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		<title>The Effect Of Catalase Concentration On The Breakdown Of Hydrogen Peroxide</title>
		<link>http://www.tallsimon.com/biology/the-effect-of-catalase-concentration-on-the-breakdown-of-hydrogen-peroxide/</link>
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		<pubDate>Tue, 01 Jul 2008 04:26:45 +0000</pubDate>
		<dc:creator>simon</dc:creator>
				<category><![CDATA[Biology]]></category>
		<category><![CDATA[enzyme catalase]]></category>
		<category><![CDATA[enzyme concentration]]></category>
		<category><![CDATA[experiment]]></category>
		<category><![CDATA[kinetic theory]]></category>
		<category><![CDATA[potato]]></category>

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The Effect Of Varying Enzyme Concentration On The Breakdown Of Hydrogen Peroxide In The Presence Of Catalase
Grade: 18/18 Words: 3000
Hypothesis – Hydrogen peroxide will breakdown to oxygen and water in the presence of Catalase.  The reaction will increase with increasing enzyme concentration when molecules of hydrogen peroxide are freely available.  However, when molecules of the [...]]]></description>
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<h2 class="style5">The Effect Of Varying Enzyme Concentration On The Breakdown Of Hydrogen Peroxide In The Presence Of Catalase</h2>
<div class="style5">Grade: 18/18 Words: 3000</div>
<p>Hypothesis – Hydrogen peroxide will breakdown to oxygen and water in the presence of Catalase.  The reaction will increase with increasing enzyme concentration when molecules of hydrogen peroxide are freely available.  However, when molecules of the substrate are in short supply, the increase in rate of reaction is limited and will have little effect.<br />
Variables – In this investigation, the variables that affect the activity of the enzyme, Catalase, were considered and controlled so that they would not disrupt the success of the experiment. <br />
i) Temperature – As temperature increases, molecules move faster (kinetic theory).  In an enzyme catalysed reaction, such as the decomposition of hydrogen peroxide, this increases the rate at which the enzyme and substrate molecules meet and therefore the rate at which the products are formed.  As the temperature continues to rise, however, the hydrogen and ionic bonds, which hold the enzyme molecules in shape, are broken.  If the molecular structure is disrupted, the enzyme ceases to function as the active site no longer accommodates the substrate.  The enzyme is denatured.<br />
To control this variable, the temperature was maintained at a fairly constant level that allowed the enzyme to work effectively (room temperature, approximately 23°C).  This was achieved by using a test tube rack and tongs to handle the apparatus so that the heat from my hands did not affect the Catalase. <br />
ii) pH – Any change in pH affects the ionic and hydrogen bonding    in an enzyme and so alters it shape.  Each enzyme has an optimum pH at which its active site best fits the substrate.  Variation either side of pH results in denaturation of the enzyme and a slower rate of reaction.<br />
In this experiment, the pH was kept constant using a pH 7 buffer, selected to maintain a pH level suited to the enzyme by being equal to the natural environment of the enzyme (potato tissue).<br />
iii) Substrate Concentration – When there is an excess of enzyme molecules, an increase in the substrate concentration, produces a corresponding increase in the rate of reaction.  If there are sufficient substrate molecules to occupy all of the enzymes’ active sites, the rate of reaction is unaffected by further increases in substrate concentration as the enzymes are unable to break down the greater quantity of substrate.<br />
To control the substrate concentration, identical quantities of the substrate were used for each reading.  To ensure that this was measured precisely, 5ml syringes were used to accurately gauge to exact quantities.<br />
iv) Inhibition – Inhibitors compete with the substrate for the active sites of the enzyme (competitive inhibitors) or attach themselves to the enzyme, altering the shape of the active site so that the substrate is unable to occupy it and the enzyme cannot function (non-competitive inhibitors).  Inhibitors therefore slow the rate of reaction.  They should not have affected this investigation, however, as none were added.<br />
v) Enzyme cofactors – cofactors are none protein substances which influence the functioning of enzymes.  They include activators that are essential for the activation of some enzymes.  Coenzymes also influence the functioning of enzymes although are not bonded to the enzyme.<br />
Unless enzyme cofactors were present in the potato tissue containing the Catalase, they were not included in this investigation and therefore would not have affected the rate of reaction and the results of this experiment.<br />
vi) Enzyme Concentration – Provided there is an excess substrate, an increase in enzyme concentration will lead to a corresponding increase in rate of reaction.  Where the substrate is in short supply (i.e. it is limiting) an increase in enzyme concentration has no effect.<br />
