Experiment Bio

Experiment 1.2: making bread using yeast in absence and presence of sugar

Problem Statement:How does the absence and presence of sugar affect the time taken for a dough to double in size?Variables:

Manipulated variable: Absence or presence of sugar Responding variable: Time taken for the dough to double in size Fixed variables: amount of flour, yeast and water and length of time taken to knead

the doughHypothesis: The dough takes time to double in size when sugar is presentMaterials: 1.5g of yeast, 5g of glucose ,100ml of warm water, 170g of flour and a piece of clothApparatus: 1000 ml beaker, 250ml conical flask, a mixing bowl and stopwatch.Technique:Measure and record the time taken for the dough to double in size with a stopwatch.Procedure:

1. 5g of glucose is mixed with 100ml of warm water in a conical flask.2. 1.5g of yeast is then dissolved in the sugar solution.3. The mixture is the left to stand for twenty minutes.4. Half the amount of flour is poured into a mixing bowl.5. Once the yeast mixture has risen and is frothy, a hole is made in the centre of the flour

and yeast mixture is poured into the hole.6. The mixture is stirred well to make the soft dough.7. The remaining flour is gradually added until the dough become difficult to stir.8. At this point, the work surface is sprinkled with flour. The dough is dumped onto the

lightly floured surface.9. the dough is kneaded for 10 minutes until it is smooth, elastic and no longer sticky.10. The dough is kneaded.11. The kneaded dough is placed in a 1000ml beaker and pressed down. Its initial volume

is recorded.12. The beaker is covered with a clean cloth and placed in a warm spot.13. The stopwatch is started and the time taken for the dough to double in size is recorded. 14. Steps 1 to 13 are repeated by preparing a new batch of dough using the same amount

of ingredients but without any sugar.Results:

DoughTime taken for the dough to double in size

(minutes)Dough with sugar 23

Dough without sugar 40

Discussions:1. The presence of sugar speed up the fermentation process. Therefore, the dough that

contains sugar doubles in size faster than the double without sugar.Conclusion:The dough takes a shorter time to double in size if sugar is present. The hypothesis is accepted.

Experiment 2.1 : Preparing and studying sides of plant cells

Materials:An onion bulb, Hydrilla sp.,iodine solution and stilled water.Apparatus:Glass slides, cover slips, a light microscope, a scalpel, forceps, a glass dropper, a mounting needle and filter papers.Technique:Temporary wet mounts of specimen are prepared, stained (using the irrigation method) and observed under the light microscope.Procedure:

1. A piece of onion scale is cut from an onion bulb using a scalpel.2. The transparent epidermis from the inner surface of the onion scale is gently peeled

off using a pair of forceps.3. The epidermis is the mounted without being folded on a slide in a drop of water.4. The specimen is covered with a cover slip at an angle of 45 degree. With the help of a

mounting needle, the cover slip is lowered slowly to ensure that air bubbles are not trapped under the cover slip.

5. The prepared slide is examined under the light microscope, first under low power, the under high power.

6. A drop of dilute iodine solution is placed onto one side of the cover slip. A filter paper is held at the opposite end of the cover slip to draw the iodine solution across the slide the cells.

7. Excess iodine is absorbed using another filter paper.8. the slide is observed again under the light microscope.9. the structure of the epidermis cells is drawn and its cellular components are labelled

accordingly. The magnification used is recorded.10. This activity is repeated with a leaf of the Hydrilla sp.

Discussions:1. Plant cells have a fixed shape.2. The iodine solution stains the nucleus brown, and the starch granules and a

chloroplasts blue-black.3. Onion cells do not contain chloroplasts.4. The left hydrilla sp. Contain chloroplasts and starch granules.5. The cellular components of plant cells as seen with a light microscope are the cell wall

, nucleus, cytoplasm and vacuoles.Conclusions:A typical plant cell has a fixed shape, a cell wall, a plasma membrane, a nucleus, cytoplasm and large vacuole. Plants cells that photosynthesis have chloroplasts.

Experiment 2.2: Preparing and studying slides of animal cells (human cheek cell)

Materials:Animal cells, methylene blue solution, iodine solution and distilled water.Apparatus:Glass slides, cover slips, a light microscope, a glass dropper, a mounting needle, filter papers and toothpicks.Procedure:

1. The blunt end of a clean toothpick is used to gently scrape the inner lining of the cheek.

2. The scrapings are mounted in a drop of water on a clean glass in slide.3. The specimen is the covered with a cover slip.4. A drop of methylene blue solution is placed onto one side of the cover. A filter paper

is then placed at the opposite side of the cover slip to draw the methylene blue solution across the specimen.

5. The slide is observed under the low-power objective lens of the light microscope and the high-power objective lens.

6. The cheek cells are drawn an the cell structures labelled accordingly.7. Steps 1 to 5 are repeated by using iodine solution to replace the methylene blue

solution.Discussions:

1. Human cheek cells do not have a fixed shape.2. The methylene blue solution stains the cytoplasm and nucleus blue, while the iodine

solution stains the nucleus brown and the cytoplasm yellow.3. The cheek cells do not contain starch granule. Therefore, the iodine solution does not

stain the cell blue-black.4. Both pant cells and animal cells have a plasma membrane and, cytoplasm and a

nucleus. However, unlike onion cells, cheek cells do not have a large central vacuole and cell walls.

Conclusions:A typical animal cell does not have a fixed shape. Like a plant cell, it has a plasma membrane, cytoplasm and a nucleus.

