• Shathurdekah Siwasubramaniam

Download from Moodle and print a picture of your gel (nobody else’s). Attach this to your write-up. If your gel shows clear documentation of each of your DNA preparations you should use this in the subsequent analysis. If (and only if) your experiments have not worked, you should use the specimen gel provided on Moodle. Measure the distance migrated from the origin for each of the size marker fragments in lane 11 and plot a standard curve of the size of the DNA markers against distance migrated. Use this to estimate the sizes (in bp) of your DNA samples;

Plasmid DNA (main band) (lane 1)= 50mm5.0kb5000bp

PCR fragment (lane 2)=85mm1.5b1500bp

Bacterial chromosomal DNA(lane 3) =40mm8kb8000bp

The PCR sample should be represented by a single fragment, why do you think this is?

The PCR sample should be represented by a single fragment, this shows the presence of the DNA fragment only, however if there are more bands present this can illustrate that there are other proteins and contaminates in the sample. Furthermore, the presence of smaller secondary bands can be as a result of where the annealing process has not allowed the primers to attach. It is important to have a single clean fragment for further DNA sequencing in order to produce an accurate outcome.

Moreover, the PCR sample should only show one band since the reaction involves amplification of the low molecular weight piece of DNA. The reaction involves the base pairing of the forward and reverse primers to the target sequence on the DNA. Heat and DNA polymerases are used to extend the annealed primers to make more of the product. This means that there is only one product/fragment present in the sample but there could be more depending on how many times it has been amplified. Since there was no restriction enzyme used in this process; the PCR sample would not have been cut at specific DNA sequences to produce several fragments.

Please attach your graph and the photograph of your gel when you hand the write-up in. You will get NO marks for this section if you forget.

Remember to annotate the picture

Experiment 2. Restriction enzymes and restriction maps

Using the standard curve from above (Experiment 1) measure the distance migrated from the origin for each of the restriction enzyme digests of the plasmid DNA (lanes 5 – 10) and estimate the sizes in bp of the fragments. Record your observations here:

What is the significance of running an uncut sample of the plasmid (lane 4)?

The uncut sample of plasmid (lane 4) is a control and is run alongside with the plasmid samples which have been cut with the several different enzymes. This is an example of positive control which gives the desired outcome given that the experiment was carried out accurately. The experimental results provides the ability for the bacteria to grow on a petri dish containing ampicilin, so the positive control is the bacteria that are known to carry the appropriate resistance marker. On my gel profile there are two bands; one thin band and a thick one. The top band can be illustrating a nicked or relaxed circle whereas the thick band shows the super coiled plasmid. The absence of any other band shows that there were no contaminations. The control allows us to distinguish whether the sample with the enzymes have been cut appropriately. This is essential to allow us to compare the conformations of the uncut plasmid with the cut plasmid samples; the method is also used to verify if the enzyme digestion has been completed. Also, the purity of the DNA can be established by examining the lane of the uncut plasm

As a result, running the uncut plasmid on the agarose gel is important because the uncut sample of the plasmid can run at differing sizes depending on the degree of super coiled they are. To analyse how the DNA travelled depends on how many fragments they have produced .i.e. one or more bands. The control uncut plasmid (with no restriction endonucleases) would have structure that is circular and super coiled therefore it would run faster on the gel (can be slow depending on the buffer used) than the cut plasmid which is linear. Running the control acts as ladder for us to determine the extent to which the plasmid has been cut in the other samples.

Use this information to make a restriction map of the plasmid. To help you we will tell you that the plasmid size is 5500bp and that it is cut once (i.e. it is linear) by HindIII; we have therefore drawn you a circle with the single HindIII cleavage site indicated at position “0” – your aim is to position the BamHI and EcoRI sites on this map relative to the HindIII site as your reference point (i.e. if you think an EcoRI site is 1800bp away from the HindIII site in one direction then mark this on the circle about one-third of the way around the circle and write next to it “1800”). It doesn’t matter which way round you do it, provided you are consistent. One more piece of advice/reassurance; estimates of DNA fragment sizes are approximate, so do not worry if your sizes do not add up exactly to 5500 – an error of 10% is acceptable.

