Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 17, Problem 17.3CC
Summary Introduction
To explain: The defects a strain of E.coli that lacks the function of RecA gene will exhibit.
Introduction: DNA refers to the genetic material of a cell. DNA is a double-stranded structure and each strand is made up of nucleotides. A
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Chapter 17 Solutions
Becker's World of the Cell (9th Edition)
Ch. 17 - The theoretical amplification accomplished by n...Ch. 17 - Bacterial replication and that in typical...Ch. 17 - Nonhomologous end-joining and synthesis-dependent...Ch. 17 - Prob. 17.3CCCh. 17 - Meselson and Stahl Revisited. For each of the...Ch. 17 - DNA Replication. Sketch a replication fork of...Ch. 17 - More DNA Replication. The following are...Ch. 17 - QUANTITATIVE Still More DNA Replication. Suppose...Ch. 17 - The Minimal Chromosome. To enable it to be...Ch. 17 - Prob. 17.6PS
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- A gene contains the sequence CGCATACGGTAC that results in the amino acid sequence arg-ile-arq- tyr. A mutation in this gene has a G inserted after the second C in the strand. How will this mutation affect the phenotype? A)This will affect the phenotype because although most of the protein will be identical, the first amino acid will be different. B)This will not affect the phenotype because only the second amino acid is different from the original protein. C)This will not affect the phenotype because the protein will be identical to the original protein. D)This will affect the phenotype because all of the amino acids after the first one will be different from the original protein.arrow_forwardThe following DNA sequences found on the sense strand belong to the same eukaryotic gene: Sequence 1: 5'-GATTCAATAAAGCTCAGATCGCTCACGTCGCGACTC-3' Sequence 2: 5'-TCCGAGGTCACTAGATACTCGTCGATCGTATAAATG-3' a) Which sequence is likely to be found upstream from the coding sequence? Justify your answer. b) Which sequence is likely to be found downstream from the coding sequence? Justify your answer. c) Which sequence will not be transcribed into an mRNA transcript? Justify your answer.arrow_forwardYou are studying the tryptophan synthetase gene that Yanofsky also examined to determine the relationship between the nucleotide sequence and the amino acid sequence of the gene. Yanofsky found a large number of mutations that affected the tryptophan synthetase gene. A) If you took this mutant E. Coli line (that has an Arginine at this location) and exposed it to a mutagen that could potentially change bases, what are the second mutations you would most likely discover that would restore the activity of the tryptophan synthetase gene and where would it be located? B) Most of the mutations that Yanofsky recovered were missense mutations. However, Yanofsky also recovered a nonsense mutation that changed amino acid number 15 into a stop codon. This codon normally encodes Lysine. Does the recovery of this mutation support the hypothesis that this Lysine residue is critical in the function of the tryptophan synthetase protein?arrow_forward
- Xeroderma pigmentosum is a genetic disease caused by an error in the nucleotide excision repair process that fixes damage to DNA by ultraviolet light. Studies have shown that it can result from mutations in any one of seven genes. What can you infer from this finding? A) There are seven genes that produce the same protein B) These seven genes are the most easily damaged by ultraviolet light. C) There are seven enzymes involved in the nucleotide excision repair process. D) These mutations have resulted from translocation of gene segments.arrow_forwardIn your laboratory, you have an F − strain of E. coli that is resistant to streptomycin and is unable to metabolize lactose, but it can metabolize glucose. Therefore, this strain can grow on media that contain glucose and streptomycin, but it cannot grow on media containing only lactose. A researcher has sent you two E. coli strains in two separate tubes. One strain, let’s call it strain A, has an F factor that carries the genes that are required for lactose metabolism. On its chromosome, it also has the genes that are required for glucose metabolism. However, it is sensitive to streptomycin. This strain can grow on media containing lactose or glucose, but it cannot grow if streptomycin is added to the media. The second strain, let’s call it strain B, is an F − strain. On its chromosome, it has thegenes that are required for lactose and glucose metabolism. StrainB is also sensitive to streptomycin. Unfortunately, when strains A and B were sent to you, the labels had fallen off the…arrow_forwardConsidering that prokaryote genomes do not have large introns, how is it possible to move a eukaryotic gene into a transformed bacterium, since they lack a spliceosome?arrow_forward
- Robert Bost and Richard Cribbs studied a strain of E. coli (araB14) that possessed a nonsense mutation in the structural gene that encodes Lribulokinase, an enzyme that allows the bacteria to metabolize the sugar arabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From the araB14 strain, they isolated some bacteria that possessed mutations that caused them to revert back to the wild type. Genetic analysis of these revertants showed that they possessed two different suppressor mutations. One suppressor mutation (R1) was linked to the original mutation in L-ribulokinase and probably occurred at the same locus. By itself, this mutation allowed the production of L-ribulokinase, but the enzyme produced was not as effective in metabolizing arabinose as the enzyme encoded by the wild-type allele. The second suppressor mutation (SuB) was not linked to the original mutation. In conjunction with the R1 mutation, SuB allowed the production of L-ribulokinase, but SuB by itself was not able…arrow_forwardRobert Bost and Richard Cribbs studied a strain of E. coli (araB14) that possessed a nonsense mutation in the structural gene that encodes Lribulokinase, an enzyme that allows the bacteria to metabolize the sugar arabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From the araB14 strain, they isolated some bacteria that possessed mutations that caused them to revert back to the wild type. Genetic analysis of these revertants showed that they possessed two different suppressor mutations. One suppressor mutation (R1) was linked to the original mutation in L-ribulokinase and probably occurred at the same locus. By itself, this mutation allowed the production of L-ribulokinase, but the enzyme produced was not as effective in metabolizing arabinose as the enzyme encoded by the wild-type allele. The second suppressor mutation (SuB) was not linked to the original mutation. In conjunction with the R1 mutation, SuB allowed the production of L-ribulokinase, but SuB by itself was not able…arrow_forwardConsider the following experiment. First, large populations of two mutant strains of Escherichia coli are mixed, each requiring a different, single amino acid. After plating them onto a minimal medium, 45 colonies grew. Which of the following may explain this result? A) The colonies may be due to back mutation (reversion). B) The colonies may be due to recombination. C) Either A or B is possible. D) Neither A nor B is possible.arrow_forward
- You have been tasked with designing a new computer algorithm to identify protein-coding genes in bacterial genomes.a) Name 3 specific criteria/patterns you will use to define whether a given sequence could contain a full-length protein-coding gene, and briefly justify your choices.b) Do you think your algorithm will work well to detect proteins encoded by the archaean Sulfolobus acidocaldarius? Why or why not?c) Do you think your algorithm will work well to detect proteins encoded in the eukaryotic yeast Saccharomyces cerevisiae? Why or why not?arrow_forwardA scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. The scientist determines the nucleotide sequence of the coding strand of the GS1 gene from strain A. It is identical to the GS1 sequence from the wild type gene except for a single change occurring approximately 1⁄3 of the way into the GS1 open reading frame. A small region of the GS1 sequence (including the site where the mutation occurs) from the wild type and mutant strains is shown below. Wild type TGTCCTCGGCCACAAGTTCTCTATC Strain A TGTCCTCGGCCACTAGTTCTCTATC How has this…arrow_forwardA scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. Sequencing of the GS1 gene from strain B shows that it is identical to the wild type gene except for a single alteration (the replacement of one nucleotide by another). How might this account for the features of the GS1 protein produced by strain B?arrow_forward
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