Laboratory+Practicals

= Contents =
 * 1) **Egg-citing Eggsplore**
 * 2) **Leek Cells Investigation**
 * 3) **Beetroot Investigation**
 * 4) **Agar Investigation**
 * 5) **Enzyme Action Investigation**
 * 6) **Yeast Investigation: Is Yeast Alive?**

1) Egg-citing Eggsplore
To find out if the internal environment of the egg changes when placed in water and syrup, through three observable changes – namely changes in length, mass and volume of the egg.
 * Aim**

a. Procedure b. Precautions c. Assumptions
 * Methods**
 * 1) Measure the length (in centimetres, cm) of Egg 1, Egg 2 and Egg 3 using the vernier calliper.
 * 2) Measure the mass (in grams, g) of Egg 1, Egg 2 and Egg 3 using the electronic balance.
 * 3) Measure the volume (in cubic centimetres, cm3) of Egg 1, Egg 2 and Egg 3 using the water displacement method. Fill the measuring cylinder up to the 100 cm3 mark, place the egg in the cylinder and subtract 100 cm3 from the new reading to obtain the volume of the egg.
 * 4) The following should be done at the same time. Place Egg 1 into the beaker containing syrup, place Egg 2 into the beaker containing water, and place Egg 3 onto a dry petri-dish.
 * 5) After 25 minutes, take Egg 1 and Egg 2 out of the beakers, and place all three eggs onto a dry paper towel.
 * 6) Repeat steps i to iii, and record the data down in a table.
 * 1) An egg was used as a control in this experiment, to ensure that the length, mass, or volume of the egg did not expand due to any factors other than the factor being investigated.
 * 1) An assumption made was that no water was absorbed into the quail egg while the volume was being measured using the water displacement method, otherwise the reading of the volume would be affected.
 * 2) Another assumption made is that the concentrations of the three eggs are about the same, so that when they are being placed into the different hypertonic and hypotonic solutions, the results are reliable.

(Time taken: 20 minutes)
 * Results**
 * |||| Length (cm) || Percentage Difference in Length (%) |||| Mass (g) || Percentage Difference in Mass (5) |||| Volume (cm3) || Percentage Difference in Volume (%) ||
 * || Before || After ||  || Before || After ||   || Before || After ||   ||
 * Egg 1 (syrup) || 3.330 || 3.330 || 0 || 12.09 || 12.00 || -0.74 || 12 || 10 || -16.67 ||
 * Egg 2 (water) || 3.035 || 3.130 || 3.13 || 11.53 || 12.63 || 9.54 || 12 || 13 || 8.33 ||
 * Egg 3 (control) || 3.190 || 3.190 || 0 || 12.17 || 12.17 || 0 || 12 || 12 || 0 ||

A. What are your predicted results? I predicted that the egg in syrup would decrease in length, mass and volume, as the syrup is more concentrated than the egg, and by osmosis, the water in the egg will move out of the egg through its semi-permeable membrane, from a region of higher solute concentration to a region of solute lower concentration. I predicted that the egg in water will increase in length, mass and volume, as the contents of the egg has a lower water potential than the water it is placed in, and by osmosis, the water from the petri-dish will move into the egg through the semi-permeable membrane, from a region of higher water potential to a region of lower water potential.

A. What are the problems that you have encountered? There was little standardisation and uniformity of the units (as well as significant figures and the like) and instruments we should use for measurement, within the class. I feel that we could have planned better in order for this investigation to be more organised. In addition, I feel that our time of 25 minutes for leaving the eggs in either water or syrup was too long. It was not practical of us to use such a timing as when the bell rang, many groups were still conducting the experiment. As such, some data will be correct for a total duration of 25 minutes of the eggs immersed in water or syrup, whereas some data will be correct for a duration of 22 minutes, 20 minutes or whatever timing the group had.

