PRACTICAL 8: INTRODUCTION TO ENZYME INHIBITION

Using the same enzyme and substrate from last week, trypsin and BAPNA, the effects of an inhibitor on the rate of the reaction were investigated. The rate of the reaction was measured again as absorbance per minute using a spectrophotometer, the concentration of substrate was increased in the different solutions.
The following table of results was obtained:

From this table it can be observed that the absorbance per minute increases for both solutions with and without inhibitor. However this is not the case for solutions D and H, this could be either a mistake when doing the calculations to obtain the absorbance per minute, or maybe the group realizing these solutions placed the cuvettes to quickly in the spectrophotometer and the reaction had not started, resulting thus in a lower value.
The graph obtained is the following:

The pink curve represents the absorption per minute of the enzyme-substrate solution without inhibitor, and the blue one represents the solution with the inhibitor. As it was mentioned before, both curves should increase continuously, however some of the results do not correspond with what was expected, but the following conclusions can be attained by observing the graph.
The absorption per minute is affected by the presence of the inhibitor, this means that less enzyme-substrate complex are found because some of the active sites of the enzymes are occupied by the inhibitor. However, this kind of inhibition, which is called competitive inhibition, can be overcome by increasing the concentration of substrate, and by doing so it would not affect eventually the maximum velocity. This is due to the fact that the enzyme-inhibitor complex is a reversible reaction and as more substrate is added the enzyme manages to overcome the effects of the inhibitor and bind to the substrate. Therefore, if substrate is increased competitive inhibition does not affect the maximum velocity of the reaction.
The case would be completely different in the case of a non competitive inhibition, in this case the inhibitor would not bind to the active site of the enzyme but to a different site called an allosteric site changing thus the structure of the enzyme, and making it unable to bind to the substrate. This situation would affect the maximum velocity of the reaction and it could not be changed by adding more substrate.

PRACTICAL 5 AND 6: PURIFICATION OF BIOMOLECULES

BIOL I 1100


In this practical Gel Permeation Chromatography, also known as size exclusion chromatography was used to separate a mixture of small and large biomolecules according to their size. In GPC the matrix contains tiny gel beads which have an open structure. The large molecules will travel faster through the gel, this is due to the fact that the small molecules can enter the gel beads which contain a gel matrix that encloses an internal solvent space whereas the large molecules cannot.
In this case the mixture contained trypsin (protein), glucose (carbohydrate), and cholesterol (lipid). Twenty 2 mL fractions were collected as the solution run through the gel together with a running buffer which keeps the beads hydrated. The larger molecules will come out first, whereas the smaller molecules will come out later because they have to travel through a bigger volume of solvent. Blue Dextran and yellow pNP were used as size markers, the blue colour came first indicating the molecules of high molecular weight and the yellow colour came last indicating those molecules with low molecular weight.
Trypsin and glucose are soluble in the running buffer solution but cholesterol is not, cholesterol should be extracted by a liquid-liquid extraction, which it wasn’t carried out. On the other hand, Protein can be detected using Bradford reagent which is a dye binding method, when protein is mixed with the Bradford reagent there is a change in colour of the reagent. The intensity of the colour, in this case blue, being proportional to the amount of protein. The enzyme glucose oxidase was used to detect the presence of sugar, again the intensity of the colour, in this case red, would indicate the presence of sugars. The following table of results was obtained when adding Bradford reagent and Glucose oxidase to twenty 100µl fractions prepared using the previously separated 2mL fractions.

The minus symbol represents no colour, whereas the + symbol represents intensity being +++ the maximum intensity. In the case of protein the colour developed was blue whereas in glucose it was red.
The results show how Trypsin which is a large molecule with a complex 3D structure, comes out at an earlier stage because it is unable to enter the beads and therefore passes through the column at the same speed as the buffer. On the other hand glucose, a monomer with low molecular weight, came out at a later stage because it is a smaller molecule which is able to enter the beads, and therefore has to travel trough a larger volume of solvent.
If a plot of the concentration of the concentration of trypsin and glucose was to be made it would look like the graph below, indicating the separation of the two components.

