Genetics - Problem Drill 08: Bacterial and Viral Genetics Question No. 1 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 1. In a transduction experiment, phage P1 is grown on a bacterial host of genotype A+ B+ C+ and the resulting lysate is used to infect a recipient strain of genotype A– B–C–. Transductants are obtained by selecting for the A+ phenotype. The genes are in an order, such that B is in the middle, and the distance between A and B is greater than the distance between B and C. Based on this information, which of the following statements is true? (A) If none of the A+ transductants were also C+, then the distance between A and C would be greater than about 100 kbp. (B) The cotransduction distance between A and B can be obtained from the Question fraction of A+ transductants that are A+ B+ C+ and A+ B+ C–. (C) It is possible that the cotransduction distance between A and C could be 0%. (D) The number of A+ transductants that are B– and C+ will be much smaller than the number of A+ transductants that are B+ and C–. (E) The number of A+B+ transductants will be greater than B+C+ transductants. A. Incorrect! From the given condition, we know the distance between A and C is greater than the distance between A and B, but there is not enough information to determine physically how many base pairs away. B. Incorrect! The distance between A and B should be calculated from the fraction of A+ transductants that are A+B- and A+B+. C. Incorrect! The distance between A and C cannot be 0% because B gene is in the middle. Feedback D. Correct! The number of A+B-C+ originate from double crossover, whose number should be much less than the single crossover A+B+C-. E. Incorrect! The distance between A and B is greater than that between B and C; therefore, the linkage between B and C is greater than that of A and B, the B+C+ transfectants will be more than the A+B+ transfectants. The key to answer this question is to understand the gene order and gene recombination in bacteria. (D)The number of A+ transductants that are B– and C+ will be much smaller than the number of A+ transductants that are B+ and C–. Solution RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 2 of10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 2. Cells of an E. coli strain that are trp lac Z met bio were mixed with cells of an E. coli strain that are trp lac Z met bio and cultured for several hours. Cells were then removed, washed, and transferred to minimal media containing lactose as the only sugar source. A few cells were able to grow on minimal media with lactose and they formed colonies. How did these few cells become trp lac Z met bio ? (A) Transformation Question (B) Transduction (C) Conjugation (D) Transposition (E) None of the above A. Incorrect! Transformation means bacteria uptake free DNA from its environment media, not from another bacteria. B. Incorrect! Transduction means bacterial cells accept genetic material via a phage infection; there is no phage involved in this case. C. Correct! Conjugation is likely what happened in this case, where two cells establish a conjugation bridge when they are close and genetic material is transferred through Feedback it. D. Incorrect! Transposition is not a genetic term. E. Incorrect! There is one correct answer above. Three types of methods enable bacterial cells to exchange or uptake genetic materials, which are transformation, transduction and conjugation. These are very useful in bacterial genetic mapping. (C)Conjugation Solution RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 3 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 3. A cross is made between two E. coli strains: Hfr arg+ bio+ leu+ x F- arg- bio- leu- (where arg is a gene for arginine biosynthesis, bio is a gene for biotin synthesis, and leu is a gene for leucine biosynthesis). Interrupted-mating studies show that arg+ enters the recipient last. Recombinants that are arg+ are selected on a defined medium containing biotin and leucine, but no arginine. The arg+ colonies are tested for the presence of the bio+ allele by replica plating onto a medium lacking only biotin; the same recombinants are tested for the presence of the leu+ allele by replica plating onto a medium lacking only leucine. The following results were obtained. Genotype Number of colonies arg+ bio+ leu+ 320 arg+ bio+ leu- 8 Question arg+ bio- leu+ 0 arg+ bio- leu- 48 The order of the genes are (starting from F factor): (A) Arg-Bio-Leu (B) Arg-Leu-Bio (C) Leu-Arg-Bio (D) Leu-Bio-Arg (E) Cannot be determined A. Correct! From the results, the most abundant one is the parental type, while the least abundant one is from double crossover. The one that changed position is in the middle. Here, the least abundant is arg+ bio- leu+; bio is the one that has changed the order and, therefore, bio is in the middle. Genetic material transfers though F factor starts from the closest to F factor. Arg+ bio- leu- results from single crossover between F factor-arg+ and another genotype bio- leu-; therefore, arg is the closest to F factor. The correct answer for this problem is Arg-Bio-Leu. B. Incorrect! Leu cannot be in the middle because it is not the one that changed position in the least abundant type. C. Incorrect! Feedback Arg cannot be in the middle either; it has not switched position in the least type. D. Incorrect! Leu cannot be the closest to F factor because Arg is, Arg+ bio- leu- results from single crossover between F factor-arg+ and another genotype bio- leu-. E. Incorrect! There is one correct answer above. Two key concepts to solve this problem: 1) Genetic material transfer though F factor starts from the closest to F factor; 2) double crossover is the least abundant type and the one whose position is changed is in the middle. (A)Arg-Bio-Leu Solution RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 4 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 4. In bacteriaphage λ you are studying three mutations that cause plaques to be either speckled, blue, or diffuse. You cross a speckled, blue, diffuse strain to wild type λ by co-infection (all bacteria are infected with both phages). You plate the resulting lysate and analyze the phenotypes of the plaques caused by the progeny phage: Phenotype Number of Plaq es Speckled, blue, diffuse 2121 Speckled, blue 3 Speckled, diffuse 344 Blue, diffuse 21 Speckled 22 Blue 349 Question Diffuse 4 Wild type 2136 Which of the following statement is NOT true? (A) The gene order is speckled – diffuse – blue. (B) The genetic distance between speckled and diffuse is 1 cM. (C) The genetic distance between blue and diffused is 14 cM. (D) The genetic distance between speckled and blue is 15 cM. (E) The double crossover frequency is 0.14%. A. Incorrect! The double crossover is the least abundant type and the one whose position is changed is in the middle. Therefore, here, diffuse is in the middle; the gene order is correct. B. Incorrect! Ignore blue and just look at speckled and diffused; the genetic distance is (3+21+22+4)/5000 = 1%. C. Incorrect! Ignore speckled and just look at blue and diffused; the genetic distance is (3+344+349+4)/5000 = 14%. D. Correct! Feedback Calculate the genetic distance the same way we did on speckled-diffused, or blue- diffused. Now, we ignore diffused and just look at speckled and blue, the genetic distance is (344+21+22+349)/5000 = 14.8%. This number is slightly smaller than 15% (the total value of speckled-diffused and diffused-blue). E. Incorrect! The double crossover rate is (3+4)/5000 = 0.14%. This is the reason that the genetic distance cannot be added up. A three-point cross can calculate double crossover, which is the reason the genetic distance at the two ends appear closer than the added up value of two regions. (D)The genetic distance between speckled and blue is 15 cM. Solution RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 5. Which of the following is NOT a cause of virus mutation in animal cells? (A) Infidelity DNA replication (B) Recombination (C) Phenotype mixing (D) Chemical reagent or Physical factors Question (E) Reassortment A. Incorrect! Viral DNA polymerase often lacks proof-reading function; therefore, the fidelity of DNA replication is compromised and mutations arise frequently. B. Incorrect! Viral genome is very complex and there are many sequence repeats which facilitate recombination; recombination is a cause of viral mutations.