BioFreaks Biochemistry Blog (v.Beta)

GGS LIVE - Site Directed Mutagenesis

2011-06-13T01:27:00.019+01:00

Hello BioFreakers!!

Today in the GGS - LIVE section the Site Directed Mutagenesis technique. Wanna see how it is done? Lets roll then.

Method: Site Directed Mutagenesis (SDM).

About: Allows for introduction of point mutations in DNA and thus sometimes in protein sequence.

What: Introduction of STOP codon in order to generate C-terminal deletion in protein X.

Site directed mutagenesis is performed on circular DNA substrate. In our study case, the cDNA of protein X will be inserted into GST vector. Cloning of the protein X was performed analogously to study case described in a different post (click here GGS LIVE - Makinga fusion protein).

To mutate protein X cDNA sequence we will need a set of primers that will be used in PCR reaction (GGS LIVE - Polymerase chain reaction (PCR)). SDM primers overlap with the target sequence, whereas regular PCR primers flank the target sequence (see cartoon below).

The idea behind SDM is that each primer will mutate a single DNA strand of the pPLAS + X cDNA plasmid giving a product of the mutated pPLAS + X cDNA. Naturally, the template will be also present in the final mixture. Tamplate is removed by DpnI, a restriction endonuclease that digest modified DNA (DNA methylation). Such modification of the DNA occurs within bacterial cells, therefore SDM product will not wear it, as it was formed in vitro. After PCR reaction and overnight digestion with DpnI, SDM reaction is analysed by agarose gel electrophoresis (see figure below).

         As you can see from the gel electroporesis analysis, desired product is present in both DpnI treated and untreated samples (R and R+D). Template +/- DpnI was used as a negative control. Additionally, amplification of 0.3kb fragment served as template control.
         You probably, wonder why SDM product and template are not observed as single band. This is due to different conformations of these plasmids. Template is supercoiled, therefore more packed and travels through gel faster and thus appears smaller on the gel. SDM product is not and therefore migrates slower.

         The R+D sample is then used to transform bacterial cells. In this process specific E.coli cells (for example Top10) uptake R+D plasmid and replicate it. This allows obtaining workable amounts of DNA, which is then send for sequencing. Several different E.coli clones are tested and when sequencing confirms mutation of pPLAS + X cDNA, DNA from correct clone is used to transform yet another specifc E.coli strain (for example BL21 pLysS).
We then express X and X-truncation proteins using these BL21 cells (see figure below). For the protein expression in E.coli tutorial, please go here GGS LIVE - Protein expression in E.coli.

As you can see, this way we can generate a mutant protein KA-CHING.

I hope you enjoyed it.

Maciek

GGSTEAM

GGS LIVE - Protein purification from E.coli

2011-06-09T09:36:00.001+01:00

Yo Yo Biofreakers,

Today we are going to have closer look at the recombinant protein purification from bacterial cells.

Method: Purification of recombinant protein.

About: Having an optimised protocol for protein purification is an essential tool to study properties of the protein of interest.

What: Purification of GST-tagged chicken protein previously expressed in E.coli.

Before we start we should first have a look at what the GST tag is. GST (glutathione S-transferase) is an enzyme that transfer glutathione (GSH) via slufhydryl group to differernt type of substrates (lipids, xenobiotics). GST has a high affinity towards its substrate glutathione and this property of GST is utilised to purify GST-tagged proteins. In order to recover GST-tagged protein from the complicated mixture of proteins, the fusion protein is incubated with agarose beads coupled to glutathione (GSH-agarose, see picture below), what leads to efficient precipitation of GST-fusion protein.

Ok Vamos!! Expression of the GST-X protein was previously demonstrated in different post GGS LIVE - Protein Expression in E.coli.

After a succesfull expression of the protein of interest in E.coli, we can purify it using one of the many protocols available for GST protein fusion purification. In general such protocol consists of three major steps: cell lysis and solubilisation of the GST-fusion protein (see cartoon below), recovery of the fusion from the lysate and elution of the GST fusion.

As you can see on the cartoon above, first GST-X fusion is expressed in large amount (for example 250 ml to 1 l culture) under previously optimised conditions (for protein expression see post GGS LIVE - Protein expression in E.coli). After protein expression, E.coli cells are harvested by centrifugation, superatant is removed and cell pellet is resuspended in lysis buffer of choice. Usually such buffer should have pH of around 7.0 - 8.0 to facilitate efficient interaction between GST and its substrate glutathione, protease inhibitors (such as PMSF) to prevent protein degradation. Additionally, lysis buffer should contain component taht will help lyse the cells, such as lysosyme (enzyme that degrades bacterial cell wall) or detergent (which disrupts bacterial cell wall). Cells are usually lysed at 4C rocking or mildly shaking what increases lysis efficiency. Cell lysate is then sonicated to share bacterial DNA (DNA makes lysate viscous and hard to work with) and help to solubilise proteins by breaking up protein aggregates. In the next step, cell debri is removed by high speed centrifugation. And there we go we have a lysate ready for protein purification.

As mentioned earlier, in order to recover our GST-X protein we have to mix our lysate containg the fusion with glutathione agarose beads. First beads have to be prepared (see cartoon below).

Glutathione agarose beads are first washed with the lysis buffer in order to remove storage solution (usually ethanol, which can impede binding of GST to glutathione). Then beads are mixed with lysate containing GST-X fusion and incubated at 4C in order to bind GST fusion to the beads. After binding step, beads have to be washed in order to remove unbound GST-X fusion and unspecifically bound proteins.

At this stage of purification GST-X fusion should be clean and depending on the nature of furhter experiments that we want to perform, we can either elute the fusion with glutathione (excess of the glutathione will compete and displace GST-X protein from the beads) or cleave the GST tag and release protein X (see cartoon below).

When purification is finished we can analyse our experiment by separating protein sample taken at each step of the purification by SDS-PAGE and stain proteins in gel with Coomasie dye. The results from GST-X purification are shown on the picture below.

As you can see from the Coomasie stained gel, the expression of GST-X fusion was nicely induced (UI and I samples). We can also confirm that the GST-X was present in the starting material (lysate IN sample). After incubation of the lysate with beads, most of the GST-X bound to the resin what resulted in depletion of GST-X, as observed in unbound sample (UN). After beads wash, a single band of GST-X was detected on the beads, indicating high purifty of this sample. Elution of the GST-X with glutathione recovered fusion protein from beads. Additionally, alternative elution by GST cleavage resulted in appeareance of two bands: a free X protein and GST tag.

Hopefully, you got the picture how protein purification can be performed using a GST tag as a bait.

I hope u enjoyed it.

Cu SOON!

Maciek

GGSTEAM

GGS LIVE - Protein expression in E.coli

2011-02-14T20:28:00.003Z

Dear BioFreaker,

today in the GGS LIVE section we are going to look at the recombinant protein expression in E.coli.

Method: Protein expression.

About: Protein expression in E.coli is a widely applied technique, which allows for quick and robust production of the particular protein. Such protein can be then used in different applications, such as antibody production, in vitro enzymatic and binding assays, crystallisation etc.

What: Expression of the GST-tagged chicken protein in E.coli.

          Ok, lets start. If we want to express any protein in the bacteria we have to first clone it (for more info about cloning please visit this post GGS LIVE cloning) into a plasmid which contains specific elements allowing for protein production within the prokaryotic bacteria cells. These plasmids utilise technology based on application of lac operon system (click here for more information about the lac operon system). We will not cover how this system works in this post, we will just go straight to the results:). The only thing you should know at this stage is that protein expression from the lac operon system is achieved by addition of IPTG compound. The IPTG turns on the protein production, which cannot happen without it.
            If a technology for protein X purification is established then expression of such can be performed using native protein sequence. If such protocol is not available we can express our protein as a fusion. The tag which will be added to the protein will later allow for its rapid and robust purification. There are many different tags available for various applications, such as His, GST, MBP, CBP, S-tag, FLAG, Strep and many more. In our study case we will work with GST-tagged protein. If you would like to know how to tag a protein of interest, please look at this post GGS LIVE - Making a fusion protein.
          After we cloned our DNA sequence of interest into a N-terminal GST tag containing vector, we have to transform E.coli strains with this plasmid and then find optimal conditions for expression of our fusion protein. This step has to be performed experimentally. The culture of bacterial cells is split into many batches where protein expression is performed at different conditions, such as IPTG concentration, temperature, time, different media composition etc. At the end of the experiment samples are separated by gel electrophopresis and proteins visualised by staining of the gel. Such gel staining, where protein expression at different temperatures was tested, is shown on the picture below:

Picture taken by Kliszczak M.

I have to mention that our protein of interest has size of 27 kDa and the GST tag is similarly big (28 kDa), what gives size of 55 kDa for the fusion. As you can observe on the gel above, increase of temperature during expression positively affects production of GST-X protein by E.coli cells. The 37C seem to be the best temperature for expression of fusion protein. In addition you can observe that GST-X has the predicted size of 55 kDa. After, establishment of perfect conditions, these can be used for production of our protein.

I hope you enjoyed. In the next post we will cover the purification of the GST fusion protein from E.coli.

Maciek GGS TEAM

GGS LIVE - Polymerase chain reaction (PCR)

2011-01-26T19:14:00.001Z

Today Biofreakers PCR!!

Method: Polymerase chain reaction (PCR).

About: PCR allows for amplification of a particular DNA sequence.

What: Ampliifcation of a gene X sequence from genomic DNA.

Ok lets start. Polymerase chain reaction is a commonly used technique. The theory behind it as well aas practical part of PCR are very simple. For PCR reaction we need:
- thermostabile polymerase - which can stand temperature up to 95 degrees Celsius,
- primers - short DNA oligonucleotides that determine DNA sequence to be amplified,
- DNA template - contains target DNA,
- buffer - supply perfect conditions for polymerase activity.

As I mentioned before theory of PCR reaction is very simple. Cartoon below represents major steps in PCR protocol:

The steps 2 - 4 are repeated between 15 - 30 times in order to obtain a workable amount of the DNA product. Time of the PCR experiment depends on the product size. After the PCR reaction small amount of the reaction mix (approximatelly 5-10%) is analysed by the agarose gel electrophoresis. Example of the gel is shown on the picture below:

As you can observe we did amplify two different DNA sequences. The positive control reaction was set up with primers that previously amplified the desired sequence from the same DNA template (product size 2.5 kb). As expected there is no signal in the negative control lane as negative control reaction did not contain the DNA polymerase. The band in the P lane (the experiment PCR reaction) need to have an expected size. Our gene of interest is 3.9 kb long.

As you can see this way we can very easily ampllify a DNA sequence of interest in order to study its sequence or cellular function.

I hope you enjoyed.

Maciek

GGSTEAM

GGS LIVE - Foci kinetics

2010-12-04T01:49:00.002Z

Yo, Yo, Yo Biofreak readers!!

Method: Foci kinetics.

About: Foci kinetics assay is performed in order to assess foci formation and resolution of particular protein under specific conditions.

What: Kinetics of gamma-H2AX foci formation after ionizing radiation in wild-type and mutant cells.

In our study case experiment we will investigate gamma-H2AX (histone modification) foci formation and resolution after treatment with ionizing radiation. Ionizing radiation cause DNA double strand breaks.
In response to such DNA damage H2AX histone is phosphorylated (called gamma-H2AX in phosphorylated state) to facilitate double strand break repair.
Briefly, DT40 cells were treated with IR (5 Gy), harvested and analysed by immunoflurescence at different times post-IR treatment (for immunofluorescence tutorial visit GGS LIVE - immunofluorescence). Pictures below represent cells stained for gamma-H2AX harvested at different times post-IR treatment.

As you see on the images above there is not much of signal from gamma-H2AX in untreated cells (there is small number of spontaneous DNA damage in unchallanged cells - red arrow heads indicate representative gamma-H2AX foci). After induction of double strand breaks with IR treatment H2AX histone is robustly phosphorylated (15 min timepoint) and this modification is removed with ongoing DNA repair. At this stage we have to quantify our results. We can eitehr simply score number of the foci per cell or score number of cells with more than X foci of gamma-H2AX. Plot below represents such quantification in which cells with 6 or more gamma-H2AX foci were scored as positive.

From the quantification plot you can see that both cell lines induce gamma-H2AX foci fomration with the same kinetics, indicating that this process is not affected in the mutant cell line. When we look at gamma-H2AX foci resolution, four hours post-IR treatment 50% of wild-type cells has resolved gamm-H2AX foci, where mutant cells need another 4 hours to accomplish the same task. Such results indicate that mutant cell line might have problems with repairing DNA damage caused by IR treatment.

I hope you enjoyed:)

Maciek

GGS TEAM

GGS LIVE - Immunofluorescence

2010-11-26T11:42:00.001Z

Yo Biofreakers:)

Today in the GGS LIVE section we are going to look at the immunofluorescence staining technique.

Method: Immunofluorescence.

About: Immunofluorecence is a technique that allows for detection of a particular protein within the cell.

What: Staining for gamma-H2AX and Rad51 proteins in chicken DT40 cells after ionizing radiation treatment.

Immunofluorescence technique is perfect for studying protein localization. Staining procedure is not compilcated and a representative protocol is shown on the cartoon below:

Briefly, DT40 suspension cells are harvested by centrifugation. Media is discarded and cell pellet is resuspended in 1 x PBS (phosphate buffered saline). Cells are adhered to poly-L-lysine slide, fixed and permeablised to allow access of molecules used in subsequent steps. Next, slide is incubated in blocking solution (usually a neutral protein, like 1 % bovine serum albumin solution). Blocking protein binds to sticky spots on slide, cells and within cells in order to decrease unspecific binding of the primiray and secondary antibodies (reduce unwanted background signal). After blocking, slides are incubated with primary antibodies that recognises the protein of interest. Notice that primary antibody has higher affinity to protein of interest and displace blocking protein. Slides are then washed to remove unspecifically bound antibodies. Later, secondary antibodies conjugated to a fluorophore are added and again washes are performed to remove unspecifically bound antibodies. Complex protein of interest - primary antibody - secondary antibody is formed and protein is ready to be detected and analysed by microscopy.
Lets have a look at our results.

As you can see in our study case experiment, cells treated with ionizing radiation were stained against two different proteins. Gamma-H2AX is a phosphorylated form of H2AX histone. H2AX histone is phosphorylated in proximity of DNA double strand break, appears as fast as 1-2 min after radiation and persists few hours post-treatment. Rad51 is a DNA repair protein and it accumulates at site of damage 2-4 hours after IR treatment. As you can see from our results both gamm-H2AX and Rad51 proteins form foci within nucleus (representative foci are indicated with yellow arrowheads). Both proteins nicely colocalise as seen on the overlay picture. With such immunofluorescence staining we can follow foci formation, resolution, change in protein localizatio or etc.

I hope you enjoyed:)

Maciek

GGS TEAM

GGS LIVE - Cell fractionation

2010-11-23T21:34:00.001Z

Method: Cellular fractionation.

About: Cellular fractionation is a technique that allows analysing protein localization to different cell compartments.

What: Comparison of protein X and Y localization in wild-type and mutant cell lines.

There are many protocols for cellular fractionation. Selection of a particular fractionation protocol may depend on cell type used, required fractionation resolution or protein analysed. In our study case we will look at a very basic fractionation protocol for mammallian cells. Cartoon below represents majors steps of the protocol:

Briefly, cells are incubated in hypothonic buffer to facilitate cell lysis. Cell membrane of swollen cells is mechanically disrupted by action of dounce homogeniser. At this stage cellular organelles are spun and supernatant containing cytoplasmic proteins is saved. Nuclei are incubated (usually with rocking or rotating) in high salt and detergent containing buffer to extract nuclear proteins. Again cell remaining are spun and supernatant containing nuclear proteins is saved. In the last step DNA bound proteins are extracted either by sonication (DNA sharing) or DNAase treatment (DNA digestion). After solubilization of DNA bound proteins extracts are spun and supernatant containing DNA bound proteins saved. All obtained samples can be now analysed by a method of choice. In our study case we will analyse our samples by Western blotting. Equal volume of each sample is separeted by SDS-PAGE to retaing similar loading and proteins are detected by Western blotting (for Western Blot tutorial click GGS LIVE Western Blotting). Ok, lets have a look at our results. In our experiment we have fractionated wild-type and mutant cell lines. To be sure that protocol worked perfectly, we have to first look at proteins that are known constituents of each cellular fraction. Cytoplasmic protein beta-actin, DNA binding protein Histone H3 and nuclear protein nucleophosmin are only present in cytoplasm, chromatin and nuclear fractions, respectively. Additionally all control proteins are present in Whole Cell Lysate sample. This is an additionall control which tells us if our protein of interest was present in the starting material.

As you can see analysis of protein X and Y reveals their distinct localization between wild-type and mutant cell line. When Protein X is exclusively cytoplasmic within wild-type cell line, it is also present in the nucleus of the mutant cell line. On the other hand protein Y is nuclear and partially bound to DNA in wild-type but its exclusively cytoplasmic in mutant cell line.

I hope you enjoyed it.

Maciek

GGS TEAM

GGS LIVE - Making a fusion protein

2010-11-17T21:03:00.003Z

Yo BioFreakers,

today in the GGS LIVE section we will learn how to virtually design and generate a fusion protein.

Method: protein tagging.

About: protein tagging allows to perform specific experimenents that are not possible with the endogenous protein (wild-type protein).

What: desing and generation of protein X fusion with a GFP (green fluorescent protein) tag.

There are many different tags available for protein tagging. Choosing the tag depends on the experiments that we want to perfrorm with the fusion protein. So, if we want to:
- purify the protein of interest, we would use MBP (maltose binding protein), GST (gluthatione S-transferase), FLAG or His (hexahisitidine) tags,
- easily detect our protein, we would use epitope tags like myc, V5 or HA,
- follow cellular localization of the protein, we would use fluorescent tag like GFP.
The most important is to remember that fusion protein might behave differently in vivo than wild-type protein (folidng, solubility or activity, etc may change) so it is crucial to check if the fusion is functional before we start our experiments.

Ok lets start with design of the fusion protein. In our study case we will use a sequence of a protein X (shown below) and a pEGFP-C1 and -N1 plasmids for N and C terminal tagging, respectively.

The red triplets are start and stop codons, respectively. We are also going to need information about multi cloning site (MCS) sequence of the pEGFPN1 and -C1 vectors, where DNA sequence of protein X will be inserted. Information about those is shown below.

As you see there are many restriction sites available in both plasmids. Our aim here is to find a restriction enzymes that will cut in MCS of our vectors but not in the sequence of protein X (we can do that with any cloning software, like free pDRAW32 which you can download from here). In our study case two enzymes XhoI and EcoRI are cuttining in the MCSs but not in sequence of protein X. What we have to do now is to virtually introduce XhoI and EcoRI sequences at 5' and 3' end, respectively (sequences recognised by XhoI and EcoRI endonucleases are available here XhoI and EcoRI). There is one more issue to look at before we are going to add our restriction sites (see picture below).

