Posts Tagged ‘experiment’

How The Population Of Yeast Changes Over A Number Of Days

Friday, June 6th, 2008

A Level Biology Project

Aims

This is an experiment to examine how the population of yeast alters over a number of days.

Background Information

Yeast is a single celled fungus. It reproduces asexually by budding. It respires anaerobically.

Formula: Yeast + Carbohydrate ------------> Alcohol + Carbon Dioxide

Apparatus Needed For The Experiments

  1. Yeast
  2. Cider
  3. Conical Flask
  4. Muslin
  5. Haemocytometer
  6. Colorimeter
  7. Pipette
  8. Pipette Filler
  9. Measuring Flask

Method

I am planning to use two methods to measure how the population of yeast changes over time. Method 1 uses a haemocytometer whilst method 2 uses a colorimeter to measure the number of yeast cells each day.

A haemocytometer is a microscope slide which has an etched grid on it. It consists of a 1mm² square known as a A square which is divided into 25 B squares which have an area of 0.04mm². These B squares are each divided into 16 C squares which have an area of 0.0025mm².

A colorimeter is a machine that is used to see how much light can pass through a liquid. It shows how much light is being transmitted through a sample of liquid. As the number of cells gets higher, less light will be transmitted through the sample. Special thin walled test tubes are used in the colorimeter so that they do not affect the amount of light passing through the sample.

To prepare the yeast solution do the following steps.

1. Measure out 50cm³ of cider into a conical flask using a measuring flask. Cider is used to provide food for the yeast to grow in.

2. Using a 1cm³ pipette carefully measure out 1mm³ of yeast suspension and add it to the conical flask containing the cider.

3. Mix the solution and then put a muslin cloth over the top of the conical flask which should be held in place using an elastic band.

4. Put this conical flask in an oven pre heated to 40ºC. The change in yeast population can then be measured by using a haemocytometer and by using a colorimeter.

The Haemocytometer Test

1. First set up the microscope, place the cover slip on the haemocytometer slide between the two grooves and put the slide on the microscope.

2. A drop of yeast solution should be taken from the conical flask using a dropper pipette and put next to the cover slip. The yeast solution will be drawn under the cover slip by capillary reaction.

3. Focus the microscope so that you can see the smallest type C squares.

4. Count the number of yeast cells that you can see in 10 randomly picked squares. Note down all these results in a table such as the one below.

Test Number 1 2 3 4 5 6 7 8 9 10 Average
Day 1                      
Day 2                      
Day 3                      
Day 4                      
Day 5                      
Day 6                      

This haemocytometer test should be repeated daily for about two weeks. The results in the table can then be plotted into a graph.

The Colorimeter Test

The number of cells can be measured by with a colorimeter using the following method.

1. Calibrate the colorimeter by pouring 3cm³ of plain cider into a colorimeter test tube and placing the test tube into the colorimeter. Adjust the colorimeter so that with the plain cider it is showing 100% light transmission. Take the plain cider test tube out of the machine.

2. Take the conical flask containing the yeast solution out of the oven. Using a pipette measure out 3cm³ of the yeast solution into a clean colorimeter test tube.

3. Place this test tube into the colorimeter into the machine and note down the value of light transmission into a table such as the one below.

Day 1 2 3 4 5 6 7 8 9 10
Colorimeter Reading                    

4. Pour the yeast solution back into the conical flask. Put the muslin back on the flask and return it to the 40ºC oven.

Both the haemocytometer and the colorimeter method of measuring the change in yeast cell population should be repeated at daily intervals for roughly two weeks.

A graph of Cell Numbers against time can then be plotted using the data. To convert the colorimeter readings to numbers of cells a calibration chart should be used.

Predictions

I predict that as the number of days goes up the number of yeast cells in the solution will go up. This will continue for a number of days until the rate starts to slow down and eventually it will stop going up as no new cells are being produced.

The rate will go up over time because the yeast cells reproduce asexually which means that they are able to reproduce very quickly. They also have the perfect conditions for reproducing. They are being kept at a temperature of 40ºC which is very near to their optimum living temperature. They have lots of food from the cider and they have a lot of space to reproduce in. The rate will slow down after a while because the yeast will start killing itself off with the alcohol which it produces as part of it's respiratory process. This is known as a negative feedback reaction.

Results

The Experiments - started to carry out the above experiments but after a few days decided to give up on the haemocytometer test because it was too time consuming. I carried out the colorimeter readings as written above and the results can be seen below. In order to get the number of yeast cells produced I am using a calibration curve so that by using the percentage transmission results the number of cells per mm² can be read off it.

I carried out the above experiment and these results were obtained. To make the colorimeter test as accurate as possible as well as my own results I used six other peoples sets of results and then took the average of each days results to work out how many yeast cells there are per mm² using the calibration curve.

Results Table

A graph of the log of the cell numbers against time was plotted. A log graph is used because the increase in the number of cells per mm² over the 11 days is so big that it would not be practical to have a graph with numbers that large on it. This graph shows the total cell numbers both living and dead.

Yeast Graph

Interpretation

The number of cells increased as the time went by as I had predicted. It increased as a fairly steady rate which can be seen on the Log graph. The population is able to increase at this rapid rate due to it's asexual reproduction. When the yeast cell is ready to reproduce a bud starts to grow out from it’s cell membrane which gets bigger and bigger until it is big enough to break off and become a new yeast cell. This is shown in the diagram below. [not reproduced]

These two yeast cells are then able to reproduce making four cells then eight then sixteen and so on. The yeast cells have plenty of space to grow in and lots of food from the cider. They are also being kept at 40ºC which is close to their optimum temperature for living. The number of cells does not keep doubling in this way. This is because the cells start to die through natural death and they are also killed by the alcohol which the yeast produce as part of their natural respiration process.

From day 3 to 4 the Log number increases from 6.87 to 6.95 which is an increase in cell numbers from 7413102 to 8912509. This is an increase of 20% for one day. From days 8 to 9 the Log goes up from 7.22 to 7.30 which is an increase in cell numbers from 16595869 to 19952623 which is an increase of 20% for one day. This shows that the rate of increase up to day nine is very steady.

After day nine the rate of population increase starts to slow down which is shown by the decreasing gradient on the graph. From day 10 to 11 the Log goes up from 7.37 to 7.41 which shows an increase in cell numbers from 23442288 to 25703958 which is an increase of 9.6% for one day. This is a fairly large drop in the expansion of the population of yeast. The rate of increase from day 10 to 11 is half that of the increase from day 3 to 4.

The rate has started to decrease by 9 and 10 because the yeast is starting to be killed off by the alcohol. This alcohol is produced as a by-product to the yeast's respiring. The alcohol poisons the yeast which causes it to die. This is known as a negative feedback reaction. The increase in yeast cells numbers leads to less space for them to reproduce so they are having to compete for space. The cells which don't have enough space will soon die. The yeast cells may also be running out of carbohydrates from the cider so this could lead them to die. The increase in alcohol and the lack of space starts to kill the cells off and so after 9 to 10 days the rate of increase in cell numbers slows down. If the experiment was continued for a few more days then I would expect the rate to stop as by then all the yeast cells would be dead.

The graph plotted of cell numbers does refer to the total number of cells both living and dead. If just the number of living cells was counted then the number of cells would start by rising at a slow rate as reproduction is only just beginning (Lag Phase). The numbers of yeast cells would then rise at a very fast rate before levelling off (Log Phase). By the time C is reached the alcohol levels have built up so the birth rate is equal to the death rate causing the population to remain constant. After D the numbers would then start dropping as the yeast cells die from lack of space and alcohol poisoning. This is known as the death phase. The theoretical curve is shown below. [not reproduced]

A - B Lag Phase
B - C Log Phase
C - D Stationary
D - E Death Phase

Limitations

To help make this experiment more accurate, I used seven sets of results and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:

a) It is also impossible to precisely measure out the amounts of Yeast or cider when making the solution. As the scale on the pipettes shows the volume to the nearest mm³ the volume of the solutions that I used should be correct to the nearest mm³.

b) It is also impossible to perfectly calibrate the colorimeter corectly each time. This slight variation could lead to some minor innacuracies in the results. The colorimeter was used because the haemocytometer method was taking too long to do.

c) The experiment was not carried out at the same time each day which could lead to uneven gaps between measurements.

d) The oven temperature may not have remained constant all the time causing the rate of reproduction to alter as the oven temperature changed.

