Differences between prokaryotic and eukaryotic genomes

This was written by Direncan Boyraz (Izmir Institute of Technology)

1)Genome structure: 

Prokaryotic genome has only one chromosome in the cytoplasm, but eukaryotic genome has multiple chromosomes inside a nucleus. Because of multiple number of chromosomes, eukaryotes can be haploid or diploid as difference than always diploid prokaryote. Also Prokaryotic genome own  a circular structure, but there are exceptions, besides always linear strands are present in the eukaryotic cells and linear DNA provide telomeres, non-coding DNA ends, for eukaryotic cells. The eukaryotic genome is more complex with longer gene, on the other hand, prokaryotic genome has up to 90% coding sequences than often around 3% in eukaryotic genome. Supercoiling of DNA exist in both, but patterns of them are different because of circular and linear genome. One replication site or Ori is found in prokaryotic genome and it serves quicker division rate more bidirectionally. Eukaryotes have long repeating nucleotides than rare prokaryotic genome (1).  

2)Gene Structure: 

Both of them have some common features about genes. Replication directions, DNA and RNA structures, principles of work of polymerases can be seen as examples. Eukaryotic genes have introns and exons. In mature eukaryotic mRNA, only exons exist. Also eukaryotic genes are clustered, so genes that are related to similar functions are on same region of chromosome and they are controlled independently. Operons or polycistronic mRNAs do not exist in eukaryotes and in prokaryotic gene, regulatory parts are on same place with actual gene (2,3).      

3)RNA polII-driven transcription initiation:  

In contrast, prokaryotes have simpler transcription. σ factors in prokaryotes attach to promoter with mostly RNA pol and start transcription. Different σ factors bind to different promoters for a responsible gene. In Eukaryotes, transcription initiation are quiet different and more complex. This complexity comes from supercoiled structure and regulator elements. RNA pol can’t initiate transcription. Some DNA binding proteins such as transcription factors and enhancers helps to change conformation of chromatin structure. RNA pol can bind to promoter with help of transcription factors and starts transcription (4).     


[1]Brown TA. Genomes. 2nd edition. Oxford: Wiley-Liss; 2002. Chapter 2, Genome Anatomies. Available from: https://www.ncbi.nlm.nih.gov/books/NBK21120/  

[2]Alberts B, Johnson A, Lewis J, Raff M, Roberts K, Walter P (2002).Molecular Biology of the Cell(Fourth ed.). New York: Garland Science.  

[3]Shafee, Thomas; Lowe, Rohan (2017). “Eukaryotic and prokaryotic gene structure”.WikiJournal of Medicine.4(1).doi:10.15347/wjm/2017.002  

[4]Dvir, Arik & Conaway, Joan & C Conaway, Ronald. (2003). Assays for Investigating the Mechanism of Promoter Escape by RNA Polymerase II. Methods in enzymology. 370. 733-40. 10.1016/S0076-6879(03)70059-7.  

The best model organisms for chromatin studies

This was written by Direncan Boyraz (Izmir Institute of Technology)

1) Saccharomyces cerevisiae 

Saccharomyces is one of easier organisms to be analysed for isoforms of histones and histone modifications chromatin structure studies than other complex eukaryotes; additionally Saccharomyces has common modifications with mostly all eukaryotes (1). There are some key points for this. Firstly, promoters of yeast as enhancers can be activated by becoming nucleosome-depleted. Second, nucleosomes on chromatins are well located. Distances and bordering regulatory of nucleosome provide convenience. Third, highly transcribed genes own low nucleosome occupancy because of RNA polymerase II activity with its associated factors (2).  

2) Drosophila melanogaster 

Fruit fly as a model organism has some advantage in chromatin structure analysis such as DNA methylation. DNA methylation in flies are manageable than other animal or models. We have available data for the system of it and also existence of less methylated genome than human genome  with less number of gene is useful for epigenetic studies. More restrainable number of chromosomes form an easier environment for observations. Fruit fly is one of most model organism for those experiments. Easy observable and high similarity to human make it a good model(3).  

3) Arabidopsis thaliana 

Histone modification and DNA methylation are to key for studying chromatin structure and natural changes Arabidopsis lives brought it to rapid modifications evolutionary.  Also epigenetic variation is proved with some ways such as modifications and the presence of repeated sequences or transposons within the promoters. First and most known plant on the earth is Arabidopsis, and this makes it a good model organism for that (4). 


[1] Rando, Oliver J., and Howard Y. Chang. “Genome-Wide Views of Chromatin Structure.” Annual review of biochemistry 78 (2009): 245–271. PMC. Web. 28 Feb. 2018.  


