Mark Scheme Biology Paper 2 Ujian Pra 2007

4551/2 PRA TRIAL BIOLOGY Mark Scheme CONFEDENTIAL

JABATAN PELAJARAN NEGERI MELAKA

MARK SCHEME BIOLOGY PAPER 2

PRA TRIAL SPM EXAMINATION 2007

SECTION A

QUESTION 1

Item Num.

Answer Marks Remark

1(a) (i)

(ii)

P : XylemQ : Phloem

Organ ;made up of ground tissue, epidermal tissue, mesophyll tissue and vascular tissue // consists of various types of tissues combined together to perform spesific functions.

11

11

2

2

(b) (i)

(ii)

Q / Phloem tissue composed of sieved tubes

with the end walls of each cell are perforated by pores to form sieves plates

which allow substances to pass from one cell to another.

Translocation

1

1

1

1

Max2

1

(c) (i)

R : Companion cell S : Sieve tube cell

11 2

© 2007 PRA TRIAL BIOLOGY SPM

1


Question Answer Marks Remarks

(ii) The plant will be dye; without R / companion cell ; no

energy will be provided to the sieve tube;

hence dissolve organic substances/sucrose/ cannot be transported (from leaves to the storage organ/other part of plant)

11

1 Max 2

Any 2

(d) The branch will be die; owing to a lack of organic substances

in the parts below the ring.

11 2

Reject :plant

Marks 13

QUESTION 2

Item Num.

Answer Marks Remark

2(a)(i) Interphase 1 1

(a)(ii) P1 Synthesise of protein and new cytoplasmic organellesP2 DNA replication

P3 Accumulates energy

1

1

1

Max 2

(b) Q Chromatid.

R Spindle fibers11

Max 2

( c )(i)

(c) (ii)

Stage M

Anaphase

1

1 2 marks


2


(d)1

1

D- diagramL- label plat cell

2

All 2 correct

– 2 m

1 correct – 1 m

(e) 4 1 1

(f)) Cloning / culture tissue Easier to get infected // no variation

1

1 2

12 marks

(f)10/100 X 20000 = 2000 X 10/100 = 200 kJ

11 2

12 marks

QUESTION 3

Item Num.

Answer Marks Remark

3(a)Able to mark and labeled the producer correctly

Answers: Hydrilla sp or lotus plants

Able to mark and labeled the consumer correctly

Answers: Small fish/big fish/duck/tadpole/ black beetle

1

1 2

(b)Able to state the niche of duck

Answers: Duck eat the lotus plant/ duck eat

Hydrilla sp/ Duck eat prawn

1 1


3

or

Plat cell


(c) Able to explain the role of lotus plant

Sample answers: Lotus plants is a floating plant that

able to receive sunlight to conduct photosynthesis process

They can produce rapidly by vegetative propagation

And the submerged plant die & being decomposed

1

1

1

Max 2

( d )Able to explain the benefits of this biogradable herbacides

Sample answers: The biodegradable pesticides will

only kill producer So they cannot produce rapidly by

vegetative propagation And decoposition process will stop

1

1

1

max2 marks

(e) Able to explain how the excess fertilizer pollute the freshwater pond

Sample answers:Draining of excess nutrient from the fertilizer into the pond

Encourage the rapid growth of algae The excessive growth of algae

restricts the penetration of light into the water

And when they die,decomposer used up all the dissolved oxygen/supply of oxygen in the pond decrease

1

1

1

1

Max

3

(f)Answer

10/100 X 20000 = 2000 X 10/100 = 200 kJ

11 2

12 marks


4


QUESTION 4

Item Num.

Answer Marks Remark

4(a) (i)Able to name the hormone labeled X and Y

Answers X- FSH

Y- LH

1

1 2

4(a)(ii)Able to draw the sctructure of P

Answer

1 1

4(b) Able to explain the process of cell division which produce gamete Q can result in the variation.

Sample answer Crossing over during prophase 1 Recombination of allele during

prophase 1 Behaviour of the choromosomes

during anaphase 1

1

1

1Max 2

( c )Able to convince the woman about the treatment

Sample answer The replacement hormone stimulates

ovulation Causing the graafian follicle to form

corpus luteum corpus luteum secretes progesterone

to maintain the thickness of the endometrium

this enables implantation to occur

1

1

1

1

max3 marks


5


(d) Able to explain an imbalance of hormone X & oestrogen to the disrubtion of the developing folicle

Sample answer high oestrogen level stimulate secretion of LH and it will inhibits the secretion of FSH

and stop the growth of follicle in the ovary

1

1

1

Max

3

(e) Able to explain the importance of the pituitary gland in the developing follicle process

Sample answer the pituitary gland secretes FSH to

stimulate the growth of follicle in the ovary

and follicle able to secrete oestrogen to promotes repair of endometrium

1

1

2

12 marks

QUESTION 5

Item Num.

