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Showing posts with label 3rd year practicals. Show all posts
Showing posts with label 3rd year practicals. Show all posts

Thursday, 11 February 2010

3rd year practicals - Immunochemistry

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

Friday, 29 January 2010

3rd year practical - GDH

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

Thursday, 21 January 2010

3rd year practicals - Measurment of Lactase Dehydrogenase Activity

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

Saturday, 16 January 2010

3rd year practicals!!

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

Friday, 18 December 2009

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

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

Wednesday, 16 December 2009

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

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

Tuesday, 15 December 2009

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

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

Tuesday, 8 December 2009

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

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

Monday, 7 December 2009

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

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

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

Sunday, 6 December 2009

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

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

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


Saturday, 5 December 2009

Biochemistry Exams - Sample Papers Exposed

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

Friday, 20 November 2009

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

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

Thursday, 19 November 2009

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

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

Wednesday, 18 November 2009

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

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:



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

Tuesday, 17 November 2009

Scientific prefixes - you will be laughing:)

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!!

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.

Monday, 16 November 2009

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

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:



Sunday, 15 November 2009

3rd year practicals - exam paper !!

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: