Answers to Mid Year Biology P 2 (Section B)

Question 6
a. P : Artificial active immunity. Q : Artificial passive immunity.

b. P is immunity acquired by injection of vaccine but Q is immunity obtained by injection of antiserum.
P is injection of vaccine given before infection by antigen but Q is injection of antiserum given to patients already infected.
For P the immunity acquired is slow but Q the immunity acquired is immediate after injection.
For P the period of immunity acquired is for a long time but for Q the period of immunity acquired is temporary.
For P the body produce antibody but for Q the antibody is gained from outside.
For P, more than one injections may be needed but for Q the first injection gives the immediate effects of the immunity but a second injection may be needed.
For P the additional injection i.e. the booster dose is given so that the level of immunity reaches an effective level whereas for Q, the additional injection i.e. the booster dose is given so that the level of immunity which drops quickly after the first injection is restored.
(Can answer in table form, but in full sentences)

ci) P1 : The injection of the serum (antivenom) from snakes for victims bitten by snakes.
P2 : This will neutralize the poison of the snakes.

cii) P1 : The patient is given the anti-tetanus serum.
P2 : The antibodies of the serum will prevent the infection caused by the tetanus bacteria..
P3 : This system of defence does not last long as the patient does not produce antibodies from the body cells.
P4 : The antibodies from the serum will be destroyed and excreted.

Question 7
a)F : Leaf structure P : Explanation

F1 :The upper epidermis is one cell thick and transparent (does not contain chloroplasts)
P1 :Allow sunlight to penetrate the leaf and reach the light-trapping chloroplasts in the palisade mesophyll and spongy mesophyll cells

F2:Palisade mesophyll cells are cylindrical , tightly packed in an upright arrangement directly beneath the upper epidermis // the palisade mesophyll cells have a high density of chloroplasts.
P2: Allows the cells to receive and absorb the maximum amount of sunlight to carry out photosynthesis

F3:Spongy mesophyll cells contain chloroplasts// Irregular in shape and loosely arranged with lots of air spaces in between
P3:Allows the cells absorb the sunlight to carry out photosynthesis// allows easy diffusion of water and carbon dioxide into the cells

F4:Presence of xylem which forms continuous tubes throughout the plant
P4: Transports water and mineral salts from the roots to the leaves

F5:Presence of phloem which forms continuous tubes throughout the plant
P5: Transports organic products of photosynthesis from the leaves to all parts of the plant.

F6:Presence of stomata in the lower epidermis
P6: Allows the exchange of gases ( oxygen and carbon dioxide ) between the leaf and the atmosphere

bi) P1 : Compensation point is the light intensity.
P2 : at which the rate of carbon dioxide production during respiration is equal to the rate of carbon dioxide consumption during photosynthesis.

bii) P1 : During the night, light intensity is almost nil, green plants only carry out respiration, therefore only carbon dioxide is produced and no carbon dioxide is consumed.
P2 : At dawn, when there is light, green plants start to carry out photosynthesis.
P3 : However, because of the low light intensity, the rate of photosynthesis //the rate of carbon dioxide consumption is still much lower then the rate of respiration // the rate of carbon dioxide production.
P4 : As the light intensity increases during the day, the rate of photosynthesis increases until a point when it is equal to the rate of respiration.
P5 : At that particular light intensity, the plant is said to have reached// attained the compensation point.

biii) P1 : If a plant were to constantly remain at the compensation point, the rate of sugar ( glucose) / food production is constantly equal to the sugar / glucose/ food consumption.
P2 : As a result, there is no excess sugar/ food for growth and vital processes like reproduction.
P3 : This will cause the plant to be stunted and sterile.P4 : No oxygen will be released by the plant into the atmosphere to be used by other living organisms

Question 8
a) P1 : Depletion of the ozone layer is due to the widespread use of CFC.
P2 : It is used in aerosol, industrial solvents, electronics and Freon in air conditioners.
P3 : Ultraviolet radiation strikes a CFC molecule cause the chlorine atom to break away.
P4 : Then the chlorine atom collides with an ozone molecule and combines with an oxygen atom to form chlorine monoxide and oxygen.
P5 : Then the free atom of oxygen collides with the chlorine monoxide, the two oxygen atoms form a molecule of oxygen.
P6 : The chlorine atom is released and free to destroy more ozone molecules.
P7 : The chlorine produced re-enters the cycle
P8 : When the ozone layer becomes thinner, more ultraviolet radiation reaches the Earth

The effect of excessive ultraviolet radiation on human.
P1 : Reduction of the body’s immune system.
P2 : Skin cancer.
P3 : Cataract of the eye.