I varied the enzyme concentration by altering the number of equal sized discs of potato that contain the Catalase, in the reaction.  The greater the number of discs, the greater the enzyme concentration.<br />
Apparatus –<br />
i) A manometer ii) 30ml hydrogen peroxide<br />
iii) Manometer fluid iv) 6 boiling tubes<br />
v) Tongs vi) A test tube rack<br />
vii) A potato viii) A petri dish<br />
ix) A cork borer x) Distilled water<br />
xi) A razor blade xii) A stop watch<br />
xiii) A ruler xiv) Rubber tubing<br />
xv) A marker pen xvi) A clamp<br />
xvii) A stop watch xviii) 2 5ml syringes<br />
xix) pH 7 buffer xx) A bung<br />
Procedures – Three tubes, 10mm in diameter were bored from a potato using a cork borer.  Using a razor blade and a ruler, 122 discs, 1mm thick, were cut from the tubes and placed under distilled water in a petri dish.  This prevented the potato from being contaminated or dehydrated.<br />
5ml of hydrogen peroxide and 5ml of a pH 7 buffer were then measured and added to each of six boiling tubes using a syringe.  Care was taken to view the syringes from the side to ensure the bottom of the meniscus was lined up properly with the gradations and there were no air bubbles in the syringe.  A pH buffer was added to the boiling tubes to maintain the pH at a constant level so that changes in pH as a result of the reaction would not affect the activity of the enzyme and disrupt the results.  pH 7 buffer was selected to match the natural pH of the potato tissue and therefore suit the enzyme so that it could work efficiently.<br />
One of the boiling tubes was then connected to a manometer containing manometer fluid using a bung (see diagram below).  Holding the manometer level by the bung to ensure that the fluid was at its lowest level, a mark was drawn to indicate this point using a marker pen.   A further mark was then drawn 5cm above the original, measured using a ruler.<br />
15 pieces of potato were placed in to the boiling tube using a pair of tweezers to prevent contamination.  A clamp was then placed over the rubber tubing on the bung to ensure that all of the oxygen gas released will travel up the manometer tube and not escape.  Once the clamp was closed, the stopwatch was started to record the time taken for the manometer fluid to travel to the second mark.  When this had been achieved the time was noted in a results table and the clamp opened to allow the gas to be released and the manometer fluid to return to its original level.  Once the apparatus had been reset and any air bubbles in the manometer fluid removed, a second and later third reading was taken by re-closing the clamp and measuring the time taken for sufficient gas to be released from the reaction to force the manometer fluid back up to the top mark.  By taking several readings for each enzyme concentration, it enabled me to average the results to minimise the extent of any inaccuracies.  The experiment was then repeated for different quantities of potato discs (enzyme concentration) by using different boiling tubes containing hydrogen peroxide (see table below for quantities).  I began with an enzyme concentration of 15 potato discs rather than a lower quantity to ensure that the apparatus was working correctly.<br />
 </p>
<p>Vol. pH buffer (ml) Vol. Hydrogen peroxide (ml) No.Potato discs<br />
5 5 5<br />
5 5 10<br />
5 5 15<br />
5 5 20<br />
5 5 25<br />
5 5 30<br />
5 5 35<br />
        <br />
 <br />
 <br />
 <br />
 <br />
 <br />
 <br />
 </p>
<p>Observations and Measurements – In the boiling tubes it was clear that a reaction was taking place by the observation of bubbles of oxygen gas being released creating a ‘fizzing’ in the boiling tubes.<br />
In order to decide how varying the enzyme concentration affected the decomposition of hydrogen peroxide, the rate of reaction was measured.  To do this accurately, the time taken for a specific quantity of oxygen gas (a product of the reaction) to be released was determined.  This was achieved by observing the time taken for the manometer fluid to travel between the two marked fixed points as it was forced through the manometer by the rising gas.  This was an accurate measure of how the enzyme concentration influenced the breakdown of hydrogen peroxide, as the quantity and speed of gas produced is dependant on the rate of reaction.  The marked points remained the same distance apart for each reading for different enzyme concentrations so that they could be accurately compared and the trend observed.<br />
All measurements were taken so that the stopwatch was started once the rubber tubing was sealed and the stopwatch stopped once the manometer fluid had reached the base of the highest marked point.  To judge  accurately, the point at which the fluid reached the marked line, it was examined at eye level and the measurement taken when the bottom of the meniscus was lined up to the mark.  This was  the same for every reading.<br />
