juj 2011 biologi bahan seminar guru

BAHAN SEMINAR BIOLOGI JUJ PAHANG 2011 4551

1

BAHAN SEMINAR JUJ

PROJEK JAWAB UNTUK JAYA (JUJ)

NEGERI PAHANG

TAHUN 2011

BIOLOGI


2

Page

1.1 Contents 2

1.2 Format 3 - 4

1.3 Analysis 5 - 10

1.4 Tips 11 -18

1.5 Questions - SPM 2009 Question paper (paper 2 & 3) 19 - 48

1.6 Marking scheme 49 - 63

1.7 Example of student’s answer 82 - 85

1.8 Marking Skill 86 - 89


3

1.2 BIOLOGY SPM EXAM FORMAT ( STARTING FROM 2003) SUBJECT CODE : 4551

Index Criteria Paper 1(4551/1) Paper 2(4551/2) Paper 3(4551/3)

1 Type of

instrument

Objective Test Subjective Test Written Practical

2 Type of item Objective Item

Multiple

choice

Each item followed

by four alternative

answers A, B, C or

D

Subjective Item

Section A : structured

Item

Section B :Essay

Item

Subjective Item :

Structure

Item

Open ended

Respond

(Essay )

3 Total Question 50 ( Answer all the

questions )

Section A :

5 items (Answer all

the questions –

Section B :

4 items (Answer

any two questions)

Refer to SPM 2008

format (latest)

Structure Item

1 – 2 item

(Answer any

one question)

Open ended

responds:

1 item

( Essay

written )

4 Total Marks 50 100 50

5 Responded Blacken one space at

OMR form

Write the answer in the

space provided in the

question paper

Write the answer in

the space provided

in the question

paper

6 Duration 1 hour 15 minutes 2 hours 30 minutes 1 hour 30 minutes

7 Construct

scoring

section A :

Knowlegment – 25

Section B :

Understanding – 15

Section C :

Application skill - 10

Knowlegment – 10

Understanding – 20

Application skill – 30

Analysis skill – 15

Synthesis skill – 15

Evaluation - 10

Science process

skill :

16 aspect

Max score :3

8 Item example

based on

construct

Refer Example

Instrument: Paper

4551 / 1

Refer Example instrument:

Paper 4551 / 2

Refer Example

instrument: Paper

4551 / 3

9 Marking Dichotomous

Mark: 1 or 0

Scoring is analytical based

on scoring rubric

Scoring is analytical

based on rubric at


4

level 3.

10 Context Construct from All

learning Area are

tested

Construct from All learning

Area are tested

Construct are tested

from suitable

learning area.

11 Level of

difficulties

Easy : E

Moderate : M

Hard : H

R : S : T = 3 : 1 : 1

( 25 easy item :

15 moderate item

:10 hard item )

R : S : T = 4 : 4 : 2

( 40marks easy Item:

40marks moderate item :

20marks hard item )

R : S : T = 3 : 1 : 1

( 30 easy item : 10

moderate item : 10

hard item )

Overall

R : S : T = 5 : 3 : 2

12 Adding

apparatus

Scientific calculator Scientific calculator Scientific calculator


5

1.3 Analysis of the SPM Biology Exam Questions

Analysis of the SPM Biology Questions (2006-2010)

CHAPTER 2006 2007 2008 2009 2010

P1 P 2 P 3 P1 P 2 P 3 P1 P 2 P 3 P1 P 2 P 3 P1 P2 P3 OBJ S E 1 2 OBJ S E 1 2 OB S E 1 2 OB S E 1 2 S E

F O R M

F O U R

1. Introduction of Biology

- - - - - - - - - - - - - - - - - - - - - - - - -

2. Cell Structure and Cell Organisation

2 1 - - - 4 1/2 - - - 2 ½ 1/5 - - 2 1 - - - 3 1 -

3. Movement of substance Across The Plasma Membrane

4 - 1 - 1 2 1 - - - 3 - 4/5 - 1 5 1/5 - - - 2 1 -

4. Chemical Composition Of The Cell

2 - - 1 - 1 1 - - - 3 - - - - 3 - - - 1 4 1 -

5. Cell Division

3 - - - - 2 - - - - 1 - - - 2 2/3 - - - 2 ½ -

6. Nutrition

4 1 2 - - 9 - - 1 - 8 - 1 - - 6 - 1 1 - 5 - 1

7. Respiration

2 - - - - 4 1/3 - - 1 - - - - - 6 1 - - - 6 - -

8. Dynamic Ecosystem

4 - - - - 5 - 1 - - 5 - - 1 - 3 - 1 - - 4 - -

9. Endangered Ecosystem

2 - - - - 3 - 1 1/3

- - 3 - 1 - - 4 - - - - 3 - 1


6

CHAPTER 2006 2007 2008 2009 2010

P1 P 2 P 3 P1 P 2 P 3 P1 P 2 P 3 P1 P 2 P 3 P1 P2 P3 OB S E 1 2 OB S E 1 2 OB S E 1 2 OB S E 1 2 OB S E 1 2

F O R M

F I V

E

1. Transport 5 - 1 - - 4 - - - - 6 ½ 1 - - 7 1 1 - - 3 - 1

2. Locomotion and Support

3 - - - - - - - - - 3 - - - - 1 - - - - 1 1 -

3. Coordination and Response

5 1 - - - 7 2/3

1 - - 5 1 - - - 4 - - - - 4 - 1

4. Reproduction and Growth

9 1 - - - 6 2/3

1 - - 5 1 - - - 4 1 - - - 7 - -

5. Inheritance

2 1 - - - 2 - 1 - - 2 1 - - - 2 1/3

- - - 3 - -

6. Variation

3 - - - - 3 - - - - 2 - - - - 1 - 1 - - 3 1/2

-


7

EXPERIMENTS CHECKLISTS FORM 4 (SPM 2004-2009)

No Topic 2005 2006 2007 2008 2009 2010

Q1 Q2 Q1 Q2 Q1 Q2 Q1 Q2 Q1 Q2 Q1 Q2 1 CHAPTER 3: Akt:3.1 Size of

molecule that can diffuse through a semipermeable membran

X

2 CHAPTER 3: Movement of substances across the plasma membrane Activity 3.2 : Studying osmosis using an osmometer (page 24)

3 CHAPTER 3: Movement of substances across the plasma membrane Activity 3.4 and 3.4 : Studying the effects of hypotonic ,hypertonic and isotonic solutions on animal and plant cells. (27-28)

4 CHAPTER 3: Movement of substances across the plasma membrane Activity 3.6 : Determining the concentration of an external solution which is isotonic to the cell sap of a plant. (page 30)

X X

5 CHAPTER 4: Chemical composition of the cell Activity 4.3: Studying the effects of temperature on salivary amylase activity (page 36)

X

6 CHAPTER 4: Chemical composition of the cell Activity 4.4: Studying the effects of pH on the activity of pepsin (page 39)

7 CHAPTER 4: Chemical composition of the cell Activity 4.4: Investigate the effects of pH on the breakdown of starch by amylase. (page 41)

8 CHAPTER 4: Chemical composition of the cell

X


8

Activity 4.5: Studying the effects of substrate concentration on salivary amylase activity (page 42) /(SPM : Concentration of albumen)

9 CHAPTER 4: Chemical composition of the cell Activity 4.6: Studying the effects of enzyme concentration on salivary amylase activity (page 43)

10 CHAPTER 6: Nutrition Activity 6.1: Determining the energy value in food samples. (page 61 – 62)

X

11 CHAPTER 6: Nutrition Activity 6.3: Determining the vitamin C contain in various fruit juices. (page 65 – 66)

12 CHAPTER 6: Nutrition Activity 6.8 : Studying the effects of macronutrient deficiency in plants (page 72)

13 CHAPTER 6: Nutrition Activity 6.11 Investigating the effects of light intensity on the rate of photosynthesis. (page 76)

X

X

14 CHAPTER 6: Nutrition Activity 6.11 Investigating the effects of carbon dioxide concentration on the rate of photosynthesis.