Experiment 2.3: studying the living processes of unicellular organisms (Amoeba sp and paramecium)

Materials:amoeba sp culture, paramecium sp. culture, cotton wool, sodium chloride solution and vinegar.Apparatus:Microscope slides cover slips and a light microscope.Procedure:

1. One week before the experiment, the teacher will prepare an Amoeba sp. culture and a Paramecium sp. culture.

2. Some cotton wool is spread on a slide to limit the movement of Amoeba sp.3. A drop of Amoeba sp. culture is placed on the cotton wool. The specimen is covered

with a cover slip.4. The structure of the Amoeba sp. is observed under the light microscope.5. The structure of Amoeba sp. is drawn and labelled.6. The movement of Amoeba sp. is observed.7. A drop sodium chloride solution is placed at one side of the slide. The response of the

Amoeba sp. to the sodium chloride solution is observed.8. Step 7 is repeated by replacing the sodium chloride solution with vinegar.9. Steps 2 to 7 are repeated by replacing the Amoeba sp. culture with the Paramecium sp.

culture.Discussions:

1. - Amoeba sp. does not have fixed shape. Its shape changes constantly.- Paramecium sp. has the shape of the sole of a shoe.

2. Amoeba sp. moves by temporary extending its pseudopodia. Cytoplasm flows from one area of the cell into a protruding pseudopodium and this enables it to move in the same direction. In contrast, Paramecium sp. moves by using cilia.

3. Both organisms move in the opposite direction when they come in contact with a stimulus such as cotton wool or chemicals. Both organisms are sensitive to chemicals and physical stimuli and can respond to them.

4. Contractile vacuoles of both organisms control the water balance. The contractile vacuoles periodically expand and contract to expel excess water through the plasma membrane.

Conclusion:Unicellular organisms carry out living processes such as feeding and locomotion, and respond to external stimuli.

3.1: Studying the movement of substances across a semi-permeable membrane.

Problem statement:What factor influences the diffusion of substances across a semi-permeable membrane.Variables:

Manipulated variable: size of solute molecules Responding variable: colour of solution in the Visking tubing. Fixed variables: surrounding temperature, time and volume of solutions.

Hypothesis:The diffusion of molecules through a semi-permeable membrane is based on the size of the molecules.Materials:Benedict’s solution, 1% starch suspension, iodine solution, 30% glucose solution, Visking tubing and cotton thread.Apparatus:Test tubes, beakers and a Bunsen burner.Technique:Testing for the present of starch and glucose with iodine and Benedict’s solutions respectively.Procedure:

1. A Visking tubing is soaked in water for 5minutes to soften it.2. One end of the Visking tubing is tied firmly with a piece of cotton thread to prevent

leakage.3. The Visking tubing is filled with 15ml of glucose solution and 15ml of starch

suspension.4. The other end of the Visking tubing is rinsed with distilled water.5. The outer surface of the Visking tubing is rinsed with distilled water.6. 400ml of distilled water and 15ml of iodine solution are mixed in a beaker. The colour

of the solution is recorded.7. The Visking tubing is immersed in a beaker of solution and left to stand for 40

minutes.8. After 40 minutes, the Visking tubing is transferred to a dry beaker.9. The colour of the solutions in the Visking tubing and the beaker s observed and

recorded.10. Both solutions are tested for the presence of glucose using Benedict’s solution.11. - 2ml of each solution are poured into separate test tubes and 1ml of Benedict’s

solution is than added to each test tube. - The solution is heated in a boiling water bath for about 5 minutes and the change in

colour is then recorded.Results:

ContentsInitial colour

Final colour

Benedict’s test

Visking tubing

15ml of glucose solution + 15ml of starch

suspensionColourless

Blue-Black

Positive. A bricked-red precipitate is formed.

Beaker400ml of distilled water + 15ml of iodine solution

Yellow YellowPositive. A bricked-red precipitate is formed.

Discussions:1. The results show the size of iodine solution molecules is smaller than the size of

starch molecules.a) Iodine molecules in a beaker can diffuse through the semi-permeable membrane of

the Visking tubing.b) The content of the Visking tubing turns blue-black.

2. The colour of the solution in the beaker does not change at the end of the experiment. This proves that the starch molecules are too large to diffuse through the pores on the membrane of the Visking tubing.

a) Glucose molecules can easily diffuse through the membrane of the Visking tubing and enter the solution in a beaker.

b) Glucose molecules are small enough to pass through the pores of Visking tubing. Therefore, glucose molecules are smaller than starch suspension.

c) The brick-red precipitate that is formed as a the result of the Benedict’s test shows that glucose molecules are present in a beaker.

Conclusion:The diffusion of molecules across the semi-permeable membrane(Visking tubing) is based on the size of the molecules. The hypothesis is accepted.

Experiment 3.2: Studying osmosis using an osmometer

Problem statement:What substances can diffuse through a semi-permeable membrane?Variables:

Manipulated variable: Time (in minutes) Responding variable: Increase in the level of sucrose solution Fixed variables: surrounding temperature and concentration of sucrose solution.

Hypothesis:Water molecules can diffuse through a semi-permeable membrane.Materials:30% sucrose solution, distilled water, Visking tubing and cotton thread.Apparatus:A retort stand, 25cm capillary tube, a ruler, a marker pen, scissors, 250ml beaker, a syringe and a stopwatch.Technique:Determining the increase in the height of sucrose solution with a ruler.Procedure:

1. A Visking tubing about 8cm long is cut.2. The Visking tubing is soaked in water for about 5 minutes to soften it.3. One end of the tube is tied tightly with a piece of cotton thread to form a bag.4. The Visking tubing is filled with 30% sucrose solution using a syringe.5. The other end of the Visking tubing is tied to the bottom of the capillary tube.6. The outer surface of the Visking tubing is rinsed with distilled water.7. The capillary tube is then clamped vertically to the retort stand.8. The Visking tubing is then immersed in a beaker filled with distilled water.9. The initial level of the sucrose solution in the capillary tube is marked with a marker

pen at the beginning of the experiment.10. The level of sucrose solution is marked every 40 minutes. The level of the sucrose

solution in the capillary tube of each interval is measured and recorded.

