Answers to Hands on Pg 65-68

6.6 : Effects of Macronutrient Deficiency in Plants
Aim : To study the effects of Macronutrient Deficiency in Plants
Hypothesis : A plant is healthier if it is grown in a complete culture solution (Knop's solution)
a) MV : Components of minerals in culture solution
b) RV: Condition of plant
c) FV : Volume of solution//size and type of maize seedlings//amount of sunlight//air
Observations

Boiling tube	Nutrient deficient	Effects on seedlings
A	All	No growth or a stunted and weak seedling
B	None	A healthy seedling with a large and sturdy stem and dark green leaves
C	Nitrogen	Yellow and small leaves with minimum growth
D	Phosphorus	A stunted seedling with dark green leaves and red spots
E	Potassium	A seedling with small yellow leaves and soft stem
F	Calcium	Stunted seedling with irregular shapes of leaves and dying shoots
G	Magnesium	A seedling with yellow and dying leaves
H	Sulphur	A seedling with light green leaves and stunted roots

Discussion
1. To shield the culture solution from light because algae will grow in the culture solution if light is present and compete with the maize for nutrients.
2. To supply oxygen to the roots for respiration.
3. The seedling in A is the least healthy because no nutrients was given to it. The seedling in B was healthiest because it was grown in a complete culture solution where all macronutrients were present.
Conclusion : the hypothesis is accepted. A plant is healthier if it is grown in a complete culture solution (Knop's solution)

Answers to Hands on Pg 62-65

6.5 : Movement of substances through Visking tubing
Aim : To study the movement of substances through the Visking tubing
Hypothesis : Reducing sugar molecules which are smaller than starch molecules can move across the Visking tubing.
Variables : a) MV : Contents in the Visking tubing
b) RV : Change in colour of water samples using iodine test and Benedict's test
c) FV : Temperature//volume of solution
Observation:

Boiling tube	Beginning of experiment		After 30 minutes
	Iodine test	Benedict’s test	Iodine test	Benedict’s test
A	Yellowish brown	Blue solution	Yellowish brown	Brick red precipitate
B	Yellowish brown	Blue solution	Yellowish brown	Blue solution

Discussion
1. Salivary amylase in saliva breaks down the starch into reducing sugar. The molecules of reducing sugar are smaller in size when compared with the starch molecules. Hence, the molecules of the reducing sugar move out from the Visking tubing while the starch molecules remain inside.
2. Visking tubing represent the wall of the ileum whereas the distilled water represent the blood in the blood capillary.
3. Digestion is the breaking down of complex food molecules into simpler and smaller food molecules. These smaller food molecules are small enough to move across the membrane of the ileum into the blood capillary and transported to tissues of the whole body.
Conclusion: The hypothesis is accepted. Reducing sugar molecules which are smaller than starch molecules can move across the Visking tubing.

Answers to Hands on Pg 60-62

6.4 : Digestion of protein
Aim : To study the digestion of protein
a) MV: Absence of presence of enzyme
b) RV : Condition of solution
c) FV: Temperature//enzyme concentration//volume of albumen suspension
Observations

Test tube	Condition of content
	Beginning of experiment	End of experiment
A	Cloudy	Less cloudy
B	Cloudy	Clear
C	Cloudy	Cloudy

Discussion
1. To prepare an acid medium for the action of pepsin.
2. A: Digestion of albumen is very slow because pepsin is not very active in a neutral medium.
B: Digestion of albumen is very fast because pepsin is very active in acidic medium. The cloudy suspension of albumen turns to a clear solution of amino acids.
C: No digestion of albumen as there are no enzymes.
3. Te cloudy suspension remains unchanged because pepsin is inactive in alkaline medium.
Conclusion : The hypothesis is accepted. Pepsin catalyses the hydrolysis of albumen to polypeptides in an acidic medium.

