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Two phage phenotypes are controlled by the genes a and b. In a mapping experiment, a culture of bacteria is infected simultaneously with an a-b+ strain and an a+b- strain. When plaques are analyzed, five out of 1000 have the a+b+ or a-b- phenotype. Based on the information, how far apart are genes a and b?

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In the given mapping experiment, we are ...

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Which of the following will have the LEAST influence on the efficiency of transformation in E. coli bacteria?


A) calcium chloride treatment
B) heat shock
C) electrical field
D) chilling on ice
E) using high concentrations of DNA

F) A) and C)
G) C) and D)

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Generalized transduction by phage P1 of four E. coli genes (a, b, c, and d) shows the following percentages of cotransduction: ab51%bc2%bd12%ad40%ac19%cd79%\begin{array} { l l l r } a - b & 51 \% & b - c & 2 \% \\b - d & 12 \% & a - d & 40 \% \\a - c & 19 \% & c - d & 79 \%\end{array} What is the CORRECT order of the four genes?


A) a b c d
B) b a d c
C) d a c b
D) d c a b
E) b d a c

F) B) and D)
G) B) and C)

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The human immunodeficiency virus (HIV) is what type of virus?


A) a DNA virus that can infect a type of human blood cell
B) a double-stranded RNA virus that can recombine with the genomes of similar viruses from other species
C) a single-stranded RNA virus that can undergo reverse transcription
D) a single-stranded DNA virus that uses reverse transcriptase to make a provirus
E) a double-stranded DNA virus that can infect both humans and chimpanzees simultaneously

F) B) and D)
G) B) and E)

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In order to better understand arginine biosynthesis in bacteria, a microbial geneticist might first isolate mutant bacterial strains. a. What characteristics must these mutant bacteria have? b. Outline a strategy for isolating such mutants. c. List three possible methods for mapping the genetic location of the mutations in these strains.

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a. The mutant bacterial strains must hav...

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A transformation experiment is performed with a donor strain that is resistant to four drugs, A, B, C, and D, determined by genes A, B, C, and D. The recipient strain is sensitive to all four drugs. The treated recipient-cell population is divided up and plated on media containing various combinations of the drugs. A cell that is resistant to the added drug(s) will form a colony. The table below shows the results.  Drug(s)  Added  Number of Colonies  Drug(s)  Added  Number of Colonies  None 10,000BC51 A 1,156BD49 B 1,148CD786 C 1,161ABC30 D 1,139ABD42 AB 46ACD630 AC 640BCD36 AD 942 ABCD 30\begin{array} { l l l l } \text { Drug(s) Added } & \text { Number of Colonies } & \text { Drug(s) Added } & \text { Number of Colonies } \\\hline \text { None } & 10,000 & \mathrm { BC } & 51 \\\text { A } & 1,156 & \mathrm { BD } & 49 \\\text { B } & 1,148 & \mathrm { CD } & 786 \\\text { C } & 1,161 & \mathrm { ABC } & 30 \\\text { D } & 1,139 & \mathrm { ABD } & 42 \\\text { AB } & 46 & \mathrm { ACD } & 630 \\\text { AC } & 640 & \mathrm { BCD } & 36 \\\text { AD } & 942 & \text { ABCD } & 30\end{array} One of the genes is quite distant from the other three that are tightly linked. What is the probable order of the three tightly linked genes?


A) DAC
B) ACB
C) BDC
D) DCA
E) ADC

F) C) and D)
G) None of the above

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To MOST readily demonstrate transformation of bacteria in the laboratory one could extract DNA from:


A) an auxotroph and add it to prototrophic cells.
B) arg- cells and add it to arg+ cells.
C) streps (sensitive) and add it to strepr (resistant) cells.
D) arg+ cells and add it to arg- cells.
E) both arg- and arg+ cells, mix them, and select for recombinants.

F) B) and E)
G) A) and B)

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Outline the steps involved in mapping a bacterial chromosome by cotransformation.

