria and Bacteriophages24
- Ch. 9 - DNA Structure and Analysis38
- Ch. 10 - DNA Replication29
- Ch. 11 - Chromosome Structure and DNA Sequence Organization22
- Ch. 12 - The Genetic Code and Transcription36
- Ch. 13 - Translation and Proteins27
- Ch. 14 - Gene Mutation, DNA Repair, and Transposition34
- Ch. 15 - Regulation of Gene Expression in Bacteria22
- Ch. 16 - Regulation of Gene Expression in Eukaryotes37
- Ch. 17 - Recombinant DNA Technology27
- Ch. 18 - Genomics, Bioinformatics, and Proteomics30
- Ch. 19 - The Genetics of Cancer21
- Ch. 20 - Quantitative Genetics and Multifactorial Traits37
- Ch. 21 - Population and Evolutionary Genetics45
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Non è quello che usi tu?Cambia libro di testo
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Klug 10th EditionCh. 11 - Chromosome Structure and DNA Sequence OrganizationProblema 18
Klug 10th EditionCh. 11 - Chromosome Structure and DNA Sequence OrganizationProblema 18Capitolo 11, Problema 18
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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Identify the total number of base pairs in the human haploid genome, which is given as approximately \(3 \times 10^{9}\) base pairs.
Note the total number of Alu sequences per haploid genome, which is approximately \$10^{6}$ copies.
Understand that the problem asks for the average spacing between each Alu sequence, meaning the average number of base pairs between consecutive Alu elements.
Calculate the average spacing by dividing the total number of base pairs by the total number of Alu sequences using the formula:
\(\text{Average spacing} = \frac{\text{Total base pairs}}{\text{Number of Alu sequences}}\)
Substitute the given values into the formula:
\(\text{Average spacing} = \frac{3 \times 10^{9}}{10^{6}}\)
This will give the average number of base pairs between each Alu sequence.
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Human Genome Size and Structure
The human haploid genome contains about 3 billion base pairs (3 x 10^9 bp). This total DNA length is organized into chromosomes and includes coding and non-coding regions. Understanding genome size is essential for calculating average distances between repeated sequences like Alu elements.
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06:51
Human Genome Composition
Alu Sequences and Short Interspersed Elements (SINEs)
Alu sequences are a type of SINE, approximately 282 base pairs long, repeated about one million times per haploid genome. These repetitive elements are dispersed throughout the genome and can be used as markers for genetic studies. Their abundance affects genome structure and spacing calculations.
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05:30Human Transposable Elements
Calculating Average Spacing Between Repetitive Elements
To find the average spacing between repetitive sequences, divide the total genome size by the number of repeats. This gives the average number of base pairs between each element, assuming even distribution. This concept helps estimate genomic distances between Alu sequences.
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07:18Discovery
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How many base pairs are in a molecule of phage T2 DNA 52-µm long?
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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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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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