15 CHAPTER 7: Respiration Activity 7.6: Investigating the differences between inhaled and exhaled air in terms of oxygen and carbon dioxide contents. (page 93) (page 93 – 94)

x


9

16 CHAPTER 7: Respiration Activity 7.2 : Investigating the process of anaerobic respiration in yeast (page 85)

17 CHAPTER 8 :Dynamic Ecosystem Activity 8.1 Investigating interspecific competition of plant

X

18 CHAPTER 8: Dynamic Ecosystem Activity 8.5 Investigating the distribution of plants using the quadrat sampling technique (page 111- 112) Modified (using Grid)

X

19 CHAPTER 8: Dynamic Ecosystem Activity 8.6 Estimating the population size of animals using capture, mark, release and recapture technique (page 113)

20 CHAPTER 8: Dynamic Ecosystem Activity 8.11 Studying the effects of temperature, pH, light intensity and nutrients on the activity of yeast (page 119)

X

21 CHAPTER 9: Endangered Ecosystem Activity 9.2: Investigating the level of pollution in several different sources of water (page 128 – 129)


10

EXPERIMENTS CHECKLISTS FORM 5 (SPM 2003-2009)

No Topic 2003 2004 2005 2006 2007 2008 2009 2010

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

1 CHAPTER 1:TRANSPORT To study one of the factor of affecting the rate of transpiration.

2

CHAPTER 3: COORDINATION AND RESPONSE. - To study the effect of different quantities of water intake of urine output.

3 CHAPTER : VARIATION - To investigate continuous variation and discontinuous variation in human.

4. CHAPTER 6: VARIATION - To investigate the importance of camouflage in the survival of a species

X


11

1.4 TIPS FOR EXAM 1.4.1 Objective Question – Paper1

i. Try to answer easy questions first, followed by moderate questions and students

have enough time to answer difficult questions.

ii. Don’t take more than 11/2 minutes for each question to make sure enough time

for all questions.

iii. Read the question carefully for three times to you understand what are the

questions ask.

iv. More information for each question can get from graph, table, and diagram that

given.

v. Make ( / ) for true statement, reject all destructor and guess the best answer

when you are not sure the best answer.

vi. Make sure answer all the questions and remark all the answer and make sure:

* One question only one answer.

* Deleted wrong answer completely

* Used 2B pencil.

Vii Examples of questions form for paper 1

* Remember the fact

* Making conclusion

* Application

* Observation

* Knowlegment

* Comparisons

* Identify the problem

* Calculation


12

1.4.2 Subjective Question

Encourage the students to review the essay question first (Part B Paper2 ),before

answer the structure question, this because students will have enough time to think

some facts or explaination.

Almost structure questions based on diagram, table, data, flow chart, graph that

suitable with fact, experiment or investigation. Understand all the information

given.

Time suggestion to answer Paper 2: Part A ( 90 minutes ), Part B ( 60 minutes ), for

Paper 3 : Question 1 ( 50 minutes ) and Question 2 ( 40 minutes )

Answer in one word, one number or one simple sentence

Don’t combine the right fact with the wrong fact

Follow the instruction like : Give two examples of……., so students should give

only two examples, the third example will not get the mark.

No need write in long sentence or copy again part of the question.

Answer can be in equations form, diagram, table or graph. Calculation must be

show.

Space for write the answers and mark at end of the essays or structure questions are

given will show how long the answer must be write.

Characteristics of alveolus :

Accept Reject

Thickness of alveolus is only

one cell

Alveolus is thin

Surface of alveolus is wet wet

A lot of network of blood

capillaries covering the

alveolus

A lot of blood capillaries


13

Instruction verb like justification, evaluation, give your opinion,

Students must state like ‘ I agree / I accept / I’m not agree / I’m not accept that

statement given ( 1 mark ) and followed by opinion

Draw a diagram

* No artistic

* Big (suitable size), clear,

* Label the diagram correctly and line for label can’t be cross together

* Neat and without broken lines

Draw a enzyme structure: Size and shape of the enzyme must same with the original

Comparison - Must have similarities and differences

- One characteristic must compare between two subject in one

sentence

- Separate sentence between similarities and differences

- If answer in table, must write in full sentence

Write chemical equation :

* In word form

[ / ] Glucose + oxygen Carbon dioxide + water + energy

[ X ] Glucose + oxygen CO2 + H2O + energy

* In chemical form

[ / ] C6H12O6 + 6O2 6CO2 + 6H2O + energy


14

Draw the hybrid cross (Inheritance)

* Has key

* Label the schema diagram - Parental Genotype

- Parental Gamete

- F1 Genotype

- F1 Phenotype

Male gamete and female gamete are fertilization

* Reject combine / attach

Function of mitochondrion – Generate / provide energy

- Reject : Supply / give energy

Don’t copy again part of the question because this is not get any mark.

1.4.3 Paper 3

1.4.3.1 Question 1

i) Measuring using number

Measure / record the data using apparatus that given in the experiment / question with

the correct unit

Example : Record scale / thermometer reading, stop watch, ruler, measuring

cylinder, syringe, burette with the correct units ( if not given)

ii) Observing

Making observation based on the experiment given not on the theory. What can

observe / see only – from data, table, scale of apparatus

Example : State changes in color

State increase of thermometer reading

State changes in time

State changes in volume ( end of experiment )

• State the VALUE OF MV & RV

• The observation that can be making inference


15

iii) Making Inferences - Making initial conclusion / cause based on observation

- Inferences must be correspond with the observation ( inference (i) correspond with

observation (i) , inference (ii) correspond with observation (ii)

- Must infer MV & RV

If wrong / reject observation automatic inference will reject / wrong

iv) Controlling Variables

- Able to state all the variables, controlled, responding and manipulated variables

correctly and method to handle variable correctly.

- Must state PARAMETER like volume, temperature, mass, time, length

- State that apparatus using to get the result for responding and controlled variables.

Variable Method to handle variable correctly

Manipulated variable:

Variables that are changed in the

experiment

Examples:

Temperature of water bath, mass of

food, concentration of sucrose

solution, type of fruits

Change in mass/concentration / water

Or used different mass/ concentration /

type of food

Example :

Used different mass of food

Used 30% sucrose solution, 5%

sucrose solution 10% sucrose

solution

Replace papaya juice with orange

juice

Change the concentration of

albumen

Responding variable:

Variable that are measure after

experiment / result

Example

i) Final length of potato strip,

ii) Final temperature of water,

iii) Rate of transpiration

iv) Rate of enzyme reaction

Must state the apparatus or state the

formula using

Example :

i) Measure and record the final length of

potato strip using ruler

ii) Measure and record the final

temperature of water using

thermometer

iii) Calculate the rate of transpiration using

formula : distance divided by time

iv) Calculate the rate of enzyme reaction

using formula concentration of

albumen dived by time

Controlled variable:

Variable that constant during

Must state the PARAMETER and VALUE


16

experiment

Example:

Initial temperature of water, volume

of water, concentration of starch,

type of enzyme

and APPERATUS

Example :

Fix the temperature at 370C using

thermometer Fix volume of water at 20ml using

measuring cylinder

Fix concentration of starch at 10%

Fix type of enzyme is pepsin

v) Making hypothesis

Make a statement of hypothesis by relating the manipulated variable (MV) with

the responding variable (RV) and showing the specific relationship (H).

vi) Communication Presenting the data in certain form like table, graph, chart or diagram.