Effect on plants.
P1 : Reduction of the rate growth therefore reducing crop yields.

Effect on aquatic organism.
P1 : Death of plankton, reduce food supply to aquatic organism, fisherman’ catch is reduced.

Steps to overcome this problem
P1 : Reduce or stop using CFC or chlorine-based products.
P2 : Replace CFC with HCFC.
P3 : Use wrapping papers instead of polystyrene boxes.
P4 : Patch up the holes in the ozone layer by firing frozen ozone balls into the atmosphere.

bi) P1 : Increase in carbon dioxide concentration traps heat.
P2 : Increase in global atmospheric temperature make the atmosphere warmer / cause a rise in Earth’s temperature / global warming.
P3 : Melting of polar ice / rise in sea level.
P4 : This phenomenon is called green house effect

bii) F1 : Releases a lot of carbon dioxide in the air forms a layer of carbon dioxide in the atmosphere.
P1 : Traps heat / infra-red light / long-wave radiation.
P2: And causes global warming / green house effect / rise in atmospheric temperature.

F2 : Releases heavy smoke which results in the formation of haze / smog
P3 : Reduce light intensity for photosynthesis.
P4 : Less yield / productivity of the crop plants.
P5 :smog / haze prevents vision and results in the air accidents / difficulties during landing of aeroplanes.
P6 : Smoke will also cause health hazards / lung diseases / eye problems.
F3 :Destroys flora / fauna.
P7 : Causes substantial loss of herbs for medical purposes / timber.
P8 : Loss of habitat / gene pool / biodiversity.
P9 : Lack of water catchment area.
P10 : Loss of nutrient / expose to flash flood / landslide / soil erosion / water pollution

Question 9
ai) P1 : When a person touches a hot object, the sensory receptor at the fingers will be stimulated, triggering a nerve impulse.
P2 : The impulse is then sent to the spinal cord through the afferent neurone.
P3 : In the spinal cord, the impulse is transmitted from the afferent neurone to the interneurone and finally to the brain.
P4: The efferent neurone transmits the impulse to the effector so that the fingers can be pulled away from the hot object immediately.

aii)
-For Nervous system, message is carried in the form of nerve impulse but for endocrine system message is carried by hormones.
- For Nervous system, message is via nerve fibres but hormone is carried via the bloodstream.
- For Nervous system, speed of transmission is rapid but for endocrine system speed of transmission is slow.
-For Nervous system, area of response is limited to places with a nerve supply but for endocrine system, area of response is all over the body –one hormone may affect several target organs at the same time. (Can be in table form)

bi) P1 : The process of maintaining the balance of the physical factors of the body
P2 : like temperature, blood pressure, osmotic pressure of the cell fluids and other factors involving the pH of blood and the glucose concentration of the internal environment

bii) P1 : Information of abnormal conditions is sent to a control centre.
P2 : The effectors act to reverse the change.

biii) P1 : Insulin
P2 : and glucagon are the hormones regulating sugar in the blood.
P3 : If the blood sugar level is higher than normal, more insulin is secreted into the blood stream and transported to the liver.
P4 : In the liver, insulin stimulates the liver cells to convert excess glucose to glycogen and fats.
P5 : This causes the level of sugar to fall and return to normal.
P6 : If the blood sugar level is lower than normal, more glucagon is secreted into the bloodstream and transported to the liver.
P7 : In the liver, glucagon stimulates the liver cells to convert glycogen to glucose.
P8 : This causes the level of glucose to rise and return to normal.