Results:Time (minutes) 0 10 20 30 40Increase in height of sucrose solution from the initial level (mm)

0 30 60 90 108

Discussions:1. The results show the level of sucrose solution increases with time. This is because

water molecules from outside diffusion into the Visking tubing.2. Diffusion takes place because the size of water molecules is smaller than the size of

the microscopic pores on the Visking tubing.3. The sucrose solution cannot diffuse into the beaker because the size of the of the

sucrose molecules is much larger than that of the pores of the Visking tubing. As a result, the sucrose solution in the Visking tubing graduallly become diluted.

Conclusion:Small molecules like water can easily diffuse through a semi-permeable membrane such as the Visking tubing. The increase in the level of the sucrose solution in the capillary tube is due to the process of osmosis. The hypothesis is accepted.

Experiment 3.3: Studying the effect of hypotonic, hypertonic and isotonic solutions on plant cells

Materials: An onion bulb, 0.5 M and 1.0 M sucrose solutions, distilled water and filter paper.Apparatus:A light microscope, microscope slides, cover slips, a pair of forceps and a mounting needle.Procedure:

1. Four slides are prepared and labelled A, B, C and D.2. The epidermal layer (skin) from the inner surface of an onion cells leaf is peeled off

gently using a pair of forceps.3. The epidermis is mounted in a drop of distilled water on slide A.4. The slide is examined under the microscope. The epidermis cells are drawn and

labelled.5. A drop of 0.5 M sucrose solution is placed on one side of the cover slip. The solution

is allowed to spread out beneath the cover slip by placing a filter paper on the other side of the cover slip.

6. The epidermal cells are the observed under a light microscope. This is slide B. 7. The epidermal cells are drawn and labelled. 8. step 2 to 5 are repeated by substituting 0.5 M sucrose solution with 1.0 M sucrose

solution (slide C)9. The epidermal cells that were soaked in 1.0 M sucrose solution are then transferred

into a drop of distilled water (slide D)10. The cells are observed again under the microscope, drawn and labelled.

Observations and Discussions:

Slide Observation (appearance

of plant cells)Discussion

A ( plant cells in distilled water)

The cells appear turgid. The vacuoles swell up into a larger size.

The distilled water is hypotonic to the cell sap of the epidermal cells.

Net inflow of water by osmosis into the vacuoles causes them to swell.

At this point, the cells are highly turgid. The vacuoles and cytoplasm exert an outward force against the plasma membrane and cell wall. The cells do not burst because of their rigid cell walls.

B ( plant cell in 0.5 M sucrose

solution)

The shape and structure of the cells remain the same.

0.5 M sucrose solution is isotonic to the cell sap.

The cells do not lose or gain water.

C ( pant cells in 1.0 M sucrose

solution)

The epidermal cells look flaccid, the vacuoles lose water and become smaller. The cytoplasm shrinks and the plasma membrane pulls away from the cell wall.

1.0 M sucrose solution is hypertonic to the cell sap of the epidermal cells.

There is a net outflow of water form the vacuoles by osmosis.

The plant cell has plasmolysed

Slide Observation Discussion D ( plant cells

that are transferred from

the 1.0 M sucrose solution

into distilled water

The cells revert to their normal shape and structure.

When the plasmolysed cells are Placed in the distilled water, water

diffuses into the cells by osmosis. The vacuoles swell and the cells

become turgid without being damaged

Conclusion:1. Plant cells become turgid in a hypotonic solution.2. Plant cells undergo plasmolysis in a hypertonic solution.3. In an isotonic solution, plant cells neither gain nor lose water.

Experiment 3.4: Determining the concentration of an external solution which is isotonic to the cell sap of plant

Problem statement:What is the concentration of an external solution which is isotonic to the cell sap of a plant cells?Variables:

Manipulated variable: Concentration of sucrose solutions. Responding variable: Mass of strips of potato Fixed variables: Surrounding temperature and time

Hypothesis:The concentration of the solution which is isotonic to the cell sap of the plant cells has no effect on the mass and size of plant cells.Materials:A freshly cut potato, distilled water, 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M and 0.6 M sucrose solution.Apparatus:A razor blade or a sharp scalpel, a cork borer, Petri dishes, forceps, a ruler, 50ml beaker, an electronic balance and tissue paper.Technique:Measuring the mass of potato strips with an electronic balance and determining the percentage difference in mass.Procedure:

1. Seven Petri dishes are prepared and labelled A, B, C, D, E, F and G.2. Each beaker is filled respectively with the following solutions:

Petri dish A: Distilled water Petri dish B: 0.1 M sucrose solutionPetri dish C: 0.2 M sucrose solutionPetri dish D: 0.3 M sucrose solutionPetri dish E: 0.4 M sucrose solutionPetri dish F: 0.5 M sucrose solutionPetri dish G: 0.6 M sucrose solution

3. A medium sized cork borer is pushed through a large potato.4. The potato tissue is removed from the cork borer.5. It is then cut into a cylindrical strip 50 mm long.6. Steps 3 to 5 are placed to prepare another six cylindrical strips of the same length.7. Each strip is wiped dry with a piece of tissue paper.8. The mass of each potato strips is weighed and recorded. 9. Each strip of potato strip is the placed in a Petri dish.10. The strips of potato strips must be covered complete with the solution.11. After soaking for an hour, each strip is removed from its Petri dish and wiped dry.

The mass of each potato strip is weighed again and recorded.12. The results are recorded in a table. A graph of the percentage difference in mass

against the concentration of the sucrose solution is drawn.