Anomolies

The plotted results on the graph produce a straight line of best fit to begin with which then goes into a curve of slightly decreasing gradient. It is a very smooth graph so no anomalies are present.

Extension Work

This experiment could be improved in a number of ways. It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

The test should be carried out at the same time each day to ensure that the length of time between each measurement is the same.

A test that could differentiate between living and dead cells would be very useful because then just the numbers of living cells could be counted. This would tell us how the population of living cells alters over time rather than at the moment where we can only find out how the total number of cells changed over time.

Different sources of carbohydrate apart from cider could be used to see if this affects the rate that the yeast numbers increase.

Different sized flasks could be used as this could increase or decrease competition between the yeast cells depending on the size. This would tell us how big a factor the competition between the cells is.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.


The Effect Of Substrate Concentration On The Activity Of The Enzyme Catalase

Thursday, June 5th, 2008

A Level Biology Project

Aims

This is an experiment to examine how the concentration of the substrate hydrogen peroxide affects the rate of reaction of the enzyme catalase.

Introduction

This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Background Information

Enzymes such as Catalase are protein molecules which are found in living cells. They are used to speed up specific reactions in the cells. They are all very specific as each enzyme just performs one particular reaction.

Catalase is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from the cells. Hydrogen Peroxide is the poisonous by-product of metabolism. Catalase speeds up the decomposition of Hydrogen Peroxide into water and oxygen as shown in the equations below.

Formula:

                       Catalase
 Hydrogen Peroxide---------------------->Water + Oxygen
                       Catalase
             2H2O2------------------->2H2O+O2

It is able to speed up the decomposition of Hydrogen Peroxide because the shape of it's active site matches the shape of the Hydrogen Peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called an anabolic reaction.

Apparatus Needed For The Experiments

  1. Gas Syringe
  2. Metal Stand
  3. Yeast Catalase
  4. Hydrogen Peroxide
  5. Test Tubes
  6. Beakers
  7. Test Tube Rack
  8. Stop Watch
  9. Pipette
  10. Pipette Filler
  11. Tap Water

Method

To test out how the concentration of hydrogen peroxide affects the rate of reaction first set up the apparatus below.

[Aparatus picture not reproduced]

1. Add 2cm3 of yeast to one test tube. Add 4cm3 of hydrogen peroxide solution at a concentration of 20% to the other test tube. Use a pipette to measure out the volumes. It is very important to accurately measure the amounts of Hydrogen Peroxide, Yeast and water to ensure a fair test.

2. Pour the hydrogen peroxide solution into the test tube containing the yeast and immediately put the gas syringe bung on the end of the test tube, at the same time start the stopwatch.

3. Bubbles should start to rise up the tube and the gas syringe will move outwards, as soon as the gas syringe passes the 30cm3 mark stop the stopwatch and note the elapsed time down to the nearest 1/10th of a second.

4. Repeat the experiment with hydrogen peroxide concentrations of 16%, 12%, 10%, 8%, 4% and 0%. The 0% concentration of hydrogen peroxide solution is done as a control solution to show that at 0% concentration no reaction occurs. The different concentrations of Hydrogen Peroxide are made by adding tap water to the 20% Hydrogen Peroxide in the correct amounts. The table below shows what amounts of Hydrogen Peroxide and water are needed to make the solutions.

Concentration Of Hydrogen Peroxide

Volume Of Hydrogen Peroxide (cm3)

Volume Of Water (cm3)

20% 4 0
16% 3.2 0.8
12% 2.4 1.6
10% 2 2
8% 1.6 2.4
4% 0.8 3.2
0% 0 4

5. Repeat all the tests at least three times so that an average can be obtained. Repeating the experiments several times will help to produce better and more accurate results as any inaccuracies in one experiment should be compensated for by the other experiments. Note all the results in a table such as the one below.

Hydrogen Peroxide Concentration 0% 4% 8% 10% 12% 16% 20%
Time Taken (Test 1)  
Time Taken (Test 2)
Time Taken (Test 3)
Average of the Tests
Rate  

The rate can then be worked out by

Rate=30/Average Time

This gives the rate in cm3 of oxygen produced per second, this is because I am timing how long it takes to produce 30cm3 of oxygen. From these results a graph can be plotted with concentration on the x-axis and time taken on the y-axis.

I am using yeast catalase as opposed to catalase from apples, potatoes or liver because it is easier to get the desired amount of yeast catalase by simply measuring it off. To obtain catalase from a substance such as potato would involve crushing it and with that method you would never be sure of the concentration of the catalase. If the catalase was used up then another potato would have to be crushed and this could produce catalase of a totally different concentration which would lead to inaccuracies in the experiment making this an unfair test.

To ensure this is a fair test all the variables except for the concentration of Hydrogen Peroxide must be kept the same for all the experiments. Variables that must not be altered include:-

Temperature, yeast concentration, type of yeast, batch of yeast, volume of yeast, volume of hydrogen peroxide, air pressure and humidity.

When measuring the volumes of Hydrogen Peroxide, Yeast and Water the measurement should be taken by looking at the scale at an angle of 90 degrees to it to avoid any parallax error.

Predictions

I predict that as the substrate concentration increases, the rate of reaction will go up at a directly proportional rate until the solution becomes saturated with the substrate hydrogen peroxide. When this saturation point is reached, then adding extra substrate will make no difference.

The rate steadily increases when more substrate is added because more of the active sites of the enzyme are being used which results in more reactions so the required amount of oxygen is made more quickly. Once the amount of substrate molecules added exceeds the number of active sites available then the rate of reaction will no longer go up. This is because the maximum number of reactions are being done at once so any extra substrate molecules have to wait until some of the active sites become available.

Results

I carried out the above experiment and these results were obtained.

Hydrogen Peroxide Concentration 0% 4% 8% 10% 12% 16% 20%
Time Taken (Test 1)   47.3 18.4 17.3 14.5 10.6 9.7
Time Taken (Test 2)   43.3 19 16.7 14.9 11.2 10
Time Taken (Test 3)   52.2 17.2 18.5 11.2 8.6 7.8
Average of the Tests   47.6 18.2 17.5 13.5 10.1 9.2
Rate=30/Average (Cm3/second) 0 0.63 1.65 1.71 2.22 2.97 3.26

All the times are in seconds. The average results are all written down to one decimal place because although the stopwatch gives results to two decimal places it is impossible to get accurate times to two decimal places due to the fact that our reaction times are not fast enough to stop the stopwatch precisely. I then worked out the rates of the reactions with the equation

Rate=30/Average Time

From these rates I was able to plot a graph of the rate of reaction against concentration of Hydrogen Peroxide.

Yeast Graph

Interpretation

When the concentration of Hydrogen Peroxide is increased, the rate of reaction increases at a directly proportional rate until the concentration of Hydrogen Peroxide reaches about 16%. If you double the concentration of Hydrogen Peroxide then the rate of reaction doubles as well. When the concentration is doubled from 8-16% the rate goes up from 1.65-2.97 Cm3 Oxygen produced per second, which is an increase of 1.8 times. I would expect the rate to increase two times if the Hydrogen Peroxide concentration is increased two times because there are twice as many substrate molecules which can join onto the enzymes active sites. The reason that the number is less than two times could be put down to the fact that at 16% the Enzyme's active sites may already be close to being saturated with Hydrogen Peroxide. There may also be some experimental error which causes the inaccuracies.

After 16% the increase in the rate of reaction slows down. This is shown by the gradient of the graph going down. At this point virtually all the active sites are occupied so the active sites are said to be saturated with Hydrogen Peroxide. Increasing the Hydrogen Peroxide Concentration after the point of saturation has been reached will not cause the rate of reaction to go up any more. All the active sites are being used so any extra Hydrogen Peroxide molecules will have to wait until an active site becomes available.

The theoretical maximum rate of reaction is when all the sites are being used but in reality this theoretical maximum is never reached due to the fact that not all the active sites are being used all the time. The substrate molecules need time to join onto the enzyme and to leave it so the maximum rate achieved is always slightly below the theoretical maximum. The time taken to fit into and leave the active site is the limiting factor in the rate of reaction.

The diagram below shows what happens.

[not reproduced]

Limitations

To help make this experiment more accurate, I repeated it three times and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables except for the concentration of Hydrogen Peroxide the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:

a) There is a slight delay between pouring the Hydrogen Peroxide into the yeast, putting the bung on and starting the stopwatch. This will slightly affect all the results but as I carried out all the three steps in the same way for all the experiments it should not make any difference to the overall result.

b) It is also impossible to precisely measure out the amounts of Hydrogen Peroxide, Yeast and Water each time. As the scale on the pipettes shows the volume to the nearest mm3 the volume of the solutions that I used should be correct to the nearest mm3. The volume of gas in the test tube to start with is slightly affected by the amount which the bung is pushed down each time, if the bung is pushed down further then the volume in the tube will be less so the 30cm3 of gas is reached faster.

c) Due to the fairly slow speed of our reactions it is only possible to measure the time of the reaction to the nearest 0.1 second even though the stopwatch shows the measurements to the nearest 0.01 second.

Anomolies

The plotted results on the graph produce a straight line of best fit to begin with which then goes into a curve of steadily decreasing gradient. The only anomalies are the results at 8% and 10%. The result at 8% is slightly above the line of best fit and the 10% result is slightly below it. This is probably due to an experimental error involving one of the factors mentioned above.

Extension Work

This experiment could be improved in a number of ways. It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

Using more concentrations of Hydrogen Peroxide would have produced a better looking graph and I would have liked to use concentrations higher than 20% to extend the graph so that the maximum possible rate of reaction could be reached.

The problem of the delay between pouring in the Hydrogen Peroxide, bunging the test tube and starting the stopwatch could have been limited by getting another person to start the stopwatch when the hydrogen peroxide was poured into the tube.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.


Investigation Into The Blowfly Larvae’s Response To Light

Wednesday, June 4th, 2008

A Level Biology Project

Aims

I plan to investigate how blowfly larvae react to light.

Introduction

This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Background Information

The way in which animals respond to external factors is known as behaviour. As animals become bigger their behaviour becomes more complex. This is because larger animals tend to have a more complex nervous and hormonal system. Two of the types of behaviour are called taxis and kinesis.

Taxis is a movement whose direction is determined by the direction of the stimulus. Some animals may move towards certain stimuli such as the smell of food. Others may move away from a stimuli such as excessive heat.

Kinesis is a behaviour pattern in which the animal changes it’s rate of movement in relation to the intensity of the stimulus. For instance if an animal is used to living in a humid environment then it’s activity will increase in a non-humid environment, this will help them to find a new humid environment as quickly as possible.

Apparatus Needed For The Experiments

  1. 20 Blowfly larvae
  2. Perspex Tray and lid
  3. Lamp
  4. Stopwatch
  5. Ruler
  6. Felt tip
  7. Paper
  8. Petri dish

Method

I am planning to use the following method for my experiment.

1. The apparatus is set up as shown in the picture below.

Diagram Of The Apparatus [not reproduced]

2. A piece of marked paper (an example of which is shown below) is put in the plastic dish. Each of the positive and negative sectors should have angles of 120° and each of the neutral sectors should have an angle of 60°. The end marked “positive” is positioned nearest to the light.

Paper Layout

3. The plastic tray should be 10cm away from the front of the lamp. The lamp should be shining at a shallow angle along the tray.

4. A pencil is spun to find a random direction. The maggot is then placed in the centre of the plastic dish facing in the direction of the pencil.

5. The dish’s cover is placed on the tray.

6. As soon as the maggot’s head leaves the paper’s inner circle the stopwatch is started. The position of the maggot’s head is then marked with a small ‘a’ on the plastic cover every 5 seconds until the maggot leaves the outer circle.

7. The maggot is taken out of the tray and placed in a spare petri dish so that it is not used again. This is because a maggot that has already been used may react differently if it is used again as it’s behaviour may be affected by it’s previous experience in the tray.

8. The distance between each ‘a’ is measured and noted down in a table such as the one below. The sector (positive, negative or neutral) that the maggot left the outer sector is noted in the table. The distance between each of the points and the front of the lamp is then measured and noted in another similar table.

Time
Experiment 5 10 15 20 25 Sector
1            
2            
3            

The rate in cm/second for each 5 second interval is then worked out with the following equation

Speed=Distance / Time

and put into a table such as the one above.

8. A new sheet of paper is then placed in the perspex tray with the ‘positive’ end pointing towards the light. This is to make sure that the maggot’s behaviour is not affected by following the trails of the other maggots. The experiment is then repeated another 19 times.

Using the results I plan to plot a graph of distance from the lamp against speed. This will allow me to se how the speed of the maggot alters as it’s distance from the lamp changes.

The Chi² test will be used to see if the differences in the maggot’s direction of movement are significant.

I will use the following method to do the Chi² test. The Observed value is the number of maggots that crossed each sector. The expected value is the number of maggots that I would expect to cross each sector according to the null hypothesis.

Direction Observed Expected O - E (O - E)² (O - E)² / E
Positive  
Negative
Neutral  
Total:  

If the total of the three “(O - E)² / E” values is bigger than 3.84 then I can be 95% confident that there is a significant difference between the results.

In order to make this experiment as accurate as possible a number of steps must be taken.

  • The experiment should be carried out in darkness with only the light from the bench lamp reaching the plastic tray. All the rooms light should be turned off and the window’s blinds should be closed.
  • Each maggot should only be used once to ensure that their behaviour is not affected by their previous experience in the perspex tray.
  • A fresh sheet of paper is used each time to ensure that the maggots can not follow the trails of any previous maggots.
  • The lamp should be at the same height and distance away from the tray for each experiment.
  • The distance should be measured from the front of the lamp to the plastic tray.
  • Each maggot should be positioned on the paper facing a random direction to ensure that it does not just move in a straight line in the direction that it is positioned.
  • When marking the position of the maggot the head of the mark should be placed at the maggots head to ensure consistency throughout the experiment. The eye should be directly above the maggot to avoid parallax error when putting the mark on the perspex.

Although taking these steps will make the experiment more accurate, it’s accuracy is still limited by several factors.

  • The accuracy of each measurement is limited to the nearest 0.5 mm the ruler is graduated in mm.
  • It is impossible to manage to mark the positions exactly every 5 seconds. Some of the time intervals may be slightly bigger than others.

Predictions

I predict that the maggots will move away from the light source. This is because bright lights could kill them from exposure to ultra violet radiation, or by drying the maggots out.

If the maggots move away from the light then they will avoid the harmful ultra violet light and will avoid drying out. Staying in a darker area will make the maggots less visible to predators which will increase their chances of survival. If they move away from the light then their movement pattern is called negative phototaxis.

I predict that their movements will be faster if they are nearer to the light. By moving faster when they are near the light they will be able to get away from the light quickly. Once they are a reasonable distance away I predict that they will slow down as they are no longer in danger of being killed by overexposure to the light.

If the maggots were to stay in the sunlight then there is a chance that they could dry out and so die.

Null Hypothesis As their is the same angle for each of the three sectors (positive, negative and neutral) an equal amount of maggots should leave each sector.

33.3% of maggots should leave by the positive sector.
33.3% of maggots should leave by the negative sector.
33.3% of maggots should leave by the neutral sector.

The Experiment - The experiment was carried out as stated in the plan. As the maggots moved fairly quickly I had to use 2 second intervals between each mark instead of 5 second intervals.

Results

These are the results from my experiments.

Result Table 1

Result Table 2

Interpretation

I will analyse the results for the experiment. As I predicted the maggots moved away from the light as soon as they were placed in the plastic container. The maggot’s behaviour is therefore negative phototaxis. 15 of the maggots left the outer circle through the negative sector, showing that the maggots move away from the light. The Chi² test for the experiment is shown below.