Can we use an egg for your exploration? Is an egg a cell? Yes, we can use an egg for this exploration, and yes, an egg is a single cell. Based on information online, the Ostrich egg is in fact the largest cell on Earth. Prior to this experiment, I had always thought an egg was made up of multiple cells. I did some research online regarding a chicken egg as a cell, and learnt the following information: It has half (haploid) the DNA as a regular cell, and does not become a full (diploid) cell until it is fertilized. However, since man makes the definitions of structures, not nature, science has made the egg or ovum of three or more types by definition. With or without the yolk and albumen, the poultry ovum is the largest living cell. One of the egg (ovum) definitions does make the Ostrich Egg a one cell unit. As one source aptly puts it, "Like so many other definitions in science, there is not a complete universally accepted agreement." Regardless, there is only one cell in the unfertilized chicken egg when it is laid.
 * Discussion**

How is the drawing of your cell similar and different from a chicken egg? Similarities: Both the cell I drew and the chicken egg are separated from their external environment by a membrane. Both also contain much nutrients (in a cell, macromolecules, and in a chicken egg, fat-soluble vitamins as well as essential fatty acids). In addition, both have a semi-permeable membrane. Differences: The cell I drew is microscopic and cannot be seen without the aid of a microscope, whereas a chicken egg is macroscopic and is large enough to be seen with the naked eye. The chicken egg, unlike the cell, has a shell which is made up of calcium carbonate. This explains why it ‘disappears’ when placed in vinegar – vinegar reacts with calcium carbonate (or limestone) to produce carbon dioxide.

What is the function of the shell? Does it allow any substances to pass through? There are several layers in an egg shell, each with a different function. The outmost layer protects the embryo and allows the passages of certain gases so that the embryo can breathe.

How would you remove the shell, yet maintain a barrier that separates the egg content and the external environment? As mentioned earlier, the calcium carbonate egg shell reacts with vinegar to produce carbon dioxide.

2) Leek Cells Investigation
Topic: Homeostasis

**Aim** To estimate the approximate osmotic concentration of the leek stem cells

**Apparatus and Materials**
 * Razor blade/scalpel
 * White tile
 * 5 Petri Dishes
 * Forceps
 * Distilled water
 * 1%, 2%, 4%, 7%, 10% salt solution
 * 2 Chinese leeks (Allium tubersosum)
 * Stop watch
 * Vernier calliper
 * Thread
 * Labels
 * Weighing scale

**Procedure**
 * 1) Using the razor scalpel, cut the 2 leeks longitudinally to obtain 4 equal quarters for each leek. 8 strips will be obtained – discard 2 stalks as only 6 will be needed for this investigation.
 * 2) Weigh the stalks using the weighing scale and measure their lengths using the vernier calliper and thread.
 * 3) Using forceps, place one strip in a Petri dish of distilled water, and one each in 1%, 2%, 4%, 7%, 10% salt solution. The Petri dishes should be labelled respectively.
 * 4) After 17 minutes, use the forceps to remove the stalks from the Petri dishes.
 * 5) Measure the weight and length of the stalks.
 * 6) Draw the results in a table and graph.

**Predictions** Distilled water: The leek in this solution will become turgid and increase in length and mass by around 3%, as the distilled water is hypotonic to the contents of the cell sap in the leek stalk. 1% salt solution: The leek in this solution will become turgid and increase in length and mass by around 1-2% as the 1% salt solution is still hypotonic to the contents of the cell sap in the leek stalk, but less so than the distilled water. 2% salt solution: The leek stalk in this solution will become slightly flaccid and its mass and length will not change much as I predict this 2% salt solution is around the same concentration as the contents of the cell sap. 4% salt solution: I predict that the mass and length of the leek will decrease by around 4-5% as the solution is hypertonic to the cell sap, and the leek will become flaccid. 7% salt solution: I predict that the leek will become very flaccid, and the mass and length will decrease by around 6-7%. 10% salt solution: I predict that the leek will become extremely flaccid, and the mass and length will increase by around 9-10%.

**Results** 1) Table
 * Leek |||| Length (cm) || Percentage Difference (%) |||| Mass (g) || Percentage Difference ||  ||
 * ^  || Initial || After 17 minutes ||^   || Initial || After 17 mins ||^   ||^   ||
 * Distilled water || 6.16 || 6.61 || ++7.31 || 0.28 || 0.29 || -3.57 || Very turgid ||
 * 1% salt solution || 6.07 || 6.21 || ++2.31 || 0.29 || 0.34 || ++17.24 || Turgid ||
 * 2% salt solution || 6.395 || 6.05 || -5.39 || 0.27 || 0.28 || ++3.70 || Slightly flaccid ||
 * 4% salt solution || 6.10 || 5.83 || -4.43 || 0.39 || 0.41 || ++5.13 || Flaccid ||
 * 7% salt solution || 6.35 || 5.92 || -6.77 || 0.25 || 0.30 || ++20 || Very flaccid ||
 * 10% salt solution || 6.24 || 6.00 || -03.85 || 0.20 || 0.18 || -10 || Extremely flaccid ||