[2] Rando Lab Biochemistry and Molecular Pharmacology, Umass Medical School, Chromatin Structure and Function,  


[3] Lyko F., Beisel C., Marhold J., Paro R. (2006) Epigenetic Regulation in Drosophila. In: Doerfler W., Böhm P. (eds) DNA Methylation: Development, Genetic Disease and Cancer. Current Topics in Microbiology and Immunology, vol 310. Springer, Berlin, Heidelberg  

[4] Turck, F. and Coupland, G. (2014), Natural Variation in Epigenetic Gene Regulation and Its Effects on Plant Developmental Traits. Evolution, 68: 620–631. doi:10.1111/evo.12286  

Computational Biology Exam Sample

1) What is the difference between application and program? Explain in one (1) sentence max. (5
2) Why would we prefer command line based tool over graphical user interface (GUI) based tool?
Explain in two (2) sentences max. (10 pts)
3) Why is the Basic Local Alignment Search Tool called a heuristic algorithm? Explain in one (1)
sentence max. (5 pts)
4) What is the difference between database search and de novo search in mass spectrometry based
search tools ? Explain in one (1) sentence max. (5 pts)

Exam Questions about Computational Biology

1) GFF (General feature format) files are widely used in data annotation, describing genes and
other molecules. Find the example GFF formatted file in your cms (Q1.gff).
a) Search and print every line that contains the word “overlap” using find command in cmd
(10 pts)
b) Redirecting (pipeline) the output of the problem (a) to the input of a new find command,
find all the lines that contain the word “chrXII” (10 pts)
c) Redirect the output of problem (b) to a new file named “Answer1.txt” (5 pts)
d) Put all the command lines that you used in this question to a file named
“Answer1_cmdline.txt”. Send both of the files.
2) Find the “blast_exam” folder in your cms.
a) Index the database file (7 pts)
b) Blast the query to database using “blastp” but use a switch to filter out all the alignments
that has identity below “95.3” percent (the number “95.3” is a float value) (13 pts)
(Hint: Help is always given on command line prompt to those who ask for it)
c) Put all the command lines that you used in this question to a file named
“Answer2_cmdline.txt”. Send this file and screenshots.
3) Find the “xtandem.zip” file in your cms. Inside the “src” folder, you have “exam_spectra.mgf”
and “exam_protein.fasta” files. Don’t move any file.
a) Index the database file (from “src” folder). (5 pts)
b) Fix “taxonomy.xml” and “input.xml” files in such a way that you can search
“exam_spectra.mgf” in “exam_protein.fasta” (mgf and indexed fasta files will be used
from src folder) (18 pts)
c) Set “fragment monoisotopic mass error” option to 0.7 and run xtandem program (7 pts)
d) Put all the command lines that you used in this question to a file named
“Answer3_cmdline.txt”. Send “Answer3_cmdline.txt”, “default_input.xml”,
“taxonomy.xml”, “input.xml” and screenshot files.
Please put all your files and answers in separate folders (e.g. folder “a1”, “a2”…)
then zip them into one file with the name “yourname_yourstudentid”.zip.

ComputationalComputational Biology Exam Folder

Differences between Prokaryotes and Eukaryotes

Both prokaryotes and eukaryotic genomes are DNA genomes. Prokaryotic genome is usually found on the cytoplasm and consists of only one chromosome, while the eukaryotic one consists of multiple chromosomes and most of the eukaryotic genome is found on the nucleus. Prokaryotic genome is less complex compared to the eukaryotic one which generally consists of longer genes. The coding sequences compart nearly the 90% of the prokaryotic genome and only the 3% of the eukaryotic genome. The prokaryotic genome tends to be circular, while the eukaryotic genomes are generally linear. The prokaryotic genomes are always haploid, but the eukaryotes can be either diploid or haploid. Compared to the eukaryotic genomes, the prokaryotic genomes have a high rate of protein coding genes. Furthermore, the eukaryotic genome has hundreds of rRNA genes compared to the prokaryotic ones that have 1-10 rRNA genes. Eukaryotic genomes have a higher rate of repeated sequences, end telomeres and introns compared to the prokaryotic genomes that rarely consist of those parts. Also origin of replications usually are numerous on eukaryotes but found only once in prokaryotes.[1]. Both eukaryotic and prokaryotic gene structures include similar sequence elements necessary for gene expression. They both involve regulatory elements such as promoters, silenceres and enhancers usually on their ends and contain ORI. Also the reading strands of both eukaryotic and prokaryotic strands is on the 5′-3′ direction. Eukaryotic genes differ from prokaryotic genes because introns are absent in prokaryotic genes and in prokaryotic genes there is no such a phenomenon such as alternative splicing. Also eukaryotic genes consist of more regulatory regions compared to the prokaryotic ones. In prokaryotes the genes can sometimes organize themselves in a polycistronic operon. Riboswitches also are some regulatory regions of the prokaryotic genes that lack on the eukaryotic genes. The main difference of the RNA polymerase II assisted transcription initiation is that in prokaryotes only DNA containing promoter and polymerase II are needed, while in eukaryotes several transcription initiation factors and proteins more are needed, such as TF, TBP, TFIID, TAF[2].