Data handling – The data obtained from this investigation has been recorded in a table showing the time, enzyme concentration and rate of reaction.  This means that the results of the experiment are presented in a clear and orderly fashion that allows patterns in the results to become more obvious.<br />
The rate of reaction was calculated by dividing 1000 by the time taken for the quantity of gas to be produced from the reaction.  By calculating the rate of reaction instead of merely using the time readings, the quicker reactions will be represented as a greater value for the rate of reaction rather than a small time value.  This makes the graph more clear and easier to analyse.<br />
Patterns within the results collected from the experiment, are best shown on a graph.  This is because overall trends between the enzyme concentration and rate of reaction can be portrayed more effectively and become more obvious.<br />
Limitations and Precautions – In this investigation, I measured the rate of reaction with enzyme concentrations of between 0 and 35 units (potato discs).  At 0, there should be no reaction as there will be no substrate, however, I included it to act as a control.  This will show that it is the variable, enzyme concentration that is being measured.<br />
I decided to vary the enzyme concentration by varying the number of potato discs.  However, although the enzyme, Catalase, occurs in the potato tissue, I did not know the exact quantity and certain discs might have more Catalase than others.  This could be a major limitation in this investigation.  I have tried to compensate for this, however, by taking multiple readings for each enzyme concentration so that inaccuracies are minimised once averaged.<br />
As a precaution, I have limited my contact with the boiling tubes, as my body heat will raise the temperature, increasing the rate of reaction or expanding the gas inside the test tube moving the manometer fluid.<br />
I also monitored the temperature using a thermometer to ensure that it remained constant and not disrupt the results of the experiment by affecting the activity of the Catalase.<br />
A pH buffer was used to maintain a consistent pH level in the boiling tubes. This way there was no variation in pH that might have resulted in an increase or decrease in the rate of reaction.<br />
A major limitation of this investigation was the time. It meant that only 8 different enzyme concentrations could be measured at intervals of 5 units or potato discs.  This means that only very general, overall trends can be identified across the results.  Patterns between these values can only be approximated and are not necessarily accurate. </p>
<p>Safety – Laboratory coats were worn during the investigation to prevent chemicals from spoiling clothes.  Care was also taken whilst handling the chemicals as hydrogen peroxide is corrosive and the manometer fluid is permanently staining.  Whilst using the razor blades, care was also taken to hold them by the handle and not the blade to prevent an accident occurring.<br />
 <br />
Results – The rate at which hydrogen peroxide was broken down to water and oxygen in the presence of Catalase:<br />
The graph “The decomposition of hydrogen peroxide in the presence of potato catalase Chart 2” shows the rate of reaction up to an enzyme concentration of 25.  Up to this point the line of best fit is a straight line through the origin. This shows that without the enzyme, catalase, present no reaction takes place.  It also indicates that the enzyme concentration is directly proportional to the rate of reaction for the decomposition of hydrogen peroxide in the presence of catalase (the rate of reaction increases with increasing enzyme concentration).         <br />
The other graph, “The activity of potato catalase with differing enzyme concentrations Chart 1”, shows how the rate of reaction varies with differing enzyme concentrations over the whole range that I experimented with.  After an enzyme concentration of 25 potato discs, the line of best fit is no longer a straight line and begins to level off.  The enzyme concentration is no longer proportional to the rate of reaction, and the increases in the rate of reaction reduce dramatically. <br />
Conclusion – The reaction was fastest at an enzyme concentration of 35 potato discs.  At this enzyme concentration there were the greatest number of free active sites available to the substrate molecules so that they could be broken down. <br />
The rate increased steadily from 0 up to a concentration of 25 and slowed beyond this point to give a “maximum level”.  It appears that at this “maximum level”, increasing the enzyme concentration had little effect and other factors such as substrate concentration were limiting the reaction and prevented any further increases in the rate of reaction.<br />