2) Line graph on graph paper (individual and class data) 
 * Discussion

1) At which concentrations are the salt solution isotonic to the leek cells? Explain how you arrive at your answer. ** 2.6 %. On the graph (class data), I took the corresponding x-axis (salt concentration) values of percentage change 0% on the y-axis. There were two values, 1.6% from the length line and 3.6% from the mass line. I used the average of these two values as the answer to this question, which is (1.6 + 3.6)/2 = 2.6%.  2) Explain the gain or loss in the length of the stems. When the stems are placed in a solution hypertonic to its cell contents, the length of the stems decrease, because water moves from a region of higher water potential (the leek cells) through a selectively permeable membrane to a region of lower water potential (the solution) by osmosis. When the stems are placed in a solution hypertonic to its cell contents, the length of the stems increase, because water moves from a region of higher water potential (the solution) to a region of lower water potential (the leek cells) by osmosis. When the stems are placed in a solution isotonic to its cell contents, there is no net movement of water from the internal to external environment.

3) Instead of measuring the change in length of the stem, is there another variable you can measure as an indication to osmosis? Mass (g) can be measured as an indication to osmosis.

4) Suggest a way to improve this experiment. More qualitative data could have been collected. Also, the experiment could have been repeated 2 more times to ensure the results are more reliable. The individual data collected was quite inconsistent. The approximate osmotic concentration of leek cells is 2.6%. **
 * Conclusion

3) Beetroot Investigation
Aim To explore the factors which affect the movement of materials in and out of the cells, namely surface area and change in external environment - temperature and presence of alcohol


 * Apparatus/Materials**

o Razor blade/ scalpel o White tile (for cutting) o Ruler o 2 petri dishes o 5 test tubes & rack o Thermometer o Labels o Stop watch o Forceps o 2 Beakers o Distilled water o Supply of hot water o 25%, 50% alcohol o 3 droppers (water, 25%, 50% alcohol) o Beetroot (Beta vulgaris) o Spectrophotometer (Data logger) (Introduction and Procedure are on the worksheet)

For tube A, the water is observed to be a very faint pink, almost colourless, and the value of light transmission is 76.5%. This shows that only very little red pigment from the beetroot discs diffused out to the water. For test tubes D and E, very little pigment from the beetroot diffused out to the water as well. However, in test tubes D and E, the value of light transmission are 79% and 71.5% respectively. This shows that in increasing amount of red pigment diffused out to water, it is test tube E, D, followed by A.
 * Results**
 * **Tube** || **Content** || **Value of light transmission/%** || **Observations** ||
 * A || Water || 76.5 || Very faint red, almost colourless ||
 * B || 25% alcohol || 51 || Slightly red ||
 * C || 50% alcohol || 36.5 || Red ||
 * D || Hot water || 79 || Very faint red, almost colourless ||
 * E || Water (chopped up beetroot) || 71.5 || Faint red ||
 * Analysis**

For test tube B, the 25% alcohol turned slightly red and the value of light transmission is 51%. This shows that a significant amount of red pigment diffused out of the beetroot to the alcohol that made the light get blocked by 51%. For test tube C, the 50% alcohol was observed to be red, and the value of light transmission is 36.5%. The large amount of light blocked shows that a lot of red pigment diffused out of the beetroot to the alcohol.