Results:

SolutionPetri dish

Mass of the strip of potato (g) Difference

in massPercentage difference

in mass (%)

Texture and

appearanceOriginal mass

Final mass

Distilled water

A 1.40 1.80 0.40 0.4/1.4 x 100 = 28.6 Firm

0.1 M sucrose solution

B 1.40 1.75 0.35 0.35/1.4 x 100 = 25.0 Firm

0.2 M sucrose solution

C 1.40 1.70 0.30 0.30/1.4 x100 = 21.4 Firm

0.3 M sucrose

D 1.40 1.60 0.20 0.2/1.4 x 100 = 14.3 Firm

0.4 M sucrose solution

E 1.30 1.40 0.10 0.1/1.3 x 100 = 7.7 Firm

0.5 M sucrose solution

F 1.40 1.20 -0.20 -0.20/1.4 x 100 = -14.3 Soft

0.6 M sucrose solution

G 1.50 1.20 -0.30 -0.3/1.5 x 100 = -20.0 Soft

Percentage difference in the mass of potato strips against the concentration of sucrose solution

-30

-20

-10

0

10

20

30

40

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

concentration of sucrose solution (M)

pe

rce

nta

ge

dif

fere

nc

e

in m

as

s (

%)

Discussion:1. The solutions in Petri dishes A to E are hypotonic to the cell sap of the strips of

potato. Water diffuses into the cells by osmosis. This causes the mass of the strips of potato to increase.

2. The point where the graph cut the x-axis indicates that there is no difference in the percentage of the mass of the potato strip. This means that the concentration of sucrose solution (0.43 M) at this point is isotonic to the cell sap of the plant tissue. Therefore, the concentration of the cell sap of the strips of potato is isotonic to 0.43 M sucrose solution.

Conclusions:Based on the graph the concentration of the cell sap of the potato tissue is 0.43 M.

Experiment 4.1: Studying the effects of temperature on the activity of salivary amylase

Problem statement:What are the effects of temperature on the activity of salivary amylase on starch?Materials:

Manipulated variable: temperature of the medium Responding variable: rate of enzymatic reaction Fixed variables: Volume of saliva (enzyme), volume and concentration of starch

suspension, and pH of the medium.Hypothesis:The rate of the activity of salivary amylase on starch increases with the increases in temperature until it reaches the optimum temperature of 37 degree.Materials:1% starch suspension, saliva, iodine solution, ice and distilled water.Apparatus:Beakers, test tubes, syringes, droppers, glass rods, white tiles with grooves, thermometers, a Bunsen burner, tripod stand, a test-tube rack, a wire gauze and a stopwatch.Technique:Test for the presence of starch using the iodine test and record the time taken for the hydrolysis of starch to be completed with a stopwatch.Procedure:

1. Saliva collected in a beaker is diluted with an equal volume of distilled water.2. 5ml of 1% starch suspension is added to each test tube, labelled A1, B1, C1, D1 & E1

respectively with the aid of the syringe.3. 2ml of the saliva each is added to another set of the set of the test tubes, labelled A2,

B2, C2, D2 and E2 respectively using a syringe.4. Test tubes A1 and A2, B1 and B2, C1 and C2, D1 and D2, E1 and E2, are immersed

respectively into five different water baths with temperatures which are kept constant at 0°C, 28°C, 37°C, 45°C, and 60°C.

5. The test tubes are left for five minutes.6. A drop of dilute iodine solution is placed into each groove of the white tile.7. After five minutes, the starch suspension is placed into each groove of the white tile.8. The mixture is stirred using a glass rod. A stopwatch is activated and the time is

recorded as zero minute.9. A drop of mixture from test tube B is immediately placed in the first groove of the

tile containing the iodine solution.10. The iodine test is repeated every minute for 10 minutes. The dropper is rinsed in a

beaker of distilled water after each sampling.11. the time taken for the hydrolysis of starch to be completed is recorded, that is, when

the mixture no longer turns blue-black in colour when tested with the iodine solution.12. The test tubes containing the mixture are kept in their respective water baths

throughout the experiment.13. Steps 7 to 10 are repeated for test tubes C, E, G and I.14. Thermometers are used to ensure that the temperatures remain constant throughout

the experiment.15. The results are recorded in the following table and a graph showing the rate of

enzymatic reaction, 1/t against temperature is plotted (°C) is plotted.

Test tubeTemperature

(°C)Time taken for the hydrolysis of starch to be completed (minutes)

Rate of enzymatic reaction 1/t (minutes)

A2 0 Not completed after ten minutes 0B2 28 2 0.50C2 37 1 1.00D2 48 3 0.33E2 60 Not complete after 10 minutes 0

Rate of enzymatic reaction 1/t against temperature

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60 70

temperature (°C)

rate

of

enzy

mic

rea

ctio

n (

1/t)

Discussions:1. Starch is hydrolysed by salivary amylase to a reducing sugar. (maltose)2. At 0°C, the enzyme is not active. Therefore, the salivary amylase cannot hydrolyse the

starch.3. The optimum temperature for the activity of salivary amylase is 37°c, this is because

the hydrolysis of the starch is completed in the shortest period.4. Beyond the optimum temperature, the rate of enzymatic reaction decreases ad ceases

altogether. This is because the enzyme has denatured due to the high temperature.

Conclusions:Changes in temperature affect the activity of salivary amylase on starch. Salivary amylase is inactive at 0°C and denatured at 60°C. The rate of reaction catalysed by salivary amylase is highest at 37°C, which is the optimum temperature. The hypothesis is accepted.

Experiment 4.2: Studying the effects of pH on the activity of pepsin.