It is important to remove start codon of the protein when tagging it on the N-terminus to avoid an expression of untagged form of the protein. You have to remember that promoter will drive expression of any open reading frame that is downstream of it. It is also crucial to remove stop codon when we tagging protein on the C-terminus to prevent premature termination and allow expression of a fusion protein. Including above information we have:

Now we put this sequence into our plasmid and we get this:

Lets have a look at our constructs now. We are going to focus on the reading frame at the moment. In the case of N-terminal tagging our reading frame is determined by the two last codons of GFP tag. To get the right fusion DNA sequence of protein X has to be in the reading frame with the GFP tag. The reading frame is indicated by the horizontal brackets (each triplet codes for one amino acid). As you can see stop codon of the protein X (indicated with the red colour) is not in reading frame with GFP tag. To shift the reading frame we have to add two extra nuclotides to our protein X just between the XhoI restriction site and the coding sequence of protein X. Similar situation takes place with the C-terminal tagging. You can see that now start codon is not in frame with GFP tag and addition of a single nuclotide between EcoRI restriction site and protein sequence will rescue that problem (see the pictures below).

It is important to remember that addition of extra nuclotides to our sequence may result in introduction of the stop codon. We have to check our sequence before we proceed further. If everything is ready we should obtain this:

As you can see now, protein X sequence is in the frame with the GFP tag in both cases. Now the protein sequence including the XhoI / EcoRI restriction sites and extra nucleotides can be used to design primers for cloning of the protein X DNA. Such DNA then will be sequenced to check for potential mutations and if correct subcloned into pEGFPN1 or -C1 plasmids. Such construct can be later used for expression and localization studies of the protein X.

I hope u enjoyed it.

Maciek

GGS TEAM

GGS LIVE - Immuno and co-immunoprecipitation

2010-11-05T12:38:00.004Z

Whatsup you all?

GGS LIVE sections presents today an immunoprecipitation method:)

Method: Immunoprecipitation.

About: Protein precipitation using specific antibodies is a technique that allows for enrichment of a particular protein and components associated with this protein (it is a type of protein purification).

What: Immunoprecipitation using anti-myc antibody from DT40 cells stably expressing 3myc tagged protein.

This technique can be used to purify the protein of interest in order to use it in a specific assay or to identify its potential interacting partners. For example an enzyme protein kinase can be immunoprecipitated and subsequently used in the kinase assay. The cartoon below show immunoprecipitation steps.

Briefly, cell lysate is prepared by cell lysis in IP buffer of choice (the composition of the buffer might be different from protein to protein analysed and may have to be found empirically). Cell lysate is also sonicated to share the DNA that may interfere with the immunoprecipitation process and also to solubilize proteins. Next, cell membrane and debri is spun down and protein concentration is determined. On this stage the lysate is ready to go. We also should prepare our beads and antibody. The Protein A or G beads are washed three times with lysis buffer (aka IP buffer) to remove beads storage solution (usually ethanol which can interfere with the precedure). Washed beads are mixed with antibodies. After coupling, beads are mixed with previously prepared cell lysate and this is our immunoprecipitation step. After precipitation beads + antibodies + protein of interest + partner proteins are washed with lysis buffer (or other buffers) to remove unspecifically bound material. Finally, precipitated proteins are eluted and analysed by Western blotting (for the Western blotting tutorial visit this post GGS LIVE - Western Blotting). Ok lets see the results now...

On the top panel we see bands representing the protein against which the antibody was used (immunoprecipitated protein). Bottom panel shows co-precipitated protein (a partner protein of the myc-tagged protein of interest). The lanes are: input lane, representing starting material, unbound lane showing material that left after the immunoprecipitation and finally the IP lane which represents the precipitated proteins fraction. The control IP has to be performed so we are sure that the protein we are enriching for is specifically pulled down. One can imagine that protein of interest might bind to other part of the antibody (that is common for anti-myc or control antibody) or bind unspecifcally for example to the beads. Simply saying the signal form our protein should be present only in the experiment IP but not in the control IP. As you can see this is correct for our myc tagged protein which is present in the starting material, it is depleted in unbound fraction and there is an obvious enrichment of it on the beads (IP lane). In the case of the control IP again it is present in the input lane but it is absent in the IP lane (which is perfect:).

The same samples were analysed for the presence of a partner protein that is known to interact with our myc-tagged target. You can clearly see that the interacting partner is also present in all the samples beside the control IP what suggests a that it is interacting with the myc-tagged protein. Of course novel interactions have to be confirmed with reciprocal experiment.

I hope u enjoyed and understand that immunoprecipitation is a very powerful cell biology technique:)

Maciek

GGS TEAM

GGS LIVE - How to measure DNA synthesis in vivo?

2010-11-02T23:21:00.001Z

Hello Biofreakers,

Today in the GGS LIVE section we are going to look at techniques that allow to observe cells that replicate DNA (are in the S-phase of the cell cycle).

Method: Labelling and detection of replicating (aka. S-phase) cells.

About: Both methods that will be discussed (flow cythometry and immunofluorescence) use specific derivative of nucleotide that is incorporated into DNA during the S-phase of cell cycle. Later this modified nuclotide can be detected and S-phase cells analysed.

What: Labelling of chicken DT40 S-phase cells.

One of the most common nucleotide derivative used in cell biology is a bromodeoxyuridine (BrdU, for structure and other applications go here Meet the chickens - sister chromatid labelling). BrdU pulse will result in its incorporation into cells DNA. In our case study such cells will be analysed by flow cytometry and immunofluorescence. This techniques are very different but sample preparation is very similar, see picture below.

And this one:)

Similarly like in the Western and Southern Blotting, antibody used here is conjugated to a moiety that gives a specific signal, in this case antibody is conjugated to a green fluorophore called FITC. The signal from fluorophore can be detected in many different ways. In our study case, cells designated for flow cytometry will be analysed with flow cytometer:) and ones on the slide will be analysed with fluorescent microscope. Ok lets check the results of our experiment.

Lets look first at the panel A where S-phase labelled cells are analysed by flow cytometry (for those that are not familiar with flow cytometry, please visit this post GGS LIVE - flow cytometry). The first histogram in the panel A represents a DNA content profile. On the x-axis we have a DNA content. The first big peak are cells in G1 phase of the cell cyle with 2N DNA content, second smaller peak are cells in G2/M phase of cell cycle with 4N DNA content and finally all cells between those two peaks are cells in S-phase with DNA content between 2N-4N. On the y-axis we have counts (the BrdU incporporation is only relevant for the other two dot plots). So you can easily see that the most number of the cells is in the G1 phase of the cell cycle followed by S-phase and G2/M cells. This histogram is often called one dimensional cell cycle plot as it measure only DNA content. The other two dot plots are often called two dimensional as we measure DNA content and BrdU incorporation. What you can see on the first dot plot is exatly the same as on the first histogram but represented in a slightly different way. Now, there is no peaks but each dot represents a single cell. Similarly the first bunch of dots (cells:) is population of G1 phase cells and the other bunch is population of G2/M cells. Everyhting between them are S-phase cells. Look know on the last dot plot. The situation is similar but these cells were pulsed with BrdU and what you can see know is that S-phase cells did shiftet to upper values on the y-axis (which is a measure of the fluorescence that is directly proportional to the antibody that recognise BrdU). This picture is often called a horseshoe.

On the panel B we can see pictures of cells. On the first panel we see DNA that was labelled with DAPI (blue flurophore), in the middle we see a replication foci that were lebelled with BrdU and detected with antibodies. On the last picture we see an overaly of the two, which clearly indicates that replication foci are localised to DNA (what makes sense:).

Both techniques are very useful in analysis of replicating cells. The flow cytometry allows for a very robust and quick analysis of S-phase cells (be aware that flow cytometry allows for quantification of the cells in each phase of cell cycle). We can use this technique to monitor S-phase cells for example after treatment with different drugs. Immunofluorecence method does not have power of numbers but allows detecting of for example other proteins colocalization to replication foci (if protein X will be labelled red we can see if it colocalizes with replication foci). Both techniques are very powerful and are commonly used in each cell biology laboratory.

I hope u enjoyed:)

Maciek

GGS TEAM

GGS LIVE - Counting cells

2010-09-14T12:52:00.001+01:00

Hello BioFreaks Reader,

Today in GGS LIVE section we are having some basic Tissue Culture technique:

Method: Determination of cell number or cell density in the culture.

About: Determination of cell number is essential in performing experiments where appropriate number of cells is required in order to sustain uniform conditions during the experiment.

What: Determination of cell number in suspension cell culture.

The most common technique to determine cell number in culture is to count single cells using a microscope and a hemacytometer (see picture below).

To count suspension cells we need to first mixed them well to homogenise the entire culture. After mixing 10microliteres is taken and loaded onto a hemocytometer as shown on pictures below.

To count cells using hemocytometer we need to hava a closer look at it. Under magnification smooth area of hemocytometer looks like this:

The lilquid that occupy area of a centre square (here indicated with red border) which contains 25 smaller squares has volume of 0.1microliter. After loading cell onto hemocytometer we should see something like this:

The red boarder again indicates the central square that we are going to score. Additionally red arrowheads inducate three represenatative cells. We should count cells at least twice and take an average of two as the final result. For example: count one 93 cells
                                      count two 87 cells
                                             total 180 cells
                                        average 90 cells

To get a number of cells in a 1ml of culture we need to multiply our result by 10 000. Why? because volume scored is 0.1microliter and that is 10 000 less tham 1ml, so in 1ml there is 10 000 more cells than in the volume we had investigated (if you have problem with prefixes and calculations visit this post Scientific prefixes - you will be laughing). So in our study case we have 90 x 10 000 = 900 000 cells in 1ml.

I hope you enjoyed it.

Maciek GGSTEAM

GGS LIVE - Clonogenic Survival Assay

2010-09-13T23:14:00.000+01:00

Yo Nerds,

Method: Clonogenic Survival Assay (CSA).

About: Clonogenic survival assay is a long term cell viability assay in which ability of a single cell to form a colony (proliferation capacity) is scored.

What: Clonogenic survival assay after DNA damage induced by Drug X using two different cell lines: wild-type and DNA repair defficient mutant.

Clonogenic survival assay is performed differently for suspension and adherent cells. As I am working with suspension cell at the moment, I will explain only how to perform CSA for suspension cells.

Around 5 x 10^5 cells is enough to perform clonogenic survival assay with four different doses of Drug X.
If we perform clonogenic survival assay for the first time using a mutant cell line (or using a particular drug for the first time) it is important to plate different number of cells for each dose as it is hard to predict sensitivity, for example:

It is also important to seed cells in triplicate. This is because we want to base our result on a three different numbers what makes it more accurate (not on a single number but on avarage of three). When we have our cells and numbers ready we have to prepare plates. Different plates can be used for this purpose but in our lab we are using the triple vent petri dishes (see on the picture further on).

Special media has to be used in the case of suspension cells, in our lab we are using a methylcellulose based media. Methylcellulose is added to the media as a thickener so our "mobile" suspension cells do not move after we plate them. This prevents a single cell to give rise to multiple colonies. You can imagine that if media is not thick colony could split resulting in extra number of colonies at the end of the experiment. Once the media is placed in dishes we can start plating cells. For that purpose a dilution method is used, as show on the picture below:

We prepare our cells so we have a 10^5 cells per ml and we place them in 24 well plate (picture above).
We treat this cells with Drug X at doses 0, 2, 4 and 8 mg/ml (in the 0 dose we treat cells with the solvent used to dissolve the Drug X only). We incubate the cells at 37 degrees Celsius for specific time and after that time we dilute cells twice each time by factor of 10 (as on the picture above). This way we obtain three cell solutions in which we have 100 000, 10 000 and 1000 cells in total per ml. This means that if we use 100ul from each solution we will plate 10 000, 1000 and 100 cells onto dish, respectively. After the cells are plated we incubate them for 7-14 days depending on how fast cells grow and give visible colonies (see picture below).

Picture taken by Kliszczak M.

Three representative colonies are indicated with red arrowheads. We count colonies and then we calculate the survival according to the table shown below:

After the calculations we plot the % of relative survival against the doses of Drug X and we obtain a plot like on the picture below:

From this experiment we can conclude that mutant cell line is more sensitive to Drug X than wild-type cells and that this sensitivity is dose dependent. We have to repeat this experiment at least three times to obtain reliable result. Remember that all experimental steps (cell number, drug doses, time of cell treatment) has to be optimized and each time performed in the same way to minimize the errors.

I hope u enjoyed it:)

Maciek GGSTEAM

GGS LIVE - Flow cytometry

2010-05-10T22:28:00.000+01:00

Yo all,

welcome again in the Biochemistry Method section. Today we will cover a Flow Cytometry Technique.

Method: Flow cytometry

About: simply saying flow cytometry is a technique where properties of microscopic particles (such as cells or chromosomes) is examined. Flow cytometry is a very powerful method because it can measure many particles in the same time (up to thousands). Additionally, flow cytometry analysis is multiparametric, so different properties of particles can be measured at once.

What: Analysis of cell cycles distribution in different populations of cells.

How: By flow cytometry.

Ok, so let's start with some short background on how the flow cytometer looks like (see picture below)

and how it works ... (what is inside? :) picture below)

Of course this is a simplified scheme:). What you can see on the scheme is:
- a tube with your sample (in our case cells are the particles examined),
- sample collector,
- source of liquids,
- mirrors,
- laser,
- detectors,
- and waster container.

To simplify how the flow cytometer works we can say that particles are sucked from tube, enter machine flow, where they are exposed to laser and finish in waste container. Depending on particles they either adsorb or alter path of the light. This is monitored by different detectors installed in the flow cytometer. Two most important are:
- forward scatter detector - which is placed in line with the light beam and gives us information about the size of cells (particles),
- side scatter - which is placed perpendicularly to light beam and gives us information about surface of cells (particles) for example their roughness etc.
Additional detectors are installed in the flow cytometer which are able to detect other features of cells/particles (for example they can measure fluorescnce of a chemical compund bound to cell membrane giving us idea how many cells do contain such modified membrane).

In our case study we will look at cell DNA content. Ok let's start. To perform flow cytometry we need to first prepare cells. Our experimental design is as follow:
- one control sample (untreated cells, also called unsynchronous population),
- and three treated samples:

nocodazole treated cells (nocodazole is microtubule depolymerizing agent causing cells to stall at the G2/M boarded). For the cell cycle tutorial please visit this post Theory is fundamental - Cell cycle.
hydroxy urea treated cells (HU- hydroxy urea ribonucleotide reductase inhibitor. Inhibition of this enzyme leads to depletion of DNA synthesis substrates - deoxyribonucleotides). This drug stalls cells in or before S-phase.
Drug X treated cells.

First cells are treated with drugs for a specif time or left untreated. After incubation cells have to be fixed. Usually ethanol is used for that purpose but that may be different from protocol to protocol. After fixation all samples are treated with DNA intercalator Propidium Iodide (PI), a red fluorescent dye that stain DNA (see picture below).

After short treatment (when complex between DNA and dye is formed) cells are ready for analysis. So, we simply kick in machine, install our samples and run analysis. The control result from our experiment is shown on picture below.

The left-hand plot is a representation of our population (below you can see it as indicated with a red area), where each spot is a single event (cell or particle). The X-axis is a forward scatter (FSC), which if you remember tells us about the size of particle. If we move along it we go from the smallest cells (G1 phase cells), through S-phase, to the biggest G2 and mitotic cells. The Y-axis is a side scatter (SSC) and if you remeber it tells us about the sufrace of particle/cell. As you can imagine cells in G1 phase have different morphology than cells in S-, G2 or M-phase (that's why they have different position on Y-axis). We can alter position of our population on the dot plot using specific parameters but usually we try to place it in diagonal of dot plot box and between value 200 and 400 on X-axis.
If you have a look at the histogram plot now, you can see that on X-axis we have FL2 parameter (which is actually a fluorescence of our DNA dye) and on Y-axis counts. This plot simply tells us how many cells contains how much of fluorescence. You can imagine that cell in G1 phase, where there is a single copy of DNA, will have smaller fluorescence than a G2 cell which contains doubled amount of DNA. G1 peak is placed at 200 and G2 peak at 400. Everything in between is S-phase cells which contain amount of DNA between 1n and 2n. From this you can see that the most number of cells are in G1 phase then in S- and in G2-phase.

Now let's have a look at flow cytometry results of our treated cells.

We have already covered control experiment, so we start from Nocodazole treated sample.
Nocodazole stops cells at G2/M boarded. You can easily see that dot plot and histogram plot shifted to the right. The block of cells might not be so clear at dot plot but from histogram we can tell that all cells have been stalled in G2 phase (peak at 400). There is no G1-peak or S-phase area.
Hydroxyurea which prevents cells to enter S-phase, decreasaed number of S-phase and G2/M cells. G1 peak is now fatter what suggest that most of cells is stalled in G1 phase. This block is not nice as Nocodazole one but I hope you see difference between hydroxyurea treated and control cells.
Drug X treatment is toxic to cells. We can deffinitely say that it dimnishes S-phase cells and block them in G1. Additionally you can see a small sub-G1 peak which is actually a indication of apoptotic cells (during apoptosis- a programmed cell death, DNA of cell is fragmented and distributed to apoptotic bodies. This is why it appears as smaller, less than 1n).

This is it:) I hope you enjoy it.

Maciek GGSTEAM

Theory is fundamental - Cell cycle

2010-05-09T17:10:00.002+01:00

Hello BioFreak Reader,

Welcome in a brand new section where we will try to shortly and nicely cover fundamental topics of biochemistry. We are going to start with a Cycle cycle:)

What is cell cycle - (also known as cell division) is a set of events which main objective is to duplicate genomic material (DNA) equaly distribut it to two doughter cell. Four distinct phases can be distinguished during each cell cycle (see picture below). Those are G1, G2 (known as gap phases), S- and M-phases. S-phase (S stands for synthesis) occurs between gap phases and in this phase DNA duplication occurs. M-phase (M stands for mitosis) is the last phase of cell cycle, where cell finaly divides giving live to new cells. Each phase of cell cycle is extremely important and all phase specific events had to be finished before cell enters proceeds to the next one.

Maciek GGSTEAM

GGS LIVE - broken chromosomes

2010-05-09T13:49:00.001+01:00

Welcome you all,

today in Biochemistry Methods section we discuss how to assess chromosomal stability. Chromosome aberration is one of the methods.

Method: Chromosome Aberrations.

About: Chromosome aberration assay is a direct method to assess chromosomal abnormalities.

How: By visual examination of chromosomes.

To examine chromosomal stability of chicken DT40 cell line we first need to prepare chromosomal spreads (please have a quick look at this post Meet the chickens - chook chromosomes at hand). To prepare such chromosome spreads, we have to capture them in metaphase (subphase of cell division - mitosis). Usually drug called colcemid is used for that (look at the picture below).

Colcemid is a drug that interfere with formation of mitotic spindle by depolymerization of microtubules. This way we enrich population of cells that already condensed their DNA, which now has a form of chromosomes.
Later cells are harvested and incubated in hypothonic buffer (what cause their swelling) to make them more fragile. Next step is to fix them in such state and drop them onto slide. Cells burst/break when they are dropped, releasing chromosomes (as on the picture below or in this post Meet the chickens - chook chromosomes at hand).