**Discussion** When the beetroot was sliced, red pigment escaped from the beetroot cells damaged in the cutting process. Washing the beetroot discs thoroughly will ensure that all these red pigment from the damaged cells will be washed away. At the end of the experiment, we can be sure that the red pigment found in the water or alcohol is caused by diffusion of red pigment from the beetroot cells out to the water/alcohol, and not due to the red pigment escaping from the damaged cells. To ensure the natural variation of living things, which could cause errors, was minimised, all discs came from the same beetroot. The diameter of the beetroot was also about the same throughout the entire length, and the beetroot discs were all cut to the same length (2 mm). The control in this activity was test tube A, which contained 3 beetroot discs in 4 ml of water. It can be used to compare with the test tubes with alcohol present, with hot water, and with chopped up beetroot. The results of this experiment have to be analysed one variable at a time. The variables involved are surface area of the beetroot and change in external environment - temperature and presence of alcohol. Comparing test tube A and test tube E can help us find out the relationship between surface area of the beetroot and rate of diffusion of the red pigment out to the external environment. For test tube A, containing 3 beetroot discs in water, the water is observed to be a very faint red, almost colourless, with a value of light transmission at 76.5%. For test tube E, containing chopped up beetroot in water, the water is observed to be a faint red, with a value of light transmission of 71.5%. This shows that the greater the surface area of a beetroot, the more light is blocked, therefore a faster rate of diffusion. An explanation for this could be that if the beetroot is chopped up, the surface area to volume ratio is larger, and thus diffusion of the red pigment takes place more efficiently. Comparing test tubes A and D can help us find out the relationship between temperature of the external environment and rate of diffusion of the red pigment from the beetroot cells to the surroundings. Test tube A contains 3 beetroot discs in water at room temperature. Test tube D contains 3 beetroot discs in hot water (90 - 100 degrees Celsius). At the end of 15 minutes, the water in test tube A was a very faint red, almost colourless, with a value of light transmission at 76.5%. The water in test tube D was a very faint red, almost colourless as well, but the value of light transmission is 79%. This shows that the higher the temperature of the water, the less light blocked, therefore a slower rate of diffusion. However based on background knowledge (the kinetic particle theory learnt in Chemistry) it is known that a higher temperature causes particles to move faster as they have more kinetic energy, and thus a faster rate of diffusion. Perhaps there could have been an error in data collection, and comparing this data with data from other classmates could help make this investigation more reliable. Comparing test tubes A, B and C can help find out the effect of alcohol on beetroot cell membranes. Test tube A contains 3 beetroot discs in water, and at the end of the experiment, the water is observed to be a very faint red, almost colourless, with a value of light transmission at 76.5%. Test tubes B and C contain 3 beetroot discs each in 25% and 50% alcohol respectively. The 25% alcohol in test tube B is observed to be slightly red with a value of light transmission of 51%, while the 50% alcohol in test tube C is observed to be red with a value of light transmission of 36.5%. Looking at test tube A along with B and C, the presence of alcohol causes the amount of light transmission to decrease, therefore more light was blocked. An explanation for this could be that alcohol breaks down the cell membrane and thus the red pigment escaped out of the beetroot cells, causing more light to be blocked in the alcohol. Looking at test tubes B and C, the greater the concentration of alcohol, the more light blocked. Possibly, the greater the concentration of alcohol, either more cell membranes are broken down, or the cell membranes are broken down to a larger extent of damage, or the faster the action of breaking down cell membranes, causing more red pigment to escape, resulting in the greater amount of light blocked. Other factors I can think of that will affect the leaking of the pigment out of the beetroot cells would be the concentration of red pigment in and outside of the cell, as well as the size of the red pigment particles.

**Conclusion** The greater the surface area of beetroot, the faster the rate of diffusion of materials from the cell to an external environment. The higher the temperature of the external environment, the faster the rate of diffusion of materials from the cell to an external environment. Alcohol breaks down cell membranes, thus the materials from the cell will escape to the external environment. The greater the concentration of alcohol, the more materials will escape out of the cell.

** 4) Agar Investigation **
The aim of this investigation is to explore the relationship between surface area to volume ratio and rate of exchanging materials. Agar cubes are used to represent living cells in this investigation.
 * Aim**

**Apparatus/Materials** - Razor blade/scalpel - Stop watch - White tile - 3 Beakers - Ruler - 3 pieces of 2 X 2 X 2 cm 3 solid agar - Stirrer - Data logger with conductivity probe