Problem statement:What are the effects of pH on the activity of pepsin?Variables:

manipulated variable: pH of medium Responding variable: Rate of enzymatic reaction Fixed variables: Volume and concentration of albumen suspension, volume and

concentration of pepsin solution and temperature of mediumHypothesis:The optimum pH for the activity of pepsin is an acidic medium of pH 3.Materials:Albumen suspension, 1% pepsin solution, 0.1 M hydrochloric acid, 0.1 M sodium hydroxide solution and distilled water.Apparatus:Beakers, droppers, thermometers, test tubes, 5 ml syringes, pH paper, wire gauze, a stopwatch and a test tube rack.Technique:Observe the condition of mixtures before and after 20 minutes. Procedure:

1. An albumen suspension is prepared by mixing the egg white from an egg with 500 ml of distilled water. The suspension is bile and cooled down. Large particles are removed by using glass wool.

2. Three test tubes are prepared and labelled P, Q and R respectively.3. A syringe is used to add 5 ml of the albumen suspension to each test tube.4. the solutions below are added to the test tubes:

5. A piece of pH paper is dipped into each test tube and the pH value is determined and recorded.

6. All the test tubes are immersed in a water bath maintained at 37 °C for 20 minutes.

P: 1 ml of 0.1 M hydrochloric acid +1 ml of 1% pepsin solution.Q: 1 ml of distilled water + 1 ml of 1% pepsin solution.R: 1 ml of 0.1 M sodium hydroxide solution + 1 ml of 1% pepsin solution.

7. The conditions of the mixtures are observed at the beginning of the experiment and again after20 minutes.

8. The results are recorded in a table.

Results:

Test tube pH

Mixtures

At the beginning of the experiment

After 20 minutes

P 3 Cloudy ClearQ 7 Cloudy CloudyR 8 Cloudy Cloudy

Discussions:1. Pepsin hydrolyses albumen (protein) into polypeptides in an acidic medium. The

solution turns clear because polypeptides are soluble in water.2. The pH condition in test tube P (pH 3) is optimal for the function of pepsin. This is

because the contents of the test tube P become clear at the end of the experiment. This shows that albumen has been completely digested or hydrolysed by pepsin.

3. The content of test tubes Q and R are still cloudy at the end of the experiment. This shows that a neutral (test tube Q) and alkaline pH (test tube R) are not suitable for the activity of pepsin.

Conclusion:The activity of pepsin is affected by the pH of its medium. An acidic (pH 3) is the most suitable pH for pepsin to function efficiently. The hypothesis is accepted.

Experiment 4.3: Studying the effects of substrate concentration on the activity of salivary amylase.

Problem statement:What are the effects of substrate concentration on the activity of salivary amylase?Variables:

Manipulated variable: concentration of starch suspension. Responding variable: Time taken for the hydrolysis of starch to be completed. Fixed variables: Enzyme concentration, temperature and pH of medium.

Hypothesis:The rate of enzymatic reaction increases with the increase in substrate concentration until it reaches a maximum rate.Materials:Starch suspensions at various concentrations (0.1%, 0.2%, 0.3%, 0.4%, 0.5% and 0.6%), 0.1% amylase or saliva suspension, iodine solution and distilled water.Apparatus:5 ml syringes 1 ml syringes, test tubes, grass rods, a stopwatch, a white tile with grooves, droppers and measuring cylinders.Technique:Test for the presence of starch using the iodine test and record the time taken for the hydrolysis of starch to be completed with a stopwatch.Procedure:

1. 10 ml of 0.1% salivary amylase is prepared.2. Six test tubes, labelled A to F are prepared.3. 4 ml of starch suspensions of various concentrations are poured into the following test

tubes using different syringes.A: 0.1% starch suspensionB: 0.2% starch suspensionC: 0.3% starch suspensionD: 0.4% starch suspensionE: 0.5% starch suspensionF: 0.6% starch suspension

4. The test tubes are immersed in a water bath at 37°C.5. Drops of iodine solution are added separately onto the grooves of the white tile.6. 1 ml of 0.1% amylase is added to test tube A using a syringe.7. The stopwatch is activated immediately. The contents are stirred with a glass rod. A

drop of the mixture is tested with the iodine solution on the white tile. This step is repeated at 30-second intervals until the mixture stops turning blue-black in colour when tested with iodine solution. The time taken for hydrolysis of starch to be completed is recorded.

8. Steps 6 and 7 are repeated with test tubes B, C, D, E and F. at every sampling, the dropper must be rinsed with clean distilled water.

9. The results are recorded in the table below.

Results:

Test tubeConcentration of starch

suspension (%)

Time taken for the hydrolysis of starch to

be completed

Rate of reaction = substrate concentration/time (%

minutes)(seconds) (minutes)

A 0.1 240 4.0 0.025B 0.2 240 4.0 0.050C 0.3 240 4.0 0.075D 0.4 240 4.0 0.100E 0.5 300 5.0 0.100F 0.6 360 6.0 0.100

The effects of the substrate concentration on the activity of salivary amylase

0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

concentration of starch suspension (%)

rate

of

reac

tio

n (

% m

inu

te)

Discussions:

1. The graft shows that the rate of reaction increases with the increase in the concentration of starch suspension until a maximum point. At this point, a further in starch concentration does not increase the rate of reaction anymore.

2. The graft shows that the enzyme reaches its saturation point at 0.4% of substrate concentration.

3. At this substrate concentration, the rate of reaction does not increase even thought the substrate concentration is increased because the concentration of the enzyme has become a limiting factor.

Experiment 4.4: Studying the effects of enzyme concentration on the activity of salivary amylase.