Picture taken by Kliszczak M.

As you can see, breaks in chromosomes can be easily seen and scored (red arrow heads). Usually cells are left untraeted (to score spontaneous aberrations) and treated with ionizing radiation (IR) or with drug of interest. Chromosomal aberrations are then scored and compared to untreated sample.

I hope you enjoy it.

Maciek GGSTEAM

Meet the chickens - sister chromatid labelling

2010-05-09T10:08:00.002+01:00

Yo all,

welcome again in Meet the chickens section. Today a specific labelling of chromosomes (actually sister chromatids) will be presented.

As you probably remember from previous post (Meet the chickens - chook chromosomes at hand), there is a way to examine karyotype (set of chromosomes) of chicken DT40 cell line (or any other cell line:). In this post we will combine this technique with a labelling procedure. This protocol will give us oportunity to see differential staining of sister chromatids.

Ok lets go.

In this protocol a nucleotide derivative, called BrdU (5-bromo deoxyuridine) is used to labell DNA for two cell cycles (see the picture below).

Simply saying, cells are treated with BrdU, which is incorporated to DNA during S-phase (DNA synthesis phase of cell cycle). It is important that labelling is performed for exactly two cell cycles. This is because after two cell cycles (two subsequent S-phases) one of sister chromatids is complately substituted with BrdU where other one is half substituted half not. This feature of labelling can be now used to distinguish between sister chromatids (they simply have different physical and chemical properties).

First cells are stopped in metaphase (cell division phase) using a specific drug (usually colcemid a microtubule depolymerysing agent) that prevents mitotic spindle formation. After that, cells are swollen (to make them fragile) and fixed (to fix their state:). Further, cells are droped onto slide to open them and release chromosome spreads. Next few steps leads to differential staining of sister chromatids (see picture below).

In the first step, cells are treated with a DNA intercalating dye called Hoechst 33258 (this is a fluorophore that adsorbs UVC light at 258nm). Intercalating means that it gets in between the DNA base pairs stacks (see picture below)

Red bars represent an intercalating agent bound to DNA/

Picture taken from http://en.wikipedia.org/wiki/File:DNA_intercalation.jpeg

As you remember one of the chromatids is complately BrdU substituted. BrdU contains bromide and occupy more space than a regular nucleotide (remember that bromide atom is way more bigger than carbon, hydrogen or nitrogen). Additionally BrdU nucleotide is light sensitive. Hoeachst intercalation occurs only in monosubstituted chromatid but not in disubstituted one (there is not enough space for Hoechst intercalation, because of two bromide atoms).

Further, slides are exposed to UVC light. Hoechst 33258 adsorbs the UVC light protecitng monosubstituted chromatid from UV light degradation. Subsequently DNA is stained with another dye called Giemsa stain. Degraded DNA do not stain with Giemsa dye revealing a differential effect as on picture below. Additionally to UVC light effect it is believed that different set of proteins are bound to BrdU disubstituted chromatid comparing to monosubstituted (what leads to differential staining).

Picture taken by Kliszczak M.

As you can see on the picture above mono- and disubstituted chromatid are indicated with blue and red arrowheads, respectively. Monosubstituted chromatid appears darker (as a result of Giemsa staining) than disubstituted (pale colour, no staining).

This technique is being used to assay chromosomal stability. Any DNA damage that occurs in the cell and is repaired by recombination (exchange of DNA information between the sister chromatids), will be visualized by this technique. You can see such events on the picture above. Two of them are indicated with black arrowheads. Some gene mutations (for example Rad54 or Rad51 - these genes code for proteins involved in DNA recombination pathways) leads to decreased number of Sister Chromatid Exchanges (SCE). Usually a specific drug (in this case Mitomycin C) is used to induced SCE's. After induction SCE's frequencies between Wild Type and mutant strains are comapred.

I hope you enjoy it.

Maciek GGSTEAM

Western Blot in Pictures

2010-05-05T23:33:00.001+01:00

Hello Biofreak,

I do not have to explain what this setion is going to be about:)

Enjoy!!

!!To download the WESTERN BLOT TUTORIAL IN PICTURES click on the link!!

MACIEK GGS

GGS LIVE - Gene Targeting

2010-04-26T21:04:00.218+01:00

Welcome again Dear BioFreak Reader,

It has been a long time:) But GGS is back:)

Today we are going to look at Gene Targeting Method.

Are you ready?? Let's go!

Method: Gene Targeting

About: Simply saying, gene targeting is a method that allows to remove/alter a specific DNA sequence from the genome of cell.

What: Knockout of exemplary gene A in order to study its function.

How: By generation of a targeting vectors and their subsequent application.

Before we start, we need to know a little bit more about gene targeting, so:

Gene targeting - is a multistep process that allows to delete or mutate a gen of interest. Gene targeting can be either permanent or conditional. In the former a specific DNA sequence is deleted, where in the latter such sequence is retained but can be deactivated upon specific conditions.

Ok, lets start the fun. We are going to split the process into few steps:

1. mapping of genomic locus of exemplary gene (will called it gene A, from now on),
2. design the targeting strategy,
3. and finally screening and detection of knockout bearing cells.

1. Mapping of gene A genomic locus. To disrupt gene A we need to gather a genomic DNA sequence bearing the locus of interest. To map the genomic locus of gene A we can use either its cDNA (complementary DNA which is simply just a coding sequence of the gene) or mRNA sequence. If such information is not available in database (for example NCBI database) it has to be first obtained by cloning gene A and its sequencing. But let's imagine that a coding sequence of gene A is known. We use such sequence to screen the genomic database by performing BLAST search against database (see picture below).

Cartoon above represents a genomic locus of an exemplary gene A, aligned with its cDNA. The roman and arabic numbers show exons (coding sequence) and introns, respectively (notice that cDNA do not contain introns:). As you can see by performing such database search (BLAST) we are able to map a genomic localization of specific gene. The real BLAST result is shown on the figure below.

You can see here two exemplary bits of the cDNA sequence aligned against the genomic DNA. Notice that Query and Subject sequences are cDNA and genomic locus, respectively. The top one corresponds to the first exon (I) and the bottom to the last exon (V). Such search gives us two information:

- genomic localization of gene A (top table, gene A is present on chromosome 2),

- chromosomal position of gene A - notice that top bit, an exon I (Query position 1) is located at a genomic position 5834742 (this is were our coding sequence starts in the genome). The same for the exon V, which ends at genomic position 5839788. We know that there is 5 exons in total, so we can estimate the size of gene A locus which is:

Size of gene A genomic locus = 5839788 - 5834742 = 5046bp = 5.05kb

In this case our genomic locus is relatively small and we will be able to knockout whole sequence of gene A. Some gene loci are longer and different approaches might applied for their disruption (I am working with 122kb locus which is around 24 times more than our exemplary gene A:). In our study case we will try to knockout complete sequence of gene A.

2. Design of targeting strategy. Strategy includes:

- desing of targeting vector (this will be used to disrupt gene A),

- desing of probes, (these will be used to detect knockout cells).

For generation of the aboves we will use genomic locus sequence of gene A. Additionaly we will need a DNA sequence, roughly about 10kb up- and downstream of gene A locus. We are going to use these flanking sequences to generate "arms" of targeting vector and probes (see picture below)

5' and 3' arms will be cloned into a specific plasmid (click here for Cloning tutorial) and have to be identical (homological) to sequences up- and downstream of gene A, respectively (as shown on figure above). Between arms a non-homological sequence to gene A is going to be inserted. Usually such sequence codes for a resistance to a specific drug, which later can be used to select cells that upteken and integrated targeting vector into their DNA.

Notice also that three HindIII restriction sites are present in the DNA region of interest. Two outside the gene A (up- and downstream to gene A) and one inside the gene A sequence. When gene A will be replaced by restriction cassette, the central HindIII site will not be present anymore. In this way a digestion pattern of genomic DNA will change. If locus is not targeted it will show wild type like pattern; 5'probe will detect 5.5kb band and 3'probe a 6.5kb one. But when targeting event takes place both probes should recognize a band of 11.0kb (that depends on restriction cassette size). To detect such change Southern Blot will be used.

Notice also that 5' and 3' probes will hybridize to DNA regions outside arms (upstream to 5' arm for 5' probe and downstream to 3' arm for 3' probe). It is important that probes are localized in the region outside arms. This is because in some cases our targeting vectors will be randomly integrated into genomic DNA (will not target gene A). If probes would hybridize to an arm region, they would pick up also those random integration events which are not of interest. This way we make our detection step easier

3. Screening and detection of knockout cells. When strategy is ready and targeting vectors are generated, we put them into action:). Wild type cells are transfected with a targeting vector and usually left around 24h to let them recover from stress. Next day cells are treated with a specific drug, depending on the cassette used and plated into 96 well plate. Conditions here are specifically optimized such approximatelly one cell that uptaken the DNA should be present in each well. This, and only this cell will grow and give a rise to a colony (other cells should die as they are not resistant to the drug). After few days single colonies are picked and expanded, so we have enough cells to extract DNA and freeze other half of cells for the time of screening. In the next step DNA from each clone (usually from 20-100 clones per transfection) is isolated and digested with an restriction enzyme (in our case HindIII). For the Southern Blot technique tutorial click here. An exemplary result is shown on the picture below.

As you can see clones 1, 2, 3, 4, 5, 6 and 8 are targeted as the digest pattern of genomic DNA has changed comparing to WT lane and non-targeted C7 clone. In this case 5'probe was used. We observe a correct band shift from 5.5kb to 11.0 kb. Notice also that there still is a WT band in each of the targeted clones. This is because there are two copies of gene A in each cell (there are usually two alleles of a gene). Clones 1, 2, 3, 4, 5, 6 and 8 are called heterozygotes for gene A (one targeted allele one WT allele). To obtain a full knockout we need to perform another round of a targeting on one of those heterozygotes using another targeting vector (bearing different resistance cassette, we cannot use the same vector because heterozygotes are already resistant to the drug, so we would not be able to select against the knockout cells). In this case wild type band will dissapear and only targeted band will be present.

I hope you enjoyed it:)

CyA Soon

Maciek GGS

Biochemistry Related Videos !!

2010-02-19T22:44:00.004Z

Dear BioFreak,

If you would like to see some cool movies about basic Biochemistry Techniques, I recommend my friend Osin's Youtube channel.

Oisin Keely - Biochemistry Techniques

You will find there videos about how to:

- use standard biochemistry tools (pipettes, spectophotometers),

- perform biochemistry techniques (affinity chromatography),

- build molecular models,

- cite papers and labell your figures correctly,

- and more (biochemistry lectures and interviews)!!!!

Maciek GGSTEAM

3rd year practicals - Immunochemistry

2010-02-11T21:06:00.000Z

Hello Students,

This week we have been exploring a world of immunochemistry:). We have been analysing two Unknown samples (U1 and U2) for presence of progesterone. Our aim in this practical was to estimate concentration of the progesterone in the those samples.

Ok let's begin.

You suppose to download your results from the blackboard and they should look roughly like this, see Table 1 below (remeber!! those numbers were made up by, my just for presentation purposes).

Using the Table 1 you calculate following values and you put them in Table 2 (noticed that obvious outliners where not included in the calculations - red highlights):

Now, we have to plot the B/B0 against the progesterone concentration, like on the plot below:

To get our Unknown samples concentrtion you read them manualy from the plot. So for example:

Unknown 2 = ~50pg

This is it:) I hope you enjoyed it:)

CyA Next Time:)

Maciek GGSTEAM

Amino acids - the handy way:)

2010-02-06T20:14:00.010Z

Yo, yo, yo you all!!

I have not been updating this section for a long time:) I have another idea, but I have been strugling to present it for a while. You judge it yourself and leave the comment:)

Enjoy:)

Ok. Before we start there is some rules to pick up:) LOL:) See the description and picture below.

1. Your wrist represents backbone of amino acid and it is always the same.

2. First carbon atom will be always placed in the area of your palm, including first knuckle of each finger.

3. Next atoms (carbon, oxygen, nitrogen or sulfur) will occupy finger knuckles and tips (like shown for thumb and small finger).

4. We will bend rules a little in bigger structures:)

5. Be aware that some bond lengths and angles are not real:)

6. Hands photographed from top.

Ok Lets gO!

First four structures:)

Short explenation (SE): Gly - just writst, Ala - Fist, Ser - your hand looks like S, Cys - Central finger .

It is getting better....

SE: Pro - for Pentagon and looks like letter P as well, Thr - Looks like T, Val - Looks like leter V, Ile - Fingers looks like I and L.

Next four....

SE: Leu, Asp and Asn are exactly the same only atom and number of bonds change, Met -first picture you MET two hands.

More ....

SE: Glu, Gln are the same (again only atoms and number of bonds change), Lys - is Long and Arg- is as long as Lys but splits at the end.

Two more ...

SE: I have no suggestions:)

And last two:)

SE: His - looks like house (nitrogens on the same knuckles) and Trp - house with pool (nitrogen at finger tips:)

If you want to download the Amino Acid Handies (AAHandies) please visit the link below:

Amino Acids are easy - the handy way (AAHandies).

I hope that will work for some of you:)

CyA SooN

Maciek GGSTEAM

3rd year practical - GDH

2010-01-29T17:28:00.004Z

Hello Students,

This week we have been working with Glutamate Dehydrogenase. We have been assaying 7 different samples for GDH activity . The protocol was exactly the same as last week, so I will not include any of that in this lab report. Just go back to the last post and you can find there how to prepare the lab report.

The only difference is, that this time you also have to calculate the Specific Enzyme Activity U/mg. To get the specific activity of an enzyme we need to know what is the Enzyme Activity (U) which you calculate from the plots you got. Additionally, you need to know amount of the enzyme.

From Blackboard (provided by Peter Creighton) we know that:

Using the Table above we can calcuate our Specific Activity which is simply:

Enzyme Activity (U) / Protein Amount (microgram) = Specific Enzyme Activity (U/mg)

To get the Protein Amount we need to go back to the protocol and see how much of the sample we have used. For example if we have used 500microlitres = 0.5ml of the sample, the amount of the protein is:

Protein Concentration = Amount of the protein (mg) / Volume (ml)

Amount of the protein (mg) = Volume (ml) x Protein Concetnration (mg/ml)

Remember to use appropriate units!!!!!!!!!!!!!!!!!! If you do not do so, your results will not make any sense:)!!!!!If you have problems with convertion of units/prefixes click the link below and visit this post

Convertion of Scientific Prefixes!!

Ok that would be it for this week:) Remember to pay attention to your calculations!

Maciek GGSTEAM

3rd year practicals - Measurment of Lactase Dehydrogenase Activity

2010-01-21T20:28:00.004Z

Hello Student,

today we will work with the Lactase Dehydrogenase (LDH). In our experiment we had been assaying activity of the Lactase Dehydrogenase in 7 different samples: homogenate, Supernatant and Pellet 1, Supernatant and Pellet 2 and Supernatant and Pellet 3.
To measure the LDH activity we were looking at its substrate NADH (see the picture below). Maximum adsorbance for NADH is 340nm. Beacause NADH is used in this reaction by the LDH we expect that adsorbance will decrease with time of experiment.

To prepare your lab report you need to determine the LDH activity in each sample provided and discuss on the results obtained.

To determine the LDH activity you will need your "fancy" adsorbance plots that you have obtained during the practical (see the picture below).

From the adsorbance plot you have to read out the change of the adsorbance that happened over one minute. To perform that we ... (see the picture below).

In the first step we tick out six vertical boxes (each of them is "worth" 10s). As you see adsorbance change over time. To get the change of the adsorbance draw a line down from the top of the curve (1min time point) to the bottom of it (the zero time point). The sector on the x-axis is your change in adsorbance over one minute. Remember that each of the small boxes is equal to 0.1.

When you read out change in adsorbance/min for each sample we construct table like this, where you put your result (see below).

To calculate the Enzyme Activity we are going to use a Lambert Beer Law. We know that enzyme activity is described as the turn over of the amount of substrate over time.Additionally, from the Lambert Beer Law we know that adsorbance of compont X is proportional to its concentration (see below).

After you get the Enzyme Activity of each sample you put it in the Table and discuss your results:) Remember to give the lab reports till tomorrow evening (if someone need more time for that please leave them till Sunday evening or bring it to me monday).

That would be it:) Not so bad ha?

Till Next Time NUIG:)

Cya Soon

Maciek GGSTEAM

3rd year practicals!!

2010-01-16T10:42:00.001Z

Hello Students,

Welcome in the second semester:) Second semester practicals for 3rd year Biochemistry, start next Monday the 18th and Wednesday the 20th January.

Cya Soon and Have a Nice Day

Maciek GGSTEAM

GGS LIVE - Southern Blotting

2010-01-12T00:10:00.002Z

Welcome again in the Biochemistry Methods Section,

Let me introduce today:

Method: Cold Southern Blot

About: Simply sayin Southern Blot is a method were a presence of specific DNA sequence (or its lack) can be detected in the DNA sample of interest. Using the Southern Blot technique we can detect abnormalities/mutations in the whole genomes (karyotypes) like deletions or insertions and many more.

What: Detection of a specific DNA sequence (gene) in the samples form three different cell lines.

How: By cold southern blot technique where non-radioactive DNA probe is used. Major steps of the protocol include: DNA digestion, agarose gel electrophoresis, hybridization with a specifically designed DNA probe and target DNA sequence detection.

Ok so lets start:) Before we go please have a look at these two cool animations about the southern blotting
Southern Blot and Southern Blot 2.

In our case study we will look for presence of a specific gene sequence in three specimens. Each specimen is a different cell line and we will isolate and screen their DNA to look for presence of exemplary gene A.

First thing to do before we start our analysis is to design a probe that will identify DNA sequence of interest. Our probe is of course a short DNA sequence that is complementary to the target sequence. To design the probe we need to know the sequence of the gene A. In this case best probe to use, is a DNA sequence that is a part of our gene A (look at the picture below).

After we design the probe on the computer/paper we have to synthetize it. We obtain the probe by PCR reaction (polymerase chain reaction). To detect the probe we have to label it with a moiety or chemical group that later will be used for detection. We can either use a radioactive or non-radioactive labelling approach for that purpose. In former one, nucleotides bearing a radioactive isotope of phosphorus are used. In the latter, nucleotides bearing unique chemical group/moiety are utilized, which later during the detection step are recognized by specific antibodies

At this stage we have our probe ready and now it is time to prepare DNA samples. We isolate DNA from our specimens and in the next step we have to digest it. We need to digest our DNA in order to release the sequence of interest. Imagine, if we do not digest DNA it will be very hard (almost impossible) to work with such bulk DNA molecule. Additionally, digestion can give us information how big is the DNA fragment that bears our sequence of interest (of course that fragment will be different for various restriction enzymes).

In our study case we will use a BamHI enzyme to fragmentarize our DNA sample. After DNA digestion we separate the DNA fragments by agarose gel electrophoresis, see the picture below.

The "smeary like thing" in lanes Sample A-C, is our digested DNA. Genomic DNA digest result in releasing DNA fragments of various sizes (ranging from very small ones to big). Remember that genomic DNA is a huge molecule that posseses many BamHI sites.

