Concept explainers
For bacteria that are
a. Describe the state of the F factor.
b. Which of these cells are donors? Which is the recipient?
c. Which of these donors can convert exconjugants to a donor state?
d. Which of these donors can transfer a donor gene to exconjugants?
e. Describe the results of conjugation (i.e., changes in the recipient and the exconjugant) that allow detection of the state of the F factor in a donor strain.
f. Describe a “partial diploid” and how it originates.
To review:
For bacterial strains
Description for the state of the F factor.
Identification and explanation of donor and recipient cells.
Identification and explanation of donor cells that can convert exoconjugants to donor state.
Identification of the donor that can transfer the donor gene to exoconjugants.
Description of results of conjugation that allow detection of the state of the F factor in the donor strain.
Description on partial diploids.
Introduction:
Conjugation is the process of transfer of the DNA from one donor bacterium to a recipient bacterium. It is also known as lateral gene transfer. Conjugation is always occurring between the donor cell and the recipient cell, it does not occur between two donor cells. The exoconjugant cell is the recipient cell that has a modification in their genome by receiving the DNA from a donor cell in the process of conjugation. During conjugation, bacteria come in contact with the help of sex pilus. The bacterium that has sex pilus is the donor - male bacterium, and the bacterium that lacks sex pilus is the recipient - female bacterium.
The cell that has the ability to donate their DNA is called
Explanation of Solution
The
Hfr (high frequency recombination)
Donor cell is a cell that donates factor/s to another cell by conjugation. Only that bacterial cell which already has the
The
Hfr and
At the time of conjugation, bacterium comes in contact with the help of sex pilus. The bacterium that has sex pilus is the donor - male bacterium, and the bacterium that lacks sex pilus is the recipient - female bacterium. Conjugation tube is formed between the donor and recipient, and F sex factor is transferred from
A bacterium that has two copies of some of its genes (not all) is termed as partially diploid. In a bacterium, one copy of the desired gene is present on the bacterial genome, and another copy may be fused with the plasmid. Now, the bacterial cell possesses one complete set of genes and identical copy of portion of the genome.
Example of formation of partial diploid:
Conclusions:
The bacterium that has plasmid is
The
Hfr and
A bacterium that has two copies of some of its genes (not all) is termed as partially diploid.
Want to see more full solutions like this?
Chapter 6 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- Conjugation between an Hfr cell and F- cell does not usually result in conversion of exconjugants to the donor state. Occasionally however, the result of this conjugation is two Hfr cells. Explain how this occurs.arrow_forwardWhen plasmids are isolated from bacterial cells, they may existin a number of forms.a. List the different forms that may be found.b. Which do you think would migrate the fastest and farthest in anelectrophoresis experiment and why?arrow_forwardI need hand written solution only otherwise I will down votearrow_forward
- When Avery and his colleagues had obtained what was concluded to be the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, RNase, and DNase, followed in each case by the assay for retention or loss of transforming ability. What were the purpose and results of these experiments? What conclusions were drawn?arrow_forwardExamine the table below for the pathway with precursors / products A, B, C, D, and E and mutant enzymes 1, 2, 3, and 4. Use the information given to draw the pathway. (+ indicates growth, 0 indicates no growth) A B C D E 1 + 0 0 0 + 2 + + 0 0 + 3 + + 0 + + 4 0 0 0 0 + Later a new mutation (5) is found that fails to complement mutant 2. Between what two compounds on the pathway is this new mutant located?arrow_forwardSelect all that apply from the following to each transformation process : The following are White non-glowing colonies White glowing colonies Blue non-glowing colonies Blue glowing colonies None of the above The transformation are: a. LB + kanamycin plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain . b. LB + kanamycin plates that have been spread with the E. coli cells transformed with water, what phenotype of the colonies do you expect to obtain. c. LB plates that have been spread with the E. coli cells transformed with your pHSG298, what phenotype of the colonies do you expect to obtain. d. LB + kanamycin plates that have been spread with the E. coli cells transformed with PHSG298, what phenotype of the colonies do you expect to obtain. e. LB plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain. f. LB plates…arrow_forward
- A transduction experiment was carried out to map the his and leu genes, which are involved in amino acid biosynthesis, of E. coli using a his+ leu+ donor and a his- leu- recipient. his+ leu- transductants will grow on agar media with: OA lacking histidine, containing leucine O B. containing histidine, containing leucine OC containing histidine, lacking leucine O D.a and b are correct O E. a, b and c are correctarrow_forwardhe bands indicated by arrow "A" (to the left of the western blot) most likely represent: A. PX-HA lacking any post-translational modifications. B. dephosphorylated PX-HA. C. phosphorylated PX-HA. D. PX-HA with its N-terminal signal sequence still attached. E. glycosylated PX-HA.arrow_forwardIdentify the most mistaken (wrong) choice: a) Transcription machinery and an enhancer can bind to the chromosome at the same time. b) Organic matters may interfere with heat treatment of bacterial growth control. ( c) Nitrocellulose can be used to filter out microorganisms from a liquid solution. d) Time to kill a bacterial culture is not proportional to the number of microbes in the culturearrow_forward
- Some explanation for this image.arrow_forwardA researcher discovered Suramin as a potent and selective inhibitor of Mycobacterium tuberculosis RecA protein and the SOS response. What would be the outcome if the action of RecA was inhibited during the SOS response? O LexA would not autolyse, and therefore the transcription of DNA repair genes would not occur; the bacteria would likely not survive. O LexA would autolyse, and therefore the transcription of repair genes would not occur; that would be lethal to the bacteria. LexA would autolyse, and therefore the transcription of DNA repair genes would occur; and the bacteria will be susceptible to antibiotics. LexA would not autolyse, and therefore the transcription of repair genes would occur; essentially that would be lethal to the bacteria.arrow_forwardSeveral common antibiotics affect some strains of bacteria's ability to carry out transcription and/or translation. For example: Rifamycin inhibits prokaryotic RNA polymerase Chloramphenicol blocks the transfer of the peptide from the P to A site. a) For each of these drugs, identify at what point it could affect the process of DNA->RNA->protein. Be as specific as possible. b) Why do you think these drugs kill bacteria but spare animal cells? (Hint: remember bacteria are prokaryotes)arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning