Answer to Written Practical Question (Hands on Pg 40)

1a)

Set	Time taken (min)
A	20
B	10
C	3
D	6
E	30

b)(i) Observation 1-The time taken for complete hydrolysis of starch in set C/at 37oC is 3 minutes.
Observation 2-The time taken for complete hydrolysis of starch in set E/at 50oC is 30 minutes.
(ii) Inference 1 -The time taken for complete hydrolysis of starch in set C is fastest because 37oC is the optimum temperature for enzyme activity.
Inference 1 - The time taken for complete hydrolysis of starch in set E is the slowest because at too high temperature, that is 50oC, enzyme are denatured.

c)

Variable	Method to handle the variable
Manipulated variable Temperature	5 sets of apparatus using water baths of different temperatures, 15oC, 25oC, 37oC, 45oC and 50oC are set up.
Responding variable Time taken taken for complete hydrolysis of starch	Measure and record the taken for complete hydrolysis of starch using a stopwatch
Fixed variable pH// enzyme concentration// substrate concentration	Fix the pH to be the neutral throughout the experiment Fix 1ml of saliva throughout the experiment Fix 5 ml of starch throughout the experiment

d) The higher the temperature, the higher the rate of enzyme reaction until an optimum temperature is reached.
e)

Temperature (oC)	Time taken for hydrolysis of starch (minutes)	Rate of reaction (min -1)
15	20	0.05
25	10	0.10
37	3	0.33
45	6	0.17
50	30	0.03

f) An increase in temperature will also increase the rate of reaction until the optimum temperature of 37oC is reached. The rate of reaction decreases after 37oC.
g)An enzyme/amylase hydrolyses starch into a product that make the iodine solution remains unchanged/yellow, and the rate of reaction is affected by temperature.
h)The rate of reaction will be zero because all enzymes are denatured at high temperature of 70oC.

ANSWERS TO PG 38-40

Section A
1. a) Saliva amylase breaks down starch into maltose.
b)(ii) The rate of reactions in sets P and Q are slow because enzymes are less active at low temperatures. The rate of reaction in set R is the highest because it is in its optimum temperature of enzme activity. Starch is hydrolysed very quickly. There is no breakdown of starch in set S as enzymes are denatured after they have been boiled.
c) The reaction in set R will stop as salivary amylase is not active in acidic medium.
d) Increase the temperature to 40oC.
e)(i) Wash the clothes in water of temperatures between 37oC and 40oC.
(ii) A temperature range of between 37oC and 40oC is an optimum temperature for enzymes to break down stains.

2. a) To study the effect of pH on enzyme activity.
b) A: Clear solution
B: Cloudy suspension
C: Cloudy suspension
D: Clear solution
c) In test tube A, enzyme pepsin breaks down albumen because pepsin works effectively in an acidic medium. Neutral medium is not suitable for the action of pepsin as shown in test tube B. There is no breakdown of proteins in test tube C as there are no enzymes present.
d) It is because a temperature of 37oC is an optimum temperature for enzyme activity.
e) There is no albumen/substrate, so no reaction takes place.

Section B
3a)

Characteristic	Explanation
Enzymes are biological catalysts and enzyme reactions are reversible	.Enzymes speeds up the rate of metabolism in the body. · Metabolic reactions are reversible · Enzymes remain unchanged at the end of reaction
Enzyme action is highly specific	· Acts on specific substrate only · Has an active site which is complimentary to certain substrate molecule only · Acts on ‘lock and key hypothesis’
Enzymes are sensitive to pH	· Normally function best at neutral eg amylase · Some enzymes work best in acidic condition. For example, pepsin · Some enzymes work best in alkaline condition. For example, trypsin.
Enzymes are sensitive to temperature	· Work best at temperatures between 35oC and 40oC. · Enzymes are inactive at low temperatures. · They are denatured at high temperatures.

b)

Acidity or alkalinity of solutions affect rate of reaction.

Each enzyme has an optimum pH so as to function at its maximum rate.

For example, pepsin at pH 3, salivary amylase at pH 7, trypsin at pH 8.5

Change in pH alters the charges on the active sites of an enzyme.

Alterations in the ionic charges also changes configuration of substrate.

Excess hydrogen ions attach to active sites of enzymes when pH is low.

Substrate is not able to bind to form enzyme-substrate complex.

Free hydroxyl ions attach to active sites when pH is high.

Extreme changes in pH alter the structures of enzymes and substrates.

4a) An enzyme has a specific three-dimension shape. As it is a protein molecule, the polypeptide chain fold to form an active site. Active site of enzyme complements the structure of the substrate it acts on. Substrate molecule fits into active site of enzyme like a key fits into a lock. Substrate is represented by the key while the lock represents the enzyme. When substrate fits into active site of enzyme, an enzyme-substrate complex is formed. Hydrogen and ionic bonds hold the enzyme-substrate complex. Reaction takes place to convert the substrate into products. The products have different shapes. The product then leaves the active sites.

b)Food processing industry
In meat industry, protease is used to tenderise meat.
In fish industry, protease is used to remove the skin of fish.
In dairy industry, lipase is used for ripening of cheese, rennin coagulates milk in the making of cheese, lactase is used to hydrolyse lactose.
In baking industry, amylase is used to breakdown starch in the making of bread, glucose oxidase is used for the stability of bread dough.
In brewing industry,zymase is used to hydrolyse sugar into ethanol.
In seaweed products, cellulase is used to hydrolyse cellulose to extract agar-agar from seaweeds.