Table - Column and row with correct title and units ( manipulated and

responding variable)

- Sufficient and systematic data (observational data )

Graph - Both axes labeled with correct units (1m)

- Uniform scale

- All points plotted correctly (1m)

- Smooth curve and correct shape (1m)

Chart - Title of the chart

- Both axes labeled with correct units

- Uniform scale

- Bars plotted correctly

- Correct shape

Diagram - No artistic

- Big (suitable size), clear,

- Label the diagram correctly and line for label can’t be cross together

- Neat and without broken lines

Calculation - Work out accurate calculation

- Wright formula

- Replacement with correct data

- Answer with correct unit


17

vii) Interpreting Data

- Based on the communicating data, able to state correctly the

relationship between the variables

- Support with theory

viii) Relationship between space and time

- Quantity and time (concentration, volume)

- Relationship between manipulated / responding variable with time

- Support with theory

ix) Predicting

Give once value that may be true base on the trend / data before and support by

Theory

x) Defining by operation

- Base on experiment, refer observation

- Including data, color, or time

- Refer to RV , HP

- Can’t base on theory

xi) Classifying

Can group the answer base on the certain character


18

1.4.3.2 Question 2 ( NEW FORMAT)

i) Problem statement (01) 3M - In question form.

- Relationship between manipulated and responding variable

- End of sentence has question mark (?)

iii) Hypothesis (02) 3M

Make a statement of hypothesis by relating the manipulated variable with the

responding variable and showing the specific relationship.

iv) Variables (03) 3M

- Manipulated variable - 1m

- Responding variable -1m

- Controlling variable - 1m

v) List of apparatus and materials (04) 3M

Don’t separate between apparatus and materials

vii) Experimental Procedure or method (05) 3M

List down the complete and correct technique used based on the following

criteria:

K1 : Technique of assembling the apparatus and materials to carry out

the experiment

K2 : Technique of fixing the constant variable

K3 : Technique of changing the manipulated variable

K4 : Technique of measuring the responding variables

K5 : Technique of taking precautions to increase accuracy State precautionary

in the experiment

Scoring :

K’s Score

5K 3M ,

3-4K 2M

2K 1M

1K 0M

viii) Presentation of data (06) 3M

- Title of column and row with correct unit

(manipulated and responding variable) 1m

- List Manipulated Variable correctly 1m


19

1.5 PAPER 2 – SPM 2009

Section A

[60 marks]

Answer all questions in this section


20


21


22


23


24


25


26


27


28


29


30


31


32


33


34


35


36


37


38


39


40


41


42

1.5.2 PAPER 3

1. An experiment was carried out to investigate the effect of different duration of activity

on percentages of carbon dioxide in exhaled air.

In this experiment, the student rest for 10 minute and his exhaled air is collected to

analyse the carbon dioxide content in the air sample.

The experiment is repeated with the same students after running on the spot 1 minutes,

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.

Table 1shows the reading of sample air column for all the four activities carried out by the

student.

Activity Before using potassium hydroxide solution

Initial

reading for

all

activities

Length of air column

Table 1

10 cm


43

Table 2 shows the result of the length of the air column after being treated with potassium

hydroxide solution.

Activity After being treated with potassium hydroxide solution

Resting 0

minutes


1 minutes

running on the

sport


2 minutes

running on the

sport


3 minutes

running on the

sport


Table 2

cm

cmcm

cm

cm

cm

cm

cm

cm

cm

cm


44

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

[3 marks]

1(a)

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

Observation 1:

.……………………………………………………………………….

………………………………………………………………………..

Observation 2:

………………………………………………………………………...

………………………………………………………………………....

[3 marks]

(ii) State the inferences from the observations in 1( b) (i).

Inference from observation 1:

………………………………………………………………………….

………………………………………………………………………….

Inference from observation 2 :

………………………………………………………………………….

………………………………………………………………………….

[3 marks]

1(b )(ii)

For

examiner’s

use

1(b)(i)


45

(c) Complete Table 3 based on this experiment.

For

Examiner’s

Use

Variable Method to handle the variable

Manipulated variable

………………………………

………………………………

………………………………

………………………………………….

………………………………………….

……………………………………….....

Responding variable

………………………………

………………………………

……………………………….

………………………………………….

………………………………………….

…………………………………………..

Controlled variable

………………………………

………………………………

……………………………….

…………………………………………..

…………………………………………..

………………………………………….

Table 3

[3 marks]

1(c)

(d) State the hypothesis is for this experiment.

………………………………………………………………………………….

………………………………………………………………………………….

………………………………………………………………………………….

………………………………………………………………………………….

[3 marks]

1(d)


46

(e) (i) Construct a table and record all the data collected in this experiment.

Your table should have the following titles:

Type of activity

Initial and final readings of the lengths of air column

Percentage of final readings of the lengths of air column

Use the formula :

Percentage of =

Carbon dioxide

[3marks]

(e) (ii) Use the graph paper provided answer this part of the question.

Using the data in 1(e)(i) , draw a line graph of percentage of carbon dioxide against

the time of activity .

[3 marks]

1(e)(ii)

For

Examiner’s

Use

1(e)(i)

Initial length of

Air column

Final length of

Air column

Initial length of air column

- x 100%



47

Graph of the percentage of carbon dioxide against the time of activity

Percentage of carbon dioxide (%)

Time (min)


48

(f) Based on the graph in 1(e)(ii), explain the relationship between the time of activity and

the percentage of carbon dioxide in the air sample. .

……………………………………………………………………………..……………

…………………………………………………………………………...………….…..

…………………………………………………………………………………………..

[3 marks]

(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.

……………………………………………………………………………………..…..

……………………………………………………………………………………..…..

……………………………………………………………………………………..…..

[3 marks]

(h) State the operational definition for exhale air.

……………………………………………………………….…………………………..

…………………………………………………………….……………………………..

………………………………………………………….………………………………..

[3 marks]

1(f) 1(h)

For

Examiner’s

Use

1(g)

1(g)

1(i)


49

(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

Table 4

[3 marks]

1(i)

Total

For

Examiner’s

Use

1(g)

Beaker Boiling tube Exhaled air sample Water

Potassium hydroxide solution Rubber tube J-Tube


50

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 or method

Presentation of data

[17 marks]

END OF QUESTION PAPER


51

No Mark Scheme Sub

Mark

Total

Mark

1 (a)(i)

(ii)

(b)

(c)

(d)

Able to label P and S

Answer

P: plasma membrane

S: Rough endoplasmic reticulum

Able to explain the function of the chromosomes

Sample answers

P1: Chromosomes carry genetic information

P2: Which determines the characteristics and function of the cell/

For synthesis of enzymes/ protein

Able to explain what will happen to the production of extra

cellular enzyme if olgi apparatus and S are absent.

