2004 SPM Sample answers

a) An athlete has just completed a 400 metre track event. His leg muscles are painful due to muscle fatigue. The athlete then carries out the following actions.

· Wears a track suit

· Takes a few long deep breaths

· Walks freely as a ‘cooling down’ activity.

Explain why the athlete carries out these actions.

b) Human and fish respiratory systems are adapted to function in their respective habitats. Compare and explain the adaptations.

Answer

a) The athlete wears a track suit so that there is no sudden decrease in temperature in his muscles.

A sudden decrease in temperature will cause the accumulation of lactic acid and muscles pain/cramp.

He draws a deep breath several times to oxidize lactic acid that has been accumulated and pays back the oxygen debt.

He walks freely so that his muscles take time to oxidize 1/6 of lactic acid to carbon dioxide, water and energy and 5/6 of the lactic acid is converted back to glucose.

b)

	Human	Fish
Organ	Human’s respiratory structure is the lungs	Fish’s respiratory structure is the gills
Habitat	Human lives on land	Fish lives in the water
Function	Lungs do not need a medium to function	Gills need flow of water through the gills for gaseous exchange
Adaptation	Human lungs have a lot of alveoli and covered with a large network of blood capillaries. Gaseous exchange occur by diffusion through the alveoli	Water is drank in and flows through the gills for absorption of oxygen to blood capillaries which are cross linked with the filaments of gills. Then water flows out through the operculum and absorbs carbon dioxide from the blood capillaries to the water to be released
Characteristics	Total surface area is big with 700 million alveoli/ many alveoli to increase total surface area for diffusion Thin alveolar wall which helps in the diffusion of gases, The wall of the alveolus is moist whereby respiratory gases can dissolve in it, The alveolus is well supplied by blood capillaries to facilitate diffusion and transportation of the gases.	Has thin epithelial wall of gill filaments to facilitate diffusion of gases. Many plate like projections (lamella) at gill filaments to increase the total surface area for diffusion A network of blood capillaries in the gill filaments to facilitate diffusion and transportation of the gases.

SPM 2010 Biology Paper 1 Answers

Biology Paper 1

1.B 2.B 3.C 4.D 5.A 6.C 7.B 8.C 9.B 10.C
11.B 12.D 13.B 14.D 15.C 16.A 17.C 18.B 19.B 20.A
21.D 22.A 23.A 24.D 25.A 26.B 27.A 28.D 29.C 30.C
31.C 32.A 33.D 34.A 35.C 36.B 37.C 38.D 39.B 40.D
41.A 42.A 43.C 44.A 45.D 46.A 47.A 48.A 49.D 50.B

2010 Biology Paper 3- Question 2

Organisms in an environment compete with each other for the same basic needs in their survival. Competition between individuals of different species is called interspecific competition and competition between individuals of the same species is called intraspecific competition. The effect of the competition is shown in their growth such as the height, size and dry mass.

Based on the above information, plan a laboratory experiment to study the effect of interspecific competition between maize and paddy plants on their growth.

The planning of your experiment must include the following aspects:

• Problem statement
• Hypothesis
• Variables
• List of apparatus and materials
• Experimental procedure
• Presentation of data

Sample answer

Problem statement (3 marks)	What is the effect of interspecific competition between maize and paddy plants on their growth?
Hypothesis (3 marks)	Maize plant grow taller than paddy plant// The growth of maize plant is less when it is grown with paddy plant than when it is grown alone.
Variables (3 marks)	Manipulated variable : Type of plant// Maize and paddy plants (only one) Responding variable : Growth of plant//dry mass//height//size (only one) Constant variable : Distance between each seedling//amount of water//intensity of sunlight//nutrient//temperature//duration (only one)
List of apparatus and materials (3 marks)	Apparatus : Tray, oven, balance and ruler Material : Paddy, maize, soil, water
Experimental procedure (3 marks)	1. 3 trays : X is planted with paddy, Y planted with maize and Z with maize and paddy. 2. The seedling trays are filled with equal amount of garden soil. 3. Initial dry mass/any other growth parameter of the seedlings are measured. 4. 5-30 paddy/maize seedlings are planted at 5 cm apart. 5. The seedlings are watered everyday. 6. The trays are weeded. (precaution) 7. After 7-30 days, 5-10 paddy seedlings are removed from tray X and the roots are cleaned (precaution) and then heated/dried in an oven at 105oC. 8. The average dry mass of seedlings is measured using a balance and recorded. 9. Steps 7-8 are repeated with maize seedlings in tray Y and maize and paddy seedlings in tray Z. 10.All data are recorded in a table.
Presentation of data (2 marks)	Types of plants(Tray) Initial dry mass (g) Final dry mass (g) Difference in dry mass (g) Paddy (X) Maize (Y) Paddy (Z) Maize (Z)