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Cotransformation is a technique used to map bacterial chromosomes by identifying the relative positions of genes on the chromosome. The steps involved in mapping a bacterial chromosome by cotransformation are as follows: 1. Selecting the genes of interest: The first step in cotransformation mapping is to select the genes that will be used as markers for the mapping process. These genes should be easily identifiable and have known locations on the bacterial chromosome. 2. Creating a donor DNA library: Next, a donor DNA library is created by isolating DNA from the bacterial strain of interest and fragmenting it into smaller pieces. These DNA fragments will contain the genes of interest and will be used to transform the recipient bacterial strain. 3. Transforming the recipient strain: The donor DNA fragments are then introduced into the recipient bacterial strain through a process of transformation, such as electroporation or conjugation. This results in the incorporation of the donor DNA fragments into the recipient bacterial chromosome. 4. Selecting for cotransformants: The transformed recipient bacterial strain is then plated on selective media that allows for the growth of only those bacteria that have acquired both the donor DNA fragments and the recipient chromosome. These bacteria are known as cotransformants. 5. Analyzing the cotransformants: The cotransformants are then analyzed to determine the order and relative positions of the genes of interest on the bacterial chromosome. This can be done through various genetic and molecular techniques, such as PCR, DNA sequencing, or genetic mapping. 6. Mapping the bacterial chromosome: By analyzing the cotransformants, the relative positions of the genes of interest on the bacterial chromosome can be determined. This information can then be used to create a genetic map of the bacterial chromosome, showing the locations of the genes and their distances from each other. Overall, cotransformation is a powerful technique for mapping bacterial chromosomes and can provide valuable insights into the organization and structure of bacterial genomes.

Which explanation is one of the reasons for the ability of the influenza virus to evolve rapidly?


A) Its single-stranded DNA genome is sensitive to environmental factors that create mutations.
B) The enzyme that replicates its RNA genome is prone to make mistakes that result in mutations.
C) Its ability to integrate into the host genome allows it to remain in host cells for long periods of time.
D) It can convert its RNA genome into a DNA copy by using the enzyme reverse transcriptase.
E) It has the ability to replicate its genome both within the host cell and outside of it.

F) C) and D)
G) B) and C)

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When the F factor integrates into the E. coli chromosome, the result is an _____ strain.


A) Hfr
B) F-
C) F+
D) F´
E) F+/-

F) C) and D)
G) A) and B)

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What are plasmids and what purposes do they serve?

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Plasmids are small, circular, double-str...

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Which of the following statements concerning the role of reverse transcriptase in the life cycle of certain RNA viruses is FALSE?


A) It is produced from the pol gene.
B) It makes DNA from an RNA template.
C) It is essential for producing a provirus.
D) It is part of the life cycle of retroviruses.
E) It is very accurate and leads to a low mutation rate.

F) A) and E)
G) None of the above

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Which of the following statements is FALSE?


A) When Hfr cells are crossed with F- cells, most of the recipient cells remain F-.
B) The F plasmid or factor is capable of integrating into the bacterial chromosome at several different sites.
C) F+ cells are normally capable of undergoing conjugation with F- and Hfr recipient cells.
D) F+ cells can convert F- cells to F+ upon conjugation.
E) Hfr cells are capable of transferring part of the bacterial chromosome to recipient cells upon conjugation.

F) A) and B)
G) B) and E)

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Bacterial strains that can produce all the necessary compounds and therefore grow on minimal media are called:


A) autotrophs.
B) heterotrophs.
C) prototrophs.
D) omnitrophs.
E) auxotrophs.

F) C) and D)
G) A) and B)

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C

You are mapping five bacterial genes using cotransformation frequencies. The genes are designated A, B, C, D, and E. You perform a series of crosses and measure the percent cotransformation between all pairs of genes. The results are shown in the table below. Percent contransformation between gene pairs ABCDEAB55C1432D806522E4525434\begin{array} { l l l l l l } & A & B & C & D & E \\ A & - & & & & \\ B & 55 & - & & & \\ C & 14 & 32 & - & & \\ D & 80 & 65 & 22 & - & \\ E & 45 & 25 & 4 & 34 & - \end{array} What is the CORRECT order for the five genes?