Answers to Hands on Pg 57-60

6.3 : Digestion of starch

Aim : To study the digestion of starch
Variables : a) MV : Absence or presence of starch
b) RV : Benedict solution turns brick red-Presence of reducing sugar
c) FV : Temperature//concentration of starch
Results:

Test	Content	Observation
Iodine test	Starch suspension	Dark blue
	Saliva	Yellowish brown
Benedict’s test	Starch suspension	Blue solution
	Saliva	Blue solution
Test	Test tube	Observation
Iodine test	A	Yellowish brown
	B	Dark blue
	C	Dark blue
Benedict’s test	A	Brick red precipitate
	B	Blue solution
	C	Blue solution

Discussion

1. To make sure that there is no starch and reducing sugar in the saliva and no reducing sugar in the starch.
2. Salivary amylase
3. Maltose
4. In test tube A, the salivary amylase breaks down the starch into soluble maltose, which is a reducing sugar. In test tube C, there is no breakdown of starch as the enzyme has been denatured after boiling.
5. There is no breakdown of starch because there is no enzyme in the distilled water.
Conclusion: Hypothesis is accepted. Salivary amylase in the saliva breaks down starch into reducing sugar.

Answers to Hands on Pg 74

AIM	To show that light is needed for photosynthesis
PROBLEM STATEMENT	Is light needed for photosynthesis?
HYPOTHESIS	Plant needs light for photosynthesis.
VARIABLES	Manipulated: Presence or absence of sunlight Responding: Presence or absence of starch Fixed: Temperature, light intensity, concentration of carbon dioxide
MATERIAL & APPARATUS	Materials: Ethanol, iodine solution, aluminium foil and paper clips Apparatus: Beaker, boiling tubes, Bunsen burner, tripod stand, wire gauze and white tile
TECHNIQUE	Test for starch in leaf using iodine solution and draw the diagram of the leaf to show the part of the leaf that turned dark blue.
PROCEDURE	1. Keep a plant in the dark for two days. 2. Choose a healthy leaf on the plant and cover part of a leaf with the aluminum foil as shown. 3. The plant was exposed to sunlight for 6 hours. 4. Light the Bunsen burner and boil the water. When the water had boiled add the leaf. 5. Wait one minute for the leaf to boil (this is to get rid of the waterproof layer and break the open cells and make it soft). 6. Turn off the Bunsen burner (for safety reasons, we are going to use ethanol), and take out the leaf. 7. Put the leaf in a boiling tube and cover with ethanol. 8. Be careful when using ethanol because its highly flammable. 9. Put the tube of ethanol plus leaf into the beaker of hot water. 10. Dip the leaf back into the hot water so it can get the ethanol off. 11. Spread the leaf out on a tile. 12. Add about five drops of iodine on to the leaf and observe after about two minutes after the iodine had soaked in. 13. Draw the diagram of the leaf to show the part of leaf that turned dark blue. 14. Repeat steps 4-13 on another leaf (not covered) on the same plant.
RESULT	Diagram after experiment
CONCLUSION	Hypothesis is accepted. Light is needed for photosynthesis.

Answers to Hands on Pg 72

Section B

3a)

(a) Waxed cuticle

• Reduces loss of water through transpiration.

Upper epidermis

• Transparent thin layer of cells which allows sunlight to reach the mesophyll layer.

Palisade mesophyll cells

• Elongated cells perpendicular to the surface.
• They are densely packed and rich in chloroplasts to allow maximum absorption of sunlight.

Spongy mesophyll cells

• Irregular in shape, loosely arranged and contain chloroplasts.
• Large air spaces between cells allow gaseous exchange between the cells and the surrounding air.

Stoma

• Regulate the passage of oxygen, carbon dioxide and water vapour across the surface of the leaf.

Guard cells

• Contain chloroplasts.
• Control the closing and opening of the stoma.

Vascular bundle

• Xylem is used for the transport of water and mineral salt.
• Phloem is used for the transport of soluble food materials from the leaf to other parts of the plant.

b) Photosynthesis involves light reaction and dark reaction.