**Procedure** Table 1: Comparison of rate of conductivity change for different surface area to volume ratios Table 2: Results for 2 X 2 X 2 cm 3 agar cube, 8 1 X 1 X 1 cm 3 agar cubes and 64 0.5 X 0.5 X 0.5 cm 3 agar cubes
 * 1) Keep 1 agar cube intact. Cut the other two into eight 1 X 1 X 1 cm 3 cubes and 64 0.5 X 0.5 X 0.5 cm 3 agar cubes respectively.
 * 2) Set up the data logger. (Detailed instructions on worksheet)
 * 3) Prepare a beaker with 200 ml of tap water.
 * 4) Place the conductivity probe in the water.
 * 5) Place the largest agar block into the beaker and click the ‘run’ icon.
 * 6) At the same time, start the stopwatch and time for 2 minutes.
 * 7) Stir the water with the rod continuously and gently while the results are recorded by the data logger.
 * 8) After 2 minutes, press the red ‘hand’ button.
 * 9) Repeat this for the other two agar blocks.
 * Results**
 * **No. of pieces of agar cubes** || **Length (cm)** || **Surface Area (cm** **2** **)** || **Volume (cm** **3** **)** || **Surface area to volume ratio** || **Rate of conductivity change** ||
 * 1 || 2 || 24 || 8 || 3:1 || 0.75 ||
 * 8 || 1 || 48 || 8 || 6:1 || 1.16 ||
 * 64 || 0.5 || 96 || 8 || 12:1 || 2.16 ||
 * 2 X 2 X 2 cm 3 agar cube |||| 8 1 X 1 X 1 cm 3 agar cubes |||| 64 0.5 X 0.5 X 0.5 cm 3 agar cubes ||
 * Time (s) || Average (Conductivity I/O-1) (mS) || Time(s) || Average Conductivity I/O-1(mS) || Time(s) || Average Conductivity I/O-1(mS) ||
 * 0 || 0.08 || 0 || 0 || 0 || 0.48 ||
 * 10 || 0.08 || 10 || 1.02 || 10 || 0.86 ||
 * 20 || 0.08 || 20 || 0.75 || 20 || 1.12 ||
 * 30 || 0.13 || 30 || 0.77 || 30 || 1.54 ||
 * 40 || 0.22 || 40 || 0.76 || 40 || 1.34 ||
 * 50 || 0.35 || 50 || 0.89 || 50 || 1.46 ||
 * 60 || 0.46 || 60 || 0.93 || 60 || 1.59 ||
 * 70 || 0.58 || 70 || 0.95 || 70 || 1.73 ||
 * 80 || 0.65 || 80 || 0.97 || 80 || 1.79 ||
 * 90 || 0.69 || 90 || 1.04 || 90 || 1.9 ||
 * 100 || 0.70 || 100 || 1.08 || 100 || 1.98 ||
 * 110 || 0.74 || 110 || 1.13 || 110 || 2.23 ||
 * 120 || 0.75 || 120 || 1.16 || 120 || 2.16 ||

The average rate of conductivity change from the lowest to highest is: 0.5 X 0.5 X 0.5 cm 3 agar cubes followed by 8 1 X 1 X 1 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cubes, then lastly the 2 X 2 X 2 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cube. Respectively, their rates of conductivity change are 2.16, 1.16 and 0.75. As the surface area to volume ratio decreases, the rates of conductivity change decreases. Likewise, as the surface area to volume ratio increases, the rates of conductivity change increases. The rates of conductivity change are representative of the amount of NaCl in the beaker of 200 ml of tap water, and thus are representative of the rate of diffusion of NaCl from the agar cube(s) into the beaker of water. The greater the value of the rate of conductivity change, the greater the amount of NaCl in the beaker of water and thus the faster the rate of diffusion of NaCl into the beaker of water. **
 * Analysis