2010 Biology Paper 3 - Question 1(Continuation)

1(e(i) Construct a table and record all the data collected in the experiment. Your table should have the following titles:

• Type of activity
• Initial and final readings of the lengths of air column
• Percentage of carbon dioxide in the air sample

Type of activity	Initial readings of the lengths of air column (cm)	Final readings of the lengths of air column (cm)	Percentage of carbon dioxide in the air sample (%)
Resting 0 minute	10.0	9.9	1
1 minute running on the spot	10.0	9.8	2
2 minute running on the spot	10.0	9.7	3
3 minute running on the spot	10.0	9.6	4

(e) (ii) Use the graph paper provided on page 9 to answer this question. Using the data in l(e)(i), draw a line graph of the percentage of carbon dioxide against the time of activity.

Uniform scales on both axes : 1 mark

4 points plotted correctly : 1 mark

Join all 4 points correctly : 1 mark

f) Based on the line graph in l(e)(ii), explain the relationship between the time of activity and the percentage of carbon dioxide in the air sample.

When the time of activity increases, the percentage of CO2 released increases because more energy is required, so respiration takes place at a faster rate.

g) This experiment is repeated on the same student but the exhaled air is collected 10 minutes after each activity. Predict the percentage of carbon dioxide released. Explain your prediction.

1% //decrease because respiration went back to normal rate, ie same as resting. Less energy is produced/needed.

h) State the operational definition for exhaled air.

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

(i) The following list is part of the materials and apparatus used in this experiment.

Complete Table 4 based on the list given above.

Material	Apparatus
Exhaled air sample Water Potassium hydroxide solution	Beaker Boiling tube Rubber tube J-Tube

2010 Biology Paper 3 - Question 1

Q1. An experiment was carried out to investigate the effect of different duration of activity on the percentage of carbon dioxide in exhaled air.

In this experiment, a student rests for 0 minute and his exhaled air is collected to analyse the carbon dioxide content in the air sample.

The experiment is repeated with the same student after running on the spot for 1 minute, 2 minutes and 3 minutes.

The exhaled air is collected immediately after each activity. In this experiment, a J-tube is used to analyse the carbon dioxide content in the exhaled air. Potassium hydroxide solution is used to absorb carbon dioxide in exhaled air.

Sample answers

(a) Record the lengths of air column in the four boxes provided in Table 2.

Activity	Length of air column (cm)
Resting 0 minute	9.9
1 minute running on the spot	9.8
2 minute running on the spot	9.7
3 minute running on the spot	9.6

(b) (i) State two different observations made from Table 2.

Observation 1: The length of air column during resting time is 9.9 cm

Observation 2: The length of air column after running on the spot for 3 min is 9.6 cm

(ii) State the inferences which correspond to the observation in 1(b) (i).

Inference from observation 1: During resting, the least CO2 is released and absorbed by potassium hydroxide solution

Inference from observation 2: During running on the spot for 3 min, the most CO2 is released and absorbed by potassium hydroxide solution

c) Complete Table 3 based on this experiment.

Variable	Method to handle the variable
Manipulated variable Time of activity // Type of Activity	Use different time of activity, ie 0 minute, 1 minute, 2 minutes and 3 minutes // Resting and running on the spot
Responding variable Length of air column	Measure and record the length of air column using a metre rule.
Constant variable Same student Initial length of air column	Fix the same student to carry out all the activities Fix the length of air column to be 10 cm.

(d) State the hypothesis for this experiment.

As the time of activity increases, the length of air column decreases/CO2 released increases

To be continued......