Light reaction

• Chlorophyll absorbs light energy.
• Light energy causes the chloropyhll to emit electrons.
• Light energy also splits the water molecules, into H⁺ ions and OH^- ions.
• OH^- ions combine to produce water and oxygen.

24OH^- - 24e --->12H₂O + 6O₂

• H⁺ ions accepts an electron to form a hydrogen atom which is used in the dark reaction.

Dark reaction

• Hydrogen atom reduces CO₂ to form CH₂O
• 6 units of CH₂O combine to form 1 unit of glucose, C₆H₁₂O_6.

4. a)

Bread is a carbohydrate which consists of starch.

Starch is digested into glucose.

Half-boiled egg contains proteins which are digested into amino acids.

Glucose and amino acids are absorbed into the villi of the ileum.

Glucose and amino acids enter the blood capillaries of the villi and are transported by the

hepatic portal vein to the liver.

From liver, glucose and amino acids are transported to body cells.

Glucose is oxidised to release energy.

Excess glucose is converted to glycogen and stored in the liver.

Amino acids are used for synthesis of new protoplasm in the body cells.

Amino acids are used for synthesis of hormones, enzymes and plasma proteins.

Excess amino acids are converted to urea and excreted through the kidneys.

Margarine contains lipids.

Lipids are digested into fatty acids and glycerol.

Fatty acids and glycerol are absorbed into lacteals in the villi.

Glycerol and fatty acids are transported in the lymphatic system before entering the blood circulatory system.

Glycerol and fatty acids recombine to form lipids.

Lipids are used as components of the plasma membranes.

Excess lipids are stored as fat in the adipose tissue.

b)

Rice is starch.

Starch is a complex molecule which consists of polysaccharide chains.

The size of starch molecule is too big to move across the intestinal wall.

Starch needs to be broken down into smaller and simpler units.

These smaller units (monosaccharides) are able to move across the intestinal wall into the blood stream.

Salt, which is also known as sodium chloride, dissolves in water to form sodium ions and chloride ions.

Sodium ions and chloride ions are small enough to pass through the intestinal wall to be absorbed into the blood stream.

Therefore, salt does not need to be digested.

Answers to Hands on Pg 71-72

Section A

1. a) i) Daily protein need = 20/7.5 = 2.67 gkg^-1 body mass

ii) Daily protein need = 70/55 = 1.27 gkg^-1 body mass

b) A child needs more protein to build new cells and tissues for growth. A teenage boy at the age of 17 has almost reached puberty. Hence, less protein is needed because his growth rate is slower.

c) Meat and soya beans contain all the essential amino acids needed by our body. However, vegetables do not contain all the essential amino acids.

d) Vitamin D is needed for the absorption of calcium and phosphorus. A growing child needs more calcium and phosphorus for the formation of strong bones and teeth. Hence, a child needs more vitamin D. As for the teenage girl, she needs less vit D because her bones and teeth are almost completely formed.

e) A teenager needs more iron for the production of haemoglobin to transport oxygen because he/she is more active and bigger in size as compared to a child. A teenage girl will need more iron than a teenage boy because she loses blood through menstruation.

2. a) i) To study the need of light for photosynthesis

ii) To study the need of carbon dioxide for photosynthesis.

b)i) To prevent the covered part of the leaf from receiving sunlight.

ii) Solution X is potassium hydroxide solution and is used to absorb carbon dioxide from the air around the leaf.

c) i) Dark blue on the part of the leaf which is not covered with black paper.

ii) The whole leaf gives a yellowish brown colour.

d) No photosynthesis takes place on the covered part of leaf P as it does not receive any light. Leaf Q shows the absence of starch because there is no carbon dioxide for photosynthesis to take place.

e) The plant needs to be put in a dark room for 24 hours to ensure the leaves of the plant do not contain starch at the start of the experiment.