A) A B C D E
B) C E A B D
C) D A B E C
D) B E A C D
E) E A D B C

F) C) and E)
G) A) and E)

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Both retroviruses and lysogenic bacteriophages employ a mechanism that allows them to be replicated and passed from cell to cell without producing viruses. What is the common mechanism that these two very different viruses use?

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Both retroviruses and lysogenic bacteriophages share a common mechanism known as integration, which allows them to incorporate their genetic material into the host cell's genome. This process enables the viral genetic information to be replicated and passed from cell to cell without the immediate production of new virus particles. For retroviruses, this process begins when the virus infects a host cell and uses its reverse transcriptase enzyme to convert its RNA genome into DNA. This newly synthesized viral DNA is then integrated into the host cell's DNA with the help of an enzyme called integrase. Once integrated, the viral DNA, now referred to as a provirus, can be replicated along with the host cell's DNA during cell division. The provirus can remain dormant within the host genome for an extended period, and the host cell can function normally until certain triggers lead to the activation of the viral genes, resulting in the production of new viruses. Lysogenic bacteriophages, also known as temperate phages, have a similar mechanism. After infecting a bacterial cell, the phage DNA is incorporated into the bacterial chromosome at a specific site. The integrated phage DNA, known as a prophage, is then passively replicated along with the host's genome as the bacterial cell divides. The prophage can remain latent within the host genome, and the bacterium continues to live and reproduce normally. However, under certain conditions, such as stress or exposure to UV light, the prophage can be induced to exit the lysogenic cycle and enter the lytic cycle, where it becomes active, replicates, and produces new phage particles that can lyse the host cell and infect new cells. In summary, the common mechanism used by both retroviruses and lysogenic bacteriophages is the integration of their genetic material into the host cell's genome, allowing for replication and transmission without immediately producing new viral particles. This integrated state can be maintained for long periods, contributing to the persistence and spread of these viruses within a population of cells.

(a) What is an F´ plasmid and how is it formed? (b) Explain how an F´ can be used to construct a bacterial strain that is partially diploid. (c) Explain how partial diploid strains can be used to assess interactions between different alleles (e.g., lac + and lac-).

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(a) An F´ plasmid is a type of plasmid f...

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A virulent bacteriophage is used to infect a prototrophic bacterial culture. Phages are collected from the culture and are used to infect a new bacterial strain that has several auxotrophies. After infection, rare prototrophs are found: met+leu +0met+pro+1met+his+0pro+leu+2pro+his+0leu+his+1\begin{array} { l l } \mathrm { met } ^ { + } \text {leu } ^ { + } & 0 \\\mathrm { met } ^ { + } \mathrm { pro } ^ { + } & 1 \\\mathrm { met } ^ { + } \mathrm { his } ^ { + } & 0 \\\mathrm { pro } ^ { + } \mathrm { leu } ^ { + } & 2 \\\mathrm { pro } ^ { + } \mathrm { his } ^ { + } & 0 \\\mathrm { leu } ^ { + } \mathrm { his } ^ { + } & 1\end{array} How are the auxotrophy genes organized on the bacterial chromosome?

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The data provided suggests that a virule...

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HIV has a high mutation rate. What causes this, and how might this be advantageous to the virus?

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HIV has a high mutation rate due to the ...

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The table below shows the results of interrupted-mating experiments with three different Hfr strains. What is the order of the genes, starting with C?  Hfr strain  Order of transfer 1A,B,E,D,F2D,F,C,G,A3D,E,B,A,G\begin{array} { | c c | } \hline \text { Hfr strain } & \text { Order of transfer } \\\hline 1 & A , B , E , D , F \\\hline 2 & D , F , C , G , A \\\hline 3 & D , E , B , A , G \\\hline\end{array}


A) C, G, A, D, F, B, E
B) C, F, D, B, A, E, G
C) C, B, E, D, F, G, A
D) C, G, A, B, E, D, F
E) C, D, F, G, A, B, E

F) A) and D)
G) A) and C)

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