Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 9, Problem 25P
Eukaryotic genomes are replete with repetitive sequences that make genome assembly from sequence reads difficult. For example, sequences such as CTCTCTCTCT .(tandem repeats of the dinucleotide sequence CT) are found at many chromosomal locations, with variable numbers (n) of the CT repeating unit at each location. Scientists can assemble genomes despite these difficulties by using the paired-end sequencing strategy diagrammed in Fig. 9.9. In other words, they can make libraries with genomic inserts of defined size, and then sequence both ends of individual clones.
Following are 12 DNA sequence reads from six cloned fragments analyzed in a genome project. 1A and 1B represent the two end reads from clone 1, 2A and 2B the two end reads from clone 2, etc. Clones 1–4 were obtained from a library in which the genomic inserts are about 2 kb long, while the inserts in clones 5 and 6 are about 4 kb long. All of these sequences have their 5′ ends at the left and their 3′ ends at the right. To simplify your analysis, assume that these sequences together represent two genomic locations (loci; singular locus), each of which contains a (CT)n repeat, and that each of the 12 sequences overlaps with one and only one other sequence.
1A: CCGGGAACTCCTAGTGCCTGTGGCACGATCCTATCAAC
1B: AGGACTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCT
2A: GTTTTTGAGAGAGAGAGAGAGAGAGAGAGACCTGGGGG
2B: ACGTAGCTAGCTAACCGGTTAAGCGCGCATTACTTCAA
3A: CTCTCTCTCTCTCTCTCTCTCAAAAACTATGGAAATTT
3B: TAGTGATAGGTAACCCAGGTACTGCACCACCAGAAGTC
4A: GGCCGGCCGTTGTTGACGCAATCATGAATTTAATGCCG
4B: TCATGGGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGA
5A: TAGTGCCTGTGGCACGATCCTATCAACTAACGACTGCT
5B: AAGGAAAGGCCGGCCGTTGTTGACGCAATCATGAATTT
6A: CAGCAGCTAGTGATAGGTAACCCAGGTACTGCACCACC
6B: GGACTATACGTAGCTAGCTAACCGGTTAAGCGCGCATT
| a. | Diagram the two loci, showing the locations of the repetitive DNA and the relative positions and orientations of the 12 DNA sequence reads. |
| b. | If possible, indicate how many copies of the CT repeating unit reside at either locus. |
| c. | Are the data compatible with the alternative hypothesis that these clones actually represent two alleles of a single locus that differ in the number of CT repeating units? |
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Eukaryotic genomes are replete with repetitive sequences that make genome assembly from sequencereads difficult. For example, sequences such asCTCTCTCTCT . . . (tandem repeats of the dinucleotide sequence CT) are found at many chromosomallocations, with variable numbers (n) of the CT repeating unit at each location. Scientists can assemblegenomes despite these difficulties by using the pairedend sequencing strategy diagrammed in Fig. 9.9. Inother words, they can make libraries with genomicinserts of defined size, and then sequence both endsof individual clones. Following are 12 DNA sequence reads from sixcloned fragments analyzed in a genome project. 1Aand 1B represent the two end reads from clone 1, 2Aand 2B the two end reads from clone 2, etc. Clones1–4 were obtained from a library in which the genomic inserts are about 2 kb long, while the inserts inclones 5 and 6 are about 4 kb long. All of these sequences have their 5′ ends at the left and their 3′ endsat the right. To simplify…
Chapter 9 Solutions
Genetics: From Genes to Genomes
Ch. 9 - Match each of the terms in the left column to the...Ch. 9 - For each of the restriction enzymes listed below:...Ch. 9 - The calculations of the average restriction...Ch. 9 - The DNA molecule whose entire sequence follows is...Ch. 9 - Why do longer DNA molecules move more slowly than...Ch. 9 - Agarose gels with different average pore sizes are...Ch. 9 - The following picture shows the ethidium...Ch. 9 - The linear bacteriophage genomic DNA has at each...Ch. 9 - Consider a partial restriction digestion, in which...Ch. 9 - The text stated that molecular biologists have...
Ch. 9 - a. What is the purpose of molecular cloning? b....Ch. 9 - a. DNA polymerase b. RNA polymerase c. A...Ch. 9 - Is it possible that two different restriction...Ch. 9 - A plasmid vector pBS281 is cleaved by the enzyme...Ch. 9 - A recombinant DNA molecule is constructed using a...Ch. 9 - Suppose you are using a plasmid cloning vector...Ch. 9 - Prob. 17PCh. 9 - The lacZ gene from E. coli encodes the enzyme...Ch. 9 - Your undergraduate research advisor has assigned...Ch. 9 - Which of the enzymes from the following list would...Ch. 9 - You use the primer 5 GCCTCGAATCGGGTACC 3 to...Ch. 9 - a. To make a genomic library useful for sequencing...Ch. 9 - Problem 15 showed part of the sequence of the...Ch. 9 - Eukaryotic genomes are replete with repetitive...
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