Problem Drill 08: Bacterial and Viral Genetics Q

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.

C. Correct! Phenotype mixing is a phenomenon in which the coat protein may be mixed up by two different viral strains; however, it does not refer to mixing of DNA and, therefore, it is not a cause of mutation (which refers to genetic material only).

Feedback D. Incorrect! Chemical reagent and some physical conditions can cause DNA mutation.

E. Incorrect! Some viral genome is composed of segments of DNA and, when co-infection occurs, different strains may uptake DNA fragment from another strain, leading to mutation phenotype.

Mutation is very common in viral strains; they may be a result of recombination between viral and host genome. And by reassortment: coinfection of cells by viruses with segmented genomes can produce progeny with different combinations of the parental genome segments.

(C)Phenotype mixing

Solution

RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 6 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. 6. Which of the following statements about binary fission is true?

(A) Prokaryotes don’t reproduce sexually. (B) Prokaryotes can reproduce asexually and sexually. (C) In binary fission, spindle formation does occur. (D) Binary fission leads to the formation of two daughter cells without the Question process of cytokinesis. (E) Prokaryotes contain many linear chromosomes.

A. Correct! Prokaryotes don’t reproduce sexually.

B. Incorrect! Prokaryotes don’t reproduce sexually.

C. Incorrect! In binary fission, spindle formation does not occur.

Feedback D. Incorrect! Binary fission includes the process of cytokinesis.

E. Incorrect! Prokaryotes contain a single circular chromosome.

Binary fission is “asexual” reproduction. One cell divides into two cells of the same size. The process results in the reproduction of a living cell into two equal or near equal parts. Prokaryotes contain a single circular chromosome. Prokaryotes do not reproduce sexually. Prokaryotes reproduce by binary fission. In binary fission, spindle formation does not occur in mitosis. The chromosome is attached to the inside of the plasma membrane. No other proteins are found on the chromosome.

(A) Prokaryotes don't reproduce sexually.

Solution

RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 7 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. 7. Transformation is ______.

(A) The genetic alteration of a cell by taking up and expressing foreign RNA. (B) The genetic alteration of a cell by releasing foreign DNA into the extracellular environment. (C) When a bacteria cell takes up foreign DNA and, if done successfully, would be known as competent.

Question (D) When a bacteria cell takes up viral DNA and, if done successfully, would be known as competent. (E) When a bacterium takes up foreign DNA and automatically replicates it, along with its own DNA.

A. Incorrect! Transformation is the genetic alteration of a cell by taking up and expressing foreign DNA.

B. Incorrect! Transformation is the genetic alteration of a cell by taking up and expressing foreign DNA.

C. Correct! Transformation is when a bacteria cell takes up foreign DNA and, if done successfully, would be known as competent.

Feedback D. Incorrect! Transformation is when a bacteria cell takes up foreign DNA and, if done successfully, would be known as competent.

E. Incorrect! When a bacterium takes up foreign DNA and if it has an origin of replication recognized by the host bacteria, it will replicate it along with its own DNA.

In transformation, cells pick up pieces of DNA released from other bacteria and use it to acquire new abilities. Cells can pick up pieces of DNA, which are released by live bacteria. If the foreign DNA has an origin of replication recognized by the bacteria, it will be replicated along with their own DNA. Bacteria which are able to uptake DNA are called "competent”.

(C)When a bacteria cell takes up foreign DNA and, if done successfully, Solution would be known as competent.

Question No. 8 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. 8. Which of the following statements about a plasmid is not true?

(A) Conjugative plasmids have tra-genes, which perform the process of conjugation. (B) Non-conjugative plasmids cannot initiate conjugation by themselves. (C) Two classes of plasmids are degredative and counter conjugative. Question (D) Every plasmid has at least one origin of replication. (E) The plasmids of most bacteria are linear.

A. Incorrect! Conjugative plasmids do have tra-genes, which perform the process of conjugation.

B. Incorrect! This is true; non-conjugative plasmids cannot initiate conjugation by themselves.

C. Correct! While there is a class of plasmids known as degredative, counter conjugative is not one.

Feedback D. Incorrect! This is true; every plasmid has at least one origin of replication to enable them to be copied independently from chromosomal DNA.

E. Incorrect! The plasmids of most bacteria are circular.

One way of grouping plasmids is by their ability to transfer to other bacteria. Conjugative plasmids have tra-genes, which perform the process of conjugation, the sexual transfer of plasmids to another bacterium. Non-conjugative plasmids are not able to initiate conjugation so can only be transferred with the help of conjugative plasmids. Another way to classify plasmids is by function. There are five main classes: Fertility F-plasmids, Resistance-(R)plasmids, Col-plasmids, Degradative plasmids and Virulence plasmids.

(C)Two classes of plasmids are degredative and counter conjugative.

Solution

Question No. 9 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. 9. A virus ______.

(A) Is either made of DNA or RNA. (B) Is always made up of DNA. (C) Is known as a bacteriophage and infects immune cells in humans. (D) Has a capsid on the inside to facilitate attachment to the host cell. Question (E) Contains nucleic acid, either DNA or RNA on the outside of the capsid coat.

A. Correct! A virus is either made of DNA or RNA.

B. Incorrect! A virus is either made of DNA or RNA.

C. Incorrect! A virus, known as a bacteriophage, infects bacteria.

Feedback D. Incorrect! A virus has a capsid on the outside to facilitate attachment to the host cell.

E. Incorrect! A virus contains nucleic acid, either DNA or RNA inside the capsid coat.

Virus – an “organism” whose genetic material may be either DNA or RNA surrounded by a protein coat. Viruses take over other cells to make more viruses. There is still much debate on whether viruses should be considered living. Bacteriophages are viruses that infect bacteria. Capsid: The protein coat which acts like a membrane and allows viruses to attach to host cells. Genetic Material: The DNA or RNA that is the blueprint to make more viruses.

(A)Is either made of DNA or RNA.

Solution

Question No. 10 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. 10. Which of the following statements about the life cycle of a virus is true?

(A) Viruses have two life cycles: lysogenic and retroactive. (B) One of the viruses life cycle is the lytotonic phase. (C) In the lysogenic cycle, the viral DNA is incorporated into the host genome and is copied during DNA replication. (D) In the lysogenic cycle, the viral DNA is incorporated into the host genome, Question and then the cell is then lysed or ruptured to release copies of the virus. (E) The lysogenic cycle and the lytic cycle can continue simultaneously in viruses.

A. Incorrect! Viruses have the following two paths for their life cycle: lysogenic and lytic.

B. Incorrect! Viruses have the following two paths for their life cycle: lysogenic and lytic.

C. Correct! In the lysogenic cycle, the viral DNA is incorporated into the host genome and is copied during DNA replication.

Feedback D. Incorrect! In the lysogenic cycle, the viral DNA is incorporated into the host genome and is copied during DNA replication.

E. Incorrect! While the lysogenic cycle can lead to the lytic cycle in some circumstances, they do not occur simultaneously.

Viruses can follow two life “paths”. These two alternative paths are called “lytic cycle” or “lysogenic cycle”. In the lytic cycle, the virus infects the host and replicates. Upon completion of replication and new virion formation, the host cell is ruptured (lysed) and viron released. In the lysogenic cycle, the viral DNA incorporates into the host’s DNA and is copied along with the host DNA replication.

(C)In the lysogenic cycle, the viral DNA is incorporated into the host genome and is copied during DNA replication. Solution