PRACTICAL 7:THE KINETIC BEHAVIOUR OF ENZYMES

BIOL I 100-3

In this practical it was studied the effect of changing the amount of enzyme and substrate in an enzyme catalysed reaction. The enzyme used was trypsin, and BAPNA, which is an artificial substrate molecule, was used as the substrate.
The rate of the enzyme catalyzed reaction is expressed as amount of substrate converted to product per minute, in this practical a spectrophotometer was used, the amount of UV radiation absorbed is equal to the amount of substrate produced. Therefore, the unit of rate used is the change in absorbance per minute Abs/min. This will be calculated by subtracting the absorbance at 30s (A1) from the Absorbance at 90s (A2). It is important that our A1 is a value is in the first seconds of reaction (15 or 30 s) because this way the initial velocity will be obtained, the initial velocity is the instant before all of the substrate is converted into product and the curve levels off
The following table of results was obtained when the concentration of BAPNA was increased:

A plot could be drawn for each of the concentrations, time would be on the X axis and the absorbance on the Y axis, this way a curve would be obtained for each of the concentrations and the initial velocity would be calculated for each curve by drawing the tangent line for each curve and then drawing a vertical line through the point of initial velocity in the top most curve, wherever this vertical line intersects the other curves would be the initial velocity. These values would be used on the Y axis on a second graph together with the substrate concentration in the X axis to give a concentration plot. This would probably be a more accurate way of obtaining the initial velocity. However, for simplicity because otherwise 5 graphs would be needed, the Abs/min was calculated subtracting the absorbance value at 90s and the absorbance value at 30s, thus obtaining the initial velocity. The following concentration plot was obtained when the amount of substrate was increased:

It can be observed how the velocity increases with the increase in substrate, but as the substrate becomes larger the increase in rate becomes smaller. The substrate an enzyme have to collide forming an enzyme-substrate complex to form a product, at high substrate concentrations more enzyme-complex are found which increases the initial velocity. However, if the substrate concentration is increased at this point when most of the enzyme is present as enzyme-complex state, the reaction rate will level off as it reaches a maximal velocity.
The following table was obtained when the concentration of enzyme was increased:

The negative values obtained in the first measurements were probably due to the fact that the cuvette was introduced too quickly in the spectrophotometer and the reaction had not started.


The following concentration plot was obtained using the data on the table:

It can be observed how the initial velocity increases briefly to start decreasing rapidly. This is due to the fact that the same amount of substrate is present, but more enzyme is added, therefore, more enzyme-substrate complex are formed and the substrate is transformed into product rapidly. However, once the substrate is over only enzymes are present, thus the rate of reaction starts to decrease.

PRACTICAL 4: SEPARATION OF LIPIDS VIA THIN LAYER CHROMATOGRAPHY.

BIOL I 100-4

In this practical the lipids present in egg yolk were separated through absorption chromatography using a TLC plate. The concept is similar to paper chromatography, a plate containing a thin hydrophilic layer called a matrix is used, and a non polar solvent is allowed to move up the plate through capillary action. Thus more polar lipids will move more slowly than non polar lipids.
The egg yolk solution was compared against six standards; cholesterol, cholesterol palmitate, trioleine, linoleic acid, phosphatidylcholine, and phosphatidylethanolamine.
The result obtained was that egg yolk contains trioleine, cholesterol palmitate and phosphatidylethonolamine, the latter being the most polar of the three elements hardly moving up the TLC because of the presence of several polar groups; phosphate groups, amine groups and several oxygens , and cholesterol palmitate being the most non polar.

PRACTICAL 3: IDENTIFICATION OF THE SEQUENCE OF A POLYPEPTIDE

BIOL I 100-3

In this practical the composition of several unknown dipeptides was found using paper chromatography. This is a partition chromatography technique where cellulose in the form of paper acts as a hydrophilic support medium or liquid stationary phase where those aminoacids with polar side chains will interact. A second mobile phase containing water and a non polar solvent was used, the paper was introduced in a chromatography tank where the non polar solvent moved up the paper by capillary action, those amino acids with non polar side chains will move up the paper whereas, as it was said before, those with polar chains will interact with the hydrophilic stationary phase moving up the paper slightly.
The native unknown dipeptide and a hydrolised form of it were compared to six standard aminoacids to find out which were the components of the dipeptide. The hydrolised form will be separated into its constituent peptide residues. The results obtained were:


In the case of dipeptide HA it can be seen that the aminoacid which moves further up is leucine which has a non polar chain:

On the other hand tryphtophane is slightly more hydrophilic because of the presence of an amine group on its side chain:

Overall, amino acids side chains are very important to determine the properties of a protein affecting the way they fold and their structure.