Sample answers

P1: The production of (extracellular) enzyme is incomplete//no

Production of enzyme

P2: (Without S) the synthesized protein cannot be transported to

Golgi apparatus

P3: (Without golgi apparatus,) the protein cannot be modified

P4: Protein cannot be sorted

P5: Enzyme cannot be packaged

P6: Enzyme/ protein cannot be transported out of plasma membrane

Any 4

Able to explain why the sperm cells contain more mitochondria

Answer

P1: Mitochondria generate energy

P2: Sperm need more energy to propel/ swim toward the uterus/

Fallopian tube

Able to explain how lysosomes eliminate damaged organelles in

the cells

Sample answers

P1: Lysosomes secretes lysozyme/ hydrolytic enzymes

P2: that digest/ hydrolysed the damaged organelles

1+1

1+1

4

1+1

1+1

2

2

4

2

2

12


52

2 (a)

(b)(i)

(ii)

(c)

(d)

Able to name the processes involved in the movement of X and Y

Answer

Process of X: Active transport

Process of Y: Diffusion/ passive transport

Able to name X and Y

Answer

X: Any examples of ions/ Calcium ion/ potassium ion/ sodium ion/

iodine

Y: Oxygen

Able to describe the movement of Y

Sample answers

P1: Y diffuses from alveolus to blood capillary

P2: From higher concentration to lower concentration

Able to explain the condition of the plant cell after being

immersed in the solution

Sample answers

P1: The cell flaccid/ plasmolysed

P2: The cell is hypotonic

P3: Water diffuse out from the cell

P4: By osmosis

P5: The plasma membrane moves away from the cell wall

Any 3

Able to state one advantage and two disadvantages of method

used

Sample answers

Advantage

A1: Last longer

A2: Less spoilage of food

Disadvantages

D1: Less nutrient content

D2: Change in colour of food

D3: Change in taste of food

1+1

1+1

1+1

3

1+2

2

2

2

3

3

12


53

3 (a)(i)

(ii)

(b)(i)

(ii)

(c)

(d)(i)

(ii)

(iii)

Able to name the cell division

Answer

Meiosis I/ Meiosis

Able to arrange the stages of cell division in the correct sequence

Answer

R, P, S, Q

Able to explain the chromosomal behavior in stage R

Sample answer

P1: pairing of homologous chromosomes

P2: Bivalent/ tetrad formed

P3: Exchanged of genetic material

(Any 2)

Able to state one importance of the chromosomal behavior

Sample answers

(causes) variation

Able to explain how zygote is formed

Sample answers

P1: Sperm is haploid and ovum is (also) haploid

P2: Sperm fused/ fertilized with ovum

P3: To form diploid zygote

(Any 2)

Able to state the number of chromosome in the offspring

Answer

45

Able to name the genetic disease suffered by the offspring

Answer

Turner’s Syndrome

Able to give reason for the disease

Answer

Lack of one X chromosome

1

1

2

1

2

1

1

1

1

1

2

1

2

1

1

1


54

(iv)

Able to explain how radioactive rays causes this genetic disease

Sample answers

P1: Radioactive rays is a mutagen

P2: Causes (chromosomal) mutation

P3: Lead to non-disjunction of X chromosome

(Any 2)

2

2

12

4 (a)

(b)

(c)

(d)

(e)(i)

Able to name the level of organization in protein structure X and

Y

Answers

Protein X: primary structure

Protein Y: secondary structure

Able to describe the structure of protein X

Sample answers

P1: consists of one polypeptide chain

P2: sequence of amino acids

Able to explain how the products are formed

Sample answers

P1: Hydrolysis take place

P2: peptide bond is broken down

P3: to form dipeptides/ peptides

(Any 2)

Able to state why animal proteins are first class protein

Sample answers

Animal protein contain all the essential amino acids

Able to explain the effects of high temperature at 65C on silk

garment

Sample answers

P1: Protein denatured

P2: Structure of protein change

P3: Shape of the protein change

1+1

1+1

2

1

3

2

2

2

1

3


55

(ii) Able to suggest two ways to maintain the quality of the silk

Sample answers

P1: Wash in cold/ warm water// do not wash in water in high

temperature

P2: Do not dry the silk under direct sunlight

P3: Dry cleaning

(Any 2)

2

2

12

5 Able to draw triceps muscle that is involved in the movement

Answer

Drawing criteria

M: Triceps muscle is thinner than biceps

T: 1 tendon at scapula, 1 at the humerus and 1 at ulna

Able to state the characteristic of P which helps in the movement

Sample answer

Inelastic// strong

Able to explain the action of the muscles which cause the

movement

Sample answers

P1: antagonistic action// triceps contract, biceps relax

P2: transmitted a force to the ulna by tendon

P3: pull ulna downward/ forearm straightens

Able to explain the health problem faced by an old person when

tissue X is impaired

Sample answers

1+1

1

3

3

2

1

3

3


56

P1: Difficulty in movement/walking

P2: Pain/ swelling/ stiffness at the joints

P3: Due to wear ant tear of X/ cartilage// X become thinner

P4: Osteoarthritis/ arthritis

(Any 3)

Able to explain why an athlete must do a warming up exercise

before starting an event

Sample answers

P1: To increase temperature of the body/ muscles

P2: Enabling more efficient use of energy// increase the production

of ATP

P3: Increase blood circulation// increase the heart beat// supply

more

Oxygen

P4: Prevent muscles injuries/ muscles cramp

(Any 3)

3

3

12

6 (a)

(i)

Able to explain why pituitary gland is a ductless gland and a

master gland

Sample answers

P1: Pituitary gland secretes hormones

P2: (Directly) into the blood stream

P3: Which transport the hormones to the target organ

P4: (Pituitary gland secretes hormones) which stimulate other

endocrine glands (to secretes their hormones)// any example:

P5: Specific examples; pituitary gland secretes FSH to ovary

causing ovary secretes oestrogen

(Any 4)

Max:

4


57

(a)

(ii)

(b)

Able to explain the role of the pituitary gland in regulating the

blood osmotic pressure when intake of water is too little

Sample answers

P1: (When intake of water is too little) blood osmotic pressure/

concentration is higher (than the normal range)/ increase

P2: Osmoreceptor/ hypothalamus detects the increase/ the change

(in the blood osmotic pressure)

P3: And produce (electrical) impulse

P4: Which stimulates pituitary gland

P5: to release / secretes (more) ADH/ antidiuretic hormone

P6: ADH increase the permeability

P7: of the collecting ducts / distal convulated tubule

P8: More water is reabsorbed (from the tubules into the blood

capillary)

P9: (Resulting in) blood osmotic pressure decrease / back to normal

P10: Urine produced is less/ more concentrated

(Any 8)

Able to explain the transmission of impulse across Q

Sample answers

P1: Q is synapse/ synaptic cleft

P2: When (electrical) impulse reaches the synaptic knob/ terminal/

terminal axon/ dendrite/ presynaptic membrane

P3: It triggers/ stimulates/ causes the (synaptic) vesicles

P4: To release neutrotransmitter

P5: Mitochondrion (in the synaptic terminal)produces energy/ ATP

P6: For active transport/ transmission of the impulse// synthesis of

Neurotransmitter

Max:

8

Max:

8


58

P7: (the neurotransmitter) diffuse across synaptic cleft/ synapse/ Q

P8: To the dendrite/ neurone R/ post synaptic membrane

P9: Transmission of information/ (across Q) is in the form of

chemicals// the transmission of information across Q involves

conversion impulse from electrical impulse to the form of

chemicals impulse

P10: Reachin R, the transmission is in the form of an electrical

impulse// reconversion of impulse in the form of chemicals

(back) to electrical impulse is formed at R

(Any 8)

20

7 (a)

(b)

Able to explain what happen to the tree after one month

Sample answers

P1: The part of the stem above the ring swells/ bigger

P2: Because the organic food substances / glucose (synthesized

during photosynthesis)/ accumulates at this part of the stem

P3: Food cannot be transported below the ring/ downwards

P4: Because the phloem is removed

(Any 4)