Problem statement:What are the effects of enzyme concentration on the activity of salivary amylase?Hypothesis:The rate of enzymatic reaction increases with the increase in enzyme concentration as long as there are no other factors limiting the rate of reaction.Variables:

Manipulate variable: Concentration of enzyme Responding variable: Time taken for the hydrolysis of starch to be completed. Fixed variables: substrate concentration, temperature and pH of medium

Materials:1% starch suspension, 0.8% amylase or saliva suspension, iodine solution and distilled water.Apparatus:5 ml syringes, 1mlsyringes, test tubes, glass rods, a iodine solution and distilled water.Technique:Test for the presence of starch using iodine test and record the time taken for the hydrolysis of starch to be completed with a stopwatch.Procedure:

1. Six test tubes labelled A to F are prepared.2. The test tubes contain the following mixtures:A: 0.5 ml of 0.8% amylase + 2.5 ml distilled waterB: 1.0 ml of 0.8% amylase + 2.0 ml distilled waterC: 1.5 ml of 0.8% amylase + 1.5 ml distilled waterD: 2.0 ml of 0.8% amylase + 1.0 ml distilled waterE: 2.5 ml of 0.8% amylase + 0.5 ml distilled waterF: 3.0 ml of 0.8% amylase 3. Test tubes A to F are immersed in a water bath set at 37°C.4. Meanwhile, drops of iodine solution are added separately onto the grooves of the

white tile.5. 4 ml of 1% starch suspension is added to test tube A using a syringe. Immediately,

the stopwatch is activated and the time is recorded as 0 minutes.6. The mixture in the test tube is stirred using a glass rod. A small amount of the

mixture is removed and tested with iodine solution on the white tile.7. The iodine test is repeated at 30-second intervals until the mixture does not turn

blue-black when tested with iodine solution. The time taken for the hydrolysis of starch to be completed is recorded.

8. Steps 4 to 7 are repeated with the contents of test tubes B, C, D, E and F.9. All results are recorded and tabulated in a table.10. The rate of enzymatic reaction (1/time) is then calculated. A graph of the rate

enzymatic reaction, 1/t against the enzyme concentration is plotted.

Results:

Test tubeConcentration of amylase (%)

Time taken for the hydrolysis of starch to be completed Rate of enzymatic reaction

(1/t) ( minutes-1)(seconds) (minutes)

A 0.17 330 5.5 0.18B 0.33 150 2.5 0.40C 0.50 90 1.5 0.67D 0.67 60 1.0 1.00E 0.83 60 1.0 1.00F 1.00 60 1.0 1.00

Rate of reaction against enzyme concentration

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

concentration of amylase (%)

rate

of

reac

tio

n (

min

ute

-1)

Discussions:1. A higher concentration of the enzyme contains more enzyme molecules to hydrolyse the

starch molecules. Hence, it takes less time for the hydrolysis of starch to be completed.2. From the graph, it can be seen that the rate o reaction increases when enzyme

concentration is is reached. A further increase in enzyme concentration will not increase the rate of reaction. This is because the substrate concentration has become the limiting factor.

Conclusion:The rate of reaction increases with the increase in enzyme concentration until a certain concentration of enzyme is achived. The hypothesis is accepted.

Experiment 5.1: Preparing and observing a slide of the onion root tip to identify different stages of mitosis.

Problem statement:Can all the different stages of mitosis be observed in the meristem of an onion cell? Hypothesis:All the different stages of mitosis are identifiable in a prepared wet mount of an onion root tip, seen through a light microscope.Apparatus:A microscope, a microscope slides, a cover slip, two pins, a scalpel, a pair of forceps, one Petri dish and an electric hotplate.Materials:Acetic orcein stain, a few pieces of blotting paper and three days old root tips from an onion bulb.Technique:Onion root tip are removed, stained and observed through a microscope.Procedure:

1. The terminal 2 mm of a young onion root tip is removed.2. The root tip is put in a Petri dish containing acidified acetic orcein stain.3. The Petri dish is placed on a warm hotplate for five minutes.4. A drop of acetic orcein stain is put on a slide and the root tip is placed in the stain.5. A fine needle is used to break up the root tip and it is covered with a cover slip.6. The slide is placed on a flat surface, it is covered with a blotting paper and is press

down firmly over the cover slip.7. The mount is viewed under low and high power of a microscope and the cells

showing the different phases of mitosis are observed.8. The nuclei of various phases of mitosis are recorded and labelled.

Results:

Conclusion:All the different stages of mitosis are identifiable in a prepared wet mount of an onion root tip, seen through a light microscope. The hypothesis is accepted.

Experiment 6.1: Determining the energy value in food samples.

Problem statement:What is the energy value of different food samples?Hypothesis:The energy values of different food samples are different.Materials:Distilled water, a peanut (whole), plasticine and cotton wool.Apparatus:A boiling tube, a thermometer (0-100°C),a retort stand, a pin (5-8 cm long), a shield, a measuring cylinder, a Bunsen burner and an electronic balance.Procedure:

1. One peanut is weighted and is weight is recorded.2. 20 ml of water is poured into a boiling tube.3. The boiling tube is clamped to a retort stand.4. The initial temperature (t1 °C) of the water in the boiling tube is recorded.5. The peanut is spiked firmly to the end of a pin which is mounted on some plasticine.6. The apparatus set-up is placed in a shield.7. The peanut is ignited by holding it to the flame of a Bunsen burner. The burning

peanut is immediately placed beneath the boiling tube to heat the water.8. The water is stirred gently with the thermometer.9. The final temperature, that is the highest temperature (t2 °C), is recorded as soon as the

peanut has stopped burning.10. The energy value of the peanut is calculated using the following formula:

Energy of peanut = 4.2 (J g-1 °C-1) × mass of water (g) × increase in temperature (t2 - t1) ÷ ( mass of peanut (g) × 1000)

(kJ g-1)

11. The results are tabulated as follows.

Results:

Heat released by 1 g of burning peanut (energy value) = (4.2 × 20 × 37) 0.5 × 1000 = 6.2 kJ g-1.Discussions:

1. When the peanut is burnt, heat energy is release.a) The assumption made in this investigation is that the water absorbs all the

energy that is released from the oxidation of the food sample.b) In reality, not all the energy released by the peanut is used to heat the

water. Hence, the result obtained in this investigation is less than the actual result.

c) This is because some of the energy is lost in the form of heat and light when the food burns. Some energy is also absorbed by the boiling tube.