Discussion – The results of this investigation are as I predicted in the hypothesis:  “The reaction will increase with increasing enzyme concentration when molecules of hydrogen peroxide are freely available.  However, when molecules of the substrate are in short supply, the increase in rate of reaction is limited and will have little effect”.  The reasons for this are that there are number of variables that influence the decomposition of hydrogen peroxide in the presence of Catalase.  Some of which can be classified as limiting factors i.e. the reaction is dependant or “limited” by their availability, to be able to function effectively; these include enzyme concentration, temperature and substrate concentration. All of these factors are required for an efficient reaction to take place, even when one is freely available the reaction can still be limited by the availability of the others.  When I increased the enzyme concentration, it meant that there were more free active sites for the substrate molecules so that a greater quantity of substrate molecules could be broken down into products.  However past a certain point, which in my investigation was at an enzyme concentration of 25 potato discs, there were many free active sites but insufficient substrate molecules to occupy them.  Increasing the enzyme concentration further without increasing the substrate concentration has no effect on the rate of reaction which eventually will remain constant.<br />
From the line of best fit on the graph “Chart 1”, it is clear that some of the points do not exactly fit.  They are anomalies.  Although they have only slight inaccuracies, they are an indicator of possible errors in the investigation.  These may have occurred in either the measurement of the quantities of the enzyme and substrate or the measurement of the time taken for the manometer fluid to rise five centimetres up the manometer tube.  Another possibility was that fluctuations in temperature caused the rate of reaction to increase or the gas inside the boiling tube to expand, forcing the fluid to rise up the manometer tube.  Although minimal contact was made with the apparatus during the investigation, slight undetected variations in the room temperature may have led to inaccuracies.<br />
The precision of this experiment, generally, was very limited since insufficient readings were taken.  Although the range of enzyme concentrations taken was large, the difference in enzyme concentration between each reading was too great to distinguish a value between them.  For example, the rate of reaction at an enzyme concentration of 15 potato discs was 35 + or – 4.  This results in an error of uncertainty of 11%<br />
The shape of the graph is as I predicted showing that as enzyme concentration increases so does the rate of reaction.  This is because at a greater enzyme concentration, there are more free active sites available for the substrate and so more products can be made in a shorter length of time.  However, it is not possible to take precise readings from the graph between the plotted points since insufficient readings were taken.  To be able to do this, intermediate enzyme concentrations would have to be measured so that the shape of the graph would be more exact.<br />
Suggestions and Improvements &#8211; To create a more accurate experiment in the future, several precautions or alterations can be made:<br />
 <br />
· Instead of using potato discs that have slight variations in size, and volume of catalase, as a source for the enzyme, a 1 molar solution of the enzyme could have been diluted to create different concentrations.  This way the concentrations can be measured far more accurately reducing the chances of errors in the investigation. <br />
· In this experiment 8 enzyme concentrations were considered.  However, although there was a large range, insufficient intermediate measurements were made creating gaps between the measurements where guess work is needed to predict the rate of reaction at these points e.g. point A on graph “Chart 2”. In a future investigation, a far greater number of enzyme concentrations between those already recorded should be tested reducing the extent of any anomalies on a graph where the line of best fit must be drawn.<br />
· In this investigation each reading was repeated so that an average rate of reaction for each enzyme concentration could be calculated.  This could be improved by repeating the reading more frequently thus reducing the extent of any anomalies further, once averaged.</p>
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		<title>Osmolar concentration of a potato</title>
		<link>http://www.tallsimon.com/biology/osmolar-concentration-of-a-potato/</link>
		<comments>http://www.tallsimon.com/biology/osmolar-concentration-of-a-potato/#comments</comments>
		<pubDate>Sun, 11 May 2008 04:52:14 +0000</pubDate>
		<dc:creator>simon</dc:creator>
				<category><![CDATA[Biology]]></category>
		<category><![CDATA[cell]]></category>
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		<category><![CDATA[graph]]></category>
		<category><![CDATA[molar]]></category>
		<category><![CDATA[osmosis]]></category>
		<category><![CDATA[potato]]></category>
		<category><![CDATA[solute]]></category>
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		<description><![CDATA[Presented with raw potato how would you carry out an experiment to determine the concentration of the potato cell in molar strength? 