Answers To Mid Year Biology Paper 3 - Question 2

Problem Statement
Which fruit juice contains the highest concentration/percentage of Vitamin C?

Objective of investigation
To determine the vitamin C content in pineapple, lime and guava (take from the question).

Hypothesis
Lime juice contains higher concentration/percentage of vitamin C compared to pineapple juice and guava 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

Material & Apparatus
Materials : freshly prepared pineapple, lime and guava 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 concentration of vitamin C in fruit juice by using formula :
Volume of 0.1% ascorbic acid to decolorise 1 ml DCPIP X 1.0 (mgcm-3)
volume of fruit juice needed to to decolorise 1 ml DCPIP


or Calculate percentage of vitamin C in fruit juice by using formula :
Volume of 0.1% ascorbic acid to decolorise 1 ml DCPIP X 0.1%
volume of fruit juice needed to to decolorise 1 ml DCPIP


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 is recorded.
7. Steps 1-6 are repeated using fresh guava, pineapple and lime juice.
8. The volume of fruit juices required to decolourise the DCPIP solution in each fruit juice is recorded in a table.
9. The percentage and the concentration of vitamin C in each fruit juice are calculated.


Presentation of Data

Solution/ fruit juice	Volume of solution/fruit juice needed to decolourise 1 ml of DCPIP solution (ml)	Percentage of vitamin C (%)	Concentration of vitamin C (mg cm -3)
0.1% ascorbic acid	Table Cell	Table Cell	Table Cell
Pineapple juice	Table Cell	Table Cell	Table Cell
Lime juice	Table Cell	Table Cell	Table Cell
Guava juice	Table Cell	Table Cell	Table Cell

Conclusion
Lime juice contains higher concentration/percentage of vitamin C compared to pineapple juice and guava juice. Hypothesis is accepted.

Answers to Mid Year Biology Paper 3 – Question 1

1a. P – 7.5 cm
Q – 6.5 cm
R – 5.0 cm
S – 4.0 cm

bi)1.The final length of the potato strip immersed in 0% sucrose solution is 7.5 cm.
2.The final length of the potato strip immersed in 10% sucrose solution is 4.0 cm.

bii) 1. For the 0% sucrose solution, the increase in length is caused by the diffusion of water molecule into the cell sap by osmosis.
2. For the 10% sucrose solution, the decrease in length is caused by the diffusion of water molecule from the cell sap by osmosis.

c. Manipulated variable : Concentration of sucrose solution
Method to handle :Use different concentration of sucrose solutions:0%, 3%, 5%, 10%

Responding variable: Final length of potato strip/change in length of potato strips
Method to handle : Measure and record the final length of the potato strip using the given scale

Constant variable : Time of immersion/type of potato/size of potato strip/ volume of sucrose solution
Method to handle : Fix the time for the immersion of the potato strips at 30 min for all the expt/ Fix the type of potato to be the same / fix the diameter of the potato strips by using the same cork borer for all the strips/ Fix the same volume of sucrose solution at 20ml for all the expt.

d. The higher the concentration of sucrose solution, the shorter the length of the potato strips / inversely.

e. 1. Able to state all titles with units correctly
2. Able to record all the data correctly
3. Able to calculate and record % change correctly

f. Able to draw : 1. X – axis and Y – axis with correct scales and unit
2. correct transfer of points
3. correct shape of curve

g. Concentration of sucrose solution which is isotonic to the cell sap is 5%.
From the graph, the point where the graph cuts the X-axis indicates the concentration of sucrose solution that does not cause any change to the length of the potato strips.

h. The potato strip will be shorter than 4.0cm,maybe 3.5cm. This is because more water molecules diffuse out from the cell sap to the hypertonic solution outside.

i. Isotonic concentration is the concentration of sucrose solution that causes no change in the length of potato strips that are immersed in it.

j. 0% - Hypotonic solution
3% - Hypotonic solution
5% - Isotonic solution
10% - Hypertonic solution

OPERATIONAL DEFINITION

2006

Based on this experiment, what can you deduce about this enzyme?
( Pepsin on albumin)
This enzyme/pepsin hydrolysed/digest/act on albumen suspension which is cloudy to become clear and the rate of enzyme/pepsin reaction is affected by the concentration of substrate/albumen

2007 Based on the results of this experiment, what can be deduced about photosynthesis? Photosynthesis is a process where Hydrilla sp. in sodium hydrogen carbonate solution produces bubbles and the number of bubbles released is influenced by light intensity

2008
State the operational definition for population distribution of Pleurococcus sp.
Population distribution is defined as the total surface area covered by Pleurococcus sp. in a 10cm X 10cm grid at different light intensity/is affected by light intensity.