2 – https://www.ncbi.nlm.nih.gov/books/NBK9935/

Fosforilimi Oksidativ dhe Sinteza e ATP-s 

Fosforilimi oksidativ dhe fotofosforilimi sigurojnë me te shumtën e ATP-së në shumicën e gjallesave. Fosforilimi oksidativ është termi përdorur për dukurinë e reduktimit te oksigjenit në nje molekulë uji ku elektronet dhurohen nga NADH  dhe FADH2. Fosforilimi dhe fotofosforilimi janë të ngjashme nga ana mekanike për tre arsye: 

1-rrjedha e elektroneve në kanale të suportuar nga membrana 

2-potenciali elektrokimik midis membranave 

3-sinteza e ATP-së me anë të protonëve që ndodhen në gradientin e kundërt 

Pjesët përbërëse të mitokondrisë 

1-Membrana e jashtme-e përshkrueshme nga molekulat e vogla dhe jonet 

2-Membrana e brendshme-e pakalueshme nga jonet ose molekulet e vogla duke përfshirë dhe jonin hidrid. Ajo përmban transportuese elektronë që janë përgjegjës për procesin e respirimit, transportuesë të tjerë, sintazen ATP, translokatorin ADP-ATP 

3-Matriksi-Përmban kompleksin e piruvatit dehidrogjenues, enzimat për ciklin e acidit citrik, enzimat për beta-oksidimin e acideve të yndyrshëm, ATP, ADP, jonet e kalciumit, kaliumit. 

Fosforilimi oksidativ fillon me hyrjen e elektronëve në zinxhirin respirativ. Enzimat dehidrogjenues mbledhin elektronë nga ciklet katabolike dhe i udhëzojnë ato drejt NAD, NADP ose FAD që njihen ndryshe si pranueset universale të elektronëve. Enzima dehidrogjenuese e lidhur me NAD shërben për të hequr dy atome hidrogjeni prej substrateve të tyre. Njëri nga këto hidrogjene trasmetohet drejt NAd+ kurse tjetri ngelet në gjendje të lirë si jon hidrid. NADH mbledh elektrone nga ciklet katabolite kurse, NADPH mundëson reaksionet anabolike me elektrone.  

 Flavoproteinat pë rmbajnë një nukleotidë flavine (FMN ose FAD). Nje nukleotidë flavine e oksiduar pranon një  elektron dhe në kë të më nyrë prodhon FADH+ ose FMNH+ ose mund të pranojë dy elektrone pë r të prodhuar FADH2 ose FMNH2. Potenciali i reduktimit ë sht i varur ndaj lidhjes me proteinat e ndë rlidhura me të .  

Kalimi i elektronave në pë r transmetuesit që pë rmbajnë membranë .  

Transferimi i elektronë ve në  fosforilimin oksidativ mund të ndodhë në pë rmjet transfertë s direkte të elektronë ve, transfertë s si atom hidrogjeni ose si jon hidrid.  

Molekulat që transferojnë elektrone: 





Proteinat që  pë rmbajnë hekur dhe sulfur 



Janë  koenzima të  lidhura me yndyra të tretshme Q benzokinone dhe pë rmbajnë   një zinxhir anë sor isoprenoid. Ato pranojnë 1 ose 2 elektrone. Mund të shpë rndajë edhe elektrone edhe protone që kanë një rol të rë ndë sishë m në pompë n protonike.  


Ë shtë e aftë pë r të kapur dritë n e diellit në pë rmjet grupit heme që pë rmban, si a, b apo c. Kofaktorë t a apo b lidhen në një më nyrë kompakte me proteinat, kurse grupet heme të llojit c lidhen në më nyrë  kovalente në pë rmjet lidhjeve Cys. Citokromat e llojit a ose b janë poteina integrale kurse ato të llojit c mund të kalojnë lehtë sisht pasi janë të tretshme.  