In the next step we need to transfer the DNA from the gel to the membrane. Membrane is easier to manipulate, because it is less fragile that gel (southern blot membranes are positively charged to attract negatively charged DNA). Just before the transfer gel has to be prepared. Usually it is incubated with different bufferes which remove traces of proteins (residual proteins could interfere with subsequential steps), denaturate and nick DNA. All of that facilitate transfer and further detection of target DNA. There are different ways of the DNA transfer and one of them is the capillary transfer, see the picture below.

After the transfer is finished, DNA is crosslinked to the membrane with UV light (during the crosslink a chemical bond is formed between DNA and a membrane).

Crosslinked membrane is incubated overnight with the probe solution, at a specific temperature which depends on the probe sequence. Hybridizaion is usually performed in hybridizator where membrane is placed in a special tube (see the picture below).

Alternatively membrane can be sealed in the plastic bag containing a probe solution.

After the overnight hybridization mambrane is washed several times to remove unbound/excess probe. In the Cold Southern Blot, probe is detected using specific antibody. Because of that, we need to block the membrane before incubating it with the antibody solution. Blocking prevents antibody to stick/bind to unspecific sites on the membrane. When the membrane is blocked, we can add the antibody solution. Silimarly to western blot a complex between the DNA, probe and antibody is formed during that step (see the picture below and the link to Western Blot technique). Again membrane is washed several times, but this time excess of the antibody is removed.

Antibody used in Southern blot is conjugated to enzyme that catalyze a reaction that emits photon. When the "sandwich complex" is formed, substrate for antibody conjugated enzyme is then applied to the membrane.

Membrane is then sealed in a plastic bag and placed in the cassette. Next, photographic film is applied and cassette is closed and left for several hours. After that time film is developed (see the picture below).

Above you can see a result of our experiment. Our Southern Blot shows that cell lines A and C but not B contains the gene A (target sequence). Additionally we can see that fragment size that bears the target sequence is identicall in specimens A, C and its size can be estimated around 1.8kb (remember that this fragment is specific for BamHI digest and it will be different for other restriction enzyme).

As you see the Southern blot technique is able to detect presence of exemplary gene A.

In the next post about the Gene Targeting I will expose how we can use the Southern blot method in other way.

I hope you enjoy it:)

CyA Soon

Maciek

GGSTEAM

Welcome in 2010!!

2010-01-04T22:05:00.000Z

Welcome you all,

hello, hello, hello in 2010. Happy, Happy new year:) How are you? I hope you are fine and this year will be a very good to all of you.

I have just come back from the christmas holidays and I am ready to work again:) I charged my batteries and here we go BioFreaks. This year, there will be more and more attractions at the BioFreaks Blog. Few knew sections will appear and I hope you will enjoy them all:)

Again Happy New Year and CyA SooN:)

Maciek GGSTEAM

MaRRy MaRRy Christmas and a HaPPy NeW Year!!!!

2009-12-22T15:58:00.000Z

All you that visitied my BioFreaks Blogg during its short life:) I wish you all the best!!

GGSTEAM

Amino acids - the chemist way of life 3D STRUCTURES !!!

2009-12-18T22:48:00.009Z

Hello, Hello, Hello,

It has been a long time since I update this section:) I do not have to much but always something:) Actually, this section is one of the most visited by you:)

So lets go:)

Today we have a 3d representations of 20 aminoacid structures. Check this out!

Do not forget to leave a comment.

Be Well,

Maciek GGSTEAM

Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I - Question 5

2009-12-18T20:09:00.005Z

Yo, yo, yo!

I said I am back:) This evening together with Question 4 we expose Question 5 of the Sample Exam Paper for 3rd year Undenominated Science (to go to paper click here).

Ok, let's go team.

Question 5 For a competitive inhibitor of an enzyme:

(a) Draw a Lineweaver-Burk (double reciprocal) plot for the enzymatic reaction
in the presence and in the absence of the inhibitor
(b) Explain why the intersection of the lines, plus or minus inhibitor, occurs where
it does
(c) If the inhibitor is present at a concentration equal to its Ki, by how much does
the slope of the plot increase?

Part A) and B) Lineweaver-Burk is a double reciprocal of Michaelis-Menten plot, where [1/V] vs 1/[S] values are plotted. Picture below:

OK, now depending on the nature of inhibitor Km (Michaelis constant, a specific substrate concentration where V = 0.5Vmax) and Vmax (maximum velocity) values behave differently. When there is a presence of:

- competetive inhibitor, Vmax do not change but Km increase. This is because of the competition between inhibitor and substrate for active site of the enzyme. If more of the substrate has to be added to achieve the same result, it means that Km has to increase as well. We can observe that at the L-B plot, see below:

- non-competetive inhibitor interacts with enzyme in a difference place than active site. Its action changes conformation of the protein, what leads to deacrease in substrate processing. In this case adding more substrate will not increase velocity. Km stays constant Vmax decrease, see below.

There are also other inhibitors: reversible, irreversible, mixed inhibitions etc. Their effect on the Km and Vmax might be highly specific or a combination of the above.

Part C) Hmm it is a shame but I am not sure about the answer. I tried to solve it this way:

Remember about the red text:).

CYA SOON HAHAHA:)

MACIEK GGS TEAM

Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I - Question 4

2009-12-18T15:09:00.000Z

Hello Students,

I am sorry I have stopped posting on the Exam Questions, but I am back :)

Today Question 4 from the 3rd Year Undenominated Science Paper (to see which one click here Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I.)

Question 4 Describe how proteins recognise DNA using bZip and zinc finger motifs as examples.

Proteins can interact with DNA by charge interaction (DNA negative, proteins positive), hydrogen bonding (between protein and nucleotides in the major or minor groove) or by wrapping/enclosing DNA, etc.

Zing finger motif is a part of the protein where neigbouring cysteines and histidines fold into a zinc-binding pocket (see here zing-finger motif binding to zinc or below).

Picture taken from http://en.wikipedia.org/wiki/File:Zinc_finger_rendered.png

Presence of the zinc atoms are essential for proper protein function and conformation. On the next two picture you can easily see how binding of the zinc atoms allow protein to accommodate shape fitting into DNA double helix:

Picture taken from http://www.rcsb.org/pdb/explore/explore.do?job=graphics&pdbId=1TF6&page=0&pid=1560961164203

Accesion number 1TF6

And here DNA binding domain of Oestrogen Receptor.

bZIP motif is a basic region of the leucin zipper, a part of the leucine zipper DNA binding domain. See picture below. Remember that leucine zippers are helical and always (similarly like the zinc fingers) act together. Two zipper proteins dimerise and interact with DNA.

Picture taken from http://bssv01.lancs.ac.uk/ads/BIOS336/336L7.html

One face of each helix is hydrophobic (leucine rich region) and the other one is hydrophilic (charged aminoacids). The basic part of the dimer is interacting with DNA.

Picture taken from http://www.rcsb.org/pdb/explore/explore.do?job=graphics&pdbId=1JUN&page=0&pid=14814961579302

Access number 1JUN

Examples: zinc finger domain TFIIIA and leucine zippers C/EBP.

I hop you enjoy it:)

Maciek GGSTEAM

GGS LIVE - Live Cell Imaging

2009-12-16T23:19:00.052Z

Welcome all in the Biochemistry Methods Section,

Method: Live Cell Imaging

About: Simply saying Live Cell Imaging is a technique which utilize advanced microscopy methods to analyse, observe and record behaviour of the living organisms, cells, organelles and even single proteins. Whole cell organisms (and organelles) are usually visualized with phase contrast and differential interference contrast (DIC) techniques which allow to investigate theri optical properties, dynamics, and morphology. When single proteins are investigated, a fluorescent fusion of the protein of interest has to be expressed in the cell to visualize it (fluorescent and confocal microscopy).

What: U2OS cells (osteosarcoma) expressing RFP-H2B (Red Fluorescent Protein - Histone 2B fusion, DNA visualization) and GFP-centrin (Green Fluorescent Protein - Centrin fusion, Centrosome visuzalization).

How: Live cell imaging using a Delta Vision Microscope. Cells were followed for 24h.

What you can see on the picture below is a few U2OS cells expressing histone H2B fused to RFP (red coulour) and centrin fused to GFP (green colour, see picture below).

Histone H2B and centrin are DNA binding/packing and centrosome proteins, respectively (for more information click here Histones, DNA packing proteins and here Fluorescent Micriscopy and Centrosome Staining). DNA will appears as a round shape like structure where centrosome is just a distinct spot (you may see it only in some of the cells, because not all of them express GFP-centrin or centrosome not always stay in focus of the camera). You will see few cells dividing (see below, an example of cell that will divide during the film is indicated with blue frame).

And now the movie:

Movie taken by Helen Dodson, PhD

I hope you enjoy it:)

Have a nice day,

Maciek GGSTEAM

Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I - Question 3

2009-12-16T17:04:00.001Z

Yo, yo, yo, yo you all,

Today we continue with the 2007/2008 3rd Year Undenominated Science Exam Paper Semester I which you can find here Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I.

Question 3 Many hormones regulate gene transcription in eukaryotic cells. Explain, using examples, how this is achieved by lipid soluble and lipid insoluble hormones.

This question is easy and I bet you know why it is easy. It is easy because we have already answered this question. It has appeared in the previous paper we have exposed:) To see the answer go here Hormone triggered gene expression.

You only have to mention the difference between water soluble and water insoluble hormones.
Water insoluble - hormones (like the one in the animation here McGraw Hill - Mechanism of Steroid Hormone Action) require protein carriers which are necessary for their transport (remember, blood and cytoplasm are water solutions). Notice that water insoluble hormones do not have any problems with passing through the cell membrane (because their hydrophobic - partially water insoluble).
Of course there are also water insoluble hormones that interact with the membrane receptros on the target cells. Binding to the receptor changes its conformation what leads to signal transduction into the cell, where appropriate gene transcription activators/deactivators are switch on/off.
Either scenario leads to transcription of essential genes and production of proteins, what leads to hormone induced reaction.
Water soluble - hormones do not have any problems with traveling through the blood as they dissolve in it very easily. But they do have problems entering the cell (passing the membrane). As the membrane is highly hydrophobic there is no way water soluble molecule can pass it. In this case water soluble receptors have to interact with the cell surface receptors which transduce the signal further within the cell. Similarly as above appropriate transcription factors are activated what leads to hormone specific response.

Examples:
Water insoluble - cholesterol, estrogen (female hormone), testosteron (male hormone).
Water soluble -epinephrine, dopamine.

To sum up see the picture below:

Picture taken from http://www.biologie.uni-hamburg.de/b-online/library/falk/Endocrine/endocrine.htm

I hope you enjoy it:)

Have a nice day and studying,

Maciek GGSTEAM

Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I - Question 1 and 2

2009-12-15T21:26:00.002Z

Hello Students,

Welcome again in the Biochemistry Exams Section. Today we will start to work with another Biochemistry Exam Sample Paper which you can find here Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I.

Ok let's go team.

Question 1. The replication of eukaryotic chromosomes requires the cooperation of multiple
proteins. Discuss the activities of polymerases (DNA and RNA) in this process.

As you see Question 1 from this paper is exactly the same as on the paper we have already, previously exposed. To see the answer to the question go here DNA replication enzymes. If this question did show up last year and two years ago, we can conclude that a similar question can came up this year as well. So it is essential you get familiar with it:)

Ok that was easy:).

Question 2. Explain and discuss the term “junk DNA” OR Write an essay on protein folding and processing in eukaryotic cells and outline the role that chaperone proteins play in these events.

We will start with the essay part. As you probably noticed, we have already answered this question as well. Last paper had a similar problem to solve but this time you are asked to write an essay about protein folding and processing. Just go here Protein Folding, Processing ang Targeting (Part A of the question 4) and use the infromation provided to write nice essay.

Ok, let's explain what the junk DNA is?

Junk DNA - simply saying, it is a DNA which have no known function. About 95% of the human genome is considered as a junk DNA (about 5% encodes for the proteins, RNA's etc).
The best example of the junk DNA are introns. As you know most of the genes are composed of exons (protein coding sequence) and introns (non-coding sequence). After gene transcription (before protein synthesis), splicing machinery remove introns and join exons together to form a protein coding sequence. Because introns are removed in this process they are designated as junk DNA.
Pseudogenes are another example of the junk DNA. Pseudogene is a DNA sequence which is similar (sometimes almost identical) to another gene but it is never expressed (protein is never produced from this sequence). So this is like a copy of a particular gene but it is never used.

Why we actually need a junk DNA? We need it because we actually do not comprehend its function yet:) I am sure that these functions will be revealed in the future.
Remember that genomes are dynamic structures and junk DNA might be essential for them. Imagine that junk DNA might interfere with gene transcription (activation or deactivation). It is possible that junk DNA form complexes with DNA what is essential for genome metabolism or it is a platform for processes that regulates genome maintenance (etc).

This is it:) Soon next questions:)

Have a nice night:)

Maciek GGSTEAM

Biochemistry Exams - 2008/2009 3rd year Undenominated Science Semester I - Question 6

2009-12-08T19:47:00.001Z

Hi ya all,

Today we carry on with the 3rd year Biochemistry Exam Paper which you can find here: 2008/2009 3rd year Undenominated Science Exam Paper Semester I.

Question 6. Describe the main structural features of glycoproteins and proteoglycans and, by giving examples, outline some of the key biological functions of these glycoconjugates.

Super short background:)

Glycoproteins - are proteins which carry oligosacharide (sugar) chains covalently attached to aminoacid sidechains (N-glycosylation at Asparagine and O-glycosylation on Serine, Threonine, hydroxylysine and hydroxyproline).

Picture taken from http://www.ideacenter.org/stuff/contentmgr/files/e27b080d92450837e43d44bf73780847/misc/glycoprotein.jpg

Process of sugar attachment to protein is called glycosylation. Glycosylation is a example of either post-translational (after protein being synthesized) or cotranslational (during protein synthesis) modification.

Most of the glycoproteins are targeted to the cell membrane where they are responsible for cell to cell interactions. The best example of the glycosylated proteins are mucins. Mucins are proteins found on the surface of epithelial cells of the respiratory and digestive tracts. They play very important protective role. They are part of the mucus that prevents from antigens like bacteria, viruses or other particles. They play similar role in the digestive tract where they protect epithelial cells from harmful action of acids or proteolytic enzymes.

Proteoglycans - are also glycoproteins but they can carry form one to few long chains of sugars (glycosaminoglycans). Proteoglycans are usually negatively charged as the sugar residues contain acid moieties.

Picture taken from http://medinfo.ufl.edu/pa/chuck/summer/handouts/images/gag.jpg

One of the proteoglycan function is to "fill" spaced between the cells (extracellular matrix). They are responsible for strength and flexibility of the connective tissues. It is also involved in cell to cell interactions and cell signaling pathways (they regulate movement of molecules through extracellular matrix). Because proteoglycans bear negative charge their also involved in sequestering metal ions (like sodium, potasium or calcium) and water.

Good example of such proteoglycan is a aggrecan which is a major component of the cartilage. Aggrecans bears very high negative charge which function is to regulate osmotic potential of the cartilage. This gives cartilage its strength and flexibility.

Different examples of glycosylated proteins:

- immunoglobulins (anitbodies),

- glycoproteins are responsible for different blood types,

- Glycoprotein-41 and glycoprotein-120 are HIV viral coat proteins and play essential role in the HIV infection of T4 helper cells.

That would be it:)

I hope it was helpful:)

Maciek GGSTEAM

Biochemistry Exams - 2008/2009 3rd year Undenominated Science Semester I - Question 5

2009-12-07T22:51:00.001Z

Welcome Again in the Biochemistry Exam Exposed Section,

Today we will expose Question 5 from this exam paper 2008/2009 3rd year Undenominated Science Exam Paper Semester I.

Question 5 sounds... “Modest elevations of homocysteine have multiple causes, including low levels of folic acid, vitamins B6 and B12”. Expand on the metabolic bases to this statement.

Short, short background;

homocysteine - is a homologue of aminoacid cysteine differing by an additional methylene (-CH₂-) group.

Check the figure below.

Homocysteine is synthesized from methionine (figure above) by removing the terminal methylene group (CH3-). On the other hand homocysteine can be recycled to give cysteine or methionine.

Answer to Question 5 lies in the methabolic pathways that deal with homocysteine degradation. So, biosynthesis of homocysteine (see figure below) starts from the methionine which first is conjugated with ATP to form S-adenosyl methionine. In the second step methyl group which is bonded to sulphur atom is transfered to acceptor molecule (norepinephrine which is converted to epinephrine). In the third step adenosine (ATP residue) is hydrolized what gives homocysteine.

Probably at this stage some of you wonder why the methyl group from the methionine, is simply removed to give the homocysteine. It requires addition of adenosine first beacuse it makes the methyl group higly labile. In the case of methionine, the methyl group is too stable to be transfered.

Homocysteine can be transformed back to methionine or to cysteine. The former requires THF (tetrahydrofolate) a folic acid or or cyanocobalamin (vitamine B12) and latter pyridoxine (vitamine B6) as enzyme coffactors. So simply saying if there is no folic acid, vitamin B12 and B6 available in the body, enzymes that requires those molecules for proper function cannot fulfill their roles. If they are not able to catalyse reactions mentioned above homocysteine levels will raise.

You can also include in the answer what are the consequences of such elevation.

It is known that increased homocysteine levels cause cardiovascular diseases. How this can happen? Scientists believe that homocysteine simply degrades proteins like collagen and elastin (artery proteins) because it competes with the cystein to form disulfide bonds.

This is it:)

Question 6 soon.

Maciek GGSTEAM

Biochemistry Exams - 2008/2009 3rd year Undenominated Science Semester I - Question 4

2009-12-07T19:38:00.002Z

Hello,

It is time for Question 4.

We are refering to paper that can be found here: 2008/2009 3rd year Undenominated Science Exam Paper Semester I

Question 4 EITHER complete both parts A and B below (each part is worth equal marks)
A. Name two different types of post-translational modifications, specify the
amino acids involved and the chemical effect of the modification. Describe
how each modification enables biochemical function using examples.
B. Explain how the repeating chemical structure of the polypeptide
backbone gives rise to protein secondary structure.
OR
Write an essay on the processing and targeting of newly synthesized
proteins in eukaryotic cells.

This one is a little bit longer but still very easy:)

Let's taka care of the Part A first with a short background:
As you know proteins can be post-translationaly modified. Those modifications affect protein function, stability, activity, binding affinity etc.
List of protein modifications is still growing:
- phosphorylation (phosphate group attachment),
- methylation (methyl group attachment),
- acetylation (acetyl group attachment),
- ubiquitination (attachment of ubiquitin peptide),
- sumoylation (attachment of SUMO peptide),
- argininylation (attachment of arginine),
- ADP-rybosylation (attachment of ADP-ribose),
- citrulination (deamination of arginine),
- myristoylation (atachment of myristic acid group),
- hydroxylation (attachment of hydroxyl group),
- prenylation (attachment of prenyl groups),
- palmitoylation (attachment of palmitic acid group),
- glycosylation (attachment of sugar residues),
- and others.
Remember that also:
- proteolytic protein cleveage is a post-translational modification.