Direction Observed Expected O - E (O - E)² (O - E)² / E
Positive 1 6.67 -5.67 32.11 4.82
Negative 15 6.67 8.33 69.44 10.42
Neutral 4 6.67 -2.67 7.11 1.07
Total: 16.3

As the total is bigger than 3.84 I can be 95% sure that there is a significant difference between these and the results as predicted in the null hypothesis. I can therefore reject the null hypothesis. Maggots will move away from the light in an experiment such as this. For example maggot 17 heads away from the light at a speed of 1.3cm/s. By the time it has crossed the outer sector it has slowed down to 1.1cm/s. 15 of the 20 maggots display similar behaviour and move away from the light and exit the circle by the negative sector.

They have light stimuli on either side of their head. By moving their head they can detect where the stimuli is coming from. If the maggot detects that the light is stronger on the left hand side of it’s head then it will turn right so that it is moving away from the light. If the light is stronger on the right hand side of it’s head then it will turn left. This helps the maggot to move away from the light. If the light intensity is the same on both sides of it’s head then it will head in a straight line.

This could explain why maggot 18 headed towards the light. When it was placed in the tray it may have been pointing towards the light so it was receiving equal light on both stimuli. It therefore moved forwards in a straight line towards the light. It did speed up as it got closer to the light as it must have been trying to get away from the light as quickly as possible. As the light was of equal strength on both sides of it’s head it was unable to tell that it was heading towards the light.

By looking at the graph of the speed of movement against the distance from the lamp it can be seen that the majority of the lines move downwards and to the right. This shows that as the maggots get further away from the light their speed decreases. This maggots may therefore be showing photokinesis. More experiments would need to be done to confirm this. Kinesis is a reaction to the intensity of the stimulus and not the direction so photokinesis could be tested by using a light whose brightness can be altered.

Many of the maggots moved slowly during their first few seconds in the tray. This is probably due to them getting their bearings and working out where the light is coming from before moving off in the opposite direction.

Secondary Data

Here is some secondary data which I will use to compare with the data that I obtained.

Secondary Data

By carrying out the Chi² test this data produces similar results to my own.

Direction Observed Expected O - E (O - E)² (O - E)² / E
Positive 0 3.33 -3.33 11.11 3.33
Negative 9 3.33 5.67 32.11 9.63
Neutral 1 3.33 -2.33 5.44 1.63
Total: 14.6

These results confirm that we can be 95% sure that the difference in direction is significant.

By looking at the data and the graph it can be seen that 9 of the 10 maggots head away from light. 7 of them slow down as they get further away from the light. None of the maggots head towards the light. This data produces the same conclusion as my data and confirms that my results are correct.

The maggots show negative phototaxis behaviour. The maggots use the sensors on either side of their head to work out where the light is coming from. They then turn away from the light and head towards the dark to get away from the light. They do this to avoid the harmful ultra violet light which can harm them. By performing this negative phototaxis behaviour the maggots can increase their chance of survival.

Limitations

The accuracy of this experiment is limited by a number of factors.

a) It was impossible to get rid of all background light as other similar experiments were being carried out in the same room.

b) The accuracy of each measurement is limited to the nearest 0.5 mm the ruler’s are graduated in mm.

c) I had no way of knowing how old the maggots were. The maggot’s response may alter with age which could explain why some maggots did not show the expected response.

d) As the bench lamp was only 10cm away from the perspex tray it may have caused a heating effect on the tray. The maggot’s may therefore have reacted to the heat as well as the light. This could have been prevented by placing the lamp further away.

Extension Work

This experiment could be improved in a number of ways.

1) It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

2) Each person should have carried out their experiments in a different room to cut out all background light.

3) A perspex screen could have been placed between the light and the perspex tray to reduce any heating effect the light may have on the maggots.

4) Different ages of maggots could have been tested to see if the maggots response to the stimuli varies with age.

5) Different intensities of light could be used to see if the maggots react differently from one light intensity to another.

6) Different colours of light could be used to see if this has an effect on the maggots.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.


Factors Affecting The Rate Of Photosynthesis

Tuesday, June 3rd, 2008

A Level Biology Project

Aims

I plan to investigate how different factors affect the rate of photosynthesis. I will be changing the levels of light and CO2 and then measuring the photosynthetic rate.

Introduction

This is a A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Background Information

Yellow flower The rate of photosynthesis is affected by a number of factors including light levels, temperature, availability of water, and availability of nutrients. If the conditions that the plant needs are improved the rate of photosynthesis should increase.

The maximum rate of photosynthesis will be constrained by a limiting factor. This factor will prevent the rate of photosynthesis from rising above a certain level even if other conditions needed for photosynthesis are improved. This limiting factor will control the maximum possible rate of the photosynthetic reaction.

For instance, increasing the temperature from 10ºC to 20ºC could double the rate of photosynthesis as the plant's enzymes will be closer to their optimum working temperature. As the temperature is increased, molecules in the cells will be moving at a faster rate due to kinetic theory. If the temperature is raised above a certain level, the rate of photosynthesis will drop as the plant's enzymes are denatured. They will therefore be more likely to join onto the enzymes and react.

The amount of water available to the plant will affect the rate of photosynthesis. If the plant does not have enough water, the plant's stomata will shut and the plant will be deprived of CO². It is difficult in normal lab conditions to prove that water directly affects photosynthesis unless a heavy isotope is used to trace the path of water.

Chlorophyll is needed for photosynthesis. This can be proved by studying a variegated leaf. It is however very difficult to study how different levels of chlorophyll in the plant will affect it's photosynthesis rate. This is because in a variegated leaf the cells either contain chlorophyll or they don't.

Carbon dioxide concentration will directly affect the rate of photosynthesis as it is used in the photosynthesis reaction. It is also easy to change the amount of carbon dioxide that the plant receives.

Light is also directly used in the photosynthesis reaction and is easy to change in normal lab conditions. Carbon Dioxide and Light are the factors that I will change in the experiment as they are easy to change and measure.

Apparatus Needed For The Experiments

  1. Elodea
  2. 20mm² syringe
  3. Capillary tubing
  4. Stand
  5. Stopwatch
  6. Ruler
  7. NaHCO³ Solution
  8. Bench lamp
  9. Distilled water

Method

I could measure the decrease in the substances needed for photosynthesis, such as how much the amount of CO2 decreases over time. This is however difficult in normal lab conditions. I will instead measure how one of the products of photosynthesis (oxygen) increases over time. I am planning to use the following method for my experiment.

  1. The apparatus is set up as below with the syringe full of the 0.01M solution of NaHCO3 solution. Two marks 10cm apart are made on the capillary tubing.
  2. The syringe is placed 0.05m away from the lamp.
  3. Using the syringe plunger the meniscus of the NaHCO3 is set so that it is level with the first mark.
  4. A stopwatch is then started. The meniscus should gradually move down the capillary tube as the elodea produces oxygen as a by-product of photosynthesis. As the oxygen is produced it increases the pressure in the syringe and so the meniscus is pushed down the tube.
  5. When the meniscus reaches the level of the bottom mark the stopwatch should be stopped and the time should be noted in a table such as the one below.
Molarity of NaHCO3
Light Intensity 1/d² (m) 0.00
(Distilled water)
0.01 0.02 0.05 0.07 0.1
400            
278            
204            
156            
100            
25            
11            
4            

The light intensities have been worked out using the following equation

Light Intensity = 1 / Distance² (m)

6. Using the same piece of elodea and the same distance between the lamp and the syringe the experiment (steps 1 to 5) should be repeated for the other concentration of NaHCO3.
7. The experiment (steps 1 to 6) should then be repeated at each different distance between the syringe and the light for all the NaHCO3 concentrations. The remaining distances are 0.05m, 0.06m, 0.07m, 0.08m, 0.1m, 0.2m, 0.3m, and 0.5m.
8. The entire experiment should then be repeated three times in order to obtain more accurate data and to get rid of any anomalies that may occur in a single experiment.