Zinxhiri respirativ i mitokondrisë  

Kahu i procesit synon drejt rritjes së potencialit reduktues.  

NADH-Q-citokroma b-citokroma c1-citokroma c-citokroma a-citokroma a3-oksigjeni 

 Komplekset multienzimatikë  që luajnë rol në  transfertë n e elektronë ve 

Kompleksi I: NADH-Jubikuinone 

Deri në 42 zinxhirë anë sorë , duke pë rfshirë  flavoproteina si dhe të paktë n 6 qendra me hekur dhe sulfur.  


IMViC Tests and Catalase Test

The aim of this experiment is to determine kind of Enterobacteriaceae bacteria that are found in the tubes labelled as α and β by using IMViC tests.
In microbiology, kind of bacteria in Enterobacteriaceae of determination is done by using some methods. The rapid and useful test is IMViC. For checking danger or safety of water or food, this test is so important. (1)
Enterobacteriaceae is a huge family in bacteria. They have Gram-Negative bacteria including harmless and pathogenic kinds. Coliforms are found in this family. Coliforms are separated according to fermenting lactose with gases and acid. Coliforms are rod-shaped, Gram-Negative, non-spore forming bacteria. They mostly live in aquatic environment and fecal contamination is considered with coliforms. Because of that, they are so important as indicator for these purpose. They are used for pollutions or determining pathogenic organisms. For example, Escherichia coli is one of the most known coliform like Klebsiella pneumonia and Enterobacter aerogenes. Coliforms are obtained by using IMViC test. IMViC include some serial test to determine species of coliform. Each letter except “i” refers to first letter of tests. (1, 2 and 3)
Indole test is first test and used for obtaining present of indole amino acid. Some coliforms use tryptophan as substrate and they produce indole after reaction. Kovac’s reagent, so dangerous with air, is used for that. The reagent reacts with presence of indole and gives pink/red colour on top layer of the tube. E.coli is one of bacteria giving positive result for indole test. Many Bacillus gives negative result like this experiment. (4, 5 and 6)
Methyl-red or shortly MR test is based on pH value in solution. It acts like an indicator giving red colour with acidic solution. Glycoses is first step for generating energy for living organisms. At end of glycoses, pyruvates or pyruvic acid are produced from glucose. Naturally pyruvic acid makes acidic. Bacteria convert them to more stable and safer acidic forms such as formic acid, acetic acid or lactic acid. Eventually acidic media causes change of colour with reagent. E.coli forms acidic media and it causes positive or red colour at MR test. B.subtilis demonstrates negative result or yellow-brown colour. (4, 5 and 6)
VP test or Voger-Proskauer test is the other step of IMViC. VP is for determined existence of acetone in bacterial culture. Some bacteria live by using glucose and turning into acetylmethylcarbinol, shortly acetone. Two solutions are added for that. First one is alpha-naphthol. It binds to acetone and the second solution, potassium hydroxide forms cherry red colour if it is positive. Negative is to see yellow-brown colour for this test. B.subtilis is in positive group; however same test gives negative with E.coli. (4, 5 and 6)
Citrate Utilization test is a test to control bacteria use whether citrate or not as energy-carbon source. When bacteria are added into a media that have sodium citrate and pH indicator like bromothymol blue as basic test materials, citrase enzyme in bacteria breaks down citrate to oxaloacetate and acetate. Oxaloacetate becomes to pyruvate and CO2. On the other hand, after some reaction with acetate, sodium citrate and ammonium salts cause changes in alkaline pH value. These changes forms colour media to blue from green if it is positive like in B.subtilis. Negative result doesn’t demonstrate any growth on media. E.coli can be given an example for negative. (4, 5 and 6)
Catalyst test is one of simple tests. Some bacteria can break down H2O2 to water for producing energy. In this test, H2O2 is dropped onto cells directly and if bubble is observed, it means test is positive. For E.coli, it is variable, but for B.subtilis, it is positive. (4, 5 and 6)
Material and Methods:
Indole Production Test,
0.3 ml of Kovac’s reagent was taken and transmitted into the sample tubes α and β with Tryptone in hood. After a while, colours of samples were observed.
Methyl-Red Test,