To answer the Part A we need to discuss only two modifications. In my opinion the easiest two to describe would be protein phosphorylation and protein modification by proteolytic cleveage. So:
Protein phosphorylation - is a protein modification where phosphate group is attached to a serine or threonine residues (rarely to tyrosine residue) by a protein kinases (removal of the phosphate groups is mediated by phosphatases). See figure below:

Very important to remember is that phosphate group bears negative charge. Modification of protein by phosphorylation cause conformational change in the target protein (negative charge on the phosphate group will attract positively and repell negatively charged parts of the protein). Conformational change of the protein can affect its activity, stability, binding affinity and others. Imagine that enzyme active site of protein X is not accesible (let say it is hindered by a part of the protein) but upon the phosphorylation its conformation changes, active site becomes opened and substrate can bind. See figure below.

ATM protein is a good example of activation by phosphrylation. ATM protein is a kinase which in unperturbet cell cycle is present as a dimer and stays inactive. ATM is activated by ionizing radiation (IR) which cause DNA strand brakes. As a part of the DNA damage response each ATM protein phosphorylates its partner within the dimer (autophosphorylation). Upon phosphorylation ATM protein is released from the dimer and kinase is able to modifiy other protiens through phosphorylation.

Protein Cleaveage is easy to explain as well. There is a lot of proteins that are cleaved by proteases in order to became active. You know that most of the proteolytic enzymes like trypsin or chymotrypsin are produced as zymogens (not-active enzymes). After zymogen secretion, they are cleaved and became active.
Remember that in general most of the protein modifications work in a simillar way. They change protein conformation (you can easily imagine that if you chop out a piece of the protein that it will change its structure) what leads to change in protein activity, stability etc etc.

In Part B we need to explain how protein aminoacid backbone affects protein secondary structure. As you know there is only two major types of protein secondary structure:
- alfa-helis,
- beta-strand.
See fiigure:

Picture taken from http://www.nature.com/horizon/proteinfolding/background/images/importance_f3.gif

Both alpha-helix and beta-strands are stabilized by hydrogen bonds formed between hydrogen of the backbone alpha-amino group and oxygen from the backbone carboxyl group. Depending on the aminoacid sequence protein can either accomodate beta-strand or alpha-helix conformation. Remember that alph-helices and beta-strands can occure simultaneously in the same protein (one part of the protein folds this way the other one folds differently).

This is it:)

In this question you either answet part A and B or you have to write a short assay about: processing and targeting of newly synthesized proteins in eukaryotic cells.
Of course I am not goind to write an assay here beacuse that does not make any sense but I will put an essential information that you should include in your assay (write about). This part of the question is actually very similar to Part A because protein processing and targeting is achieved by protein modification.

Ok so let's start. You have to remember that after synthesis protein has to be:
- folded. In this process proteins called cheparones play a crucial role. Proteins in the presence of cheparones fold properly and efficiently (Have a look at this animation - Willey Interactive - Protein Folding). Unfolded protein is useless for the cell and therefore it is degraded.

- targeted. Proteins contain a signal sequence which is specific for each cell compartment. They are often called sorting signals and these signals interact with specific receptors, either on the target organelle or a carrier protein. For example there is a nuclear location sequence as well as a membrane location sequence etc.
Remember that proteins can be targeted to the different locations by a post-translational modifications as well (to see what I mean see the top of that post:). Good example of such is a SUMOylation process. Protein which was modified by attachment of SUMO peptide is usually targeted to the nucleus.
Different examples are myristoylation and prenylation modifications which target proteins to membranes (it makes sense as myristic acid and prenyl group are hydrophobic molecules).
Notice also that proteins which suppose to be targeted to the cytoplasm usually do not bear any location signal (those are proteins like enzymes, structural proteins, translational machinery protiens).

To sum up please see this animation Protein Targeting.

- processed. Some proteins have to be modified before they became active or fulfill their function (see the Part A of that question as well). For example proteolytic enzymes have to be cleaved at specific sites to became catalytically active (like trypsin). Other proteins have to bele phosphorylated, methylated or glycosylated to perform their function.
You can imagine protein processing as a set of essential protein modifications/alterations processes which lead to protein activation, localization etc.

Essential animations that will help you to see what is going on:)

Protein Modification
Protein Trafficking

Ufff that was a loooooong one:)

CyA, next time question 5.

Maciek GGSTEAM

Biochemistry Exams - 2008/2009 3rd year Undenominated Science Semester I - Question 3

2009-12-06T15:07:00.010Z

Whatsuuuup ya all??

Right here right now question 3:)

To see sample exam paper go to: 2008/2009 3rd year Undenominated Science Exam Paper Semester I

Question 3
Gene expression can be regulated by activating transcription factor activity.
Explain with reference to hormone-inducible gene expression in eukaryotic cells.

First short background:
transcription factor - is a protein that bind to specific DNA sequences in order to control production of mRNA from DNA. Transcription factors act either alone or with other proteins to promote (as an activator), or block (as a repressor) recruitment of RNA polymerase (enzyme which produce mRNA out of DNA) to specific genes.

Picture taken from http://en.wikipedia.org/wiki/File:TATA-binding_protein.png

Transcription factors enhance or silence gene transcription in response to different stimuli. On the figure above you can see a transcription factor (blue) bound to DNA (red). Notice that DNA bends upon transcription factor binding. This conformational change of DNA interfere with RNA polymerase binding, what result in either shutting down or enhancment of gene transcription.
Remember that there are transcription factors for different sets of genes.
For example hormones can induce gene transcription through binding to hormone-inducible transcription factors. Before we go further please have a look at this animation McGraw Hill - Mechanism of steroid hormone action.

As the animation explain, when hormone molecules enters the cell they bind to the transcription factors. In the nucleus hormone-transcription factor complex binds to specific DNA sites. That interaction is responsible for inducing or shutting down gene transcription. Without the hormone molecule, transcription factor does not bind to DNA and because of that it does not induce gene transcription.
Similar events occure when other types of transcription factors are activiated but mechanism might be different.

Be aware that activation of transcripation factors is not the only way of regulating gene transcription. For more details please see those animations:
McGraw Hill - Control of gene expression in eukaryotes
McGraw Hill - Transcription complex and enhencers

To sum up, what you have to remember is:
- transcription factors - bind to DNA and enhances or silences gene expression,
- transcription factors are activated by different stimuli and either activate or silence gene transcription,
- gene expression might be as well regulated in different manner (mRNA modification and translation, etc),

This is it. Question 4 soon:)

Maciek GGSTEAM

Biochemistry Exams - 2008/2009 3rd year Undenominated Science Semester I - Question 2

2009-12-06T11:58:00.003Z

Hello Again,

After this short invitation/background text:) Question 2 from 3rd year Undenominated science exam paper will be exposed (info about the paper can be found here 2008/2009 3rd year Undenominated Science Semester I).

Ok here we go !!

Question 2 Discuss the role of histone proteins in packaging the genome and regulating its function.

Short background, histones are:
- small, alkaline proteins (positively charged beacause of high number of lysines and arginines in the sequence)
- there are five classes (H1/H5, H2A, H2B, H3, H4)
- found in nucleous
- interact with other histones and DNA

Histones interact to form an octamer structure which is called a nucleosome scfaffold. It is made of H2A/H2B tetramers and another tetrameric H3/H4 subunit. Because histones are positively charged (a lot of Lys and Arg in the sequence) the octamer easily interact (ionic interaction) with the DNA (which is negatively charged). DNA/octamer complex is called a nucleosome (see figure below).

Picture taken from Nucleosome Structure.

Nucleosomes allow DNA compaction necessary to fit genomes of eukaryotes inside cell nuclei.

Remember that each group (H1/H5, H2A, H2B, H3 and H4) contain many histone isoforms and all of them can form octamer structure. Depending on that structure DNA wrapped around it may have different properties.

Keep also in mind that histones like other proteins can be post-translationaly modified. There are different modifications detected at histones: methylation, acetylation, phosphorylation, ubiquitination, sumoylation, citrulination and ADP-ribosylation. Each modification and combinations of them have an impact on histone structure and function. Histone modifications affect:

- DNA replication,

- DNA compaction (euchromatin and heterochromatin)

- gene transcription,

- DNA repair - when DNA is damaged histones are modified to recruit DNA repair proteins,

- DNA/protein complexes - their assembly or disassembly,

- their stability - when histones have to be removed, their modification can target them for degradation and when they are needed they are targeted to DNA,

- and more.

Before you finish have a look at theis animation Biology Animations - Histone modification.

I hope you enjoy.

Soon next question:)

Maciek GGSTEAM

Biochemistry Exams - Sample Papers Exposed

2009-12-05T14:23:00.003Z

Hi Students,

Welcome in the new section of the BioFreaks Biochemistry Blog. In this section we will work with the previous years exam papers. Simply as that:) Today we start with 2008/2009 3rd year undenominated science paper which you can find here:

2008/2009 3rd year Undenominated Science Semester I

If this is a wrong paper please let me know:)

Ok. Let's go!!

Question 1. Discuss the role of polymerases (DNA and RNA) in eukaryotic DNA replication.

Before we go into the answer to that question please have a quick look at this animation McGraw Hill - DNA replication fork, which actually is great overview of the DNA replication process.

To answer the question you need to remember that in DNA replication process are required:
- DNA polymerases (Polymerase III and I) - DNA synthesis
- RNA polymerase (called primase) - RNA primer synthesis on lagging strand
- helicases - DNA unwinding
- DNA ligase - strand ligation
- other proteins - which for example enhances fidelity and procesivity of the DNA replication process or those that monitor replication fork stability.

You all know well that each DNA strand has two different ends (5' end with phosphate group and 3' end with the hydroxyl group). DNA polymerases are able to elongate the new DNA strand only in 5' to 3' direction. Just remember that 3'OH end is the only one which can attack the 5'TriPhosphate on the newly incoroporated nucleotide to form a bond (never the other way around, see the figure below).

The other feature of the DNA polymerases you have to keep in mind is that they cannot start DNA replication without primer (DNA must be at least partially double stranded to be a substrate for DNA polymerases).

Because of those reasons one strand is processed continuesly and the other one in fragments (called Okazaki fragments). Once the primer is added by RNA polymarase leading strand (continues one) can be relplicated. For the lagging strand (discountinous one) it looks different but it is still easy:). It starts the same way, RNA polymerase add primer, DNA Polymerase III extend the primer, then it is replaced by DNA Polymerase I (remember that RNA polymerase add U opposite to A not T, so primer must be replaced) and then short DNA fragments are ligated (fused) by DNA ligase. And the story starts again till whole DNA molecule is replicated.

To sum up:
- we need both DNA and RNA polymerases to replicate DNA,
- leading strand - RNA Pol adds primer, DNA Pol III replicate the strand,
- lagging strand - again RNA Pol adds primers but this time DNA Pol III and I cooparate to replicate the strand which at the end is fused by DNA ligase.

It is a good idea to put a simple fugire together with your discussion as it always indicated you understand what is happening on the molecular level. Start with the figure like the one below and then discuss the question. Another adventage of having the picture is that you focus will easilly focus on what you have to write about.

Picture taken from McGraw Hill animations

http://highered.mcgraw-hill.com/olc/dl/120076/micro04.swf

That would be it for the Question 1. If you have any problems with the answer or there is something not clear do not hesitate to contact me by leaving the comment or at m.kliszczak1@nuigalway.ie.

Cya soon:) Question 2 is coming:)

Maciek GGSTEAM

Meet the chickens - Fluorescnet Microscopy and Centrosome Proteins

2009-12-02T11:27:00.002Z

Hello Again in the Meet the Chickens section,

Today we will have a look at the DT40 cell line centrosomes but first I will give you a short background.
Centrosome is large cytoplasmic organelle. Its core is compsed of mother and doughter cetrioles which is surrounded by a pericentriolar matrix (which compose of dozens of proteins). The doughter centriole is always formed de novo and the other one comes from the previous cell cycle from the mother cell (mother centriole).
During the cell cycle, centrosome duplicates and just before the cell division there are two (and there should be only two) centrosomes in each cell. Higher number of centrosomes can be dangerous for the cell as division of the genetic material could be impaired (look at this animation McGraw Hill - Mitosis and Cytokinesis). Abnormal genetic material division can lead to gain or loss of chromosomes. Such situation may lead to pathogenic state (for example cancer).
Centrosomes are often called microtubule organizing centres. What it means is that microtubules nucleate at centrosome providing connection between chromosomes and centrosomes. When microtubules attachement to chromosomes is completed they can be pulled to opposite poles of the cell which are designated by position of the centrosomes (see picture below).

Centrosomes are involved not only in cell division. They also serve as a platform for different proteins and processes. Many proteins have been shown to localize to centrosomes and for some of them this localization is necessary for their activation or proper function.

We can visualize centrosomes in the cell using specific antibodies which recognize proteins associated with them. A list of the proteins which localize to centrosomes is still growing. The well known centrosomal proteins are: centrin, gamma-tubulin, pericentrin, Kizuna, Nedd1 and more. On the picture below you can see a example of such staining.

Picture taken by Kliszczak M.

On the photos above you can cell in late stage of mitosis (anaphase). Each separate channel is represented in gray scale.

DNA was stained with DAPI and on the merge photo it appears as blue stained. You can see that DNA is already sepeareted to opposite poles of the cell.

On the Centrosome layer which appear green on the merge picture you can see a GFP fusion of centrosonal protein (Green Fluorescent Protein) which is known to localize to centrosomes (notice the two red circles). The cell has a lot of the green background which is common when fluorescent protein is expressed in the cell (GFP signal).

On the red channel (Microtubule) the protein which builds up the microtubules is stained. You can see the microtubules connecting opposite poles of the cell. Notice that at the poles microtubule staining is very bright. Those big blobs are actually centrosomes where microtubules nucleate. Notice also that those microtubules centres do colocalize with the green spots (GFP fusion of centrosomal protein).

Merge image represents all channel superimposed together. Microtubule staining is not very visibile at this photo, beside at the poles where you can also see a GFP fusion protein. On the merge picture you can clearly see that DNA is pulled to the opposite poles of the cell which are designated by centrosomes.

Scheme below represent another example of centrosomal protein staining. Where gamma-tubulin and another centrosomal protein are visualised.

Picture taken by Kliszczak M.

Again the DNA stained with DAPI appers blue on the merge photo. In this case there are three different cells. One which is not dividing (an interphase cell) and the other which just finished the division (two cells on the right hand side, you can tell it because DNA is sitll compacted).
On the green and red channels you can see centrosomal proteins which appear as bright spots.
Merge channel talks for himself:)

As you probably noticed the Flurorecent Microscopy is a powerful tool which allows to visualize a specific protein, protein complex or even an organelle. With this technique we can monitor protein behaviour. For example if the protein X is a cytpolasmic protein but during mitosis or after, let say stress response (like DNA damage, or heat shock response) it is targeted to nucleus or chromatin (DNA) we can easily detect that shift.
The ultimate technique now for visualizing protein of interest is live imaging (which will appear in on this blog soon:). Using a specially designed microscopes we can follow live cell which express a fluorescent fusion (our protein + fluorescent tag) of our protein of interest and in real time its localization.

I hope you enjoy todays post:)

Have a nice day.

Maciek GGSTEAM

GGS LIVE - Western Blotting

2009-11-29T15:55:00.220Z

Method: Protein Electrphoresis and Western Blotting

About: Simply saying protein electrophoresis is a technique that allows to separate mixture of proteins by either their size or charge. After protein separation specific antibodies are used to detect and analyse protein of interest (western blotting). Before you go further please have a look at this animation about the

protein electrphoresis Biosolutions Blog - Protein Electrophoresis.

What: Detection of protein A, B and C in human protein extracts.

How: SDS-PAGE (sodium dodecylsulfate polyacrylamide gel electrophoresis) is a form of protein electrophoresis where detergent (SDS - to see the structure click here SDS structure) is used to give all proteins a negative charge so they migrate in the same direction in the polyacrylamide gel. Proteins migration in the polyacrylamide gel is similar to DNA migration in the agarose gel. Concentration of the polyacrylamide (usually expressed as %) defines gel pores size. Small protein easily get through the pores and migrate faster where bigger proteins do not and migrate slower.

In this experiment we are going to analyse five different human cell lines (called clone A, B, C, D, E) for presence of protein A, B and C. Before we separate proteins by gel electrophoresis we need to prepare our protein extracts. There are different ways of protein extraction but most of them contain cell lysis, debri (cell membranes etc) and DNA separation and protein concentration quantification steps. Example of a such extraction is presented on the figure below:

After protein extraction and concentration quantification from Clones A to E we prepare samples for gel electrophoresis. We take the same amount of the protein for each extract. What I mean here is that for example we take 30microg of the proteins from each of the Clones A-E. Suplement that with the sample buffer which contains:
- reducing agent, usually beta-mercaptoethanol (to reduce disulfide bonds in proteins - denaturation),
- glycerol, which makes our samples heavier and easier to load on the gel,
- dye, usually bromophenol blue (makes our sample loading visible),
- dergent, a sodium dodecylsulfate (denaturates proteins).

After sample preparation, proteins are boiled (at 95 degrees Celsius for 3-5min) and loaded onto gel, like you can see in the figure below (remember that ladder, a protein reference mixture has to be loaded as well).

Dark blue bands which you see here represents proteins (in the real life, we would not see proteins separating) and light blue bands are ladder proteins. If we are using a prestained marker we would see ladder bands on the gel separating in real time:). After the electrophoresis, proteins are transfered to a membrane beacause working with gel in subsequent steps would be very hard and chellenging (polyacrylamide gels are very fragile). There are different types of membranes (for example PVDF - polivinyldiene fluoride, or nitrocellulose) and all of them have high affinity to proteins. In this step membrane is placed on the top of the gel and squeezed together in a special cassette. Again current is applied and proteins migrate from the gel and stick to the membrane (see figure below).

Notice that position of the proteins on the membrane is the same as on the gel before the transfer. Observe also that other binding sites on the membrane are still unoccupied. Because membrane has high affinity to proteins we need to block those sites. Blocking of this sites is necessary as in subsequent detection step we will use antibodies (which are also proteins) and those could stick to the membrane giving us confusing background signal (see figure below). Usually for blocking we use source of neutral protein like low fat milk, BSA (bovine serum albumin) or FBS (fetal bovine serum).

After blocking membrane can be cut into pieces and then each piece is incubated with the specific primary antibody which recognize the protein of interest (usually overnigth, at 4 degrees Celsius) and then with secondary antibody (1hr, at room temperature) that recognize the former one. Usually secondary antibody is conjugated to another molecule which is capable of producing signal that latter is detected. In this case it is a Horse Reddish Peroxidase (HRP) an enztme, which catalyse reaction and one of the products of that reaction is a photon. Complex Protein-Primary Antibody-Secondary Antibody (called a sandwich complex) is formed and a substrate of HRP is added. This way we can visualize the protein of interest (see figure below).

Membrane is placed in a special cassette and photographic film is placed on it. Film is exposed in the place where the protein of interest is present. Developing the exposed film reveal presence of the protein (see figure below).