PRACTICAL 2: Investigating the Chemical Properties of Sugars.

BIOL I 100 – 2

PRACTICAL 2: Investigating the Chemical Properties of Sugars.

The aim of this practical was to test the property of sugars to act as reducing agents. Benedict’s reagent was used to test this property. In this solution Cu2+ ions are reduced to Cu+ by certain sugars changing its colour.
Two monosaccharides sugars, fructose and glucose were used, and to disaccharides lactose and sucrose. Solutions as 0.25% and 1% solutions in water were mixed witn Benedict’s reagent and left in warm water for five minutes. The following table of results was obtained.



As it can be seen the two monosaccharides sugars, fructose and glucose, are clearly reducing agents, this is due to the fact that it is the aldehyde group which gives sugars its reducing properties and in the case of monosaccharides the aldehyde group is free to react as a reducing agent, especially at 1%, this reducing capacity is decreased in a lower concentration. On the other hand the two disaccharides, lactose and fructose, have very little reducing properties in the case of lactose and none in the case of sucrose.This is due to the fact that it is the aldehyde group which gives sugars its reducing properties.In the case of lactose we find two glucose rings joined by glycosidic bond forming an acetal (one carbon joined to two oxygens), the ring with the acetal is non reducing but the other ring still has a hemiacetal (carbon joined to two oxygens one of the oxygens being attached to hydrogen) and is capable of acting as a reducing agent though the reaction is weak.
In the case of sucrose we observe that it is non reducing in both concentrations, this is due to the fact that the glycosidic bond is formed between the reducing ends of both glucose and fructose, therefore both rings are locked and unable to act as reducing agents.
Therefore, Benedict’s reagent could be used as a mean of testing the amount of reducing sugar in a solution as it would result in a different colour depending on the concentration of sugar, and whether it is monosaccharide or disaccharide.

PRACTICAL I: Physical Properties of biomolecules

BIOL I 100-1

PRACTICAL I: Physical Properties of biomolecules.

In this practical the polarity, solubility in water, of different biomolecules was tested. A 1% solution of each compound was prepared and their solubility was tested in ; heptane, acetone, and distilled water. Compounds were classified as soluble, insoluble or partially soluble in each of the solvent solutions. The following table of results was obtained;

It can be observed that cholesterol, which is a lipid, is partially soluble in water. By looking at its molecular structure it could be assumed that it is hydrophobic, however is an amphiphilic compound possessing both hydrophilic and lipophilic properties. This explains the way cholesterol is transported in the blood stream within lipoproteins which are molecules with a water soluble outward face and a lipid soluble inward face. Cholesterol resulted to be insoluble in heptane, this is due again to the amphiphilic nature of the cholesterol molecule, heptane is a long chain of hydrocarbons whereas acetone a carbonyl group with a negatively charged oxygen and a short chain of hydrocarbons, this makes it a more suitable substance to act as a dissolvent for an amphiphilic compound. On the other hand it can be seen in the table of results that cholesterol ester is insoluble in water, but still soluble in acetone. However, it could be supposed by looking at its structure that it is partially soluble in water like cholesterol because of the presence of an extra hydroxyl group, and amide group and a sulfonyl group.
Palmitic acid and taurodeoxycholic acid are also amphiphilic both having a long chain of hydrocarbons and a hydrophilic tail.
Valproic acid, D+Glucose and caffeine are clearly hydrophilic. Leo-Trp dipeptide is soluble in water because proteins have all their polar groups to the outside being thus able to form hydrogen bonds with water.
Finally, the dye mix is clearly hydrophobic being soluble only in the organic compounds.
Overall the more polar groups a molecule has the more soluble it is, and the more non polar the less soluble it is. It is the overall ratio of polar to unpolar groups which is determinant of solubility.