2. To obtain an accurate result,a) The food must be oxidise completely by making sure the flame does not

extinguish too quickly.b) A shield is used to enclose the boiling rube to reduce heat loss to the

surrounding.c) The distance between the food and the boiling tube must not be too far.

3. The class of food with the highest energy value are lipids.

Conclusions:The energy value of the peanut is 6.2kJ g-1.

Mass of peanut (g) 0.5Mass of water (g)(assuming 1 ml of water weights 1 g)

20

Initial temperature of water. (t1 °C) 29Final temperature of burning (t2 °C) 66Increase in temperature (t2 - t1 °C) 37

Experiment 6.2: Testing the presence of starch, reducing sugar, non-reducing sugars, proteins and lipids in food samples.

Problem statement:Does starch, reducing sugar, non-reducing sugars, proteins and lipids presence in food samples.Materials:Iodine solution, Benedict’s solution, sodium hydrogen carbonate solution, 20% sodium hydroxide solution, hydrochloric acid, 1% copper (II)sulphate solution and food smples as follows;A : Starch suspensionB: 5% glucose solutionC: 1% sucrose solutionD: Albumen suspensionE: vegetable oilApparatus:Test tubes holders, 250 ml beakers, a Bunsen burner, a dropper, wire gauze, a tripod stand and filter paper.Procedure:1. Five samples of food labelled A, B, C, D and E are prepared.2. The following food tests are carried out to determining the presence of starch, reducing

sugars, non-reducing sugars, proteins and lipids in these food samples.

Test for Reagent Procedure Observation ConclusionStarch Iodine

solution1) 2 ml of each food

sample is poured into test tubes.

2) Three drops of iodine solution are added to each food sample.

3) Any change in each mixture is observed and recorded.

Food sample A turn blue0black. The other food samples remain brownish-yellow.

Food sample A contain starch.

Reducing sugar

Benedict’s sugar

1) 2 ml of each food sample is pored into a test tube.

2) 1 ml of Benedict’s solution is added to each food sample.

3) The mixture is shaken and then heated by placing the test tube in a boiling water bath.

4) Any change in colour in each mixture is observed.

A brick-red precipitate is formed in food sample B. the other food samples remain blue in colour.

Food sample B contains reducing sugar.

Test for Reagent Procedure Observation Conclusion Non-reducing sugar

Dilute hydrochloric acid and Benedict’s solution

1) 2 ml of each food sample is poured into a boiling tube. A few drops of dilute hydrochloric acid are then added to each food sample.

2) Each mixture is then heated in a boiling water bath for about five minutes.

3) The boiling tubes is removed and each of the mixtures is cooled under a running tap.

4) The acid in each sample is neutralised by adding sodium hydrogen carbonate powder until effervescences stops.

5) Benedict’s test is then carried out to test for the presence of non-reducing sugar in the mixture.

6) Any colour change in the mixture is observed.

A brick-red precipitate is formed in food sample C. Food samples A, D and E remain the same

Food samples C contains non-recucing sugar.

Protein 20% sodium hydroxide solution and 1% copper (II) sulphate solution ( Biuret’s test)

1) 2 ml of each food sample is poured into a test tube.

2) 1 ml of 20% sodium hydroxide solution is the added to each of the food samples and shaken well.

3)A few drops of 1% copper (II) sulphate solution are then added slowly to each of the mixtures.

4)Each mixture is shaken well and allowed to stand. Any colour change in the food sample is observed.

5)Any colour change in the food sample is observed.

Food sample D turns purple. The other food samples turn blue in colour.

Food sample D contain protein.

Lipid 1) A small amount of each of the food samples is rubbed on a piece of filter paper.

2)The filter paper is then dried.

3)The filter paper is held against the light.

4)All observations are recorded.

Sample E shows a translucent mark. The filter papers on the other food samples remain opaque.

Food sample E contains lipid.

Discussions:1) When a reducing sugar is heated with Benedict’s solution, the reducing sugar reduces

the blue copper (II) sulphate in Benedict’s solution to form a brick-red precipitate of copper (I) oxide.

a) A brick-red precipitate indicate the present of reducing-sugar.b) The final colour of the solution depends on how much of the precipitate is formed.

Therefore, the colour gives an indication of how much reducing sugar is present.c) A brick-red precipitate indicates that a large amount of reducing sugar is present. If

the original pale blue colour of the solution remains, this indicates that reducing sugar is not present.

2) The different food samples are first heated with dilute hydrochloric acid to hydrolyse the non-reducing sugar to its constituent monosaccharides which are reducing sugar.

Conclusions:Food sample A contains starch, food sample B contains reducing sugar, food sample C contains non-reducing sugar, food sample D contains protein and food sample E contains lipid.

Experiment 6.3: Determining the vitamin C content in various fruit juices.

Problem statement:Do different types of fruit juices contain similar amounts of vitamin C?Variables:

Manipulated variable: Types of fruit juices Responding variables: Volume of fruit juice needed to decolourise DCPIP solution. Fixed variables: Volume of DCPIP solution and standard concentration of ascorbic

acid solution.Hypothesis:Lime juice contains a higher concentration of vitamin C compared to pineapple juice and orange juice.Materials:1.0% dichlorophenolindophenol solution (DCPIP), 0.1% ascorbic acid solution, freshly prepared lime juice, pineapple juice and orange juice.Apparatus:Specimen tubes, a syringe (1 ml), syringes ( 5 ml) with needles, beakers ( 50 ml), gauze cloth and a knife.Technique:Record the volume of fruit juice needed to decolourise 1 ml of DCPIP solution with a graduated syringes.Procedure:

1. 1 ml of 0.1% DCPIP solution is placed in a specimen tube using 1 ml syringe.2. The 5 ml syringe is filled with 0.1% ascorbic acid solution.3. The needle of the syringe is placed into the DCPIP solutions .4. The ascorbic acid solution is added drop by drop with the needle of the syringe.

