Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 6, Problem 23ESP
Summary Introduction
To determine: The conclusion to the linkage experiments conducted on the mutants.
Introduction: The mutation is the change in the
Summary Introduction
To determine: The significance of part B of the experiment.
Introduction: Binary fission is the means of asexual reproduction opted by prokaryotes like bacteria to multiply and produce identical cells. The process is faster than the sexual reproduction and therefore, helps the bacteria to populate within a shorter period.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
In studies of the amino acid sequence of wild-type and mutant forms of tryptophan synthetase in E. coli, the following changes have been observed: Determine a set of triplet codes in which only a single-nucleotide change produces each amino acid change.
In the procedure shown, why was it necessary to link thecoding sequence for the A or B chains to the sequence forβ-galactosidase? How were the A or B chains separated fromβ-galactosidase after the fusion protein was synthesized in E. coli?
From Table 20-3, would you expect the noncoding mutation g4205a to be fixed before or after the coding mutation G238S in a population of bacteria evolving resistance to the antibiotic cefotaxime? Give at least tworeasons for your answer.
Chapter 6 Solutions
Concepts of Genetics (12th Edition)
Ch. 6 - When the interrupted mating technique was used...Ch. 6 - In a transformation experiment involving a...Ch. 6 - In complementation studies of the rII locus of...Ch. 6 - A 4-month-old infant had been running a moderate...Ch. 6 - Prob. 2CSCh. 6 - Prob. 3CSCh. 6 - Prob. 4CSCh. 6 - HOW DO WE KNOW? In this chapter, we have focused...Ch. 6 - Review the Chapter Concepts list on p. 123. Many...Ch. 6 - With respect to F+ and F bacterial matings, answer...
Ch. 6 - List all major differences between (a) the F+ F...Ch. 6 - Describe the basis for chromosome mapping in the...Ch. 6 - In general, when recombination experiments are...Ch. 6 - Why are the recombinants produced from an Hfr F...Ch. 6 - Describe the origin of F bacteria and merozygotes.Ch. 6 - In a transformation experiment, donor DNA was...Ch. 6 - Describe the role of heteroduplex formation during...Ch. 6 - Explain the observations that led Zinder and...Ch. 6 - Prob. 12PDQCh. 6 - Two theoretical genetic strains of a virus (abc...Ch. 6 - The bacteriophage genome consists of many genes...Ch. 6 - If a single bacteriophage infects one E. coli cell...Ch. 6 - A phage-infected bacterial culture was subjected...Ch. 6 - In recombination studies of the rII locus in phage...Ch. 6 - In an analysis of rII mutants, complementation...Ch. 6 - If further testing of the mutations in Problem 18...Ch. 6 - Using mutants 2 and 3 from Problem 19, following...Ch. 6 - During the analysis of seven rII mutations in...Ch. 6 - In studies of recombination between mutants 1 and...Ch. 6 - Prob. 23ESPCh. 6 - An Hfr strain is used to map three genes in an...Ch. 6 - A plaque assay is performed beginning with 1 mL of...Ch. 6 - In a cotransformation experiment, using various...Ch. 6 - For the experiment in Problem 26, another gene, g,...Ch. 6 - Bacterial conjugation, mediated mainly by...Ch. 6 - A study was conducted in an attempt to determine...
Knowledge Booster
Similar questions
- You have constructed a recombinant vector harbouring pyruvate kinase. Subsequently, youwish to introduce your recombinant vector into Arabidopsis plants, since the enzyme isessential for the production of oil in developing seeds of Arabidopsis thaliana and oil crops.Explain how could you produce a transgenic Arabidopsis plant harbouring your syntheticgene by Agrobaterium tumefaciens-mediated transformation?arrow_forwardThe genetic alteration responsible for sickle-cell anemia in humans involves: a transition mutation from A to G, substituting glutamic acid for valine in a-globin a transversion mutation from T to A, substituting valine for glutamic acid in b-globin a transition mutation from T to C, substituting valine for glutamic acid in b-globin a transversion mutation from G to C, substituting glutamic acid for valine in a-globin a frameshift mutation of one ATC codon, removing glutamic acid from b-globinarrow_forwardIn site-directed mutagenesis experiments of an enzyme, scientists altered an aspartate residue to glutamate, lysine, phenylalanine, or valine. Which substitution is expected to have the least effect on enzymatic acitivity? Group of answer choices Glutamate Valine Lysine Phenylalaninearrow_forward
- 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? Why or why not?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_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
- A mutant has no activity for the enzyme isocitrate lyase.Does this result prove that the mutation is in the geneencoding isocitrate lyase?arrow_forwardAs part of a project investigating potential new drug targets in the fight against malaria, you are seeking to clone the gene for a protein from the malaria parasite Plasmodium falciparum. You wish to express this protein in BL21 (DE3) cells, a standard laboratory strain of Escherichia coli. After purification of your protein, you run an SDS-PAGE gel and notice that the major band has lower molecular weight than expected, so you fear you are getting a truncated version. 1. What technique could you use to confirm that you are obtaining a shortened version of your intended protein? explainarrow_forwardAs part of a project investigating potential new drug targets in the fight against malaria, you are seeking to clone the gene for a protein from the malaria parasite Plasmodium falciparum. You wish to express this protein in BL21 (DE3) cells, a standard laboratory strain of Escherichia coli. After purification of your protein, you run an SDS-PAGE gel and notice that the major band has lower molecular weight than expected, so you fear you are getting a truncated version. (a) Give TWO possible causes of your protein becoming truncated. explainarrow_forward
- After mutagenesis of wild type Vibrio fisheri, you isolate two different mutant strains (A and B) that, unlike the wild type cells, fail to luminesce when grown to high density in a flask with appropriate medium. Curiously, however, when you inoculate both mutant strains in the same flask, you observe that the mixed (A+B) culture begins to emit light after growing dense. a) What gene/functions are likely affected in each of the two mutants? b) How does this explain their phenotypes?arrow_forwardHigashi et al. (1986) found that three critical mutations were found in one of the two genes coding for 21-hydroxylase to render the gene nonfunctional. The following is a partial DNA sequence from one of the 21-hydroxylase genes where one of the three most common mutations occurs. For the following piece of normal DNA from the 21-hydroxylase gene, fill in the corresponding mRNA codons, and amino acids, and then do the same for the mutated sequence below it. GCT GAC GTC CTC CTC normal DNA sequence normal mRNA sequence normal amino acid sequencearrow_forwardThe consensus sequences for the -35 and -10 sites are shown below. Explain what the numbers next to the nucleotides mean. If you were going to mutate two of these nucleotides in order to eliminate the function of one or both of these sites which nucleotides would you mutate and why?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)
Biology
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Cengage Learning