After developing, proteins appear on the film as different size bands. As you remember in our experiment we have been analysing five different human cell lines (Clones A-E) for presence of the proteins A, B and C. Notice that for each experiment, membrenes were incubated with antibodies specific for protein of interest and control protein. The control blot is performed because it tells us that we have loaded the proteins onto gel and and additionally, it tells us if the amounts are equall between the samples.

Lets have a look on our results.

The membrane on the left hand size show analysis of the protein A. Firstly look at the control protein. You can see that the amount loaded is equall everywhere. You can clearly see that protein A is present in all of the analysed cell lines and additionally that levels of protein A between Clones A-E is the same.

The middle blot show the same for the protein B (in this experiment we analysed only Clones A-D). In this case protein B is present in Clones A, B and C but not in the Clone D. We can be quite sure about that conclusion because we did loaded proteins as the control protein is present in that sample.

The right hand size membrane show analysis of protein C. And similarly you can see that proteins are loaded equally and that protein C is present in all of the samples analysed. On this blot we can additionally see that protein C is being phosphorylated in vivo (what means within cell). Protein phosphorylation can be detected by Western Blotting because usually phospho forms of proteins migrate differentially than unphoshorylated forms. In this case phospho form of protein C migrates slower and appears as a shifted band above the unphosphoryalted form of the protein. Conclusion that can be draw from that blot is: protein C is present in the Clonses A-E and additionally that protein C exist as a phosphoform in the Clones C-E.

Remember that analysis of the blots must be performed only in the presence of appropriate controls.

That is it:)

I hope you like it:)

Maciek GGSTEAM

GGS LIVE - DNA digest

2009-11-25T19:38:00.007Z

Method: DNA digest

About: Simply saying DNA digestion is cutting the DNA into pieces (before you go further have a quick look at this animation about restriction endonucleases McGraw Hill Animations - Restriction Endonucleases). What are the purposes of DNA digestion? They can be different like analytical (to see if the DNA is what we thing it is), preparative (to extract specific DNA sequence) or others (for example southern blot, cell transfection or DNA based assays).

What: Digestion of the plasmid DNA with either single, double or no cutting enzyme.

How: By incubation of the digestion reaction containing DNA with appropriate restriction enzyme.

In our experiment we are going digest our DNA sample with two different enzymes. One of them (BamHI) cuts only once in our sequence but the other restriction enzyme (HindIII) digest our DNA at two different sites. As a control for the experiment we will incubate DNA sample reaction containing no enzyme. Our DNA is a plasmid DNA of around 5,5kb in size (see picture below):

Regular digestion reaction is composed of:
                              x microl of DNA
                              5microl of 10x Buffer
                              5microl of 10xBSA
                              0,5 - 1,0microl of enzyme
                              Up to 50microl of high pure water
where:
DNA - our sample (amount of the DNA used depends on assay)
Buffer - delivers appropriate conditions for the enzyme
BSA - presence of other proteins (Bovine Serum Albumin) increase enzyme stability
Enzyme - no comments:)
Water - no need to comment:)

Usually we incubate the reaction at 37 degrees C for ... (time of incubation depends on amount of DNA or enzyme used). After the digestion an agarose gel electrophoresis is performed to see if the digestion occured (see below).

As you see at the picture above as expected, single cut and double cut of our DNA results in single 5,5kb band or two 2,5kb and 3,0kb bands, respectively. Remember that plasmid DNA can exist in different conformations, relaxed or supercoiled. Supercoiled form of DNA (in which plasmid DNA usually exists) is more compacted and that is way uncut DNA appears as a smaller band on the agarose gel (it migrates faster than expected from its size).

This is it:)

I hope you enjoy it:)

Maciek

GGS Team

GGS LIVE - Cloning

2009-11-22T13:39:00.014Z

Method: Cloning

About: What is cloning you ask! I see cloning as set of methods which allow to copy specyfic piece of DNA and then to use it as we wish.

What: Amplification of three different cDNAs (cDNA is a protein coding part of the gene for more informations see this amnimation McGraw Hill Animations - cDNA (complementary DNA)) by PCR and transfering them (cloning) into first transient vector and afterwards to a final (target) vector. Each of our cDNAs is around 300bp.

How: by series of DNA digestions and ligations.

Before we start please have a look at this cool animation about the cloning:) McGraw Hill Animations - Steps in Clonig a Gene

When we start cloning from the scratch first what we have to do is to amplify the desired DNA sequences using the PCR technique (about the PCR in the future Biochemistry Methods Section:). After amplification, sample of the reaction (around 10%) is analysed by agarose gel electrophoresis to check presence of the desired product. Example of such gel you can see on the Figure 1.

You can see here three different products of PCR (each around 300bp, compare to the ladder DNA) which from now on we will call S1, S2 and S3. Positive Control lane is a PCR product that previously was shown to be ampliefied (we used that as a validation of our PCR) and in the Negative Control lane no DNA tamplate was used (no PCR product is expected). After we make sure that our products are fine (and they are, all of them have expected size, PCR worked ==> Postive Control) we need to prepare another gel where we separate rest of the reaction volume (95% that left) and extract/purify the desired DNA band from the gel. Look at the Figure 2:)

After extraction purity of the DNA is analysed (again by running the gel) and then DNA is ligated/fused with transient vector. Vectors used in this step are designed to facilitate the fusion of the PCR product with the plasmid. In our case we will use a vector that is provided in linearised form, ready to go:) See Figure 3.

As you can see size of our vector (which we are going to call pXT) is around 3,0 kb (3000bp). After the ligation, bacterias (usually the E.coli) are used to multiply the plasmid so we will have a good amount to work with. After plasmid DNA isolation out of the bacterias we need to test if the ligation took place, so we perform a control DNA digest. See Figure 4.

From the different restriction enzymes available (I mean those that cut our plasmid and insert DNA) we chose those that will give us the simpliest (or the most obvious) band pattern. Example above:) I choose BamHI enzyme that cuts once in the backbone of the vector and once in our DNA of interest (all three cDNA's are similar sizes and have the single BamHI site so I did not show the other two). This way digest of the DNA should give us two bands, if the ligation took place and only one band (around 3,0kb), if the ligation failed.

After we make are sure that we have our DNA ligated into transient vector, we might either sequence the DNA to check if it is what we expect (mutations can be introduced during PCR reaction) or go to the next step (it depends on what we are going to do with the DNA after the cloning).

In case of the cDNA it is common to get it sequenced. When we get the sequence results back and everything is ok we go to the next step which is to clone the insert DNA of interest from transient vector to the final vector:) In order to do that we digest the target vector (pXA, size around 5,5kb) and cut the inserts out of the pS1/S2/S3 vectors with the same restriction enzyme mix. Remember that each restriction eznyme generates different DNA ends. To make our live easier, we design cloning in the way that we can use the same restriction enzymes to cut the vector and inserts out. This way we generate DNA ends that are compatible what increase efficiency of the ligation step. See Figure 5.

On the left side of the Figure 5 we see our target vector (size 5.5kb) and our transient vectors (pS1/S2/S3) bearing the inserts. We digest them with HindIII/XbaI enzyme mix to linearize target vector and to cut our inserts out like you see above. In the next step we have to extract the insert DNA from the gel to separate it from the backbone. Afterwards we ligate the inserts with the target vector. Then using bacterias we amplify DNA and again after isolation of plasmid DNA we set up the control digest to see if we got it ligated (this time to the final vector). Check it out on Figure 6.

This time I choose a restriction enzyme PvuI that cuts three times in the backbone of the pXA vector and either once or does not cut at all in the insert DNA. As you see depending on the insert sequence the PvuI site is in different position what results in specific band pattern for each of the vectors. Please notice that bands 1069bp and 1096bp will appear as one big band as they do not seperate greatly (their size is too similar).

This is it:) After we make sure that our plasmids are ok (we can perform another control digests) we can finally use our DNA. About that in the future Biochemistry Methods Section.

Cya Soon:)

Maciek GGS Team

3rd year practicals - exam paper - Question 6!!

2009-11-20T17:49:00.005Z

Hello You All,

Finally we reached the last Question of the 3rd year practical exam sample paper:) I hope you did enjoy the problem solutions provided and that you feel a little bit more confident about the exam. Remember to believe in your skills. That will defenitely help you to go through the exam questions smoothly:)

Aha I almost forgot, it is getting better beacause there is more to come:) In the near future you will find here (in the different section called 3rd year Biochemistry) a sample paper of the lecture exam with the solutions. Can you believe it:) Ok let's finish first the practical paper:)

All let's rollin' baybe! Question 6 Part a) This question is very simple. First two parts are about mixing stuff:) You are asked to explain how to prepare 250ml of 25% (w/v) ammonium chloride solution. Remember that:

(w/v) stands for weight to volume

(v/v) stands for volume to volume

Usually buffers or solutions are prepared with water. If you remember the density of the water is roughly 1.0kg/L = 1g/ml so it means that 250ml of water weights 250g. So mass of our solution should be around 250g. We have

mass of the ammonium chloride = 250g x 25% = 250g x 0.25 = 62.5g

So to prepare our solution we need 62.5g of the ammonium chloride and fill it up with the water to 250ml to get in total 250g of the solution. So we need to add about 250g - 62.5g = 187.5g of water = 187.5ml of water:)

In the second part you are asked to calculate what is the concentration of the solution c2 if you resuspend 60g of ammonium chloride in the 200ml of water. So:

200ml water = 200g of water = Solvent Mass

60g of ammonium chloride = Substance Mass

%Concentration C2 = (Substance Mass / Solution Mass) x 100%

where

Solution Mass = Substance Mass + Solvent Mass

We have:

%Concentration C2 = 60g/(60g + 200g) x 100% = 23%

Part b) You are asked about the same stuff here, but the author of the question wanted to complicate things a little bit more (but we will get it anyway:). So we need to prepare 250ml of the 0.4mol/L = 0.4M sodium chloride (58.4g/mol) solution which is supplemented with the 0.2% sodium azide (65.01g/mol). First we will deal with the sodium chloride. We have:

Concentration of NaCl = 0.4mol/L = 0.4M

Finval Volume = 250ml = 0.25L
Remember this simple equation: Molar Concentration = n (number of moles) / v (volume)

number moles of the NaCl required = 0.25L x 0.4mol/L = 0.1mol

We know that 1mol of the sodium chloride weights 58.4g, so how much 0.1mol weights:)

1mol - 58.4g

0.1mol - xg

x = (58.4g x 0.1mol) / mol = 5.84g mass of the NaCl required

We calculate mass of the the sodium azide in the analogous way to the part a) So,

Mass of the sodium azide = 250g x 0.2% = 250g x 0.002 = 0.5g

And again we fill it up to 250ml with water.

In the second part you are asked to calculate the concentration of sodium chloride and sodium azide if you dilute prepared solution to 1L. This is easy:) First we calculate the dilution factor:

dilution factor = final volume / starting colume = 1L / 0.25L = 4

So we know that we have diluted our solution 4 times. So:

final concentration of sodium chloride = starting concentration / 4 = 0.4M / 4 = 0.1M

final concentration of sodium azide = starting concentration / 4 = 0.2% / 4 = 0.05%

Part c) Here you need to explain what accuracy and standard deviation means.

Accuracy - (simply saying) it is clossenest of the measurment to its actual (real) value.

Standard deviation - (simply saying:) tells you how thightly different measurments are clustered around the mean in a set of the data.

Part d) In this part you need to actually calculate the standard deviation of different experiments (A1 - A9) using adsorbance as your sample readout. We know that:

where:

SD - standard deviation

Xi - value of single measurment

X with the dash = mean of all measurments in experiment

n - number of samples

The other creepy stuff in that formula means that you have to sum up (greek sigma letter) square of the difference between sample and the mean for of all the samples. To get our SD we prepare Table:

That is it:) After the SD calcilation is done you need to comment on the precision of each experiment. So first we will look what the precision is:)
Precision - which might be called reproducubility or repeatabillity as well is the degree to which repeated measurments under the same conditions show the same results.
Now we go back to our Table and compare SD of our experiments which actually say about our precission (there where measurments are precise the SD is small) and we comment. For example, the experiment A6 is the most precisie and the worts one is experiment A5 etc (of course you should put more than one sentence:)

Part e) As usually the last part is about simply calculations:

c1 = 50microM = 0.050mM = 50 000nM

c2 = 5nM
dilution factor = c1 / c2 = 50 000nM / 5nM = 10 000 [no unitis:)]

c3 = c1 diluted 500 times = c1 / 500 = 50microM / 500 = 0.1microM = 100nM = 0.0001mM

c4 = 5.84% = it means that in 100g of the solution there is 5.84g of the NaCl (100g x 5.84% = 5.84g)
n (number of moles) = mass of the substance / molecular mass
nNaCl = 5.84g / 58.4 g/mol = 0.1mol
Molar Concentration = number of moles / volume
MC of NaCl = 0.1mol / 1L = 0.1mol/L = 0.1M

We have done it:) CyA Soon:)

Remember that if you have any questions leave the comment on any post or mail me at m.kliszczak1@nuigalway.ie!!

Maciek GGS TEAM

3rd year practicals - exam paper - Question 5!!

2009-11-19T19:42:00.003Z

Hello Guys,

Today wego with the Question 5.

Check this out:)

Question 5 Part a) In this part you are asked to calculate the Vo (initial rate mmol/min/ml) of the Alkaline phosphatase. All details of the assay and obtained adosrbance are listed. To get the Vo we need: adsorbance (A410nm = 0.754), time (5min), volume of the reaction (3ml) and extinction coefficient (17500 1/Mcm). We know that:

A = e c l

where: A - adsorbance

e - extinction coefficient (1/Mcm)

c - concentration (M)

l - path length (cm)

To get the product concentration (c) we need to convert the formula:

c = A / (e x l)

Now we substitute the values:

c = 0.754 / (17500 x 1) = 4.3 x 10^-5 M = 4.3 x 10^-5 mM/ml

We know that assay was done for 5min in 3ml and 1/100 dilution of the enzyme was used. We need to inculde that data in our calculations. We have:

Vo = (c x 3ml x 100) / 5min = (4.3 x 10^-5 x 3ml x 100) / 5min = 2.58 x 10^-3 mmol/min/ml

In the Part b) you are asked to prepare Michelis - Menten and Lineweaver - Burk plots using the data provided in the Table:

To plot the Michaelis-Menten plot you use data from the Table, but to prepare the Lineweaver-Burk plot you need to calculate the reciprocals of the [S] (substrate concentration) and Vo (initial rate). After you have them you construct two plots (remember to scale as accurate as possible, in that way your results will be very accurate).

Part c) Remember that for the MM plot you read out your Vmax and Km values directly from the plot but for the second graph (LB plot) you need to do simple calculations to get the Vmax and Km. For example from the MM plot we can conclude that approximately Vmax = 5.1 mmol/min and that Km = 1.6mM. From the LB plot we have (approximately:):

y-intercept = 1/Vmax = 0.15

Vmax = 1/0.15 = 6.7mmol/min

x-intercept = -1/Km = -0.35

Km = -1/-0.35 = 2.8mM

Remember that values for Vmax and Km obtained from the MM plot are less accurate that from the LB plot. Why is that? Because linear model is easier to describe than the expotential one. I mean that it is always easier to find a linear equation that will fit to our data than the expotential one:)

Part d) Calculations again:) I will do it fast this time (if you need further assistance email me at m.kliszczak1@nuigalway.ie or look at the previous Questions from this exam paper).

c1 = 300mM = 300 000 000 = 3 x 10^8 nM

dill. factor = c1/c2 = 300 000 000 / 30 x nM / nM = 3 x 10^7 = 10 000 000 = 10^7(dilution factor no units)

c1 / dlution factor = c3 = 300 000 000nM / 15 = 20 000 000nM = 20mM

That would be it:) Tomorrow the final question:)

Cya

Maciek GGS TEAM

3rd year practicals - exam paper - Question 4!!

2009-11-18T19:16:00.005Z

Hi you all 3rd years !!

Today we continue to with the 3rd year practical sample exam paper:)

Question 4 Part a) Main reagents of the PCR reaction are:

DNA tamplate - I think I do not have to explain that:) DNA template contains the piece of the DNA that we wish to amplifiy.

Primers - short pieces of DNA (oligonucleotides, usually around 20-40 nucleotides) that specify which sequence of DNA will be amplifiied.

Buffer - usually provided as 10x. Diluted buffer provides optimum conditions (pH, salt concentration, coffactors etc) for the PCR reaction.

dNTP's - deoxynucleotidetriphosphate's (dATP, dTTP, dCTP, dGTP) are substrates in PCR reaction and are necessary to elongate the primers what result in PCR product.

MgCl2 - magensium chloride is essential reagent in the PCR reaction (sometimes it is icluded in the buffer solution but not always). MgCl2 (actually the magnesium divalent cations) stabilize the negtive charge of the dNTP's (remember that phosphate groups on nucleotides are negatively charged). This stabilization helps in alignment of the nucleotides during their incorporation to newly synthetized DNA strand. You can imagine it like that:)

This interaction is way more complicated (first it is not flat like that:). This is just to give you an idea what is happening:)

Enzyme - those are always DNA polymerases. There are different types of polymerases used for PCR purposes but all of them are thermostabile. Some of them are more error prone than others. Examples of the polymerase used: SigmaTaq, Takara LA Taq, KOD Polymerase and more.

Water - used as an essential solvent.

In the second part of the question you are asked to explain differentce between PCR and RT-PCR. RT-PCR can stand for either Real-Time PCR or Reverse-Transcriptase PCR. I will give you short definitions of both so you will know the main differences between them.
Real Time PCR is a very sensitive version of the regular PCR where real time analysis of the obtained product is possible. For that purpose fluorochromes are used to labell the arisen products what gives us possibility to quantify the amount of the product in whatever time during the reaction. Simply Real Time PCR is a quantitative version of the regular PCR.
Reverse Transcriptase PCR is a reaction in which cDNA (protein coding DNA sequence) is produced using the RNA. What it that means is that RNA is first used as a template to produce DNA (Reverse Transcriptase is an enzyme that uses RNA as a template to produce DNA) and then this DNA is used to amplify our sequence of interest.