Sample question on photosynthesis

Question :
Green plants synthesize their food through the process of photosynthesis. The chemical process of photosynthesis can be summarized as in the schematic diagram below.
a) State the meaning of photosynthesis based on the schematic diagram above.
b) Starting with water and carbon dioxide as the raw materials, describe how a green plant produces starch molecules.
ANSWER
a) A process where a green plant produces glucose which will undergo condensation to form starch from carbon dioxide and water in the presence of chlorophyll and sunlight.

b)During light reaction, chlorophyll captures light energy which excites the electrons of chlorophyll molecules to higher energy levels.
In the excited state, the electrons can leave the chlorophyll molecules.
Light energy is also used to split water molecules into hydrogen ion (H⁺) and hydroxyl ions (OH^-) (Photolysis of water).
The hydrogen ions (OH^-) then combine with the electrons released by chlorophyll to form hydrogen atom.
The energy from the excited electrons is used to form energy-rich molecules of ATP.
Hydroxyl ion loses an electron to form a hydroxyl group.
This electron is then received by chlorophyll.
The hydroxyl groups then combine to form water and oxygen gas.
During the dark reaction, the hydrogen atoms are used to fix carbon dioxide in a series of reactions catalysed by photosynthetic enzymes and caused the reduction of carbon dioxide into glucose.
The glucose monomers then undergo condensation to form starch which is temporarily stored as starch grains in the chloroplasts.

Vitamin C content in fruit juices

Please read either from Hands on Pg 55 /the sample experiment plan given below :

Objective	To determine the vitamin C content in pineapple, lime and orange juice.
Problem statement	Which fruit juice contains the highest concentration /percentage of vitamin C?
Hypothesis	Lime juice contains higher concentration/percentage of vitamin C compared to pineapple juice and orange juice.
Variables	Manipulated : Types of fruit juice
	Responding : Concentration/percentage of vitamin C in fruit juices
	Constant : Volume of fruit juices/concentration of DCPIP solution/ Standard concentration of ascorbic acid solution
Apparatus and materials	Materials : freshly prepared pineapple, lime and orange juices, 0.1% ascorbic acid solution, DCPIP solution Apparatus : 5ml syringe(with needle), 1ml syringe, specimen tubes, 5ml measuring cylinder
Technique	Measure and record the volume of fruit juice needed to decolourise 1 ml DCPIP using a syringe// Calculate the % of vitamin C in fruit juice by using formula : Vol of 0.1% ascorbic acid X 0.1 Volume of fruit juice OR Calculate the content of vitamin C in fruit juice by using formula : Vol of 0.1% ascorbic acid (mgcm-3) Volume of fruit juice
Procedure	1. 1 ml of DCPIP solution is placed in a specimen tube using a 1 ml syringe. 2. 5 ml syringe was filled with 5 ml of 0.1% ascorbic acid solution. 3. The needle of the 5 ml syringe was immersed into the DCPIP solution. 4. Ascorbic acid solution was added drop by drop until the DCPIP solution, and gently stirred with the needle of the syringe. (Do not shake the tube vigorously). 5. Ascorbic acid solution is added continuously until the DCPIP solution is decolourise. 6. The volume of ascorbic acid solution used measured using a syringe and recorded. 7. Steps 1-6 are repeated using fresh orange, pineapple and lime juice. 8. The volume of fruit juices required to decolourise the DCPIP solution in each fruit juice is measured using a syringe and recorded in a table. 9. The percentage and the concentration of vitamin C in each fruit juice are calculated using the above formula. Precaution: 1. The juice must be freshly prepared. 2. Do not shake the DCPIP too vigorously.
Recording data/ result	Solution/ fruit juice Volume of solution/fruit juice needed to decolourise 1 ml of DCPIP solution (ml) Percentage of vitamin C (%) Conc. of vitamin C (mg cm-3) 1 2 3 Average 0.1% ascorbic acid Pineapple juice Lime juice Orangejuice
Conclusion	Lime juice contains higher concentration/percentage of vitamin C compared to pineapple juice and orange juice. Hypothesis is accepted.