BIOLOGY PAPER 3 (2003-2009)

2003
Q1. To study the effect of camouflage on a predator-prey relationship
[Chapter 8]
Q2. To study the relationship between the concentration of CO₂ and the rate of photosynthesis [Chapter 6]

2004
Q1.To determine the % of vitamin C in fruit juices using DCPIP solution [Chapter 6]
Q2.To estimate the size of the bird population. [Chapter 8]

2005
Q1.To determine and compare the energy content in white bread and peanuts [Chapter 6]
Q2.To determine the concentration of the solution which will maintain the mass of plant tissues.[Chapter 3]

2006
Q1. To study the effect of concentration of albumen suspension on the rate of reaction of pepsin enzyme [Chapter 4]
Q2. To determine concentration of sucrose which is isotonic to cell sap of mango[Chapter 3]

2007
Q1.To study the effect of light intensity on the rate of photosynthesis of a Hydrilla sp. [Chapter 6]
Q2.To study the effect of temperature on the rate of anaerobic respiration in yeast [Chapter 7]

2008
Q1.To investigate the effect of light intensity on the population distribution of Pleurococcus sp. [Chapter 8]
Q2.To study the size of molecules that can diffuse through a semi permeable membrane [Chapter 3]

2009
Q1.To study the effect of light intensity on the rate of photosynthesis. [Chapter 6]
Q2.To study the effect of temperature on the rate of enzyme reaction. [Chapter 4]

2010
Q1. To investigate the effect of different duration of activity on the percentage of carbon dioxide in exhaled air.[Chapter 7]
Q2. To study the effect of interspecific competition between maize and paddy plants on their growth.[Chapter 8]

2007 Biology Paper 3-Question 2

Anaerobic respiration is the process of releasing energy from the breakdown of glucose in the absence of oxygen. Yeast can carry out anaerobic respiration in a glucose solution.

Glucose ---> Ethanol + Carbon dioxide + Energy

The rate of anaerobic respiration is affected by several factors such as temperature, pH values and nutrients.

Based on the above information and equation, design a laboratory experiment to investigate the effect of temperature on the rate of anaerobic respiration in yeast.

The planning of your experiment must include the following aspects:

§ Problem statement

§ Aim of investigation

§ Hypothesis

§ Variables

§ List of apparatus and materials

§ Technique used

§ Experimental procedure or method

§ Presentation of data

§ Conclusion

ANSWER

:1. Aim of investigation

To investigate the effect of temperature on the rate of anaerobic respiration in yeast.

2. Problem statement

What is the effect of temperature on the rate of anaerobic respiration (in yeast)?

3. Hypothesis

The higher the temperature, the higher the rate of anaerobic respiration (in yeast).

4. Variables

Manipulated variable: Temperature

Responding variable: Rate of anaerobic respiration//No of bubbles released//Time taken for lime water to turn chalky

Fixed variable: Volume/conc. of yeast suspension/glucose/pH/Time taken

5. Apparatus:

Boiling tube, Manometer/Capillary tube, Water bath//Beaker + thermometer,

Stopwatch, Marker/thread, Rubber stopper, Ruler, Rubber tubing, Retort stand, measuring cylinder

Material :

Yeast suspension / Yeast, Glucose solution, Paraffin oil (Yeast & Glucose – must have)

6. Technique

Using a ruler to record the change in height of the colored liquid/Record the number of bubbles released (after 10 minutes) using a stopwatch

//Calculating the rate of anaerobic respiration (in yeast) by using the formula :

Number of bubbles / Change in height

Time taken

//Take and record the time taken for the lime water to turn chalky using a stopwatch.

7. Procedure :

1. Set up the apparatus as shown in diagram.

2. Place the boiling tube with 15 cm³ of glucose solution in a beaker at 20^oC for 5 minutes.

3. Add 15 cm³ of yeast suspension into the boiling tube.

4. Mark the initial height of coloured liquid in manometer.

5. Start the stopwatch and mark the level of coloured liquid in the manometer after 10 minutes.

6. Record all data in a table.

7. Repeat step l, 3, 4, 5 and 6 by placing the boiling tubes in a water bath at temperatures of 30^oC, 40^oC and 50^oC.

8. Make sure all the joints of the apparatus are air tight //Repeat the experiment for each temperature to get average readings.

9. Plot a graph of rate of anaerobic respiration in yeast against temperature.

Precaution :

1. Ensure the temperature is stablised at the fixed temperature before taking the reading if the height of the coloured liquid//counting the number of bubbles released.

2. Make sure all the joints of the apparatus are air tight //Repeat the experiment for each temperature to get average readings.

8. Result: Draw a table

9. Conclusion:

Hypothesis is accepted. The higher the temperature, the higher the rate of anaerobic respiration in yeast.