Able to explain how the process of transpiration can prevent

overheating in plants

Sample answers

P1: Heat from the sun/ sunlight is absorbed by the leaves

P2: (Heat from the sun causes) water on (the external surface of) the

mesophyll cells/ leaves to evaporate

P3: Thus the air space (in the mesophyll layer) is saturated with

water vapour

P4: Outside the stomata/ leaves, the air in the atmosphere is drier/

has less water (vapour) than in the leaves/ less humid// the

Max:

4

Max:

8


59

(c)

concentration of water (vapour) in the air space is higher than

the concentration of water (vapour) in the atmosphere

P5: Sunlight/ on a hot day (stimulates) the opening of stomata

P6: Water in the xylem is drawn/ absorbed into the mesophyll cells

P7: To replace the water lost during evaporation/ transpiration

P8: Thus, the plant loses heat

P9: Causes the cooling effect on the plants/ reduce the temperature

(Any 8)

Able to explain the similarities and the differences of blood

circulatory system in X and Y

Sample answers

S: Similarities

D: Differences

S1: Both have closed circulatory system

S2: Blood flows in blood vessels

S3: Both have heart

S4: Which pump blood to body cells

S5: Both have valves in veins

S6: Blood flows in one direction only

(Min 2)

Differences

X Y

D1 Double circulation//

systemic and pulmonary

circulation

Single circulation

D2 Blood flows through heart

twice

Blood flows through heart

once

D3 Heart has 2 atria and 2 Heart has 1 atrium and 1

Max:

8


60

ventricles ventricle

D4 Has a four chambered heart Has a two chambered heart

D5 Deoxygenated blood from

the heart is pumped to the

lung

Deoxygenated blood from the

heart is pumped to the gills

D6 Oxygenated blood is

pumped from the heart to

body cells

Oxygenated blood flow from

gills to body cells

D7 Oxygenated blood has

higher pressure

Oxygenated blood has lower

pressure

D8 Gases exchange occurs at

lungs/ alveolus

Gases exchange occurs at

gills/ filaments/ lamellae

(Max 6)

20

8 (a)

Able to explain why the mangrove ecosystem has to be preserved

and conserved

Sample answers

P1: Act as (shore) protection// barrier against strong coastal winds/

waves// waves breaker

P2: To prevent soil erosion/ landslide/ flash flood/ damage from

tsunami

P3: (Cable/ prop/ knee) roots (in soft and muddy soil) traps

sediments/ soil / rubbish / garbage

P4: Thus, maintaining water quality

P5: Maintaining/ increasing biodiversity/ varieties of organisms

P6: As a site of breeding/ feeding of fauna/ aquatic animals

P7: As a habitat for fauna/ animals

P8: Conserving species/ organisms/ fauna

P9: Allows longer life span (for aquatic organisms

Max:

10


61

(b)

P10: Maintaining the population of species/ organisms// maintaining

balanced ecosystem/ food web/ food chain/ dynamic ecosystem

P11: Source of raw materials for construction industry/ piling/

furniture / boats/ houses/ production of charcoal/ tannin/ source

for research

P12: As an eco-tourisms attraction

P13: minimize climate change/ drought/ hash climate/ maintaining

the temperature

P14: Prevent/ reduce green house effect/ global warming

P15: by maintaining CO2/ O2 content in the atmosphere// reduced

CO2

(Any 10)

Able to explain the effect of an increase in carbon dioxide

concentration in the atmosphere on the ecosystem

Sample answers

P1: green house effect

P2: green house gases/ CO2 trapped/ absorb more heat in the

atmosphere

P3: (The layer of green house gases/ CO2) acts as an insulator/

barrier

P4: To prevent heat (energy) from being transmitted to space// heat

is reflected back to earth// heat cannot escape to the space

P5: Causes a rise in temperature (of the atmosphere)/ leads to

global warming

P6: Melting of glaciers/ ice sheets

P7: Cause sea level to rise/ flood/ sinking of the island

P8: Cause changes in wind direction/ sea currents

P9: Cause climatic changes/ examples: drought, typhoon

P10: High (atmospheric) temperature reduces the rate of

photosynthesis

P11: Productivity of the crops/ livestock decrease

P12: Destruction/ disruption of food chains/ food webs// cause

imbalanced ecosystem

P13: Cause extinction of species// biodiversity decrease

Max:

10

20


62

9 (a) Able to explain how steamed fish is digested in part R, S and T

in the system.

Sample answer

F : steamed fish rich in protein(P)/ and fat(L)

P1: R/ Stomach secrete pepsin.

P2: Which hydrolysed protein to peptone/ polypeptide.

P3: Equation P2

P4: In acidic medium.

L1: Fat is not hydrolised.

P5: S / Duodenum received trypsin from pancrease

P6: Peptones/ polypeptides is hydrolysed (by trypsin) to peptides.

P7: Equation P7

P8: In alkaline medium

L2 : Fat is emulsified/ change to tiny droplets// small particles.

L3 : Fat is hydrolysed by lipase to fatty acid and glycerol.

L4 : Equation L3

P9 : T/ Ileum secrete erepsin/ peptidase

P10: Peptidase is hydrolysed by erepsin/ peptidase to acid amino.

P11: Equation P10

L3 : Fat is hydrolysed by lipase

L4 : Fat is hydrolysed to fatty acid and glycerol by lipase.

(Any 10)

Able to explain the long term effect of consuming excess bread,

butter and fried chicken on a person’s health.

Sample answer

Bread (C)

C1: Bread is (rich) in carbohydrate.

C2: High glucose content in blood.

B3 / C3 : Cause high blood pressure.

Max:

10

Max:

10


63

C4: Diabetes(mellitus)/ glucosuria.

C5: Excess glucose / glycogen is converted to fats.

C6: Stored in adipose tissue/ under the skin.

B7 / C7 : Obesity.

Butter (B)

B1: Butter is (rich) in fats / lipid

B2: Result in high cholesterol level in blood.

C3 / B3 : Causes high blood pressure

B4: Lipid / cholesterol deposited on/in (inner) wall of arteries.

B4A: Lumen arteries become smaller/ blood flow become slower/

blockage/ clog.

B5: Arteriosclerosis/ angina/ stroke/ heart attack/ embolism.

B6: Excess fat is stored in adipose tissue/ under the skin.

C7/ B7 : Obesity.

Protein (P)

P1: Fried chicken is (rich) in protein.

P2: Excess amino acid is converted to urea/ ammonium compound.

P3: Liver failure/ malfunction.

P4: More urea to be removed( by kidney).

P5: Kidney failure/ malfunction.

P6: Excess uric acid will cause gout/ explain gout.

(Any 10)

20

1.6. BIOLOGY 3

Mark Scheme SPM 2010

1 (a) KB0603 – Measuring Using Numbers

Score Mark Scheme


64

3

4 tick

Able to record all 4 reading for length of air column correctly.

Activity Length of air column (cm)

Resting (0 min) 9.8 / 9.9

Running on the spot ( 1 min) 9.75 / 9.8



2

3 tick

Able to list 3 readings correctly.

1

2 tick

Able to list 2 readings correctly.