Measuring the volume of oxygen is more accurate than counting the number of bubbles produced as each bubble could be a different size. In order to make this experiment as accurate as possible a number of steps must be taken.

  • The experiment should be carried out in darkness with only the light from the bench lamp reaching the elodea.
  • The same piece of elodea should be used each time in order to make sure that each experiment is being carried out with the same leaf surface area.
  • The amount of NaHCO3 solution should be the same for each experiment. 20mm² should be used each time.
  • The lamp should be at the same height for each experiment. It should be level with the syringe each time.
  • The distance should be measured from the front of the lamp to the syringe. Although taking these steps will make the experiment more accurate, it's accuracy is still limited by several factors.
  • Some of the oxygen will be used for photosynthesis by the plant.
  • Some of the oxygen will dissolve into the water.

From these recorded times I will work out the rate of the reaction using the following equation.

Rate Of the Reaction = 1 / Time (s)

Using these rates I plan to plot a graph of the rate of reaction against light intensity.

Predictions

Light

I predict that if the light intensity increases the rate of the reaction will increase at a proportional rate until a certain level is reached, the rate of increases will then go down. Eventually a level will be reached where increasing the light intensity will have no more effect on the rate of reaction as there is some other limiting factor.

Light is needed for photosynthesis in plants. When chloroplasts in the leaf's cell are exposed to light they synthesise ATP from ADP. Oxygen is produced as a by-product of the photosynthesis reaction. Therefore increasing the concentration of light will increase the amount of ATP being synthesised from ADP and so more oxygen will be released as a by product.

NaHCO3

I predict that as the concentration of NaHCO3 increases the rate of the reaction will increase at a proportional rate. Eventually increasing the NaHCO3 concentration more will have no effect as other limiting factors will be limiting the rate of photosynthesis. Carbon dioxide is needed for the photosynthesis reaction. It is used to make the organic products of photosynthesis. If the elodea is able to absorb more CO2 then the rate of photosynthesis will increase as the plant is able to make more of the organic compounds. The plant is given CO2 in the form of NaHCO3.

Results

Pooled results from the group were used. They were taken over a 2 day period.

Molarity of NaHCO3
Light Intensity 1/d² (m) 0.00
(Distilled water)
0.01 0.02 0.05 0.07 0.1
400 3571 1666 1099 523 200 243
278 1670 5183 988 600 375 262
204 4998 4485 1175 1005 473 351
156 5590 2300 1770 1445 621 550
100 9990 3150 2900 2552 1224 645
25   4762 3984 2850 1640 1408
11   5945 4348 3780 2830 2564
4   16480 11904 5196 6578 3226

Using these results I worked out the rate

Rate Of the Reaction = 1 / Time(s) x 1000

The rate was multiplied by 1000 to make the numbers easier to handle.

Molarity of NaHCO3
Light Intensity 1/d² (m) 0.00
(Distilled water)
0.01 0.02 0.05 0.07 0.1
400 0.28 0.60 0.91 1.91 5.00 4.12
278 0.60 0.19 1.01 1.67 2.67 3.82
204 0.20 0.22 0.85 1.00 2.11 2.85
156 0.18 0.43 0.56 0.69 1.61 1.82
100 0.10 0.32 0.34 0.39 0.82 1.55
25   0.21 0.25 0.35 0.61 0.71
11   0.17 0.23 0.26 0.35 0.39
4   0.06 0.08 0.19 0.15 0.31

A graph of the rate of reaction against light intensity was drawn. It shows how the amount of CO2 and light affect the rate of photosynthesis. Lines of best fit were drawn for each CO2 concentration to make up for any inaccuracy in any individual result. The line of best fit gives a good picture of how the overall rate of reaction is affected by the light and CO2.

Rate of photosynthesis graph

Interpretation

I will analyse the results for how the amount of light and CO2 affects the rate of photosynthesis.

My prediction that the rate of photosynthesis would go up if the light intensity and NaHCO3 levels were increased proved correct. As the elodea absorbed the light and CO2 it produced oxygen gas which increased the pressure in the syringe. This pushed the air bubble in the capillary tube down. The chloroplasts produce ATP and reduce NADP to NADPH2 when exposed to light. It is at this stage of the reaction that oxygen is produced as a waste product.

As predicted when the light intensity increases so does the rate of photosynthesis. I predicted that a level would be reached where increasing the light intensity would have no more effect on the rate of reaction as there would be some other limiting factor which limits the rate of the reaction. The rate increases at a steady rate as the light intensity increases until near the end of each line where the rate of increase decreases. This is either because the photosynthesis reaction has reached it's maximum rate of reaction or another factor is limiting the rate. As 6 different CO2 concentrations were used I can see that the first five reactions are not occurring at their maximum rate as there is the 0.1M NaHCO3 rest which is occurring at a faster rate then the other 5. The photosynthesis reactions of the other five test must therefore be limited by the concentration of CO2 to the plant.

As predicted when the NaHCO3 concentration is increased the plant in able to get more CO2 which causes the rate of reaction to go up. I predicted that once the NaHCO3 had been raised above a certain level increasing the rate further would have no effect as there would be other limiting factors limiting the rate of the reaction. As the NaHCO3 concentration in the water was increased the rate of photosynthesis was able to go up. The plant therefore made more oxygen as a waste product. At a NaHCO3 concentration of 0.1M once the light intensity gets above 300 the rate of reaction slows down very quickly. This could be because photosynthesis is occurring at it's maximum possible rate or because another limiting factor is limiting the rate of reaction.

Distilled Water

With the distilled water the rate of reaction went up from 0.1 to 0.4 when the light intensity was increased from 100 to 400. This is a 4 times rise which is quite large. The curve on the graph does however level out quite soon showing that the rate is being limited by the lack of NaHCO3 in the water.

0.01M NaHCO3

At a light intensity of 4 the rate is 0.06 but this rises to 0.6 when the light intensity is brought up to 400. The curve is very shallow and levels off towards a light intensity of 350 - 400.

0.02M NaHCO3

The amount of NaHCO3 is double that of the 0.01M NaHCO3 experiment. The rate also finishes off twice that of the 0.01M experiment. This would surgest that there was a directly proportional relationship between the amount of NaHCO3 and the rate of reaction.

0.05M NaHCO3

The curve for the 0.05M NaHCO3 is steeper than the previous curves. The rate rises to 1.9 at a light intensity of 400.

0.07M NaHCO3

The 0.07M NaHCO3 test produces a line which is steeper than all the previous curves. The plant is using the extra CO2 to photosynthesise more. As the plant has more CO2 the limiting factor caused by the lack of CO2 is reduced. This test did produce a big anomaly. The rate for a light intensity of 400 is 5. By following the line of best fit I can see that this result should be more like 3.5. The elodea for this test was very close to the light source. It is possible that it had been left here for a while which caused the lamp to heat the elodea up. This would have increased the rate of reaction of the plant's enzymes which would have increased the photosynthesis rate.

0.1M NaHCO3

The 0.1M NaHCO3 produced the steepest line. Near the end of the line it looks as if the rate of reaction is hit by another limiting factor. The line goes up steadily but then between a light intensity of 300 and 400 levels off very quickly. This would surgest that at a 0.1M NaHCO3 is sufficient for the plant to photosynthesise at it's maximum rate with it's current environmental conditions. Increasing the NaHCO3 concentration after this level would therefore have no effect unless the next limiting factor was removed.

The fact that the curve levels off so quickly indicates that there is another limiting factor limiting the photosynthesis. It could be temperature. These tests are being carried out at room temperature so the temperature would have to be raised another 15ºC before the enzymes in the plant's cells were at their optimum working temperature. More tests could be done by using water that was at a higher temperature to see what effect this would have on the photosynthesis rate. It is however impossible to raise the plant's temperature without affect other factors. For instance the actual amount of oxygen released by the plant is slightly more than the readings would surgest as some of the oxygen would dissolve into the water. At a higher temperature less oxygen would be able to dissolve into the water so the readings for the photosynthesis rate could be artificially increased.

It is also possible that the photosynthetic reactions in the plant are occurring at their maximum possible rate and so can not be increased any more.