The experiment was set up to be as fair as possible. Then the potato chips were cut into 15 cylinders using a cork cutter and then cut to a length of exactly 4cm [...]]]></description>
			<content:encoded><![CDATA[<p><strong><em>Presented with raw potato how would you carry out an experiment to determine the concentration of the potato cell in molar strength?</em> </strong></p>
<p>The experiment was set up to be as fair as possible. Then the potato chips were cut into 15 cylinders using a cork cutter and then cut to a length of exactly 4cm each with a scalpel. Then three were weighed at one time on a top pan balance to find an average mass, this was done because it did not involve confusion over which potato cylinder was which after the experiment. Once these averages were found and recorded the potato chips were placed into their test tube solutions of 0.0M(distilled water), 0.25M, 0.5M, 0.75M and 1.0M solutions of sucrose. Then the contents, each containing three potato cylinders were left for 50minutes. When the time was over I removed the potato with a pair tweezers and placed them on a paper towel to try and dry of any excess water. Then I weighed the potato again together in their threes on a top pan balance and found the average mass.<br />
 </p>
<p>Prediction<br />
 </p>
<p> I believe from my past knowledge of a similar experiment involving raw eggs that the potato cylinders placed in the distilled water will be hyperosmotic, therefore the water will diffuse by osmosis into the potato from an area of high concentration down the concentration gradient. This will result in a gain of mass for the potato.<br />
 </p>
<p>    However in the 1.0M solution I think that the opposite will occur as there will be a higher concentration of water within the potato than the solution, containing lots of solute, this will make the potato hpo-osmotic and it will lose its water content as the water diffuses out.Finally, a difference between the egg experiment and the potatoes that the cells will not reach equilibrium (become iso-osmotic). This is because plants have a cellulose cell wall which cannot expand beyond a certain point as it is turgid, therefore the contents will never be able to be equal to the outside or it will not be able to give out all its water as it has a rigid structure. It is the pressure created by the cell wall which stops the cell reaching equilibrium. I think that the contents of the cell will be about 0.3M concentration as it will be in between 0.25M and 0.5MThe experiment has yielded a graph, (see attached) this graph gives me an excellent set of data. The graph shows that in the distilled water the potato gains in mass by approximently 7.4% then it gains 4.4% in the 0.25M solution. In the 0.5 M solution there is a change, the gradient of the graph decreases steeply and the potato loses mass to 4.4% then in the 0.75M solution the gradient decreases as it loses yet more mass to 10.4%.<br />
 </p>
<p> The first observation I made was after the potato cylinders were placed in their solutions I could see a difference. The ones in the 0.0M and 0.25M solutions were floating and the potatoes in the 0.5M, 0.75M and 1.0M solutions were at the bottom of the test tube, this lead me to drawing my next conclusion.<br />
 </p>
<p> The graph shows that the potato in the 0.0M solution and in the 0.25M is hyperosmotic, as I said in my prediction, this means that there is a higher water potential in the distilled water and 0.25M solution than in the potato, which has a high concentration of solutes. Therefore this is why the water diffuses by osmosis down the concentration gradient from high concentration to low concentration, resulting in the potato gaining mass.<br />
 </p>
<p> The opposite occurs in the solutions where the molarity is higher 0.5M, 0.75M and 1.0M, the potato in these is hypo-osmotic because there is a higher concentration of water inside the potato than in the solutions, therefore the water diffused out of the potatoes by osmosis, down the concentration gradient and into the solutions outside, this resulted in the loss of mass.<br />
 </p>
<p> Another important fact is that the graph makes a shape that will result in the potato not being able to take in any more water or lose any more. Like I said in my prediction the cell wall causes the pressure that prevents this from happening, therefore the cell will never become iso-osmotic like an egg.<br />
 </p>
<p> On my graph I have marked the point where the graph line crosses the place on the axis where the potato neither loses mass nor gains mass. This happens at 0.36M, therefore the contents of the potato cells in molar strength is 0.36M.<br />
 </p>
<p> This result is very close to my prediction of 0.3M and it shows that the contents of the cell are between 0.25M and 0.5M which is why in these concentrations they gained mass and lost mass.<br />
 </p>
<p>  I think that the experiment went very well, there were no odd results and they produced a good graph. However there were a few areas where there could be improvement. Firstly, when I dried off the excess water on the potato cylinders after the experiment and before I weighed them, I used a paper towel. This might have either taken some water out of the potato or it might of left some excess water on the potato. This part of the experiment is difficult to     come up with an accurate and fair method, as other ways would also lead to some slight mistakes.<br />
 </p>
<p>     Also the potato itself was not definitely from the same potato and was not exactly the same size, although I did try to cut them to 4cm each, this could have effected the amount of water gained or lost.<br />
 </p>
<p>     Another way of improving the results would have been to leave the experiment running longer, this would have enabled me to find the saturation point (when the potato can no longer take in any more water) and dehydration point (when the potato cannot lose any more water) and therefore get a more accurate result.<br />
 </p>
<p>     Finally, I could extend the experiment to a more exact level by looking at the potato cylinders under a microscope, then I would be able to see the cells in greater detail and draw some more observational results.</p>
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