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Ch. 15 - Gene Mutation, DNA Repair, and Transposition
All textbooksKlug 12th EditionCh. 15 - Gene Mutation, DNA Repair, and TranspositionProblem 23
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Chapter 15, Problem 23

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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Key Concepts

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

Alu Sequences

Alu sequences are a type of short interspersed nuclear element (SINE) found in the human genome. They are approximately 300 base pairs long and are derived from the 7SL RNA gene. Alu elements are notable for their high copy number, with around 1 million copies present in a single haploid human genome, contributing to genomic diversity and evolution.
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Haploid Genome

A haploid genome contains a single set of chromosomes, which in humans consists of about 3 billion base pairs. This is half the number of chromosomes found in diploid cells, which contain two sets. Understanding haploidy is crucial for calculating the distribution of genetic elements, such as Alu sequences, across the genome.
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Base Pair Spacing

Base pair spacing refers to the average number of base pairs that separate repetitive elements like Alu sequences within the genome. To calculate this, one can divide the total number of base pairs in the haploid genome by the number of Alu copies. This metric helps in understanding the density and distribution of these genetic elements throughout the genome.
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Related Practice
Textbook Question
It is estimated that about 0.2 percent of human mutations are due to TE insertions, and a much higher degree of mutational damage is known to occur in some other organisms. In what way might a TE insertion contribute positively to evolution?
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Describe the 'end-replication problem' in eukaryotes. How is it resolved?
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The SOS repair genes in E. coli (discussed in Chapter 15) are negatively regulated by the lexA gene product, called the LexA repressor. When a cell's DNA sustains extensive damage, the LexA repressor is inactivated by the recA gene product (RecA), and transcription of the SOS genes is increased dramatically. One of the SOS genes is the uvrA gene. You are a student studying the function of the uvrA gene product in DNA repair. You isolate a mutant strain that shows constitutive expression of the UvrA protein. Naming this mutant strain uvrAᶜ, you construct the diagram shown above in the right-hand column showing the lexA and uvrA operons: Outline a series of genetic experiments that would use partial diploid strains to determine which of the two possible mutations you have isolated.
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Textbook Question
Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. The dnaE gene encodes the α subunit of DNA polymerase III. What effect is expected from a mutation in this gene? How could the mutant strain be maintained?
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Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. The dnaQ gene encodes the ε subunit of DNA polymerase. What effect is expected from a mutation in this gene?
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Textbook Question
A fellow student considers the issues in Problem 22 and argues that there is a more straightforward, nongenetic experiment that could differentiate between the two types of mutations. The experiment requires no fancy genetics and would allow you to easily assay the products of the other SOS genes. Propose such an experiment.
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