The light is probably not a limiting factor as all but one of the curves level off before the maximum light intensity of 400 is reached. The maximum light intensity that the plants can handle is therefore just below 400.

Water will not be a limiting factor as the plants are living in water. They therefore have no stomata and absorb all their CO2 by diffusion through the leaves.

Limitations

The accuracy of this experiment is limited by a number of factors.

  1. Some of the oxygen give off is used for respiration by the plant.
  2. Some of the oxygen dissolved into the water.
  3. Some was used by small invertebrates that were found living within the pieces of elodea.
  4. The higher light intensities should be quite accurate but the smaller light intensities would be less accurate because the light spreads out. the elodea will also get background light from other experiments.
  5. The lights are also a source of heat which will affect the experiments with only a small distance between the light and the syringe. this heating could affect the results.
  6. Using the same piece of elodea for each experiment was impractical as the elodea's photosynthesis rate decreased over time. By using a different piece of elodea for each experiment did create the problem of it being impossible for each piece to have the same surface area.
  7. As the tests took place over a two day period there will be some inaccuracy caused by factors such as temperature. There was no practical way for the long tests to be kept at a totally constant temperature for the two day period and they will probably have cooled down at night and then warmed up in the day leading to a slight inaccuracy.

Extension Work

This experiment could be improved in a number of ways.

  1. It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.
  2. Each person should have done their experiments in a different room to cut out all background light.
  3. All the experiments should be done sequentially.
  4. A perspex screen could have been placed between the light and the syringe to reduce any heating effect that the light may have.
  5. The experiment could have been carried out with higher NaHCO3 to see if increasing the concentration would increase the rate of photosynthesis, or if a concentration of 0.1M NaHCO3 produces the maximum rate of photosynthetic reaction.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.


Surface Area / Volume Ratio Biology Experiment

Monday, June 2nd, 2008

A Level Biology Project

Aims

This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion in substrates and how this relates to the size and shape of living organisms.

Introduction

This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Background Information

The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive.

Single celled organisms can survive as they have a large enough surface area to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.

Apparatus Needed For The Experiments

  1. Beakers
  2. Gelatin blocks containing cresol red dye
  3. 0.1M Hydrochloric acid
  4. Stop Watch
  5. Scalpel
  6. Tile
  7. Safety glasses

Method

1. A block of gelatin which has been dyed with cresol red dye should be cut into blocks of the following sizes (mm).

5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10 x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5

The triangle is of the following dimensions. [not reproduced]

The rest of the blocks are just plain cubes or rectangular blocks.

Cresol red dye is an acid / alkali indicator dye. In the alkali conditions of the gelatin it is red or purple but when it gets exposed to acid it turns a light yellow colour.

Gelatin is used for these tests as it is permeable and so it acts like a cell. It is easy to cut into the required sizes and the hydrochloric acid can diffuse at an even rate through it.

I am not using any blocks bigger than 20 x 20 x 20 as a preliminary test found that it was only practical to use blocks of 20mm³ or less as anything bigger than this would take longer than the amount of time that we have to do the experiment.

2. A small beaker was filled with 100cm³ of 0.1 molar Hydrochloric acid. This is a sufficient volume of acid to ensure that all the block sizes are fully covered in acid when dropped into the beaker.

3. One of the blocks is dropped into this beaker and the time for all the red dye to disappear is noted in a table such as the one below.

Dimensions (mm) Surface Area Volume (mm³) Surface Area / Volume Ratio Test 1 Test 2 Test 3 Average Time
 
   

4. This test should be repeated for all the sizes of blocks three times to ensure a fair test. Fresh acid should be used for each block to ensure that this does not affect the experiment’s results.

5. The surface area / volume ratio and an average of the results can then be worked out. A graph of Time against Surface Area to Volume Ratio can then be plotted. From this graph we will be able to see how the surface area affects the time taken for the hydrochloric acid to penetrate to the centre of the cube.

Predictions

I predict that as the Surface Area / Volume Ratio increases the time taken for the hydrochloric acid to penetrate to the centre of the cube will go down. This is because a small block has a large amount of surface area compared to it’s volume so the hydrochloric acid will have a large surface area to diffuse through. A larger block has a smaller amount of surface area in relation to it’s size so it should take longer for the hydrochloric acid to diffuse into the centre of the cube. The actual rate of the hydrochloric acid diffusing through the gelatin should be the same for all the blocks but when the surface area / volume ratio goes up it will take less time for the hydrochloric acid to reach the centre of the cube.

Results

I carried out the above experiment and these results were obtained.

Dimensions (mm) Surface Area Volume (mm³) Surface Area / Volume Ratio Test 1 Test 2 Test 3 Average Time
5 x 5 x 5 150 125 1.2:1 7.02 6.57 4.53 6.16
10 x 10 x 10 600 1,000 0.6:1 10.3 23.25 15.33 16.28
15 x 15 x 15 1,350 3,375 0.4:1 29.55 30.22 23.45 28.01
20 x 20 x 20 2,400 8,000 0.3:1 53.4 32.44 58.56 48.3
10 x 10 x 2 280 200 1.4:1 0.26 0.37 1.58 1.01
10 x 15 x 5 550 750 0.73:1 7.2 10.23 10.47 9.3
20 x 5 x 5 450 500 0.9:1 3.18 2.58 4.09 3.29
10 x 10 x 10
(Triangle)
441.42 500 0.88:1 9.58 3.34 5.25 6.19

The Surface area to Volume ratio is calculated by

Surface Area To Volume Ratio = Surface Area / Volume

From these rates I was able to plot a graph of the Surface Area to Volume Ratio against time.

Interpretation

In all the blocks of gelatin the rate of penetration of the hydrochloric acid from each side would have been the same but all the blocks take different amounts of time to clear because they are different sizes. As the blocks get bigger it takes longer for the hydrochloric acid to diffuse through all the block and so clear the dye. It takes longer to reach the centre of the cube even though the rate of diffusion is the same for all the cubes.

As the volume of the blocks goes up the Surface Area / Volume ratio goes down. The larger blocks have a smaller proportion of surface area than the smaller blocks. The smallest block has 1.4mm² of surface area for every 1mm³ of volume. The largest block only has 0.3mm² of surface area for each 1mm³ of volume. This means that the hydrochloric acid is able to diffuse to the centre of the smallest block much faster than the largest block. The acid took 48 minutes to diffuse to the centre of the largest block but only 1 minute in the smallest block. A living cell would not survive if it had to wait 48 minutes for oxygen to diffuse through it so living cells need to be very small.

When the surface area to volume ratio goes down it takes longer for the hydrochloric acid to diffuse into the cube but if the ratio goes up then the hydrochloric acid diffuses more quickly into the block of gelatin. Some shapes have a larger surface area to volume ratio so the shape of the object can have an effect on the rate of diffusion.

It is important that cells have a large surface area to volume ratio so that they can get enough nutrients into the cell. They can increase their surface area by flattening and becoming longer or by having a rough surface with lots of folds of cell membrane known as villi. [picture not reproduced]

The villi vastly increase the surface area of the cell whereas the cell which is round only has a small surface area in relation to it’s volume. Both cells above have an volume of 1cm³. The cell on the left has a surface area of 3cm² but the cell on the right with villi has a surface area of 10cm². The cell membrane is made up of a lipid bi-layer with many proteins integrated into it. [picture not reproduced]

Oxygen can diffuse easily through the membrane and Carbon Dioxide and other waste products can easily dissolve out. The concentration of oxygen in the cell is always lower than outside the cell which causes the oxygen to diffuse in. Gases will always dissolve from an area of high to low pressure. The concentration of carbon dioxide outside the cell is lower than the concentration in the cell so the carbon dioxide will always dissolve out of the cell.

Single celled organisms such as amoebas have a large surface area to volume ratio because they are so small. They are able to get all the oxygen and nutrients they need by diffusion through the cell membrane.