The ascorbic acid solution is continuously added until the DCPIP solution is decolourised. The volume of ascorbic acid solution used is recorded.

5. Steps 1 to 4 are repeated using freshly squeezed lime juice, pineapple, juice and orange juice. The volume of the fruit juice required to decolourise the DCPIP solution in each case is recorded in the following table.

6. The results are tabulated. The percentage and the concentration of vitamin C in each juices are calculated using the formulae below.

Percentage of vitamin C in fruit juice = volume of 0.1% ascorbic acid solution/volume of fruit juice used × 0.1%.

Concentration of vitamin C in fruit juice = volume of 0.1% ascorbic acid solution/volume of fruit juice used × 1.0 mg cm-3

Tabulation of Results:

Solution/fruit juice

Volume of solution f fruit juice needed to decolourise

1 ml of DCPIP solution

Percentage of vitamin C in fruit juice (%)

Vitamin C concentration in fruit juice (mg cm-3)

1 2 3 Average0.1% ascorbic acid

1.0 1.0 1.0 1.0

Lime juice 3.2 3.0 3.1 3.11.0/3.1 × 0.1 = 0.032%

1.0/3.1 × 0.1 = 0.32 mg cm-3

Pineapple juice

4.9 4.9 4.8 4.91.0/4.9 × 0.1 = 0.02%

1.0/4.9 × 0.1 = 0.20 mg cm-3

Orange juice

5.0 5.3 5.2 5.21.0/5.2 × 0.1 = 0.019%

1.0/5.2 × 0.1 = 0.19 mg cm-3

Discussion:1. 0.1% ascorbic acid solution contains 0.1 g ascorbic acid in 100 ml of solution.2. The vitamin C (ascorbic acid) content in fruit juices can be determined by using the

DCPIP solution. Vitamin C reduces the blue DCPIP solution to a colourless solution.3. During the investigation, the DCPIP solution should not be shaken vigorously when

the ascorbic acid solution ant the fruit juices are being added.4. This is because oxygen from the atmosphere oxides the reduced DCPIP solution and

turn it blue again. If this happens, more fruit juice is needed to reduce the DCPIP solution. Therefore, the actual vitamin C content in the fruit juice cannot be determined accurately.

5. If the vitamin C content in a fruit juice is high, less volume of the fruit juice is required to decolourise the fixed volume of DCPIP solution.

6. During the extraction of juices from the fruits, the fruits should not be pounded too hard, as the heat produced will destroy part of the vitamin C in the juices.

Conclusion:of the fruit juices tested, lime juice has the highest ascorbic acid content of 0.32 mg cm-3. the hypothesis is accepted.

Experiment 6.2: Studying enzyme action on starch

Problem statement:how does the enzyme in saliva act on a starch food sample?Variables:

Manipulated variable: Absence or presence of salivary amylase and starch. Responding variable: Presence of reducing sugar. Fixed variables: Temperature at 37 °C, starch concentration and volume of mixture.

Hypothesis:The enzyme in saliva digests starch into a reducing sugar. Materials:distilled water, 1% starch suspension, Benedict’s solution, iodine solution and saliva.Apparatus:10 ml pipette, 500 ml beaker, test tube holders, a Bunsen burner, a thermometer, droppers, a glass rod, a white tile, a stopwatch, a tripod stand, and a wire gauze.Technique:Test for the presence of starch and reducing sugar by using iodine solution and Benedict’s solution.Procedure:

1. 1 ml of saliva is collected in test tubes labelled A and C. at the same time, 1 ml of 1% cooked starch is placed into test tubes labelled B and D. test tube E contains distilled water.

2. All the test tubes are immersed in a water bath at temperature of 37°C for five minutes.

3. At the same time, iodine solution is prepared by placing a drop into each groove on the white tile.

4. After 5 minutes, the cooked starch in test tube D is added to the saliva in test tube A.5. The stopwatch is started. Immediately thereafter, a drop of starch and saliva

suspension from test tube A is placed onto the iodine solution in the first groove of the white tile. The colour of the iodine solution is recorded.

6. Step 6 is repeated at one-minute intervals until there is no change in the colour of the iodine solution.

7. Step 6 and 7 are repeated after adding 1 ml of distilled water (from test tube E) to test tubes B and C.

8. When the mixture in all the test tubes no longer changes the colour of the iodine solution, then 1 ml of Benedict’s solution is added to test tubes A,B and C.

9. The mixture in all three test tubes is shaken and then heated by placing the test tubes in a boiling water bath. Any change in the colour of the mixture is observed and recorded.

Results:Test tube

Iodine test Benedict’s test0 minutes 1 minutes 2 minutes 3 minutes 4 minutes

A Blue-black Blue-blackBrownish-yellow

Brownish-yellow

Brownish-yellow

Brick-red precipitate

B Blue-black Blue-black Blue-black Blue-black Blue-black Blue solution

CBrownish-yellow

Brownish-yellow

Brownish-yellow

Brownish-yellow

Brownish-yellow

Blue solution

Discussions:1. Test tubes B and C act as controls in this experiment.2. All the test tubes are maintained at 37°C because this is the optimum body

temperature for the action of salivary amylase.3. Only test tube A shows the presence of a reducing sugar (brick-red precipitate is

formed when tested with Benedict’s solution). The salivary amylase in the saliva hydrolyses the cooked starch into a reducing sugar during the second minute.

Conclusion:Salivary amylase in the saliva hydrolyses the cooked starch into a reducing sugar. The hypothesis is accepted