Part b) Agarose is a polysaccharide obtained from agar. Agarose has wide range of applications and one of them is gel electrophoresis. Agarose can form gel like structures with the pores depending on the agarose concentration (usually expressed as %). Agarose electrphoresis is mainly used for the separation of the nucleic acids like DNA or RNA (but it can be used for protein separation as well). Simply nucleic acids loaded onto gel will migrate through the gel pores depending on their size. Small nucleic acids molecules will migrate faster (they go easier through the pores) than the big ones.
Ethidium Bromide (usually abbreviated as EtBr, do not confuse with ethylene bromide) is an intercalating agent. Interacts with the DNA is mediated through hydrophobic interactions between aromatic rings of EtBr and nucleotides (nitrogonous bases of nucleotides contain aromatic rings). See figure below:

Taken from http://www.madsci.org/posts/archives/1999-02/919869466.Mb.1.jpg

Ethidium Bromide is a fluorophore which have maximum adsorbtion around 300nm (it can be visualised with UV lamp and has an orange like colour). Its adsorbtion is even greater when bound to DNA. Remember that EtBr is a mutagen because its intercalation cause change in DNA structure and may interfere with DNA metabolism processes like DNA replication

Part c) In this question you are asked to estimate size of the unknown DNA molecules using the standard curve. First you have to calculate the -logRf (you can calculate the logRf as well that do not change enything) and put them together in the Table:

Then we plot the -logRf vs Molecular Size (bp). Using the standard curve you read out the molecular size of the unknown samples using the -logRf. I have used equation to calculate the sizes of the unknown samples but you have to do that manually (just remember to use as accurate scale as possible). Notice that unknown sample C gives me a negative value (which does not make any sense because DNA molecule can not have negative mass:). The reason for is that Rf of this sample is outside the range of standard curve:) Be careful and alert, it often happens that there is something tricky about the questions, lectures want to confuse you, so you have to be sure and confident about your calculations and answers:) Do not let them win:) In this case you have to indicate that Rf of this sample is outside the range and it is not possible to estimate its molecluar mass.

That is all for today.

Remember that if you have any more problems or if something is not clear or understandable, just let me know. Leave a comment or send me email to m.kliszczak1@nuigalway.ie.

Tomorrow will take care of the Question 5.

Cya

Maciek

Scientific prefixes - you will be laughing:)

2009-11-17T15:30:00.004Z

Hello you all,

Through all the lab practicals I have been demonstrating I saw that students have problems with the scientific prefixes convertion for example from kg to mg or microA to kA etc. This idea is crazy but maybe that is way it may work for some of you (I think that more crazy method you use to remember something is better beacause you get it forever:). In this post you will find a way to remember and use the scientific prefixes.

If you have any suggestions about that post, please let me know:)

Ok:) Let's go:

If this made you laugh it means it is ok:) More laughable something is, it is easier to remember:)

Greetings:)

Maciek GGS TEAM

3rd year practicals - exam paper - Question 3!!

2009-11-17T15:11:00.002Z

Hi ya all,

It is time for another Question from the 3rd year practicals sample exam paper:)

Question 3, part a) In this one you are asked to explain principle of measuring the adsorbance of protein sample at A260 and A280. If you remember from the Question 2, A260nm is used to measure concentration of DNA. A280nm is specific for the proteins because at that wavelength aromatic aminoacids adsorb the light. If you measure adosrbance of you protein sample at 260nm and 280nm, you can see if your sample is mainly proteins or proteins and DNA (you estimate purity of the sample). So what you should get is low adsorbance at 260nm and high at 280nm.

You can find all aminoacids structures, three letter codes, one letter codes etc. here:

Way to remember aminoacids

or here

Way to remember aminoacids - chemist way

or here:)

Interactive aminoacid quiz

Part b) In here you have to use the data provided in the Table (see below) to prepare standard curve and estimate concentration of unknown samples.

Using the known concentration of BSA standard and corresponding adsorbance, we construct plot concentration against the adsorbance. Remember to label the x and y axis (with the units if applicable). Using the standard curve you estimate the concentration of the Unknown samples which correspond to particular adsorbance. In the case of dilutied samples do not forget to multiply obtained concentration by the dilution factor (for example Unknown A have to be multiplied by 20 as dilution factor is 20:). I used the trendline here and the equation to estimate the unknown samples concentrations. But you probably will have to do that on the paper, so remember to scale your plot as accurate as possible:)

Part c) I already answered that question. Look up here (Question 1 part a):

Specific Activity of Enzyme

To calculate specific activity of the Unknown sample E we use the concentration estimated from part b). In my case is 20.66microg/ml, we also need Total Enzyme Activity (Activity of the sample) which is 540UI/ml.

So to get the Specific Enzyme Activity we have to divide Total Enzyme Activity (Activity of the sample) by the concentration of the sample:

Specific Enzyme Activity = 540UI/ml / 20.66microg/ml = 540/20.66 UI/ml x ml/microg

Specific Enzyme Activity = 26.14UI/microg

See also Question 1 part a from the same exam paper:

Specific Enzyme Activity

Part d) As usually part d of each question is about simple calculations. First we need to express solution

c1 = 150microg/ml in kg/L. We know that there is 1 000 000 microg in kg so we need to divide our value by 1 000 000.

c1 = 150kg / 1 000 000ml = 0.00015kg/ml

and we know that there is 1000ml in L so we need to multiply the number by 1000:

c1 = 0.00015 * 1000 kg/L = 0.15kg/L

Next you are asked to give the concentration of the solution c1 after dilution by factor of 30. So it can not be simplier:). You just divide the c1 by 30:

c2 = c1 / 30 = 150microg/ml / 30 = 5microg/ml

And finally you are asked to calculate the what is the dilution factor if you go from solution c1 = 150microg/ml to solution c3 = 15ng/ml. As you remember first we have to express the solution c1 in ng/ml or solution c3 in microg/ml to have the same units. Let's say we choose the first option. We know that there is 1000ng in microg, so we need to multiply our c1 concentration by 1000:

c1 = 150microg/ml * 1000 = 150 000 ng/ml

In the next step we need to calculate how much more solution c1 is concentrated that solution c3. To get that, which is actually our dilution factor we divide c1 / c3:

dilution factor = c1 / c3 = 150 000 / 15 ng/ml x ml/ng = 10 000

Remember that dilution factor do not have any units because all of the get reduced:)

Problems with convertion of scientific prefixes?? If yes go here Prefixes convetion - you will laugh:)

And that how it is done:)

Good luck:)

Maciek GGS TEAM

3rd year practicals - exam paper - Question 2!!

2009-11-16T23:17:00.004Z

Hello Again,

Today we will work with the Question 2 of the 3rd year practical - sample exam paper. I hope you are ready:)

Question 2 Part a) Everything you need to know is on the Figure below:

and here:

Taken from http://en.wikipedia.org/wiki/File:Nucleotides_1.svg

The main differences between DNA and RNA are listed in this Table:

Part b) In this question you are asked to explain role of the EDTA, sodium dodecylsulphate (SDS) and isopropanol in DNA extraction.

EDTA - ethylenediaminetetraacetic acid, is sequestering agent which is used to bind divalent metal ions like Mg2+, Ca2+ etc (see Figure below).

EDTA is used to "neutralize" divalent cations. In the case of the DNA isolation it is used because cations like Mg2+, Mn2+ (and others) are coffactors of enzymes that digest DNA. Simply saying, to prevent DNA degradation we inactivate the enzymes by removing the divalent ions that are necessary for their activity.

SDS - sodium dodecylsulphate is a strong detergent. As all detergents it contains a hydrophobic and hydrophilic part. See below:)

In DNA purification we use SDS for two main reasons: SDS will lyse the cells (by disrupting the cell membranes) and denaturate/precipitate proteins which in this process are dispensable. You can imagine that hydrophobic part of the SDS interacts with the hydrophobic regions of the protein what cause their precipitation.

Isopropanol - is a small molecular weight alcohol. See below:)

Isopropanol is used to precipitate the DNA because alcohols (as you probably now:) dehydrate. Just remember that alcohols form hydrogen bonds with water very easily what cause DNA solubility to drop down ==> result DNA precipitation. We can say that alcohols "bind" water molecules better than the DNA, so in the "absence" of the water molecules DNA precipitate.

Part c) In this part you are asked to calculate either DNA concentration knowing adsorbance of the sample or adsorbance of the sample knowing the DNA concentration. Look into the table:

These calculations are very easy: we know that if we masure DNA adsorbance in the cuvette of 1cm thick and the reading gives us 1.0, DNA concentration is equal to 50microg/ml. So we use simple proportion:

50microg/ml ==> 1.0 adsorbance

then

x microg/ml of DNA gives us for example (solution A diluted) 0.858

So to get the DNA concentration of diluted solution A we have to multiply the 50microg/ml by the adsorbance of the sample:) And we have:

50microg/ml x 0.858 = 42.9microg/ml

If the sample was diluted prior to adsorbance measurment you have to include the dilution factor which in case of the solution A is equal to 20:) It can not be simplier:) So neat sample A have concentration 20-times more so:

42.9microg/ml x 20 = 858microg/ml

TADA:)

For calculating the adsorbance from the concentration of the sample, we do opposite calculation:

50microg/ml ==> 1.0 adsorbance

75microg/ml (solution F) ==> x adsorbance

So to get the adsorbence we have to divide the concentration of the sample (in this case 75microg/ml) by the reference concentration (50microg/ml). And we have:

50microg/ml / 75microg/ml = 1.5 adosrbance

That is it:)

Part d) In this part you are asked to perform few simple calculations. What we have here:

solution c1 = 330kg/L and express it in microg/deciL. So we know that there is 1 000 000 microg in kg so we multiply the 330kg/L by milion:

c1 = 330kg/L * 1 000 000 = 330 000 000microg/L

In the next step we have to express the L in deciL (dL). We know that there deci is 1/10, so it means that there is 100dL in L. We have then:

c1 = 330 000 000microg/L = 330 000 000 microg/100dL = 3 300 000microg/dL

You are also asked to express the number in scientific writing, so:

c1 = 3 300 000microg/dL = 3.3microg/dL x 1 000 000 = 3.3 x 10^6 microg/dL

Further we have to calculate the dilution factor after diluting the c1 = 330kg/L solution to solution
c2 = 15mg/mL. First we have to express either c1 solution in mg/ml or c2 = kg/L. Let's choose the second option..

c2 = 15mg/ml, we know that there is 1 000 000mg in kg so we divide it by 1 000 000

c2 = 15mg/ml / 1 000 000 = 0.000015kg/ml

next we know that there is 1000 ml in L so we divide the ml by 1000

c2 = 0.000015kg/mL/1000 = 0.015kg/L

To get the dilution factor we divide c1 by c2 (starting solution by the final solution):

dilution factor = c1 / c2 = 330kg/L / 0.015kg/L = (units get reduced) = 330/0.015 = 22000 !!

Again the dilution factor do not have any units (because all of them get reduced).

In the next section of the part d) you are asked to dilute the solution c1 = 330kg/L by dilution factor 66 to obtain solution c3. So you just have to divide the 330kg/L by 66:

c3 = c1 / 66 = 330 / 66 = 5kg/L

The last one is to express concentration of the solution c4 = 4mg/microL in ng/pL. We know that there is
1 000 000 of ng in mg so again we multiply the number by 1 000 000.

c4 = 4 x 1 000 000 ng/microL = 4 000 000ng/microL

We know also that there is 1 000 000 pL in microL. We have then:

c4 = 4 000 000ng / 1 000 000 pL = 4ng/pL

If you still have problems with convertion of the scientific prefixes visit this post Prefixes convertion - you will laugh:)

That is it:)

I hope you enjoy it:)

Maciek GGS Team

3rd year practicals - exam paper !!

2009-11-15T23:53:00.003Z

Hi ya all students,

Today in the 3rd year practical section you will not find lab report tutorial but I think you will be interested as well:) In the next few posts you will find all the sample exam paper anwsers:) Everyday (since today) one question will be solved!

So lets begin:

Question 1) Part a) In this question you have to calculate different values using the data provided in the Table. This is easy busy. Like here:

To calculate the Specific Enzyme Activity (IU/mg) you have to divide the Total Enzyme Activity (IU) by the Total amount of the protein (mg). It means that you have to express how much enzyme activity is there for each mg of protein present in the extract. Actually as you look at the unit of the Specific Enzyme Activity (IU/mg) it tells everything:).
To calculate the Purification Factor you divide the Specific Enzyme Activity (of each purification method) (IU/mg) by Specific Activity of Crude extract (IU/mg).
To get the Yield (%) you express in % how much of the protein is there after purification in relation to protein content of the Crude Extract.

Part b) In this part you are asked to explain what Specific Enzyme Activity and Total Enzyme Activity are? The first one is already explained in part A (no need to do that again).
Total Enzyme Activity is (as the name suggest) total activity of the sample you have obtained. So if you examined activity of let's say 1ml sample taken from your extract and it gaves you for example 5.0IU. And you have in total 40ml of your extract/purification your total enzyme activity is 40x5.0 = 200IU.

Part c) In this part you are asked to discuss 2 factors that can influence the enzyme activity in the assay. There is a lot of them, for example:
- buffer (wrong pH, salt concentration (ionic strength), detergents, etc),
- temperature (to low, to high),
- purity of the enzyme (low purity ==> low activity),
- inhibitors,
- enzyme handling,
- other molecules essential for enzyme activity (like coffactors, ions etc),
- and more:)

Lets say you choose buffer and temperature. In you discussion you should say that each enzyme works in the optimal pH (different for different enzyme) and if the buffer is not pH properly then to high or to low pH will affect our assay (it may even cause lost of enzyme activity). It goes similar with the temperature. Again as you imagine there is temperature range/optimum in which each enzyme works. Wrong temperature used during the assay will affect our results.

Part d) In this question you have to perform simple calculations (baby ones:). First you have to express 40IU/ml as mIU/microL. So, we know that there is 1000 mIU in IU yeah? So we multiply our number by 1000, what gives us this:

40IU/ml x 1000 = 40 000mIU/ml

And we also know that there is 1000 microL in ml. We have:

40 000mIU/1000microL yeah? = 40mIU/microL

Done:) If you still have problems with the refixes convertion go here Prefixes Convertion - you will laugh

Next one is even easier. You have to dilute the solution c1 = 40IU/ml by factor 20:) So you just divide the concentration of the solution c1 by dilution factor 20:) And you have:

40IU/ml / 20 = 2IU/ml

That is it:)

Next, you have to dilute solution c3 = 15mg/ml to solution c4 = 10microg/ml and caluclate the dilution factor. So to get the dilution factor you have to simply calculate how much more solution c3 is concentrated than the solution c4. To get that you divide the c3 / c4 but first you have to express either c3 in microg/ml or c4 in mg/m. Let's say we choose the first option so:

c3 = 15mg/ml = as we know there is 1000microg in mg = 15 x 1000microg/ml

So now:

dilution factor = c3 / c4 = 15 000microg/ml / 10microg/ml

dilution factor = c3 / c4 = 1500~~microg~~/ml x ml/~~microg~~

dilution factor = 1500 and the unit:) There is no unit:) Because all units are reduced.

I hope you find these answers clear and straight. Tomorrow there we will KILL Question 2:)

Maciek GGS Team

Amino acids - interactive way:)

2009-11-12T00:34:00.002Z

Hello Again,

Welcome in the Aminoacids are easy section. Remember aminoacids are easy:) The most important is to think they are easy and rest will go smoothly:) I have just found great website where you can learn aminoacids by playing an interactive quiz. It is worth trying if you want to remember the aminoacids:)

Check this out:

http://wbiomed.curtin.edu.au/teach/biochem/tutorials/aaquiz/

Have fun:)

Maciek

GGS TEAM

3rd year biochemistry practicals - Alkaline phosphatase inhibition kinetics

2009-11-12T00:23:00.005Z

Hi Students,

This week you have been working with the Alkaline Phosphatase again but this time we included an inhibitor in our reaction. This practical is about measuring the inhibitor constant (Ki) of the beta-glycerophosphate. As I already said this practical was very similar to the previous one. You had to incubate the enzyme with different concentrations of the substrate and different concentration of the beta-glycerophosphate (our inhibitor). Because some of the work necessary to prepare the lab report was already done last week, I will be refering to the 3rd year biochemistry practical - Alkaline phosphatase post, which you can find on this site as well.

Ok, lets begin mad scientists:)

OH REMEMBER AGAIN, I DO MAKE UP ALL THOSE NUMBERS. IT IS JUST TO GIVE YOU IDEA HOW TO DO YOUR CALCULATIONS:)

First we construct Table 1 which look like this:

We use the adsorbances to calculate the Vo value using the same formula as previously:

Vo = 1000 x 3 x (Adsorbance -blank)/(18600 x 5)

Remember to use the Extinction Coefficient value of 18600 microM/cm

1. We convert our Adsorbance - blank values to Vo and we put them in another Table 2:

2. In the next step we plot the Vo (microM/min/ml) values against the substrate concentration [mM] to get the Michaelis-Menten plots, like here:

3. In the next step using the values from the Table 2 we calculate the 1/Substrate Concentration and 1/Vo (as in the previous lab report), getting the Table 3:

4. We plot the 1/Vo versus the 1/[S] (where [S] - substrate concentration) to obtain the Lineweaver-Burk plots like these here:

To get the trendlines we go right-click on the data points and choose the option Add Trendline, where in Bookmark called Type you choose Linear. Then you change the Bookmark to Options and tick the Display Equation on Chart. In the option forecast you can change the length of the trendline.

We will use the equations provided to go further with our lab report but first we have to go back to the Velocity Equation for the Competetive Inhibition:

1/V = (1/ Vmax) + (Km/Vmax) x (1 + [I]/Ki) x 1/[S]

where:

V - reaction velocity

Vmax - maximum reaction velocity

Km - Michealis Menten constant

[I] - inhibitor concentration

Ki - inhibition constant

[S] - substrate concentration

Because we have ploted our 1/V against the 1/[S] it means that we can express this equation like that:

y = c + mx

where:

y = 1/V

c = 1/Vmax

m = (Km/Vmax) x (1 + [I]/Ki)

x = 1/[S]

5. From the properties of the linear plot we know that m is our slope. Using the equations provided (by Excel:) we construct the Table 4:

Using the Table 4 we construct another plot Slope (for each [I]) against the [I], and we get this:):

Again we go and add the Trendline (with the equation) which will help us to calculate the Ki. From the theory we know that x-intercept of the Dixon plot gives us the -Ki value. So using the equation:

y = 89.082x + 1094.3

From the linear function we know that to find the x-intercept the y must equal 0, so we substitute the 0 for y:

0 = 89.082x + 1094.3

89.082x = -1094.3

x = -1094.3/89.082

x = -120.491

and we know that x-intercept = -Ki so:

-Ki = -120.491

Ki = 120.491 [mM]

6. In the next step we will try to get the Ki value in the different way. Using the equations from the Lineweaver-Burk Plots we calculate the Km (exaclty the same as previously but thhis time it is actually Kmapp because includes the inhibition effect) and we put it in the Table 5:

Then we plot the Kmapp vs the [I] and using the Excell we get the equation of the trendline again. The x-intercept is our Ki.

You should end up with two different Ki values. You compare and discuss them:)

This was easy huh?

CyA Next Time:)

Maciek GGS

3rd year students !!

2009-11-11T19:15:00.000Z

Hi ya all,

I am just working on the last practical report manual, so please be patient. It should appear here tomorrow evening (maybe today night but I am not sure if I will be able to finish it:).

Do not worry this was the last practical:) I will collect the labbooks on Monday ( I do not know what about the other demonstrators) so you have plenty of time.

CyA Soon:)

Maciek GGS

More sweet you eat, less life you got!

2009-11-06T20:46:00.004Z

Welcome again in the Sciance News Section.

