# take homologous recombination) donor strain into a recombination proficient

take out questions BI324 Genetics and Evolution Lab report of the transduction practical Mon: 22.01.2018 (noon)                                                 Questions:                                          (1) The E.

coli genome is 4.6 x 106 base pairs in length.What is the maximum separation between two genes that is possible for them tobe co-transduced (simultaneously transduced) into a donor strain.

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Please explainyour answer. (5%).            44 kbp, this is the size of DNA packagedin the virus. So this will be the length of the DNA. This is 0.957% of the E.

coli that can be transduced ((4.4×10^4)/(4.6×10^6))x100= 0.957%. The maximumseparation is the distance between the two genes on a homologous chromosome                                                (2) Suppose that you obtain the followingresults in the experiment described above:-If the culture that you used in theexperiment contained 3 x107 bacteria per millilitre, calculate the average frequency oftransduction. Explain carefully how you arrived at your answer. Do not forgetto include units.

(5%) (8+5+7+4+10)/5 = 6.8 = average of allcolonies=> 1000?l/100?l= 10 =>(3×107)/10 = 3×106 => 6.8/(3×106) = 2.27x 10-6 colonies per 100 ?l                                    (3) Why is it essential to be able to selectfor transductants in a transduction experiment? (5%) A transductant is a cell or organism thathas gone through transduction. When you select for the transductant then youcan see the bacteria that contain the gene you are looking for. Also so that you can tell the difference between Onlyat low frequencies can transduction happen.                                                             (4) Do you think that it would be possibleto transfer a drug resistance marker from a recA mutant (defective ingeneralised or homologous recombination) donor strain into a recombinationproficient recipient strain? Please explain your answer. (10 %)            Yes, it is possible.

(change) RecA is a protein that binds to 2DNA molecules and aligns homologous sequences within them. It then promotes aDNA strand-exchange reaction that creates branched DNA recombinationintermediates. It is also known that the recipient strainis not defective (only the donor strain is) therefore homologous recombinationis able to happen, therefore allowing the transfer of the marker. A recA systemis not needed as the donor strain does not take in DNA.

(change) Homologous recombination isneeded to transfer the drug resistant marker and two crossovers must occur. Inthis case the defective recA mutant is the donor so homologous recombinationand transduction can’t happen.                                                 (5) It has been observed that as the amount ofbacteriophage added to a culture increases, the number of transductionsubsequently isolated rises to a maximum level, and then declines. Provide ahypothesis to explain these results. (10%) As the number of bacteriophages (thatcontain tetracycline resistance) increases thereforeincreasing the number of transductions. When the optimum number ofbacteriophages has been added this will show a peak (highest number oftransductions) then the rate of transduction will decrease as cell lysis takesplace.

(6) It has been noticed that the difference in frequencyof transduction between two genes can vary by almost 100-fold. Offer two hypotheses to explain thisobservation. Please explain your answers carefully (15 %)                                                 The position of the enzyme is determinedby the pac sequence, therefore the closer the gene is to the pac sequence givesit a higher probability of being transduced.The frequency of transduction can varybased on the distance between the two genes, also the frequency of the gene.                                                                                                 Your practical write up should include:-                         Write your practical up by dividing itinto the following sections: – Introduction, Results and Discussion. For thiswrite up ONLY, there is no need to include experimental details. Your write upshould be pitched at a level that could be understood by one of your classmates who did not attend either the lectures or practical, but must be writtenin your own words using scientific language. I expect your Introduction tobuild upon the summary information given above (60% of marks for the practicalwrite up).

Think about how the Results can best be presented. Any figuresshould be properly labelled and accompanied by a legend. I would expect theResults and Discussion sections to be relatively brief (20% each of theremaining marks for the practical write up).                                                            Along with your write up, also hand in theanswers to the five questions below.

The marks allocated for each question arelisted, and will count for the remaining 50% of the overall mark for thepractical.                                                                                                             Introduction:(lytic cycle, lysogenic cycle,transduction, aim/objective of practical)(fix) Transduction can be describedas “enhancing the rate of evolution in bacteria.” As the number of open readingframes increase the relative percent of open reading frames involved in signaltransduction also increases. Prokaryotes have a short generation time and largepopulation sizes “One mechanism of horizontal gene transfer is transductionwhich transfers DNA (genes) from one prokaryote to another.