1 ml of methyl-red indicator was added into the samples with glucose phosphate broth and the samples were waited for colour change.
Voges-Proskauer Test,
0.9 ml of solution A and 0.3 ml of solution B of Barrit’s reagent were added into samples and they were observed.
Citrate Utilization Test,
Tubes with Simmon’s Citrate Agar were transmitted into test samples and they were waited for 48 hours at 37°C in incubator. After that, they could be observed.
Catalase Test,
3% H2O2 solution was dropped onto bacterial colonies of the test samples that were incubated at small amounts onto a plate before. The samples form bubble was observed.
Samples Indole Methyl-red Voger-Proskauer Citrate Utilization
α – + – +
β + – – –
Table 1- the results of IMViC tests for 2 strains (α and β)
Catalase Test
α +
β +
Table 2- Catalase Test for 2 strains (α and β)
α and β refer to E.coli and B.subtilis as unknown samples.
In this experiment, it is aimed to determine kinds of the samples in coliforms by using some serial tests according to metabolic activities.
According to the results of this experiment, it can be seen that α and β have different metabolic activity. For first test for IMViC, indole test was done for each sample. Indole test is bacteria use tryptophan and turns it into indole amino acid. At presence of indole in media, Kovac’s reagent gives a pink layer on top of tube by forming a complex with indole. The sample β gave positive for indole test and the bacteria in sample β uses tryptophan for its metabolism. E.coli is one of kinds of bacteria that are positive at indole test. (5,6)
At second test, methyl-red test, reagent is used for obtaining pH value in sample basically. Methyl red acts as indicator to shows colour change according to pH value. Some microorganism metabolizes glucose to pyruvic acid and it makes environment acidic, but they convert pyruvic acid to some forms such as lactic acid, acetic acid or formic acid for preventing accumulation of pyruvic acid. However environment is still acidic and methyl red can give a reaction. Positive test that bacteria produce low pH value about 4 forms red colour from yellow. Negative gives yellow with more basic environment. α and β indicate different and unexpected result for MR test. The sample α may be B.subtilis and the sample β may be E.coli by comparing with the other tests, however α and β show unexpected results. There may be experimental error. The tubes that have been labelled as α and β could have been confused before experimental process. α and β may have been confused again at observation part. These kind of experimental errors may be and also contaminated culture with the other bacterial cultures or wrong aseptic techniques may give these results. (5, 6)
VP test or Voger-Proskauer is used for identified presence of acetone in bacterial culture. At VP is applied with two solutions for reacting with acetone and giving colour. These are alpha-naphthol and potassium hydroxide. If bacteria turns glucose into acetylmethylcarbinol at digestion, alpha-naphthol reacts and potassium hydroxide gives cherry red colour for positive result. At negative result, yellow-brown or copper colour indicates. In this experiment, there is no positive result. Both cell cultures don’t give acetone after their metabolic activity as expected. E.coli and B.subtilis must have negative result after VP test. (5, 6)
Citrate utilization test is to detect use of citrate by bacteria. Some bacteria use citrate for their metabolic activity. Citrate utilization test uses reagent that can change their colour after reaction in pH value. If test is positive, a change in colour forms from green to blue. B.subtilis can cause this kind of change on media. The sample α can be seen a B.subtilis culture. E.coli culture doesn’t look like that like at β culture. (5, 6)
Catalyse test is the easiest test than the other tests. Drops of H2O2 were poured on two cultures that were found on different places of same slide. Bubbles were observed at both samples. Already forming bubbles were expected at B.subtilis. The sample α that is thought it is B.subtilis shows same results with catalyse test. On the other hand, E.coli bacteria were variable about forming bubbles. In this experiment, β sample is E.coli that can form bubble with H2O2. (5, 6)
Except MR results, samples can be identified which kind of bacteria exist in tubes. According to Table 1 and Table 2, α sample is B.subtilis and β sample is E.coli. (5,6)