2009

Based on the result of this experiment, state the operational definition for the process of photosynthesis.

The process of photosynthesis is where the aquatic plant in sodium hydrogen carbonate solution produces gas bubbles and the length of the gas bubble released is influenced by light intensity.

2010

State the operational definition for exhaled air.

Exhaled air is the air column/air sample collected in the J-tube which contains CO2 and is absorbed by potassium hydroxide. The length of air column is influenced by time of activity.

2011

Define operationally hydrolysis of starch.

Hydrolysis of starch is the process where amylase breaks down starch and is shown by the time taken for iodine solution to remain yellow. Hydrolysis of starch is affected by the pH value of the solution

Biology 2009 Paper 3

Question 2.
Situation 1
Housewife A uses warm water to wash her clothes using washing liquid which contains added enzyme. The cleaning is more effective.
Situation 2
Using the same washing liquid as in situation 1, housewife B uses cold water to wash her clothes. The cleaning is less effective.

Based on both situations, design a laboratory experiment to study the effect of temperature on the rate of enzyme reaction.
The planning of your experiment must include the following aspects:

• Problem statement
• Objective of investigation
• Hypothesis
• Variables
• List of apparatus and materials
• Technique used
• Experimental procedure
• Presentation of data
• Conclusion
[17 marks]

SAMPLE ANSWER

1. Aim of investigation
To study the effect of different temperatures on the rate of enzyme reaction on starch.

2. Problem statement
What are the effects of different temperatures on the rate of enzyme reaction on starch?

3. Hypothesis
As the temperature increases, the rate of enzyme reaction on starch increases until the optimum temperature of 37oC.

4. Variables
Manipulated variable: Temperature
Responding variable : Time taken for the hydrolysis of starch to be completed
//Rate of enzyme reaction
Fixed variable : Volume/concentration of the amylase enzyme// Volume/
concentration of cooked starch

5. Apparatus:
Test tube, beakers, thermometer, droppers, stopwatch, and white tile

Material:
Starch solution, water bath, salivary amylase, iodine solution (Salivary amylase and starch solution-must have)

6. Technique :
Record the time taken for the iodine solution to stop turning to blue-black colour/for complete hydrolysis of starch using a stop watch.

7. Procedure :
1. Measure 5ml of starch solution into a test tube.
2. Put the test tube containing starch solution in water bath at 37oC for 5 minutes.
3. Put another test tube containing 2 ml of salivary amylase in a water bath at 37oC for 5
minutes.
4. Pour the salivary amylase into starch solution.
5. Put a drop of the mixture into a drop of iodine solution at interval of 1
minute.
6. Fix the volume of starch solution at 5 ml for all the experiments.
7. Fix the volume of salivary amylase at 2 ml for all the experiments.
8. Record the time taken for the iodine solution to remain yellow/ to stop
turning to blue-black colour using a stopwatch.
9. Repeat experiment at different temperature: 20oC, 30oC, 50oC, 60oC.
Precaution :
1. Ensure the temperature is stablised at the fixed temperature before
mixing the enzyme with the substrate.
2. Start the stopwatch immediately after mixing the enzyme with the substrate.

8. Result:

Temperature(oC)	Time taken for hydrolysis of starch(s)	Rate of reaction (min-1)
20	Table Cell	Table Cell
30	Table Cell	Table Cell
37	Table Cell	Table Cell
50	Table Cell	Table Cell
60	Table Cell	Table Cell

9. Conclusion:
Hypothesis is accepted. As the temperature increases, the rate of enzyme reaction on starch increases until the optimum temperature of 37oC.