Larger organisms such as mammals have a relatively small surface area compared to their volume so they need special systems such as the lungs in order to get enough oxygen. Surface area to volume ratio is very important in lungs where a large amount of oxygen has to get into the lungs. The lungs have a very large surface area because they contain millions of sacs called alveoli which allow oxygen to diffuse into the bloodstream. By having millions of these alveoli the lungs are able to cram a very large surface area into a small space. This surface area is sufficient for all the oxygen we need to diffuse through it and to let the carbon dioxide out.

By increasing the surface area the rate of diffusion will go up.

Precautions

a) All the gelatin used should be taken from the same block to ensure that all the blocks are made up of the same materials.

b) All the tests should be done at room temperature to ensure that the blocks of gelatin do not melt.

c) The same volume of acid should be used for all the tests to ensure that the rate of diffusion can not be affected by the pressure of a larger volume of acid.

d) Safety glasses should be worn to protect your eyes from the hydrochloric acid.

Limitations

To help make this experiment more accurate, I repeated it three times for each block size and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables except for the size of the gelatin blocks the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:

a) It is also impossible to precisely measure the size of gelatin block each time. I measured the sizes to the nearest mm so the sizes of block that I used should be correct to the nearest mm.

b) When the gelatin blocks are dropped into the beakers the base of the block comes into contact with the bottom of the beaker which reduces the surface area of the block that comes into contact with the hydrochloric acid.

c) The results will be slightly inaccurate as the moment when the gelatin block has lost all it’s dye is a matter of opinion and not something that can be measured precisely.

d) Due to the fairly slow speed of our reactions it is only possible to measure the time of the reaction to the nearest 0.1 second even though the stopwatch shows the measurements to the nearest 0.01 second.

Anomolies

The graph produced shows a smooth curve with a decreasing gradient as the surface area to volume ratio goes up. The only anomaly is the result for the 5 x 5 x 5 block. The result here is higher than the curve of best fit for the graph. The results for the 5 x 5 x 5 block ranged from 4.53 to 7.02 seconds with an average of 6.15 seconds. The line of best fit for the graph suggests that the average should be around 3 seconds. The anomalous result was probably due to experimental error as a result of this being the first block size that I used in the experiment. The most likely explanation is that I was unsure of how to judge when all the dye had disappeared and as a result delayed pressing the stop button of the stop watch. As the experiment progressed with the other block sizes I probably got better at making this judgement.

Extension Work

This experiment could be improved in a number of ways.

1) It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

2) Using more shapes and sizes of gelatin block would have produced a better looking graph.

3) Variables that might affect the rate of diffusion could be investigated. The rate of diffusion may also be affected by temperature, strength of acid and volume of acid

4) The block could be suspended in the hydrochloric acid so than none of it’s surfaces are in contact with the wall of the beaker. A small cradle could be used to suspend the blocks in the acid which would mean that all six sides of the cube should be in contact with the acid. This would ensure that diffusion could occur evenly through all the sides of the cube.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.


Effects Of Varying Environmental Conditions On The Rate Of Transpiration In Leafy Shoots

Sunday, June 1st, 2008

A Level Biology Project

Aims

I plan to investigate how environmental conditions affect the transpiration of plants.

Introduction

This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.

Background Information

The Sun provides the energy to turn the water in the plants into a vapour causing it to evaporate into the leave’s internal air spaces before diffusing out of the stomata into the air. This is known as transpiration. As the water evaporates out of the top of the plant it creates a suction on the column of water below it in the xylem. The upwards force on the column of water created by transpiration and the downwards force due to gravity created a tension in the column of water.

As the upwards pull is greater than the downwards pull the column of water moves up the xylem. Cohesion tension theory tells us that it is the evaporation of water from the leaves which causes the upwards movement of water. The water molecules have a high cohesion as they are polar and so are electrically attracted to each other. They are held together by hydrogen bonds. The column of water does not tend to break as it has a very high tensile strength from the bonds.

The water is able to evaporate out of the leaf as the leaf has a high water potential and the air has a low water potential so the water molecules pass down the concentration gradient from the spongy and palisade mesophyll cells into the leave’s internal air spaces before diffusing out into the air.

Transpiration is needed to keep the cells of the spongy and palisade mesophyll cells moist as this allows carbon dioxide to dissolve before diffusing into the cells for photosynthesis. The stomata open in the day to let carbon dioxide diffuse in, and to let oxygen diffuse out as part of photosynthesis. At night photosynthesis is unable to take place due to the absence of light so the stomata are closed to reduce water loss.

Light causes potassium ions to be pumped into the guard cells which lowers their water potential and so water diffuses into the guard cells causing them to go turgid and so open. At night potassium moves out of the guard cells into the surrounding cells so the water diffuses out of the guard cells causing them to close.

In hot climates the water loss by transpiration can exceed the water uptake from the roots which causes the plants to suffer from water stress. To combat this ABA is produced by the plant which causes the rapid pumping of potassium ions out of the guard cells which closes them and so reduces the water loss by transpiration.

Apparatus Needed For The Experiments

  1. Privet shoot (used as it has many big leaves)
  2. Capillary tubing (1.13mm Diameter)
  3. Bowl of water
  4. Stand
  5. Stopwatch
  6. Ruler
  7. Vaseline

Method

For this experiment a simple potometer will be made to measure the rate of water uptake from the plant. The potometer is the easiest way of measuring this and is quite accurate as well. The experiment is limited by the fact that the potometer measures the total water uptake and not just the transpiration rate.

I am planning to use the following method for my experiment.

1. A privet shoot is cut under water in a large bowl about 1 inch up the stem. This should remove any blockages that are within the xylem when the shoot was originally cut. In order to stop the xylem vessels from being crushed when the shoot is cut, it should be cut diagonally with a razor blade. The end of the privet shoot should be left in the bowl of water that it was cut in order to prevent any air bubbles getting into the system.

2. The capillary tube should be completely filled with water by submerging it into the same bowl as the privet shoot.

3. The capillary tube should be attached to the shoot underwater being careful to make sure that there are no air bubbles in the tube.

4. The joins between the capillary tube and the shoot should be sealed with Vaseline to make sure that the system is water tight.

5. Making sure that the open end of the capillary tube remains underwater the plant can be raised out of the water and clamped above the bowl of water in an upright position.

6. The whole system should now be full of water and completely air tight. As the plant transpires it will pull water up through the tubing. This apparatus will measure the total water uptake.

7. Allow the apparatus to equilibrate for about 5 minutes. As this is the control experiment the leaves should be exposed to the maximum possible light intensity by using a bench lamp. There should also be no wind in the area that the experiment is taking place.

8. Introduce an air bubble to the system by holding the apparatus out of the water until a bubble forms.

9. Measure how far the bubble has risen up the capillary tube every thirty seconds for five minutes. When measuring the distances your eye should be at a 90º angle from the bottom of the bubble. This way parallax error can be avoided when looking at the scale on the ruler. You also must measure from the same point of bubble every time in order to ensure consistency in the measurement taking. I am planning to take all my measurements from the bottom of the bubble. The distances travelled should be noted.

Diagram Of The Potometer [not reproduced]

The test should be repeated 5 or 6 more times so that an accurate average of the rate of water uptake can be obtained. After each experiment the tubing can be refilled with water by placing the end of the tube underwater and squeezing it to force the air bubble out of the tube.

10. The air temperature of the room should also be noted. This test will give the rate of transpiration under normal inside conditions. These set of results are a control set of results against which the other results can be compared.

It is very important that the whole apparatus is air tight as if any bubbles of air were to get into the xylem then transpiration would not occur. This is because any air bubble would break the continuous column of water which is usually present in the xylem and so the water molecules below the air bubble would not feel the upwards pull from the above water molecules. The xylem would soon fill with air in a process called cavitation. The plant is cut under water an inch up it’s stem in order to get rid of the bottom part of the xylem which may have an air bubble in it. Doing this makes sure that the shoot which will be used in the experiment has water in it’s xylem.

The experiment can then be repeated but this time changing the following environmental conditions.

Wind

To measure how wind affects transpiration a hairdryer or fan that blows cold air can be used to blow the leaves. The fan has to be fairly close to the leaves but not so close that it will buffet the leaves as this may cause the stomata to close.