Draw your map here, with estimated co-ordinates:

What sequences are recognised by the restriction enzymes you have used? Refer to your lecture notes or look this up in the library.

EcoRI

HindIII

BamHI

How frequently would you expect each of these enzymes to cut DNA in a typical sample?

Both EcoRI and HindIII cuts the DNA once in a typical sample. This produces 1 linear fragment from a cut plasmid. Since the recognitions site of the EcoRI and HindIII are not directly opposite each other on the DNA molecule means the EcoRI and HindIII leaves single stranded complementary overhang on the new ends. The end has been called a ‘sticky end’ because it can be rejoined to complementary sticky ends. Not all restriction enzymes make sticky ends; some cut the two strands of DNA directly across from one another, resulting blunt ends.

The BamHI will cut the DNA in a typical sample twice. This produces two fragments from a cut plasmid. The enzyme also produces the same ‘sticky ends’. The restriction enzymes commonly used generally recognize specific DNA sequences of 4-6 base pairs in length depending on the enzyme. These recognition sites are referred to as palindromic in that the 5′-to-3′ base sequence on each of the two strands is the same.

Experiment 3: Bacterial Transformation

Count the number of colonies on each of your LB + Ampicillin plates and record this in the table below. If (and only if) your experiments have not worked, you should use the specimen results provided on Moodle. From the data calculate the transformation efficiency. The DNA concentration provided was 100g/ml. You must express the result as the number of colonies you would expect to obtain from 1g of DNA (CFU/g). Note that this is not just the number of colonies multiplied by the dilution factor! If in doubt, go back and work out exactly what you did.

What is the approximate concentration of your plasmid DNA preparation (sample 1)? Explain your working.

I plotted a graph with the numbers of colonies against the known concentrations from samples 1-5. The number of colonies found in the petri dish containing the plasmid DNA preparation was 78, so by using the line of best fit on the graph from the known concentration samples, the corresponding concentration was 10.1g.

Experiment 4. DNA sequencing

Download your sequence data from Moodle. If (and only if) your experiment has not worked, you should use the specimen result provided on Moodle. To open the file you will need “Chromas Lite” software. If your computer does not have this, download the basic free Chromas Lite software from Technelysium Pty Ltd at

www.technelysium.com.au/chromas_lite.html

Click on the file icon from the Moodle page and it will open as a Chromas file. If your samples were clean enough you will see a multi-coloured graph where each base is represented by a peak of one of four colours. The sequence will be presented in a linear form in the 5’ to 3’ direction along the top. Select “copy sequence” and “plain text” from the “edit” drop down menu. Use this as a paste in the following operation:

To establish what you have sequenced, go to http://www.ncbi.nlm.nih.gov/blast/Blast.cgi

This is a database searching tool that will compare your sequence with the millions of entries deposited over the years and will give you the best matches based on a homology search.

When you have opened the home page, choose “nucleotide blast” as the BLAST program to run.

When the screen moves to the next page, paste your sequence from the Chromas Lit .seq file in into the box marked “Enter accession number(s), gi(s), or FASTA sequence(s)”.

Scrolling down the page, look in the “Choose Search Set” section and make sure the database option is set to “other databases” and that the nucleotide collection (nr/nt) box is checked.

Scrolling down the page, look in the “Program Selection” and check the “Optimize for Somewhat similar sequences (blastn)” box.

Click on the “BLAST” box.

Your data will be matched against the databases and in a minute or so you will be given a list of hits. In the middle of the screen will be a colour image that visually shows the best hits from red (highest homology) down to black (lowest homology). What are the top three hits?

1. Homo sapiens heat shock 27kDa protein 1 (HSPB1), mRNA

2. Homo sapiens heat shock 27kDa protein 1, mRNA (cDNA clone MGC:8509 IMAGE:2822325), complete cds

3. Human mRNA for heat shock protein HSP27