0

No response or incorrect response

(b) (i) [KB0601 - Observation]

Score Mark Scheme

3

Able to state any two different observations correctly according to the criteria:

P1 : MV - Type of activity / time of activity

P2 : RV - Length of air column


65

Sample answers:

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

2. The length of air column after running on the spot for 1 min / 2 min / 3 min / is

9.8 cm / 9.7 cm / 9.6 cm

3. The length of air column during resting is highest compared to the length of air

column after running on the spot for 1 / 2 /3 min //inversely

4. The longer the time of activity, the shorter the length of air column

5. At time 0 / 1 / 2/3 minutes, the length of air column is 9.9 / 9.8 / 9.7 / 9.6 cm

2 Able to state any one observation correctly or

Able to state any two inaccurate observations

Sample answers:

1. The length of air column during rest is the highest

2. The length of air column after running on the spot for 1 min / 2 min / 3 min

decreases

3. The change in length of air column after the time 3 min / 2 min / 1 min is 0.4cm /

0.3 cm / 0.2 cm

4. At 0 minute, the length of air column is 9.8 cm

5. At time 0 min, 9.9 cm

1 Able to state observation at idea level.

Sample answers:

1. The length of air column is decreasing / 9.9cm / 9.8 cm / 9.7 cm / 9.6 cm

2. The length of air column depends on activity.

3. The time of activity increasing / 0min / 2 min / 3 min

4. The activity is running on the spot.

0 No response or wrong response.

Scoring

Score Correct Inaccurate Idea Wrong

3 2 - - -

2 1 1 - -

- 2 - -

1

1 - 1 -

- - 2 -

- 1 1 -

1 - - 1

0 - 1 - 1

- - 1 1

1 (b) (ii) [KB0604 - Making inferences]

Score Mark Scheme

Able to make two inference correctly based on the criteria Note : Inference must match observation Criteria : P1 : Infer on time / infer on type of activity

1. No / less / more activity


66

3

Any

2P’s

2. Short / long / duration / time 3. Less / more energy / respiration / exercise / running /

vigorous P2 : Infer on length – Carbon dioxide release P3 : Absorb / dissolve / diffuse by KOH solution ** Accept :P3 if P2 correct

Reject :CO2 used , breath, treated by KOH

Sample answers:

1. During resting, CO2 released is absorb by potassium hydroxide solution.

2. During resting, the least carbon dioxide is released and absorbed by potassium

hydroxide solution.

3. During running on the spot for 1/2/3 minutes the CO released is absorb by

potassium hydroxide solution.

4. During 1 / 2/ 3 minutes more / less respiration occur , CO2 is released.

5. During 1/2/3 minutes long / short duration, CO2 is released

6. The longer the duration of activity / activity, more CO2is released

2

Any

1P

Able to make one correct inference and one -two inaccurate.

Sample answers:

1. During resting, CO2 is released

2. Time / Type of activity influences the amount of CO2 released

3. During running on the spot for 1 /2 / 3 minutes CO2 is released

1 Able to state one correct inference and one-two inference at idea level.

Sample answer:

1. Exhaled air contain gas

2. Air column contain gas

3. Potassium hydroxide absorb gases


Scoring

Score Correct Inaccurate Idea Wrong

3 2 - - -

2 1 1 - -

- 2 - -

1

1 - 1 -

- - 2 -

- 1 1 -

1 - - 1

0 - 1 - 1

- - 1 1

1(c) [KB0610–Variables]


67

3

Able to state all 3 variables and the 3 methods to handle the variable correctly.

Sample Answer :

Variables Method to handle the variable correctly

Manipulated variable

Time of activity // activity //

exercise // duration of activity / time

// type of activity

Change/ varies the time of activity in 0 minute, 1

minute, 2 minutes and 3 minutes //

Use / varies different time of activity //

Running on the spot of 1 minute, 2 minutes, 3

minutes //

Running on the spot at different time

Responding variable

The length of air column //

percentage of carbon dioxide //

change in length of air column

Using meter ruler (measure) and record the length

of air column //

Calculate the percentage of CO2 by using formula:

Initial Length – Final Length

of air column of a column X 100 %


Constant variable

1. The initial length of air column

2. (Same) student

3. (Same) type of activity

4. J-tube / diameter of J-tube

5. Concentration of KOH solution.

6. Temperature

7. Time to collect air sample // air

sample

** Reject : Amount / volume of

KOH

1. Fix / used the same length of air column to be

10 cm

2. Same / fix / used one student to carry out all

activities

3. All activities were running on the spot / same

activity / fix activity

4. Same / fix used diameter of J -tube

5. Fix / same concentration of KOH solution

6. Fix at room temperature

7. Air sample collected immediately / air sample

from same student

6 ticks correctly

2

4 - 5 ticks correctly.

Reject way how to handle variable if variable is wrong.

1 Able to state 2-3 ticks correctly

0

Able to state 1 tick correctly or no response


68

1(d) [KB0611- Making Hypothesis]

Score Mark Scheme

3

Able to state a hypothesis correctly following all criteria:

P1 : Manipulated variable

P2 : Responding variable

H : Relationship

Sample answer :

1. As (the time of activity) / running on the spot increases, the length of air

column decrease // CO2 released increases / Percentage of CO2 increases

2

Able to make a hypothesis relating the manipulated variable and responding variable

inaccurately

Sample answer:

1. Activity release CO2 .

2. Activity influences affected /depends on the length of air column

1

Able to make a hypothesis at idea level

Sample answer:

1. ( CO2 )gas is released.

2. The length of air column changes

3. The longer of air column, the shorter the time of activity.

4. The length of air column is 9.9 cm

0

Not able to response or wrong response.

1(e)(i) [KB0606 – Communicating]


69

Score Mark Scheme

3

Able to construct a table and record all the data correctly

T : Able to state title with the unit correctly – 1mark

D : Able to record all the data correctly – 1 mark

C : Able to calculate and record the percentage – 1 mark

Sample answers

Type of activity / Time of activity

(min)


(cm)

Final length of air column

(cm)

Percentage of carbon

dioxide (%)

Resting for 0 minutes

( 0 minutes)

10

9.9

1

Running on the spot for 1 minute

(1 minute)

10

9.8

2


(2 minute)

10

9.7

3


(3 minute)

10

9.6

4

2 Able to record two criteria correctly

1 Able to record one criteria correctly


1 (e)(ii) [KB0612 – Relationship between space and time]


70

Score Mark Scheme

3

Able to draw the graph correctly

P (paksi) : Axes :

Uniform scales on both horizontal and vertical axes – 1 mark

T (titik ) : Points :

All to plot four points correctly - 1 mark

B ( bentuk ) : Straight line

Able to join all 4 points,( line not more than 5 small boxes from last

point)

- 1 mark

%CO2

4

3

2

1

0 1 2 3 Time (min)

2

Any two criteria correct.

1 Any one criteria correct

0

No response or wrong response.

1(f) [KB0608 – Interpreting Data]

x

x

x

x


71

Score Mark Scheme

3

R +

Any

2E’s

Able to explain the relationship between the time of activity and the percentage of carbo dioxide correctly R, E1, E2, E3 R : Directly / linearly proportional

E1 : Activity / more / less / vigorous / longer / shorter activity

E2 : More breathing / rate of breathing / rate of gases exchange

E3 : More respiration / cell respiration

E4 : More energy / ATP

Reject : R – influence , depend E - anaerobic respiration , glucose oxidise, more O2 used. Sample answer:

1. As the time increases, the percentage of CO2 released increases because

more energy is required, so respiration takes place faster

2. The time of activity is directly proportional to the percentage CO2 release,

more activity so the rate of respiration increases.

2 Able to interpret the relationship incompletely : R + 1E

Sample answer: As the time activity increases the percentage of CO2 released is high

because more activity / breathing / more respiration / more energy

1 Able to interpret the relationship at idea level

Sample answer :

As the time of activity increases, percentage release of CO2 is high.

. 0 Not able to response, inaccurate response or wrong relationship.

1(g) [KB0605 – Predicting]


72

Score Mark Scheme

3

1P +

Any

2E

Able to predict the outcome of the experiment correctly.