Scientists from University of California, San Francisco find out that even small amount of the glucose (sugar) shortens lifespam of the worms (C. elegans). Some time ago scientists discovered a set of genes that extended the lifespam of the worms. Those genes are responsible for dealing with glucose that we consume. When researches disrupted those genes in worms, they life expectancy was no longer affected by glucose in their sustenance.
What that actually means for us?? Hmm lets see:)
Firslty, from research already done on worms we know that molecular pahtways that deal with the glucose work very similarly in worms and humans.
Additionally, it has been known for decades that excessive consumption of sugar is not good for us.

SO SIMPLY WE SHOULD CUT THE CONSUPTION OF THE SUGARS AS SOON AS POSSIBLE:)

I DO NOT KNOW YOU BUT UNFORTUNATELLY FOR ME IT MEANS NO BARS, NO CHOCOLATE, NO DOUGHNUTS AAAAAAAAAA YEAH YEAH I DO NOT SEE THAT COMING:)

If you want to know more go here:

http://www.scientificamerican.com/podcast/episode.cfm?id=sugar-negates-worms-life-extending-09-11-03

Or if you want to see the original study go here:

http://download.cell.com/cell-metabolism/pdf/PIIS1550413109003027.pdf?intermediate=true

Enjoy and do not eat sweets during reading:)

Cya next time

Maciek GGS

3rd year biochemistry practicals - Alkaline phosphatase

2009-11-05T18:27:00.001Z

Hi all students,

This week you have been performing the biochemical analysis of the biocatalyst. An alkaline phosphatase was used as an example.

Practical was split into two different parts. In the first one you were measuring how much of the product was produced by different concentration of enzyme. At the end of that part you had to decide which enzyme concentration is the best for our assay (correct concentration was that which gave you less thatn 10% of the product after 5min of raction). This is because you want to measure the initial rate of enzyme activity. You want to work in the conditions were not significant concentration of product is produced beacause otherwise it may affect your results (some reaction product may be inhibitors of the enzyme). If not enough enzyme is used then detection of enzyme activity becomes too complex or just belowe the detection threshold.

In the second part you were working with the chosen enzyme concentration to estimate the initial enzyme rate for different substrate concentration. In this case there where more substrate was used more product should be produced what indicates that enzyme worked with higher speed. Then you had to use these velocity values to calculate the Km (Michealis Constant which tells you how much of the substrate you need to get half of the Vmax of the enzyme) and Vmax (which is simply the maximum substrate to product conversion velocity, a characteristic feature for each enzyme).

REMEMBER THAT ALL ADSORBNCE NUMBERS WERE MADE UP BY ME. THIS CALCULATION ARE DONE TO GIVE YOU ONLY ROUGH IDEA HOW TO DO YOUR REPORT:)

GOOD LUCK

So lets beggin the calculations.

1. First you have to orginise your Part A results into a small, cool table:

Then we have to calculate the rate of the reaction, vo at each concentration of enzyme used.

Rate (vo) = x micromol product produced/min/assay

To estimate the product concentration we use the super easy Lambert-Beer Law which says that adsorbance of the substance x is proportional to the concentration and the path length.

A = e x c x l

where A - adsorbance

c - substance concentration [mol/l = M]

l - path length [1cm]

e - molar extinction coefficient (constant, characteristic for each substance) = 18.600 [L/(mol x cm)]

From that formula we extract the c which gives you:

c = A / (e x l)

For the 1/5 enzyme dilution we have:

c = 1.581/ (18600 x 1) = 0.000085 mol/L

Now we convert the mol/L to micromol/ml. There is 10^6 (1 000 000) micromols in the mol so we multiply our result by milion and there is 1000ml in L so we have to divide the result by a 1000 as well so we have

0.000085 x 1000000/1000 = 0.000085 x 1000 = 0.085micromol/ml

Now we use our calculated concentration to estimate how much of the micromoles of the product was produced per minut, per volume used in the assay. So we have to divide our result by 5 (because there was 5min incubation) and multiply by 3 (beacause reaction took place in the 3ml volume). So we have

(0.085 micromole/ml x 3) / 5 = 0.051 micromol/min/ml

In the next step we have to calculate enzyme concentration that was used for each dilution. So for the first 1/5 dilution we have: we know that neat enzyme concentration is 1mg/ml. So in 1/5 dilution we have 5 times less which is 0.2mg/ml. And we have taken 0.5ml of the dilution to the reaction which gives us 0.2 x 0.5 = 0.1mg of the enzyme. Now to calculate the concentration of the enzyme in the reaction we have to remember that volume of the reaction was 3ml, so 0.1mg of the enzyme in the 3ml gives us 0.1mg / 3ml = 0.033mg/ml.

We construct next table:

Using this table we plot the graph (see point 2 in you lab manual) product concentration produced per min per assay volume (y-axis) against the enzyme concentration [E] (x-axis) and answer the two question which you find in the manual.

3. In the next step we need to calculate the %Conversion of the substrate to the product to check which enzyme concentration is appropriate for our assay (remember that % conversion should be less than 10%). We use the similar formula as previously. We want to know how much of the product was produced in our 3ml volume reaction in 5min. So

micromol [product] formed in 5min = (A/18600) x 1000 x 3

So for 1/5 dilution in my case we have:

micromol [product] formed in 5min = (1.581/18600) x 1000 x 3 = 0.255micromol

Now we have to express that amount as the % of the initial amount of the substrate used. So we know from the protocol that stock concentration of the substrate was 20mM = 20mmol/L = 20micromol/ml. We used 0.5ml and used that in 3ml of reaction. So we took 20micromol/ml x 0.5ml = 10micromol.

%Conversion = 100 x (micromol [product] formed in 5min) / (micromol substrate present initially)

So for the 1/5 dilution we have:

%Conversion = 100 x 0.255 / 10 = 2.55%

We perform analogous calculation for other dilutions and we answer questions from the Section 3.

We put our results in the Table:

Now we go to the Part B of the report. First we fill in the table. To calculate the Vo we use the previous formule:

Vo = (Adsorbance - blank) / (18600 x 5)

To plot plot the Michealis-Menten graph we will use (Adsorbance - blank) / 5 (where 5 is our 5min incubation). This value is proportional to the Vo, as more adsorbance is present there where Vo rate is higher:).

So you should get a plot which look like that:

Then using the last table and the Michealis-Menten plot we try to estimate the Vmax and Km values. You have to remember that using the Michelis-Menten plot it is very hard to estimate those values. Usually mathematical methods are used to describe (to get the equation) how the curve is behaving. For our purpose it is ok if you roughly estimate those values. Next plot the Lineweaver-Burk plot is more accurate. So what you do is what you see in the next graph:

You gather the Vmax by dividing the Delta A/min value by extinction coefficient (18600, Vmax unit is micromol/min) and Km data directly from MM plot and you go to the next step which is the Lineweaver-Burk plot. To construct the LB plot we will need to recalculate some of the date we have used for the MM plot.

We construct a small table again:

In this table we calculate the reciprocal values of the initial substrate concentration and (Adsorbance - blank) / min which we had calculated in the last table. Remember that the (Adsorbance - blank) / min is proportional to the Vo. When you plot the 1/[S] versus 1/(Adsorbance - blank ..... you should get a straight line plot like this.

When you look at the plot you can see that the value where the line cross the x-axis gives you -1/Km which in this case is:

-1/Km = -1.5 ==> Km = 0.66 mM

And the value where the line crosses the y-axis gives you the 1/Vmax which in this case is:

1/Vmax = 5 ==> Vmax = 0.2

Remember that we have used (Adsorbance - blank.....) value insetad of the Vo so you have to divide the obtained result by 18600 (exticntion coefficient). At this stage you have Km and Vmax estimated from the MM plot and the Km, Vmax values calculated from the LB plot. Put them into the nice table and answer the question from the manual:)

Uffff that was a long report:)

See you next week:)

GGS

GGS LIVE - Protein purification methods

2009-10-27T18:04:00.006Z

LIVE FROM THE SPOT

From my own experience, I know that when you are an undergraduate student and you have to just study about all different techniques without trying them, it is hard to comprehend anything. So I decided to create the GGS LIVE section in which I will try to bring closer as much techniques as possible by showing a real stuff (solving real problems) from the lab. This is first time I do such a thing so please be understanding:) It will get better:)

Today....

METHOD: Protein purification.

ABOUT: What is a protein purification process? In my opinion it is isolation of the specific protein out of the complex protein mixture or an enrichment of it in the outcome sample. Properties of the target protein always force the way of its purification. Simply speaking if the target protein A is a negatively charged we would use a positively charged substance X to pull it down from the mixture.

WHAT: Purification of the IgG (immunoglobulin G) fraction from rabbit serum.

HOW: Protein A - is an bacterial protein that efficiently interacts with IgG's.

METHOD SCHEME:

ANALYSIS: On each step of the purification, protein samples are collected which then are separated by a PAGE (PolyAcrylamind Gel Ecectrophoresis). After that proteins are visualized by staining with Commasie dye (Brilliant blue).

RESULTS:

What you see here is a Coomasie gel where proteins are visualized by staining with blue dye. Simply gel after electrphopresis is incubated with the dye solution and then gel is destained (gel gets stained where there is no proteins as well).

Before we start I have to tell you that IgG's like other antibodies are build up of two different sets of chains, called heavy and light chain which molecular sizes are around 50kDa and 25kDa, respectively.

Let's start then!! Simply lanes load and unbound represent a starting material (crude serum) and sample after binding, respectively. You can see that there is not much difference in proteins content between those two samples. Why is that? Because only small amount of the beads was used in this particular experiment (this is a pilot experiment:). Simply there is still so much of the protein left after binding that we do not see the difference (if greater amount of the beads was used we would see depletion of the protein in the unbound sample)

Lanes Wash 1 - 4 are simply a samples that contains proteins that did not bind to the beads. You can see that there is some of protein of interest in these samples as well. This also suggests that we beads were satureted.
.
Lane Beads is the sample that represents what was present on the beads before the elution step. You can see that a good amount of "our" protein did stick to the beads and was not washed away. Additionally you can see that most of the contamitants present in the load sample is not present in this one.

Lanes Elution 1 - 5 represents samples recovered from the beads. Again there is a good amount of our protein in thsese samples (I do not know what happend in the elution 3:). You can see that most of the protein did come off the column in two first elutions.

Lane Beads after elution represents ... I do not have to tell ya:) You can see that there is still some amount of "our" protein present on the beads which can be eluted (it was eluted but I have in on a different gel).

Last lane represents sample which simply is a pull of all fractions which then were concentrated. So this sample is our final one. You can see that we have both chains present (which is good:) and that there is not much of the contamitants.

Simply we have managed to purify the IgG's fraction from the rabbit serum.

In the next part: CLONING

CyA!

Maciek

Meet the chickens - taste of fluorescent microscopy

2009-10-25T18:07:00.004Z

Welcome again. This time in the Meet the Chickens theme we will look at the DT40 chicken cells under the microscope.

Picture taken by Kliszczak M.(NUIG)

What you can see here is just a bunch of DT40 cells where DNA is stained with DAPI (blue dye). This is a grayscale format and normally under microscope DNA would appear as bright blue. Let's take a closer look to cells indicated with the red frames.
Cell with number 2 is a regular interphase cell (interphase cell is a cell which is not in the division phase). As you see there is a lot of cells looking similar to cell number 2. Mmajor population of the DT40 cells is "sitting" in the interphase (either G1, S or G2 phase of the cell cycle).

Cell number 5 is a cell which just started to condense the DNA. Final stage of the DNA condensation are chromosomes which you could see in the previous post on the Meet the chickens section (see the post rom 16th October). What you can observe here is that shape of the DNA is no longer round like but becomes clumpy and more dense.

Cell number 3 is a cell which is in the metaphase of the mitosis where already condensed chromosomes align in the center of the cell (metaphase plate). In the next step chromosomes will be pulled to opposite poles of the cell. Final stage of that event is being captured in the cells number 4.

Cell number 4, actually cells number 4 are just finising the division. You can see that each cell line got its set of the chromosomes. In the next step chromosomes will decondensed and each of the cells will enter next cell cycle.

If you look at the cell number 1 what you will see is a abnormal mitosis where condensed chromosomes are pulled to three (instead of two) different poles. This is abnormal as the cell before the division duplicates the DNA (cell do not triplicate it) so there is not enough of the genetic material to divide it between three new cells.

Usually this kind of the abnormal cells are eliminated because they simply are not healthy cells (they contain wrong amount of the DNA).

I hope you enjoy it and start to like the chickens.

GGS

Amino acids - the chemist way of life

2009-10-21T19:14:00.003+01:00

Hello Guys,

I have found a terrible mistake in the aminoacids for chemist scheme:) But, you did not either so it means you getting older lol:)

See the new SCHEME below.

Cya,

Maciek GGSTEAM

Hi ya all,

Presenting another way of remembering the aminoacids. This scheme would probably suit people with more chemical background but this is up to you. If you find it easier to remember it is even better.

I have still few more ideas how to remember all of them but I need more time to put them together.

I hope that helps.

GGS

3rd year biochemistry practical - Protein electrophoresis

2009-10-21T18:41:00.004+01:00

Hi all students,

This week we have been used electrophoresis as a protein analysis tool. All you need to know about the polyacrylamide gel electrophoresis (PAGE) you can find in the manual or just google it:)

Before we start with this week report, I have one major comment. Please remember to put all the data in your lab reports, all the numbers, calculations, observation and so on. Do not hide any values or formulas you have used for calculations because I do not know where the data is coming from. That makes the lab report correcting very difficult what results in cutting the points (haha):). Remember about that.

So let's start. This week you have to estimate the protein size using the protein marker (ladder) as you reference. You have loaded the gel (run and staining was done for you) and you got your gel pictures from the blackboard.

This is an example of the Coomasie stained gel which I will use to show how to prepare the lab report.

Figure 1.

What you see here is small part of the bigger gel (only three wells). First lane is a ladder (protein mixture of known molecular weights, sizes indicated on the left side [kDa]), second and third lane represents an unknown protein samples which size we want to estimate. Proteins used here were purified (you can see that protein X has more contamitants than protein Y and that the amounts are different:). In your case one of the samples will be a crude bacterial extract (where you will see a lot of bands, what makes sense because there is a lot of different proteins in the cell - you do not have to estimate the size of all the bands, just comment on what you see) and the other one will be a purified protein, so you should expect only one band (size of this band you have to estimate).

First step is to measure the distance of the ladder bands from the origin (you take a ruller and go go go). What you do is what you see on the next figure:)

Figure 2.

You use the top and the bottom of the gel as your origin and stop point respectively (that do not matters a lot as it is a relative distance:) and you put your results in the table (I always try to measure the distance to the centre of the band but that is up to you. Remember only to do it exactly the same for each band)

Table 1. In the table you put the distances and you calculate the retention factor (Rf) by dividing the distance that a particular band overcomed by the stop point distance. For example for 20kDa band we have 11.90/13.08 = 0.595 (where in my case 13.08 is the stop point) . Then you calculate the Log10 of each Rf.

In the next step you draw a plot Ladder Size vs Log10(Rf) and you get this:

Using the linear regresion (that would be very accurate method), or by adding the trendline in the excell (to see how, go to the previous post on the 3rd year biochemistry practicals - Bradford) or manually (do not connect the points as you want, it is not kindergarden:), just make a straight line as close as possible to all the points on the plot) we obtain the standard curve. We will use this standard curve for our estimations.

Now we measure the distances of the bands of interest, calculate their Rf and Log10(Rf):

Table 2.

Then using the equation provided (or read manually) we estimate the molecular weight of your unknown samples.For example sample X:

y = -0.011x + 0.1016

where:

y - is your log10 of retention factor

x - is your molecular weight.

We know the retention factor so we are looking for the molecular weight. Using equation we have:

x = (y - 0.1016)/(-0.011) = (-0.66 - 0.1016)/(-0.011) = 69.2[kDa]

Molecular weight of the protein in the sample X estimated by aminoacid sequence is 69.323 (the sample X is the BSA - Bovine Serum Albumin protein). You can see that our result is very close to the real value.

You perform the analogous workflow for your lab report.

Again remember to put all the data!!!!

Have fun!!

CyA in two weeks.

Maciek

Meet the chickens - chook chromosomes at hand

2009-10-16T20:51:00.008+01:00

Ladies and Gentelman,

please welcome the chicken DT40 cell line chromosomes.

Picture taken by Kliszczak M.(NUIG)

What you can see here is a methaphase spread where macrochromosomes are visible (chromosome pairs 1, 3, 4, 5, signle sex chromosome Z and trisomic chromosome 2). Rest of the chromosomes appear as a blurry spots.

Metaphase spreads of this kind are investigated in search form any abnormalities like: chromosome breaks, chromosome loss or gain and many more.

I hope you enjoy:)

Till Next Time NUIG:)

GGS

Amino acids - ways to remember 20 structures.

2009-10-15T15:43:00.007+01:00

Hi ya all,

I have came out with this topic because I realized that most of the students (including me:) have problems with remembering 20 structures of the aminoacids. I want to make up an efficient way to remember all of them.
This scheme represents all of the strucures and a potential way to suck them in.

Red parts of the structures are the ones that are changeing from structure to structure.

If you have any suggestions leave the comment, please.

I have other ideas how to put them into your mind for good.

CyA

Maciek

3rd year biochemistry practicals - Bradford assay

2009-10-14T18:48:00.006+01:00

Hey ya all,

I know you sometimes have a problems with the calculations (do not worry me too, but we will handle anything). This week we have been using the Bradford assay to estimate protein concentration of your last week samples.

All the necessary calculations, plots and tables you need to put in your lab report are here:

Page 1

Page 2

Page 3

One more thing. Chose the most accurate standard curve and use that for the calculations. Remember to comment on your choice.

Remember to include the dilution factor of your samples, what I mean is that if you are looking at the 1/10 dilution of your sample you have to multiply your result by then to obtain your initial conncentration and so on.

To calculate the total enzyme activity you have to multiply the Normal Enzyme Avctivity by the Volume of the enzyme you had (remember to include the loss of it during the centrifugation, if it happend to you).

Have fun. CyA next week.

GGS

How to submit a question?

2009-09-30T16:25:00.000+01:00

If you want to ask about anything just add a comment to this post or simply send the question to m.kliszczak1@nuigalway.ie. I will be glad to help you (if I can of course).

CyA

Maciek

The Aim and The Vision of The BioFreaks Blog

2009-09-30T12:18:00.001+01:00

The Idea of the BioFreaks blog is to help students solving their problems with chemistry and biosciences. If you are a science student this is a perfect place for you to find different answers. If you have an exercise, an exam coming, assignment, a report to prepare or a just a simple science question do not hesitate to ask me, I will do my best to help you.

My vision of the BioFreaks blog is to help people understand biosciences.

I wish you all the best and see you soon.

Maciek

My first post ever

2009-09-29T23:30:00.001+01:00

Hello Everybody,

This is my first blog and the post ever. My name is Maciek (do not pronounce it like Magic, because I am not magic at all:) and I am from Poland. Currently studying at the NUI, Galway as a PhD student in Biochemistry/Molecular Biology area. Science, exactly chemistry and biochemistry are my passion, can't live without them lol. I started this blog to share my passion with other people. If you are interested in biosciences this is a good site for you (I hope:).
That would be it for the first time.

CyA

Maciej

GGS Team