Medical field
Trypsin is used to remove blood clots and clean wounds.

Effect of enzyme concentration on the rate of biochemical reaction

Pg 33 - 4.2

Aim : to study the effect of enzyme concentration on the rate of biochemical reaction
Hypothesis : The higher the enzyme concentration, the higher the rate of biochemical reaction until it reaches a maximum rate.
Variables : a) MV - enzyme concentration
b) RV - Time take for breakdown of starch
c) FV - pH//temperature//substrate concentration
Results :

Test tube	A	B	C	D	E	F
Enzyme concentration (%)	16.7	33.3	50.0	66.7	83.3	100.0
Time take for breakdown of starch (min)	8.0	5.0	3.5	2.5	2.0	1.5
Rate of reaction, 1/time (min-1)	0.13	0.20	0.29	0.40	0.50	0.67

Graph :
Discussion :
1. F
2. A
3. The higher the enzyme concentration, the higher the rate of reaction.

Conclusion:
The hypothesis is accepted. The higher the enzyme concentration, the higher the rate of biochemical reaction (breakdown of starch) until it reaches a maximum rate.

Effect of pH on enzyme activity

Pg. 31- 4.1
Aim : To study the effect of pH on enzyme activity
Hypothesis : Pepsin works best in an acidic condition
Variables : a) MV - pH of solution
b) RV - Time taken for content in test tube to turn clear
c) FV - Temperature//enzyme concentration//substrate concentration Observations :

Test tube	pH	Time taken for the content in test tube to turn clear (min)	Rate of reaction. 1/time (min-1)
P	3	20	0.05
Q	9	Still cloudy after 1 hour	0
R	7	Still cloudy after 1 hour	0

Discussion:
1. Enzyme activity is optimum at 37oC.
2. Pepsin breaks down albumen into peptones and polypeptides.
3. pH 3
Conclusion: The hypothesis is accepted. Pepsin works best in an acidic condition.

Written Practical (Pg 28-29)

Question 1
a(i) Observation 1 :The change in length of potato strips in 0.1 M concentration of sucrose solution is +0.25 cm.

Observation 2:The change in length of potato strips in 0.4 M concentration of sucrose solution is – 0.30 cm.

(ii) Inference 1:There is an increase in length of potato strip in 0.1M sucrose solution because this solution is hypotonic to the cell sap of the potato cells. Water diffuses into the cell sap of the potato cells by osmosis.

Inference 2:There is a decrease in length of potato strip in 0.4M sucrose solution because this solution is hypertonic to the cell sap of the potato cells. Water diffuses out of the cell sap of the potato cells by osmosis.

b) As the concentration of the sucrose solution increases, the length of potato strips decreases. The concentration of the sucrose that did not cause any change in length of the potato is isotonic to the cell sap of the potato.

c) (i) & (ii)

Concentration of sucrose solution (M)	Length of potato strips (cm)						Change in length of potato strips (cm)
	Initial			Final
	1	2	Mean	1	2	Mean
0.1	5.00	5.00	5.00	5.30	5.20	5.25	+0.25
0.2	5.00	5.00	5.00	5.10	5.10	5.10	+0.10
0.3	5.00	5.00	5.00	4.90	4.90	4.90	-0.10
0.4	5.00	5.00	5.00	4.60	4.80	4.70	-0.30

d) (i) Concentration of the sucrose solution

(ii) Length of the potato strip

(iii) Temperature// Soaking time //Type of potato//Size and shape of potato

(e) (i) Use different concentration of sucrose solution

(ii) Measure and record the initial and final length of potato strips using a ruler.

(iii) Fix the surrounding temperature to be constant // Fix the soaking time to be 30 minutes for all the experiments.

f)

g) 0.25 M. There is no change in the length of the potato strips at this concentration. This means the rate of movement of water molecules in and out of the potato cells is the same.

(h) (i) Osmosis

(ii) The cell sap of the potato (0.25M) is hypertonic to the 0.2 M sucrose solution. Water molecules from the 0.2M sucrose solution move into the vacuoles of the potato cells, causing the vacuoles to exert pressures on the cytoplasm. The cells swell and lengthens the potato strips.

i) Osmosis is movement of water molecules in or out of the potato strips which causes the change in length of potato strips which is affected by the concentration of the sucrose solution in which the potato strips are immersed.