P : Correct prediction

E : Explanation

Accept P :

1. 1% ,

2. Less than 1% / 2% / 3% / 4% // decreases // low

3. For activity 2 min

4. No activity given.

Explanation :

E1. Breathing / gas exchange back to normal rate / stable rate / Breathing normally /

condition of student back to normal

E2. Relax / resting

E3. CO2 all ready exhaled / expel / remove / get rid of

E4. (Cell) respiration become slow / low / less energy produce / needed / respiration

normal

Reject :

1. Rate of heart beat

2. Lactic acid decrease

3. Anaerobic respiration

4. O2 / oxygen debt

5. Glucose

Sample answer:

1. Percentage of CO2 will decrease, cell respiration become slow / back to normal rate /

student rest for 10 minutes, less energy produced / needed // less CO2 produced.

*** If P wrong (X) , automatic no E

2

P +

any

1E

Any two correct

1

P only

Any one correct

0 Not able to response or wrong response.

1(h) [KB0609 –Defining by Operation]


73

Score Mark Scheme

3

3P’s

Able to define operationally exhaled air based on the result of this experiment.

P1 : Exhaled air contains CO2. / exhaled air is air column in (J tube) / air sample / air

collected in (J-tube)

P2 : CO2 is absorbed / diffuse / dissolve by potassium hydroxide (in J-tube)

P3 : The length of air column / Percentage of CO2 in J-tube is influenced / affected

/depends on /by time of activity // any correct hypothesis

** Reject treated

Sample answer:

Exhaled air contain CO2 which is absorbed by potassium hydroxide solution. The length of

air column in J-tube is influenced by time of activity.

** Reject theorilitical explanation

2

2P’s

Any two correct

1

1P

Any one correct

0

No response or incorrect response.

1(i) [KB0602 – Classifying]


74

Score Mark Scheme

3 Able to list all material and apparatus in Table 4 correctly

Sample answers:

Material Apparatus

Exhaled air sample

Potassium hydroxide

solution

Water

J – Tube

Boiling tube

Rubber tube

Beaker

7 ticks

2 Able to list 4 - 6 material and apparatus in Table 4 correctly

4 – 6 ticks

1 Able to list 2-3 material and apparatus in Table 4 correctly

2-3 ticks

0 no response or incorrect response

QUESTION 2 ( NEW FORMATE)


75

PROBLEM STATEMENT (01)

No. Mark Scheme Score

2(i)

KB061201

Able to state a problem statement relating the manipulated variable

(MV) with the responding variable (RV) correctly

3

P1 : MV- different / 2 type / species / plant or organisms //

maize AND paddy plant

P2 : RV - Height / size / length / any growth parameter

H : question form and question mark (?) 2P , H

Sample answer

1. What is the effect of competition between maize and paddy

plants on growth / dry mass / height / size ?

2. Does competition between paddy and maize affect their

growth?

3. Does competition between two different species / type of

plant / organisms affect their growth?

Able to state a problem statement inaccurately

Sample answer

1. What is the effect of competition between two different

species of organisms?

2. What is the effect of interspecific competition on growth / dry

mass / height / size ?

2

P1, P2

P1, H

P2, H

Able to state a problem statement at idea level 1

Sample answer P1 / P2

1. Competition occurs between two different species.

No response or incorrect response 0

** Only H – Reject

HYPOTHESIS (02)


76


2 (ii)

KB061202

Able to state a hypothesis relating the manipulated variable to the

responding variable correctly

3

P1 : MV

P2 : RV

H : relationship

Sample answer

1. The dry mass / height / size /growth of the maize plants is less

when it is grown with paddy plants than when it is grown

alone.

2. Maize plant grow taller than paddy plant

3. Stronger type of plant / species grow taller than weaker type

of plant / species

Able to state a hypothesis inaccurately

2

Sample answer

1. The dry mass / height / size / growth of maize plant is

different when grown with paddy plants than when grown

alone.

2. Maize and paddy plant had different height / dry mass / size

3. Maize plant has higher growth.

Able to state a hypothesis at idea level

Sample answer

1. The maize plant wins in the competition

2. Longer duration duration of maize and paddy, the higher the

height

1

No response or incorrect response 0

VARIABLES (03)


77


2 (ii) Able to state all three variables correctly

KB061203 3

Sample answer

1. Manipulated :

Type / species of plant / seedling // maize and paddy

2. Responding :

Dry mass (of plant ) / growth (of plant) / height / size //

change / different of height // rate / percentage of growth

3. Fixed :

Distance between each seedling // amount of water // number

of plant //

Quantity / type of garden soil // intensity of sunlight //

Duration // size of (seedling) tray // Same species of maize /

paddy // humidity

** Reject : initial height / same condition

Able to state any two variables correctly

2

Able to state any one variables correctly

1

No response or incorrect response

0

LIST OF APPARATUS AND MATERIALS (04)


78

2(iv)

KB061205

Type of experiment Apparatus Materials Marks

Height

Tray and Ruler

Basin / seedling box

2A

Paddy, maize, soil and

water

4M

3

Height : 2A + 4M

Dry Mass : 4A + 4M

Size : 3A + 4M

Dry Mass

Tray, Ruler, Oven

and Balance

3A

4M

Size / wet mass / height

/ fresh mass

Tray, Ruler and

Balance

3A

4M

Hydroponics

Tray and Ruler

Paddy, Maize And

Nutrient solution, /

Knob solution

(water and fertilizer) /

(cotton and nutrient).

Any 2A

Paddy and Maize

+

Any 1M

2

2A + Paddy and

Maize + 1M

Any 1A

Paddy and Maize

1

1A + Paddy and

Maize

*** Reject : Box, Beaker, flowering pot / vas

PROCEDURE ( 05)


79

2 (v)

KB061204

No Description Keywords K’s

1 There seedling tray A, B and C labeled refer to a diagram //

describe with one tray with paddy, one with maize and one

mixed.

Labeled / describe 3 trays K1

2a The seedling trays are filled with equal amount of garden soil Filled with soil K1

Equal amount / value K2

2b Initial reading any growth parameter is (recorded) / taken

Example : Measure the initial height / dry mass

Initial parameter K1

3 The 5-30 paddy / maize seedlings are planted at 5 cm intervals

as shown in the diagram / in the soil / cotton wool

5 – 30 seedling // 5 cm K2

4a The seedlings are watered with the same amount of water

every day / the seedling place under the sunlight

Water / sunlight / nutrients K1

Same / equal amount K2

4b The trays are weeded weeded K5

5 After 7-30 days, (5-100) seedlings are removed (from tray A) After …days, plants,

removed/taken

K1

7 days / more or same K2

6 The (root) of seedling plant are cleaned (under running water) clean K5

7 The paddy seedlings are heated / dried at (105oC) in an oven

NOTE: not for height / size

Heated / dried in oven

Reject : under the sun

K1

8a The dry mass paddy / maize seedling are (weighted) and

recorded using electronic balance

Dry mass / height / size

instrument and record

K3

8b The average dry mass of the paddy seedling is recorded

Average result, height

Average parameter K5

9a Step 5-8 are repeated with Tray B maize seedling Repeated / mention paddy

and maize

K4

9b Maize and paddy seeding are planted alternately Label / planted alternately K1

10 All data is recorded / tabulated in a table Labeled / planted alternately K1


80

K1 : Preparation of materials and apparatus (any 4)

K2 : Operating the constant variable (any 1)

K3 : Operating the responding variable (any 1)

K4 : Operating the manipulated variable (any 1)

K5 : Steps to increase reliability of results accurately / precaution

(any 1)

O X O X

X O X O

O X O X

X O X O

A B C


2 (v)

Able to describe all the K’s

5 K

KB061204 3

Any 3-4 K

2

Any 2K

1

No response or incorrect response / 1K

0

PRESENTATION OF DATA ( 06)

O O O O

O O O O

O O O O

O O O O

X X X X

X X X X

X X X X

X X X X


81

2 (vi)

KB061203


2 (vi)

Able to present all the data with units correctly

For any growth parameter

Sample answer:

Tray Change / increase in

height of plant & unit

A(refer diagram)

B(refer diagram)

C (Paddy)

Maize

Or

Type of plant

Change / increase in

height of plant / maize

and paddy & unit

Paddy

Maize

Or

Type of plant

Dry mass / height &

unit

Difference in dry mass /

height & unit //

Percentage of growth /

Rate growth & unit

Initial

Final

Paddy

Maize

2

KB061203

Able to present a table with a least two tittles correctly

Sample answer

Type of plant

Change / increase in

height of plant & unit

1

No response or incorrect response / 1K

0

1.7 PAPER 3 (EXERCELANCE STUDENT’S ANSWER)


82

Problem statement :

How does interspecific competition between maize and paddy plants affect

their growth? Hypothesis :

The stranger species of plants will achieve greater growth is in height. The

maize plants has a greater height than the paddy plants after the

experiment?