**Discussion The precautions taken in this experiment include ensuring the agar cubes were of about the same size (for the cases of the 1 X 1 X 1 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 and 0.5 X 0.5 X 0.5 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cubes) by measuring the cut cubes using a ruler. Also, the cubes were placed into the beaker of water at the same time the ‘run’ icon was clicked. ** The results above show that the 64 0.5 X 0.5 X 0.5 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cubes had the greatest rate of conductivity change. This means that this beaker of water had the greatest amount of NaCl in it, and therefore the rate of NaCl diffusing out from the agar cube into the water was the greatest. As these cubes have the greatest surface area to volume ratio, 12:1, we can conclude that the greater the surface area to volume ratio, the faster the rate of exchanging materials. The 8 1 X 1 X 1 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cubes have a surface area to volume ratio of 6:1 and a rate of conductivity change of 1.16, which are both smaller values than for the 64 cubes. It has a smaller surface area to volume ratio, and a slower rate of exchanging materials. The largest 2 X 2 X 2 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 agar cube has the smallest surface area to volume ratio of 3:1, and the smallest rate of conductivity change of 0.75. Thus, it having the smallest surface area to volume ratio has the slowest rate of exchanging materials. As compared to the experiment done in Lower Secondary where coloured agars were soaked in acid, this experiment is more accurate. The experiment done in Lower Secondary made use primarily of qualitative data. Observations of the change of colour in agar blocks were recorded. However, this experiment makes use of qualitative data. A data logger was used in this investigation. Numbers used in data allow for a more objective view as well as more accurate and reliable results, rather than descriptive qualitative data. An earthworm has a simpler transport system as it is small in size and does not require a complicated system to transport materials in and out, and to cells in the earthworm. In earthworms, the dorsal and ventral vessels connected by 5 pairs of pulsating tubes. Gas exchange takes place through capillaries in body wall under moist, mucous-covered skin. Contrary to an earthworm, a larger organism such as the tiger would require a more sophisticated and complex transport system to get materials to all cells efficiently, as well as transporting required materials into the organism and waste materials out of the organism. Diffusion alone cannot meet the needs of the inner-most cells of larger organisms and larger organisms must have some form of internal transport system for gases. This is an adaptation of an organism to increase the efficiency of exchanging materials. Once a cell grows to a certain size, it becomes too large for the complete diffusion of needed substances throughout its cytoplasm. As observed in Table 1 above, as the agar cube “grows” bigger, from a 0.5 X 0.5 X 0.5 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 cube to a 1 X 1 X 1 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 cube to a 2 X 2 X 2 cm <span style="font-family: Arial; font-size-adjust: none; font-size: 8.7px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">3 cube, the volume increases faster than its surface area. The largest agar cube has a smaller exposed surface area although its volume is the same as the other two cubes. Therefore for a large cell, the surface area of a cell membrane is not as efficient relative to the volume of the cell. Materials are unable to be adequately exchanged in and out of the cell, and unable to be efficiently transported throughout the entire volume of the cell. The cells which are smaller are thus more efficient in exchanging needed substances. As cells need to take in nutrients and get rid of wastes by the important process of diffusion, this explains the small size of all cells.

**Conclusion** The greater the surface to volume ratio, the faster the rate of exchanging materials. Therefore, the smaller a cell is, the more efficient it is in taking in nutrients from the environment and getting rid of waste materials.

5) Enzyme Action Investigation
Investigating the effect of temperature and pH on enzyme action


 * 1. Apparatus and Materials**
 * 1) Bunsen burner, tripod stand and gauze
 * 2) 3 water baths
 * 3) Thermometer
 * 4) 12 test tubes and rack
 * 5) measuring cylinder
 * 6) 4 250 ml beakers
 * 7) 2 white tiles
 * 8) 2 droppers
 * 9) ice cubes
 * 10) distilled water
 * 11) Test tube holder
 * 12) Stop watch
 * 13) Goggles
 * 14) 2% starch solution
 * 15) 1% amylase sotution
 * 16) 1 M Hydrochloric acid solution
 * 17) 1 M Ammonium Hydroxide solution
 * 18) Iodine solution

1. Label 8 test tubes A1, A2, B1, B2, C1, C2, D1 and D2. 2. Add 5cm3 of starch solution to tubes A1, B1, C1 and D1. Add 3 cm3 distilled water to D2, and then 3 cm3 of amylase solution into tubes A2, B2 and C2. 3. Place A1 and A2 into a beaker of ice water, B1 and B2 into a water bath at 370C, and D1 and D2 into another water bath at 370C. 4. Place C1 and C2 in boiling water. 5. Leave the tubes for 10 minutes to allow the solutions in the tubes to reach the temperature of the water bath. 6. Pour the contents of A2, B2, C2 and D2 into the starch tube, A1, B1, C1 and D1, next to it. Record the time immediately. Keep the tubes A1, B1, C1 and D1, containing the mixture in the respective water bath throughout the experiment and ensure that the temperatures are maintained. 7. After 10 minutes, test the mixture in each tube with iodine solution, on a white tile. Draw table 1 with appropriate headings and record the results and conclusions.
 * 2. Procedure**
 * 2.1 Part A**