Question 2

Aim:	To study the effects of isotonic, hypotonic and hypertonic solutions on animal cell.
Problem Statement:	What is the effect of isotonic, hypotonic and hypertonic solutions on an animal cell?
Hypothesis:	An animal cell will swell and may burst when put into a hypotonic solution. An animal cell will shrink when put into a hypertonic solution. There is no change in the size and shape of the animal cell when put into an isotonic solution.
Variables:	Manipulated: Concentration of solutions Responding: Conditions of cells Fixed: Time, temperature, type of cells
Materials & Apparatus:	Materials: chicken blood (sodium citrate added to prevent it from clotting), 0.17 M sodium chloride solution, 0.50 M sodium chloride solution and distilled water. Apparatus: slides, cover slips and light microscope
Technique:	Observe and draw the blood cells as seen under a light microscope.
Procedure:	1. Label four slides as A, B, C and D. 2. Put a drop of blood on slide A and cover with a cover slip. Observe under a light microscope. 3. Put a drop of distilled water on slide B and cover with a cover slip. Put a drop of blood at the edge of the cover slip. 4. Observe the slide under a microscope when the blood is drawn into the water. 5. Observe and draw the blood cells as seen under a light microscope. 6. Repeat steps 3, 4 and 5 by using 0.17M and 0.5M sodium chloride solutions on slides C and D respectively. 7. Dispose all specimens carefully after use.
Results :	Specimen A B C D Drawing of the cell Shape /condition of cell
Conclusion:	Hypothesis is accepted. An animal cell will swell and may burst when put into a hypotonic solution. An animal cell will shrink when put into a hypertonic solution. There is no change in the size and shape of the animal cell when put into an isotonic solution.

*********The End**********

ANSWERS TO PG 26-28

Section A

1. a) Draw a graph of percentage of burst red blood cells against concentration of salt solution from Table 1.

b)(i) about 0.435 g/100cm³ (read from graph)

(ii) 0.55 g/100cm³(where the graph intersects the x- axis) No bursting of red blood cells occurs for this concentration because it is isotonic to the concentration of RBC.

c)(i) The RBC will shrink/crinkle.

(ii) At concentration more than 0.55 g/100cm³ , water molecules will diffuse out from the RBC by osmosis. As a result, RBC will shrink due to loss of water.

2. a)(i) Solution X is hypotonic to the cell sap of potato cells. Water molecules from solution X move into the vacuoles of the potato cells by osmosis. The enlarged vacuole will push against the cytoplasm, causing the cells to inflate. This causes the potato strip to lengthen.

(ii) Solution Y is isotonic to the cell sap of potato cells. The rate of movement of water molecules in and out of the cells is the same. Therefore, there is no change in length of the strip.

(iii) Solution Z is hypertonic to the cell sap of potato cells. Water molecules out from the vacuoles of the potato cells by osmosis. The cells shrinks and plasmolysis takes place.

(b)(i) Hard

(ii) Soft

(c) The use of excessive fertilizers will increase the osmotic concentration in the soil water, causing water molecules to move out from the root hairs. The plant will wilt and die.

Section B

3 a) Simple diffusion - Movement of molecules in gas or liquid from a region of high concentration to a region of lower concentration.

Facilitated diffusion - Movement of big molecules along a concentration gradient with the help of protein carriers across the plasma membrane.

Osmosis - Movement of water molecules from a region of less concentrated solution to a region of more concentrated solution across a semi-permeable membrane.

Active transport - Movement of particles across the plasma membrane against the concentration gradient with the help of protein carriers and the presence of energy from ATP.

b)

Active transport	Osmosis
Active transport needs energy	Osmosis does not need energy
Active transport involves the movements of molecules or ions against a concentration gradient.	Osmosis transport involves the movements of water molecules along a concentration gradient.
Active transport takes places through the plasma membrane of a living cell.	Osmosis takes places through a semi-permeable membrane.
Active transport needs protein carriers	Osmosis does not need protein carriers

4.(a) Plasma membrane is selectively permeable. It permits lipid-soluble molecules such as glycerol, vitamins A, D, E and K to move across. Small, uncharged molecules such as water move freely across. Large molecules such as glucose and amino acids move across the plasma membrane with the aid of carrier proteins. Larger molecules such as starch cannot move across the plasma membrane.

(b) Plasma membrane consists of phospholipids bilayer and proteins. Phospholipid molecule consists of a polar head which is hydrophilic and a pair of non-polar fatty acid tails which is hydrophobic. Two types of proteins are pore proteins and transport proteins.

Plasma membrane is semi-permeable which allows certain substances to move in and out freely. Small, uncharged molecules such as oxygen and carbon dioxide move freely through the phospholipids bilayer through simple diffusion.

Water molecules which are attracted to the hydrophilic heads of the phospholipids move across through osmosis.

Lipid-soluble molecules such as fatty acids and ethanol dissolve in the lipid bilayer and move across through simple diffusion.

Large, water-soluble molecules such as glucose and amino acids require the aid of transport proteins to move them across the plasma membrane through facilitated diffusion or active transport.

Ions such as K+ and Na+ are transported across the plasma membrane through facilitated diffusion or active transport with the help of transport proteins.

c) Vegetables soak in salt solution which is hypertonic to the cell sap of vegetable cells. Harmful insecticides or fungicides which had been sprayed on the vegetables earlier diffuse out of the cells to the salt solution. Water from the cell sap in the vacuole also diffuses out the salt solution through osmosis. The vegetables become flaccid. This action cleans the vegetables of harmful insecticides but causes the vegetables to be flaccid and soft.