[1] Received on 24 May,
MacFaddin J.F. 2000. Biochemical Tests for the Identification of Medical Bacteria, 3rd ed. Lippincott Williams & Wilkins, Philadelphia, PA, USA.
[2] Received on 24 May,
Don J. Brenner; Noel R. Krieg; James T. Staley (July 26, 2005) [1984 (Williams & Wilkins)]. George M. Garrity, ed. The Gammaproteobacteria. Bergey’s Manual of Systematic Bacteriology.  2nd ed. New York: Springer.
[3] Received on 24 May,
[4] Received on 24 May,
MacFaddin, Jean F. “Biochemical Tests for Identification of Medical Bacteria.” Williams & Wilkins, 1980, pp 173 – 183.
[5] Received on 24 May,
[6] Received on 24 May,

 16S Ribosomal RNA Sequencing and Phylogenetic Tree Construction

The purpose of this experiment is to form phylogenetic tree between five different kinds of prokaryotic organisms by using 16S rRNA analysis.
For forming phylogenetic tree, genomic DNA of each organism should be isolated and then interested regions of DNA (rRNA sequences or ribosomal RNA sequences) are amplified by PCR. For knowing these sequences, amplified DNA fragments are sequenced and phylogenetic tree is formed by using bioinformatics tools.
rRNA or ribosomal RNA is essential material that joins protein production. rRNA is produced from rRNA sequences or rDNA sequences. The importance of these rRNA sequences is to have very conserved regions and also differences that species have. These sequences are used for determined place of specie in taxonomy. (1)
There are three types of rRNA in prokaryotes. These are 5S, 16S and 23S. Without eukaryotes, 16S is often used for determination of phylogenetic tree or taxonomic places of organisms. For eukaryotes, 18S is used for that. rRNA sequences include some commonplace regions. As universal, some regions are same for all organisms. These regions are much conserved because of mutual and essential sequences. These conserved regions are named as highly conserved regions. The regions 16S includes differences is called hypervariable regions or hot-spots. These differences are used for taxonomy. (1, 2)
Size of rRNA genes is suit for bioinformatics applications and tools. Also conserved regions are universal and it helps to use universal primer for amplification of the gene. After years, researchers have a huge database about rRNA genes and it makes researches easy for them. Hypervariable and conserved regions can help by different ways for constructing phylogenetic tree. (2, 3)
Before construction of phylogenetic tree, interested rRNA genes are amplified by PCR for forming better results and working easier. Ribosomal RNA sequences are used as template DNA. Universal primers also make process easier by highly conserved regions. (4)
Phylogenetic tree is a figure that is used for showing evolutionary relativeness between living organisms. Similarity of genes is related with evolutionary origins. Figures on phylogenetic tree represent different relationships or closeness. Each node is a connection between braches and it demonstrates an ancestor. Branches are formed with differences among kinds. How far two kinds are on tree can be understood by shared nodes. In a tree, if there is a common ancestor for all organisms, this tree is named rooted tree or if there is no, it is named unrooted tree. (5)
Materials and Method:
Firstly, 16S rRNA sequence of DNA that was isolated before experiment was amplified by PCR process. After PCR components were calculated, components were mixed in an Eppendorf except enzyme. Enzymes were added into mixture and then lastly sufficient amount of water was added for completing mixture 50 µl. Finally mixture was placed into PCR machine.
Required Material Stock concentration Required concentration Volume (µl)
DNA 9.4 ng 250 ng 26.6
Primer Forward 10 µM 1 µM 5
Primer Reverse 10 µM 1 µM 5
dNTP (mix) 10 µM 200 mM 1
MgCl2 25 mM 2 mM 4
Taq Pol. 5 unit/µl 2.5 units 0.5
Buffer 10X 1X 5
H20 – – 2.9
Total: 50 µl

A PCR thermal cycle:
Denaturation (30 seconds at 950C)
Annealing (60 seconds at 680C)
Extension (5minutes at 720C)
Thermal cycle of PCR was repeated for 30 times.
Products of PCR were sequenced for rRNA analysis. Sequences were used for constructing a phylogenetic tree with help of T-Coffee (6).

This experiment was intended to make phylogenetic tree between five different organisms according to 16S rRNA sequences of them by firstly using PCR with universal primers then sequence and bioinformatics tools (6).
After constructing phylogenetic tree between five organisms, relativeness of organisms can be seen. Any common ancestor of five organisms cannot be observed on tree. This demonstrates that this phylogenetic tree is unrooted.
The nodes that have two branches from a branch on tree utilize an ancestor kind in mutual. As seen in the Figure 1, the closeness of Acinetobacter haemolyticus and Escherichia coli creatures as affinity is the closest organisms than the other organisms. The node on the right between these organisms releases that they have a common ancestor in the past. Enterococcus faecium is also close to these kinds partially. These three species seem to originate from a common ancestor. The node on the left provides this information to us about the other two organisms too. If Bacillus subtilis and Staphylococcus aureus are observed, these two kinds don’t have a close origin to each other can be seen. Also these two kinds also aren’t close to the other three kinds.
[1] Smit S, Widmann J, Knight R (2007). “Evolutionary rates vary among rRNA structural elements”. Nucleic Acids Res. 35 (10): 3339–54. Received on May 14, 2017
[2]  Woese CR, Fox GE (November 1977). “Phylogenetic structure of the prokaryotic domain: the primary kingdoms”. Proceedings of the National Academy of Sciences of the United States of America. 74 (11): 5088–90. Received on May 14, 2017
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[6] (http://www.ebi.ac.uk/Tools/msa/tcoffee/).