Variables :

Manipulated variable : Species of plant

Responding variable : Average height of plant

Fix variable : Volume of water available to the plants, distance of

plants from each other.

Materials and apparatus:

Basin, soil, water, maize, seeds, paddy seeds , meter rule.

Procedure :

1. Fill the basin with soil

2. Plant a maize seed in the soil at a depth if 1cm

3. Plant a paddy seed in the soil at a depth of 1cm at a distance of 5 cm from

the maize seed in the same row.

4. Plant another 9 maize seeds and paddy seed alternately with a distance of

5cm from each other at a depth of 1 cm.

5. Water the soil evenly and place the basin in a sunny area.

6. Water the plants every morning and evening with equally amount of water

for each plant.

7. Measure and record the average height of the maize plants and paddy

plants every week for two months using a meter rule.

8. Record all the data in a table.


Type of plant seedlings Maize Paddy

Initial height of the plant seedling(cm)

Difference in height of the plant seedling (cm)

(POTENTIAL STUDENT’S ANSWER)


83

Problem statement :

What is the effect of interspecific competition between maize and paddy

plants on their growth Hypothesis :

The interspecific competition between maize and paddy plants will cause the

plants to be shorter.

Variables :

Manipulated variable : The growth of maize and paddy on the same land or

different land

Responding variable : The height of the maize and paddy plants.

Fix variable : Fertility of soil used

Materials and apparatus:

Fertilisers, soils, seeds of maize and paddy plants, water, meter rule, 0.5m x

0.5m of land, measuring cylinder.

Procedure :

1. A seed of maize planted on a land of area 0.5m x 0.5m.

2. A seed of paddy plant is planted on a land of area 0.5m x 0.5m

3. A seed of maize plant and a seed of paddy is planted on the same land of

area 0.5m x 0.5m.

4. The plants of different land are taken care with same priority.

5. Water the plants and put fertilizers which are the same throught the

experiment and environmental condition is constant.

6. After 3 months, a ruler is used to measure the height of the plants.

7. The observation is recorded in a table.


Type of plant which planted on the land Height of

plants / m

Maize plant only

Paddy plant only

Maize plant and paddy plant


84

SECTION C ( SAMPLE STUDENT’S ANSWER) QUESTION 6

(a) (i) The pituitary gland is a ductless gland. This is because the hormones secreted by the pituitary glands are not targeted to work near the place at where it is secreted. The hormones secreted by pituitary glands are carried by the blood circulatory system to certain organs or tissue for from the pituitary gland where it works. Thus pituitary gland does not require duct as it will secrete the hormone directly into the bloodstream. The pituitary gland is known as the master gland. This is because the pituitary gland is able to control the endocrine gland. Pituitary gland secretes hormones that will stimulate other endocrine gland to produce and secretes hormones. (ii) When the intake of water is too little, the water content in the body diseases and the osmotic pressure of blood increases. This causes the body system to work in a negative feedback metabolism in order to restore the normal osmotic pressure. The change in the osmotic pressure is detected by osmoreceptors in the hypothalamus. The hypothalamus in return sends a nerves impulse to the pituitary gland. The nerve impulse sent to the pituitary gland stimulates it to produce and secrete anti-diuretic hormone (ADH) into the bloodstream. The ADH hormones will increases the permeability of the wall of the distal convoluted tubule and collecting duct towards the reabsorbption of water. Thus more water is reabsorbed into the surrounding blood capillaries, increasing the water level in the blood. (b) The dendrite of neuron R is in contact with the axon terminal of the neurone P. The junction of contact between these two neuron has a narrow gap of Q which is known as the synaps. In order, the transmission of nerve impulse occur from P to neuron R, the nerves impulse must more across the Q first. The transmission of nerve impulse in neurone are carried out by chemicals in the vessicals of the synaptic knob known as the neurotransmitter. Thus, when the nerves impulse arrives at the swelling of the axon terminal of neuron P known as the synaptic knob., the vessical in the synaptic knob releases neurotransmitter into the Q. The neurotransmitter then diffuse across Q until it touches the dendrite of neuron R. This will triggers a nerves impulse will then move towards the axon of the neuron R. This is how the transmission of a nerve impulse from neuron P to neuron R across Q occur.

EXAMPLE STUDENT’S ANSWER (PAPER 2 – SPM 2010)


85

QUESTION 7 (a) After a month, the upper part at removed part will be swollen. The growth in the

lower part is retarded. This is due to the nutrients cannot get to the lower part of

the tree. The food processed from leaves is transported to all part of tree through

the phloem. The ring of bark that has been removed contain the phloem. Therefore

the phloem is removed along the bark. The removal of phloem causes the nutrient

can’t be passed down and accumulate at the upper part of the ring. The lower part as

the phloem is removed can’t gain any nutrient and causing its growth to be retarded.

(b) During hot day, the high temperature in the environment increases the

temperature at the tree. Hence, the humidity at the environment will be lower

compare to the air space in the tree due to the high temperature. This creates a

concentration gradient at water among the tree to the environment. The water

vapours in the air space will diffuse out to the outer environment through the

stomata by osmosis down the concentration gradients. Now, when the water vapours

diffuse out to the environment, the air space become hypertonic to cell in the leaf.

Hence, water diffuse out from the surrounding cell to the air space. Then once again

the water diffuse out to the environment, hence carry away the heat in the cell. Now,

when the cell become hypertonic to the adjacent cell, the water concentration of

water to the cell that has low concentration of water. This enable the water to

transport among the cell and carry away large amount of heat as water has a large

specific heat capacity. Therefore, the transpiration prevent the trees from

overheating by removing the heat accumulate in the tree via the water vapour

through the transpiration.

( c) Blood circulatory system X and Y are both closed circulatory system. Which

mean the blood is keep in vessel throughout the circulation. Hence, X and Y has a

heart to pump the blood to all part of the body. The main blood pressure come from

pumping action of the heart. As the differences for X and Y, blood circulation system

X is closed double circulatory system, while blood circulation system Y is closed single

circulatory system. Double circulatory system is a type of blood circulation system

that let the blood passes the heart twice in a single circuit. As for single circulatory

system the blood only passes through the heart once in a single circuit. Then the

gaseous exchange in the two system is also different. The gaseous exchange in

system X occur at the lung while the gaseous exchange in system Y occur at the gill.

Apart from that the heart in system X consist of two atrium and two ventricle. The

heart in system Y only consist of one atrium and one ventricle.


86

1.8 MARKING SKILL


87


88


89

juj 2011 biologi bahan seminar guru

Documents