Humidity

The shoot will be enclosed in a transparent plastic bag. The humidity will soon increase as the water vapour which has been transpired will not be able to leave the bag and so will stay around the plant thereby increasing the humidity.

Light

The windows will be covered and the lights dimmed to make the conditions darker.

Surface Area

The surface area of the leaves of the plant can be reduced by vasolining the leaves on the top and especially to the bottom side of the leaf. The bottom side of the leaf is where the stomata are so this is where most of the water loss in the plant occurs. Applying Vaseline to the bottom surface of the leaf will block the stomata and so water vapour will be unable to leave the leaf through this leaf. The effect that this has on the rate of transpiration can be measured. The surface area of the leaf that has just been vaselined can be measured by drawing around the leaf on some graph paper. The surface area can be worked out by counting the squares. This vaselining technique can be used on more and more leaves till all the plant’s leaves are vaselined.

Each experiment should be replicated several times in order to obtain more accurate data and to get rid of any anomalies that may occur in a single experiment. I will note the results in a table for each 30 second period in all the experiments.

From these recorded distances I will work out the rate of water uptake for each 30 second period for all the experiments. Using these rates I plan to plot several graphs of the rate of water uptake against the time for various environmental conditions. I shall plot the results for the same environmental factors on the same graph so that the effect that these environmental factors has on the rate can easily be compared.

I shall also work out the volume of water taken up per minute. This can be worked out with the following equation.

Volume Of Water Uptake = p x Radius Of The Capillary Tube² x Distance Travelled

Predictions

I predict that if the wind is increased the rate of transpiration will increase. In the control test where there will be no wind a band a water vapour will be able to form in the air spaces of the leaf and around it as water transpires out. This will reduce the water potential gradient between inside the leaf and the air so the rate of transpiration will be reduced. If air is blown across the leaves by the wind or in the case of this experiment a hairdryer, this band of water vapour will be blown away and further water vapour will not be able to accumulate. This will lead to an increase in the water potential gradient between the inside of the leaf and the air and so the rate at which water transpires into the air will increase.

I predict that if the surface area of the plant’s leaves available for transpiration is halved then the rate of transpiration will be halved. This is because the plant loses almost all it’s water through it’s leaves so if it loses the ability to transpire through half of it’s leaves then the plant will only lose half as much water through transpiration. The amount of surface area that the plant has should be directly proportional to the rate of transpiration.

Results

Secondary results that were taken in January 1998 under controlled conditions were provided. The environmental conditions were measured with electronic probes. These results showed the distance that the air bubble had travelled up the capillary tube over a period of 4 minutes.

Using these results I worked out the rate of travel up the capillary tube for all the experiments. I also averaged the duplicated experiments to give me an average set of results for experiments 14 and 15, and also an average of experiments 11 and 12. I averaged experiments 1, 2 and 9 to produce control data for the rate of travel up the capillary tube against which the other results can be compared.

The sheet of secondary results is shown on the next page. The results table with my extra calculations for the rates of uptake and the averages of the duplicated experiments are shown on the page after that. [not reproduced]

Graphs of the rate of uptake against time were drawn by me for the surface area of the leaves and the wind speed.

Interpretation

I will analyse the results for the surface area experiments and the wind speed experiments.

Surface Area

The graph shows that as the surface area goes down, the rate of transpiration goes down. This would seem to prove my prediction. When the surface area of the plant’s leaves available for transpiration is reduced, the rate of transpiration is also reduced. When the surface area of the leaves is reduced from 5058mm² to 2773mm² (which is a decrease of 46%), the rate of transpiration decreases 50% which would seem to suggest a directly proportional relationship between the surface area and the rate of transpiration. However, when the surface area is reduced by 76% from (2773mm² to 659mm²) the rate of water uptake only reduces by 33%. This would suggest that perhaps the rate of transpiration is reducing at a directly proportional rate but that the rest of the water is being used for photosynthesis. When the plant had a large surface area only a small percentage of water was lost due to photosynthesis. When the plant’s surface area for transpiration was reduced the effect of photosynthesis on the water loss was a lot more noticeable.

Photosynthesis was able to continue because the leaves were still on the plant and receiving light, they must have had enough carbon dioxide left in the photosynthesising cells for photosynthesis to continue after the stomata had been blocked by the vasoline. If the plant had been left for a longer period of time then the carbon dioxide levels in the photosynthesising cells would have dropped and so the rate of photosynthesis would also have reduced.

The main anomaly with this graph is that it shows the control results, (surface area of 5997mm²) with a slower rate of water uptake than the experiment where the plant had a leaf surface area of 5058mm².

By looking at the provided raw data sheet it can be seen that the first two control experiments with a surface area of 5997mm² (1 and 2) would give an average rate of 0.9cm/minute which would make the rate higher than the experiment where the surface area was 5058mm². However it is the third control experiment (number 9) which drags the average rate down to below that of the experiment with less surface area. I believe the reason the rate for control experiment 9 being significantly lower than experiment 1 and 2 is due to experiment 9 being done straight after experiment 8 where the light level was reduced to 21%. This reduced light level will have slowed down the rate of transpiration to the extent that it had not had time to recover when experiment 9 (the third control experiment) was started. The reduced light level would have caused the stomata to close which would have resulted in less water being able to transpire out of the stomata. Therefore experiment 9 produced a lower average than that of 1 and 2.

Wind

Wind moving past the leaves caused a large increase in the rate of water uptake. When there was no wind in the control experiments the average rate of water uptake was 0.85cm/minute. When the wind was added at 0.98m/s, the transpiration rate increased to1.1cm/minute, (an increase of 23%).

As I predicted, increasing the wind increases the water uptake rate. Increasing the wind speed to 1.1m/second further increased the transpiration rate to 1.13cm/second. The rate was much slower in the control test because in the absence of wind, a band of water vapour was able to form in the internal air spaces of the leaf and around the leaf as the water vapour transpired out. This reduced the water potential gradient between the spongy and palisade mesophyll cells inside the leaf and the air so the rate of transpiration was reduced. When air was blown across the leaves by the hairdryer, this band of water vapour was blown away and further water vapour was not be able to accumulate. This led to an increase in the water potential gradient between the spongy and palisade mesophyll cells inside the leaf and the air and so the rate of transpiration into the air increased.

Limitations

a) The experiment measures the total water uptake which includes water used by the plant as well as the water lost through transpiration. Though most of the water (90 - 99%) is lost through transpiration, a small amount is used by the plant for photosynthesis and so this experiment is not able to tell us the exact rate of transpiration.

b) It is often difficult to change just one environmental condition without changing another. For example the lamps used to give the leaves the maximum possible light intensity in the experiment will also have the effect of slightly heating up the leaves. This could cause the rate of transpiration to go up.

c) These experiments were carried out one after another without leaving enough time for the plant to equilibrate after each experiment. For example, after the light intensity had been reduced to 21% in experiment 8, not enough time was allowed after it for the plants transpiration rate to rise to it’s normal level. Therefore, experiment 9 which was a control experiment produced a significantly lower average rate than the control experiments 1 and 2.

d) There was a general deterioration in the results as the experiments went on. As the experiments were carried out the water would have warmed up to room temperature. This will have caused bubbles to form in the xylem as the air started to dissolve out of it. These air bubbles could have caused a loss of cohesion tension in the leaves, which may have lead to cavitation in the xylem. This cavitation would have reduced the number of working xylem in the stem and could account for the reduced rated of uptake in later experiments.

Extension Work

This experiment could be improved in a number of ways.

1) It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

2) More time should have been allowed between experiments to allow the plant to fully equilibrate before the next experiment was started. This would have reduced the number of anomalies caused by the sudden exposure of the plant to completely different conditions when the previous effects of the last experiment’s conditions had not had time to wear off.

3) The water used in the bowl could have been allowed to warm up to room temperature before the experiment began. This would have stopped the air in the water from dissolving out and leading to cavitation in the xylem vessels.

4) A different method of carrying out the experiment could have been devised that would measure the rate of transpiration instead of the total water uptake.

Disclaimer

This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.