Antimicrobial Testing

The purpose of this experiment is to see effects of ampicillin, tetracycline and ethyl alcohol to bacterial growth by disc diffusion method and dilution method.

At disc diffusion method, effects of ampicillin and tetracycline were observed effectiveness to E.coli and differences of these effects of them were observed and measured also. If these antibiotics have effects to E.coli, inhibition zones are formed by bacteria. Diameters of these zones can give information about sensitivity of bacteria to these antibiotics. Dilution method was done for a different purpose. Effects of different concentrated EtOH can be observed by O.D measurements at 600 nm and minimum concentrations of antibiotics can be measured.

An antibiotic, antimicrobial agents, is used for kill or inhibit bacterial growth. Antibiotics are generally produced from secondary metabolites of microorganisms and it is thought that sporulation is an effect in the formation of antibiotics. Antibiotics have basic two effects to decrease growth, static and cidal effect. Static effect is inhibitory for bacteria. It decreases rate of growth by interacting enzymes, proteins or pathways of bacteria. Cidal effect is lethal side of antibiotic. Vidal processes of bacteria can be stopped by this effect and bacteria can be killed. If antibiotics are classified according to spectrum, there are two types of antibiotics. Some antibiotics have a large spectrum; it means that it can affect all wide of bacterial range. This kind of them is named as board spectrum antibiotics. Narrow spectrum antibiotics have limited range as antibiotic; it means that these kinds of antibiotics can affect some specific kinds of bacteria. (1,4)
There are two basic methods to measure effect of antibiotic according to concentrations to observe sensitivity of bacteria. Disc diffusion and dilution methods are most known. Disc diffusion method is used with a piece of paper. Antibiotics dropped onto paper passed through paper and affect bacteria. Disc diffusion method forms inhibition zones on plates. These zones can give visible source about effect of antibiotic. Largeness of zones demonstrates effectivity of it. However this method is not sufficient to show a minimum concentration for effectivity. In despite of non-visible results, dilution method can give minimum inhibitory concentration (MIC). At samples having same amounts of bacteria, different concentrated antimicrobial agents show different decreasing rate of growth. O.D measurement can give a result of decreasing bacteria on numeral values. These values can reveal MIC. The lowest concentration gives no bacterial growth; it is named minimum bactericidal concentration. (2,3)
Material and methods:
Determination of Minimum Inhibitory Concentration,
Overnight E. coli culture
LB broth
Pipettes and tips
96-well plate
Bunsen burner
At this step, EtOH that have different concentration was prepared stock solution with LB broth. E.coli was transferred later to wells. At blank solutions, E.coli was not be used but LB was used. For control groups, E.coli and LB were prepared and only LB was added to empty wells to find E.coli value without EtOH.
Firstly, 5ml of bacteria samples were diluted to 10-4 to be used for 10-4 dilution part of this experiment. Then, %2 stock solution of EtOH was taken desired volume of EtOH according to M1*V1=M2*V2 to get %0.2 solution. 100µl of EtOH having different concentrations was transmitted first 5 horizontal lines on 96-well plate and 5 lines to Line B. Next, LB broth was added to complete solution to wells poured. 100µl of diluted E.coli samples was transmitted to first and second 5-horizontal lines (Line A and B). At Line D and E, instead of E.coli, LB broth was added as blank. For control, 100µl of E.coli and 100µl of LB broth were used at same volumes at first-wells of Line G and H. Other wells of G and H were used for only 200µl of LB to find control of O.D value of E.coli.
The 96-well plate was sent for measuring O.D value at 600 nm and then waited for 24 hour in incubator at 370C. The 96-well plate was measured again after 24 hour. After calculations from data, standard deviations and graph were formed.
Disc Diffusion Method
Overnight E. coli culture
LB agar plates
Pipettes and tips
Sterile filter paper discs
Bunsen burner
Firstly, labelled agar plate for ampicillin and tetracycline were prepared. The two discs were placed, each disc to one part. 25µl of ampicillin (50µg) was dropped to side speared for ampicillin and to the other part, 25 µl of tetracycline (25 µg) was dropped aseptically onto discs. After that, distilled water was dropped too onto discs. The plate was placed into incubator for a day at 370C. Next day, Diameter of zones were measured by using ruler and recorded.

