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
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Chapter 6, Problem 18PDQ
In an analysis of rII mutants, complementation testing yielded the following results:
Predict the results of testing 2 and 3, 2 and 4, and 3 and 4 together.
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In an analysis of five rII mutants, complementation testing yielded the following results:
A fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic
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Culture #
# AztR mutants
11
12
3
4.
13
14
15
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30
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16
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18
19
20
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19
In an experiment assay, sometimes there are several dilutions involved.0.5 mL of (Bovine serum albumin) BSA is pipetted into 0.5 mL of PhosphateBuffer Solution (PBS).Then 0.2 mL of the mixture is pipetted into 0.8 mL of PBS.What is the total dilution factor for this assay?
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...
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- Below is an EMSA showing four different reactions, A-D. In each tube there is some combination of labelled DNA probe, Protein X (the protein you are studying), and an antibody for Protein X. Identify which combination of components are found in each of the four reactions and explain how you determined that based on the molecular interactions being studied and your knowledge of gel electrophoresis. It is possible that multiple lanes have the same component(s). A B C D EMSAarrow_forwardAt higher amounts of protein, the Bradford assay is not linear. Consider the plot to the right: what is the maximum amount of protein a sample could contain and still fall within a standard curve? Briefly explain your reasoning.arrow_forwardBelow is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]arrow_forward
- Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6arrow_forwardIn a cotransduction experiment involving P1, the cotransductionfrequency was 0.53. How far apart are the two genes?arrow_forwardIn a cotransformation experiment (see question 4 of More GeneticTIPS), DNA was isolated from a donor strain that was proA+ andstrC+ and sensitive to tetracycline. (The proA and strC genes conferthe ability to synthesize proline and confer streptomycin resistance,respectively.) A recipient strain is proA− and strC− and isresistant to tetracycline. After transformation, the bacteria werefirst streaked on a medium containing proline, streptomycin, andtetracycline. Colonies were then restreaked on a medium containingstreptomycin and tetracycline. (Note: Each type of medium hadcarbon and nitrogen sources for growth.) The following resultswere obtained:70 colonies grew on the medium containing proline, streptomycin,and tetracycline, but only 2 of these 70 colonies grew whenrestreaked on the medium containing streptomycin and tetracyclinebut lacking proline. If we assume the average size of the DNA fragments is 2 minutes,how far apart are these two genes?arrow_forward
- Explain the similarities and differences between the procedures used for the coarse-scale mapping and fine-scale mapping of QTLs through controlled crosses, as was accomplished for QTLs involved with tomato size.arrow_forward(a) Explain sensitivity and selectivity in terms of true and false positives. (b) What is the advantage of a Needleman-Wunsch alignment compared to a seeded alignment? (c) What does the expectation parameter mean in local alignment? What will happen if the expectation value is increased from its default value of 10 to a 100?arrow_forwardComplementation tests of recessive mutants a through f produced the data shown in the table below. "+" means wild-type phenotype; "-" means mutant phenotype. PSCO a b с 0 0 0 0 0 a e a f b + C + d e + + + + + - d e f A new mutant, g, is isolated and it fails to complement c. Which other mutant(s) would g also fail to complement? (Select all correct answers.) ++ f ++ + + -arrow_forward
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