Bacteriophage is a virus thatinfects bacteria and transfers DNA from one bacterial strain to another duringtransduction. Bacteriophages can replicate by two mechanisms: called the lyticand lysogenic cycle. The lysogenic cycle allows for the replication of the bacteriophagegenome but lysis of the cell does not occur. For transduction to occur lysismust occur so this cycle is not associated with transduction. The lytic cycle onthe other hand destroys the host cell while producing new bacteriophagesbecause of the lysis of the host cell.

In this cycle bacteriophages connect tospecific receptor sites on an E. coli’s outer surface. The phage DNA (about 44kbp)is then inserted into the cell which controls the production of phage proteinsand genes using the host’s enzymes and other objects within the cell. The new bacteriophagesthen assembly with the genes being inserted into the capsid. The packagingenzyme that helps in this stage does not differentiate between bacterial andbacteriophage DNA, so defective bacteriophages (transducing particles) are formedby bacterial genomes entering bacteriophage heads. This cycle always results inthe death of the host cell, as the cell wall breaks releasing up to 200 newlymade bacteriophages which are sent to infect healthy cells which is how thecycle starts again and continues.

Transducing particles contain thehost DNA not the phage DNA which is inside an otherwise empty bacteriophagehead. The injected bacterial DNA must 2 homologous recombination events (cross-overs)between the host genome and the injected DNA fragment for transduction (key inevolution of bacteria) to actually occur. This homologous recombination is donefor exchanging DNA so the original genomic DNA is replaced by some of theinjected DNA. If a transducing particle is not homologous to the genome then itwill be part of the recipient genome therefore will be degraded by the cell. The aim of the experiment was toshow generalized transduction, this was observed by looking at the number bacterialcolonies formed. Pac is resembled by sequences in thebacterial genome, these bacterial genomes are able to provide the point ofentry for the bacterial   Results:I decided to use the class results as I donot think that my data followed the correct pattern.

I then divided the total number of colonies by the total number ofplates counted to get the average number of colonies counted per plate. Volume of lysate (?l) Average number of colonies (transductants) counted 0 0 10 1.06 20 1.78 30 2.56 40 2.07 50 1.06               This data shows that as the volume oflysate increases so does the average number of colonies counted until 30 ?l then decreases      My results: Volume of lysate (?l) Plate number Total number of colonies (transductants) counted 0 1 (control) 0 10 2 6 20 3 1 30 4 0 40 5 0 50 6 0 This table shows that as the volume oflysate increases so does the total volume of colonies counted.

For my experiments,I used a total of 6 plates all with different concentrations.     With theclass data, many plates were used for each volume of lysate so an average wasable to be taken, while for my data only one plate was taken for each volume.The class data is more accurate as the sample size is much larger. With my data,an increase then decrease happens but the optimum volume occurs at a lowervolume compared to the class data. The general patter was the same but more ofmy data showed that no colonies were formed, this could show that generalized transductionhad not occurred.

Discussion: “why lysateaffects the number of colonies” “what could have gone better”Tofind the exact optimum volume it is important to add more values between 10 ?l and 50 ?l. An example is looking at theaverage number of colonies volumes that are  4 ?l apart. Alimiting factor would be the size of the virus.  References:·      Roca, A. and Cox, M.

(1997). RecA -an overview | ScienceDirect Topics. online Sciencedirect.com. Available at:https://www.

sciencedirect.com/topics/neuroscience/reca Accessed 21 Jan. 2018.·      Book: Reece, Jane B.

, et al. Campbell Biology. ninth ed., PearsonEducation, Inc., 2014.·      Book: Taylor, MarthaR., et al.

StudyGuide for Campbell Biology, Eleventh Edition Lisa A. Urry, Michael L. Cain,Steven A. Wasserman, Peter V.

Minorsky, Jane B. Reece. Pearson, 2017.