Tetracycline (25µg) Ampicillin (50µg)
Diameters: 20mm 24mm
Table 1.1-The diameters of zones after tetracycline and ampicillin were applied with amount of antibiotics
Time (hour) Control 0.1% EtOH 0.3% EtOH 0.5% EtOH 0.7% EtOH 1% EtOH
0 -0,005718 -0,00359 -0,02137 0,038435 -0,031213 -0,03393
24 0,892181 0,824214 0,813324 0,792511 0,797294 0,738407
Table 2.1-The results of OD values (600 nm) of E.coli diluted to 10-3 after calculations
Control 0.1% EtOH 0.3% EtOH 0.5% EtOH 0.7% EtOH 1% EtOH
Standard Deviation: 0,0714277

Table 2.2-The values of standard deviation of results at 24th hour for 10-3 diluted bacteria samples

Graph 2.3- The graph that shows O.D (600 nm) at initial and 24th hour with standard deviations at different EtOH concentrations for 10-3 diluted bacteria samples

Time (hour) Control 0.1% EtOH 0.3% EtOH 0.5% EtOH 0.7% EtOH 1% EtOH
0 -0,0110466 -0,0069583 -0,03778333 -0,03321 -0,01102333 -0,01076667
24 0,948564 0,82718067 0,745006333 0,794412 0,688869333 0,486184333
Table 3.1-The results of OD values (600 nm) of E.coli diluted to 10-4 after calculations

Control 0.1% EtOH 0.3% EtOH 0.5% EtOH 0.7% EtOH 1% EtOH
Standard Deviation 0,200969

Determination of Minimum Inhibitory Concentration,
EtOH is known as antimicrobial agent and used. Alcohols, mostly isopropyl alcohols, are often used for inhibitory or killing bacteria. EtOH does that by denaturating proteins. The presence of EtOH shows an inhibitory effect and increasing concentration of EtOH decreases bacterial growth. After a point, 70%, at concentration, EtOH begins to demonstrate killing effect or biocide on bacteria. At low concentrations, isopropyl alcohols, EtOH is an isopropyl alcohol, function as inhibitor for microorganisms. The amount of concentration of alcohol is an important for this effect. In this experiment, effect of EtOH with different concentrations (0.1%, 0.3%, 0.5%, 0.7% and 1%) was indicated.
According to graphs (2.3 and 3.3) and tables (2.1 and 3.1), inhibitory effect of EtOH can be seen obviously at different concentrations. For each data, firstly an average values of test groups, blank groups and control groups according to time and amount of dilution of bacteria. Average values are used for getting O.D values of only bacteria by subtracting. Then each value was placed onto graphs. On these graphs, changes of bacterial growth can be seen easily. However, the information must be confirmed to see clear observations. Calculated O.D values being within 0.05 interval standard deviation can give information rate of growth. Standard deviations (Table 2.2 and 3.2) were calculated to show which data is consistent. According to standard deviation, values or calculated data must be within 0.05 range to confirm consistency of experiment.
When the results of 10-3-experiment with tables (2.1 and 2.2) and graphs (2.3) was observed, they don’t give so clear results is obvious. According to standard deviation table (2.2), the values of only 0.7% and 1% EtOH can give more accurate consequences. The main reason of that may occur because of a mistake of rapid and serial filling wells. On the other, if consistent results are observed, there is a decreasing at growth. General data also with Graph 2.3 reveals that. At 10-4, except control group, values have more consistent. Values are within 0.05 standard deviation range (Table 3.2). For 10-4, the results are more consistent than 10-3. On graph, decreasing and effect of EtOH concentration can be more distinguishable. A decreasing rate can be noticeable easily. For both of them, there is decreasing on growth, but there is no a point for no growing. Or the other word, a MIC (minimum inhibitory concentration) cannot be seen. The growth is slow, but still it continuous. MBC (minimum bacterial concentration) means the lowest concentration of an antibacterial agent that can kill a kind of bacteria. 0.1% of EtOH can kill E.coli bacteria for both bacteria dilutions. 0.1% of EtOH can be MBC for E.coli bacteria because at even this lowest concentration, it can form lethal effect for bacteria.
Disc Diffusion Method,
After observations, there is a difference between 2 antibiotics. According to Table 1.1, effect of ampicillin could be seen that his effect was larger in size. 4mm difference against amount of antibiotics reveals that ampicillin is more effective on E.coli bacterial culture than tetracycline. These values provide that E.coli is quite sensitive against ampicillin. With disc diffusion method, MIC or MBC cannot be determined. Only difference between antibiotics cannot be seen according to kind of bacteria.

[1] Antibiotics (3 May, 2017),
[2] Antibiotic Sensitivity Testing Methods (2 May, 2017)
[3] Types of antibiotics (2 May, 2017)
[4] Norrell A. Stephen, Messley E. Karen, 2003 ,Microbiology Laboratory Manual:Principles and Applications,2nd Edt. page 1667