1. To each of the 4 test tubes, W, X, Y, Z, add 2 cm3 of starch solution. Place the tubes in a water bath at 370C. 2. Make up the contents of the 4 test tubes as shown in the table below. 3. Test the solution with iodine solution at the beginning of the experiment and after 10 minutes. 4. Record your observations and conclusions in table 2 below.
 * 2.2 Part B**


 * 3. Results and Discussion**
 * 3.1 Part A**
 * 3.1.1 Table 1**

Amylase solution || Ice water || Blue-black || Starch present || Amylase solution || 37 Degrees Celsius water bath || Brown || Starch absent || Amylase solution || Boiling Water || Blue-black || Starch present || Distilled water || 37 Degrees Celsius water bath || Blue-black || Starch present ||
 * **Test tube** || **Contents** || **Condition** || **Iodine Test** || **Conclusion** ||
 * **A** || Starch solution
 * **B** || Starch solution
 * **C** || Starch solution
 * **D** || Starch solution


 * 3.1.2 Discussion**

//Which tube shows that starch digestion has occurred? Give your reasons.// Tube B shows that starch digestion has occurred. Iodine drops, when in contact with the solution in test tube B, was brown. This shows that starch is absent in the solution. The absence of starch implies that the starch in the solution has been digested, into maltose. //Which of the test-tubes contained mixtures that gave a blue-black colour?// Test tubes A, C and D contained mixtures that gave a blue-black colour. //Why did the mixtures in tubes stated in (2) remain blue-black?// Starch in the mixtures A, C and D did not get digested. Therefore, the iodine drops turned blue-black when they came into contact with the starch in the mixtures. This could be due to three reasons. Firstly, as in the case of test-tube A, the temperature was too low for the enzyme amylase to catalyse the reaction of breaking down starch into maltose quickly. The enzyme action could have been slow and thus a significant amount of starch remained undigested. Secondly, as in the case of test tube C, the amylase was denatured due to an extreme temperature in the boiling water. As a result of denaturation, the active site of the enzyme could have been altered, thus the enzyme was unable to catalyse the reaction of digesting starch into maltose. Most likely, all the starch in the mixture was undigested. Thirdly, as in the case of test tube D, there was no enzyme to catalyse the digestion of starch into maltose. Thus, all the starch in the mixture remained undigested. //What purpose does the tube D1 serve?// D1 served as a control to find out what would happen to a mixture which only contained starch and no catalyst to speed up the digestion of starch. //Using your knowledge of the structures of starch, describe the test you would use to detect the presence of the substance produced as a result of the reaction// //in test tube stated in (1) above.// A Benedict's test for reducing sugars would be carried out to detect the presence of maltose, a substance produced as a result of the digestion of starch by amylase. **3.2.1 Table 2** Hydrochloric Acid Amylase solution || Blue-black || Blue-black || Starch present || Distilled water Amylase solution || Brown || Brown || Starch absent || Ammonium Hydroxide Amylase solution || Blue-black || Blue-black || Starch present || Distilled || Blue-black || Blue-black || Starch present ||
 * 3.2 Part B**
 * **Test tube** || **Contents** |||| **Iodine Test** || **Conclusion** ||
 * ||  || **Initial** || **After 10 minutes** ||   ||
 * W || Starch solution
 * X || Starch solution
 * Y || Starch solution
 * Z || Starch solution

**3.2.2 Discussion**

Test-tube X shows that starch digestion into maltose has occurred. This is due to the iodine test which revealed the absence of starch (the iodine drops were brown).This implies that the starch in the mixture had been digested into maltose, such that there was an absence of starch in the mixture because the starch had been broken down into maltose. **
 * //Which tube shows that starch digestion has occurred? Give your reasons.//

//Which of the test-tubes contained mixtures that gave a blue-black colour?// Test-tubes W, Y and Z contained mixtures that gave a blue-black colour.

//Why did the mixtures in tubes stated in (2) remain blue-black?// The mixtures in those test-tubes contained starch. In test-tube W, the presence of hydrochloric acid, which has a low pH, caused the enzyme amylase to be unable to digest the starch into maltose, thus explaining the presence of starch in the mixture. In test-tube Y, the presence of ammonium hydroxide, a base which has a high pH, created an environment unsuitable for amylase to catalyse reactions due to the high pH. In test-tube Z, there was no enzyme to catalyse the reaction, thus starch remained present.

**5. Conclusion** Extremely low temperatures cause enzymes to become inactive and thus the speed of reaction is extremely slow. Extremely high temperatures cause the denaturation of enzymes, possibly altering the active site, making the enzyme unable to bind to the substrate and catalyse reactions. A suitable temperature, such as 37 Degrees Celsius, would enable the enzyme to work optimally. Extreme pHs on the acidic and alkaline ends of the pH scale cause enzymes to be unable to catalyse reactions, possibly due to an alteration of the active site. The pH of water, 7, is a suitable pH for the enzyme to work at.
 * 5.1 Effect of temperature on enzyme action**
 * 5.2 Effect of pH on enzyme action**

** 6) Yeast Investigation: Is Yeast Alive? **
How do you determine the identity of the gas? By bubbling the gas produced into limewater, calcium hydroxide or Ca(OH)2. If a white precipitate forms, the gas is carbon dioxide.
 * Scientific Experiment to Test for Metabolism**

Leave the test tubes to stand for 5 minutes. Observe. Explain your observations. In test tube A, which contains warm yeast solution, the yeast increased in amount and there are several bubbles forming. In test tube B, which contains boiled yeast solution, there was no observable change in the amount of yeast. This is because in test tube A, the yeast was alive and could metabolise and respire, producing the gas (carbon dioxide) which explains the bubbles, and also multiplying, which explains the increase in amount of yeast. However, test tube B contained yeast that was boiled and hence dead, thus no respiration or reproduction could take place.
 * Testing the Activities of Yeast Cells

Place a thumb over the mouth of test tubes A and B and shake it vigorously. Observe and give an explanation for your observations. In test tube A, the yeast increased in amount even more than before. This is probably because mixing the yeast evenly with the solution enabled more yeast to be exposed to the glucose solution, which is food for them to respire, grow and multiply. In test tube B, there was no increase in amount of yeast, as the yeast is dead.

**Examining Yeast Cells Close Up Use a cover slip to gently cover the drop of yeast from test tube A and test tube B. Examine each slide individually under a microscope. Note the difference. For the sample taken from test tube A, the yeast cells were moving and active, as they were alive. For test tube B, however, the yeast cells were stationary and unmoving.

__Discussion__


 * 1) Discuss the results you obtained with your partner. How do you interpret your results?
 * 2) What are the evidences that prove that the yeast cells are alive?
 * One evidence is that gas bubbles are produced. This is a sign that yeast is respiring and the gas is likely to be carbon dioxide. A limewater test which shows white precipitate forming can prove this, offering further evidence. Respiration: glucose + oxygen <span style="-webkit-border-horizontal-spacing: 2px; -webkit-border-vertical-spacing: 2px; font-family: sans-serif;">→ carbon dioxide + water (and also energy)
 * <span style="-webkit-border-horizontal-spacing: 2px; -webkit-border-vertical-spacing: 2px; font-family: sans-serif,helvetica,sans-serif; font-weight: normal;">Another evidence is that the yeast increased in amount. This is a result of the yeast reproducing, causing their numbers to multiply.
 * 1) When you make bread, if you just mix flour, sugar and water, the dough does not rise, and the bread will be flat and hard. If you include yeast in the bread dough, then the dough rises and the bread is bigger and fluffier. Can you explain how the yeast helps the bread dough to rise?
 * The yeast helps the bread dough to rise because it produces carbon dioxide as a product of respiration. When yeast is placed with the flour, sugar and water, it is provided with the necessary components for respiration, glucose and oxygen (from the air). The yeast then produces carbon dioxide and water (as well as energy) during respiration. The carbon dioxide gas produced fills up the spaces between the bread dough. Since there are air spaces in the dough, it increases in size and rises.
 * 1) If we were to place a few grains of the yeast in a yeast culture, do you think you will be able to see any growth after a day or two?
 * Possibly, because yeast culture contains nutrients that allow the yeast to multiply.