Answer Marks Remark

5(a)Able to name the type of variation

Answers Figure 5 (i) : Continuos variation Figure 5 (ii) : Discontinuos variation

1

1 2

(b)Able to state the factorAnswerGenetic factors

1 1

( c )(i)

Able to name the disease

Answer THALASSEMIA

1 1 mark


6


( c )(ii)Able to state two characteristics of the above disease

Sample answer

The erythrocytes are thin & delicate Absent of one of the globin chains

11

2

(d)Able to describe the differences between the two type of variationSample answer

Body mass Blood group1. Continuous variation2. the differences are not distinctive3.show a normal distribution4. influenced by environmental factors

1.Discontinuos variation2.the differences are distinctive3.show a discrete distribution4.is not influenced by environmental factors

1

1

1

Max 3

(e) Able to explain what would happen

The blood supply in the hospital will decrease

And many patients suffer or died

1

1

Total 11marks


7


SECTION B

QUESTION 6

Item No.

Scoring Criteria Mark Remark

6(a) Able to explain the sequence of changes in the rate of heartbeat due to vigorous exercise.

Sample answer :

F1- Increased respiration produces more carbon dioxide P1- this lead to a decrease in pH

F2- receptors in aorta and carotid sinusesP2.1- detect the fall in blood pHP2.2-(this chemoreceptor) send great /more impulses P2.3- to cardiac/cardiovascular centre/ accelerator centre/ P

F3-(the cardiac centre) sends out more impulses / increased rate of impulses // no impulses from inhibitory centreP3- along the sympathetic nervesP4- to SA node , the AV node and the ventricle wall / to pacemaker // pacemaker stimulatedP5- causing the rate of the heart beat to increase/ increased rate of heartbeat P6- adrenal gland releases more adrenalin (in case of stress)

1

1

111

1

1

11

1

1 Max 10


8


6 (b) (i) Able to describe how can count pulse rate to represent the heart beat

Answer:

feel the pulse or take the pulse or find the pulse

count the pulse or count beats in artery in wrist / neck

Able to count X (ventricle contraction) 80 (answer) showing 8000 divided by 100

(indicating cardiac output divided by stroke volume)

1

1

1

1 4

(ii) Able to analyse the table 1

Sample answer

P1 – when there is change in movement, rate of contraction of left ventricle in beats per min increase / ventricle contraction heart beat faster.

P2 - this cause cardiac output in cm3 per minute also increase

P3 - pumps more blood / pumps blood faster

P4 - so more oxygenated blood can be sent to muscles.

P5 - this increase respiration cell

P6 - to generate more energy/ATP for muscle contraction

1

1

1

1

1

1 6

Total 20

QUESTION 7


9


Question Answer Marks Notes7(a)(i) Able to explain the meaning of negative

feeback mechanismSample answerE1: When the value of glucose in blood increase, E2: the corrective mechanism comes into play to reduce the value the glucose in blood to normal again

1

1

2

7(a)(ii) Able to describe how hormone X and Y regulated the blood glucose level in human

Sample answerE1: The hormones X and hormone Y are produce by the cell in the pancreas E2: Hormone X is secreted by alpha cells and hormone Y is secreted by beta cellsE3: If the blood sugar level is lover than normal, more hormone X is secreted into the blood stream.E4: Hormone X is transport by blood to the liver E5: In the liver, hormone X stimulates liver cells to convert glycogen to glucose.E6: This causes the level of glucose to rise and return to normal.E7: If the blood sugar level is higher than normal, more hormon Y is secreted into the blood stream.E8: Hormone Y is transported by blood to the liver.E9: In the liver, hormone Y stimulated the liver cells to convert glucose to glycogen and fats.E10: This cause the level of sugar fall and return to normal.

1

1

1

1

1

1

1

1

1

1max 8 8

Marks 10


10


Question Answer Marks Notes

7(b)(i)

(b)(ii)

Able to explain the meaning of homeostasis

Sample answerHomeostasis is the endocrine system and the nervous system works together to maintain optimal physical and chemical conditions in the internal environment for the cell to function optimally.

Able to state the differences between the endocrine system and the nervous system

Sample answer

Refer to Appendix A.

1

8

1

Max 8

Marks 10


11


QUESTION 8

Question Answer Marks Notes8a Able to explain the type of cross breed

that involve two pairs of characteristics-this type of cross breed is the result of breeding between two opposite pairs of characteristics

Able to state the type of inheritance-known as dyhibrid inheritance and follows Mendel’s Second Law

-G represents the allele for green colour g represents the allele for striped colour L represents the allele for short shape l represents the allele for long shape

Able to show the schematic diagram of how to get the second filial generation phenotype and the ratio. Parental GGLL x ggll genotype Parental (short green) (long striped) Phenotype

Gametes GL gl

First filial generation GgLlF1 (all short green)

F1 Self cross GgLl x GgLl

GL Gl gL gl GL Gl gL gl

Punnett square is prepared to determine the phenotypic ratio in F2 generation.

1

1

1

1

1

1

1


12


Question Answer Marks Remark

GL Gl gL gl

GL GGLL GGLl GgLL GgLl

Gl GGLl GGll GgLl Ggll

gL GgLL GgLl ggLL ggLl

gl GgLl Ggll ggLl ggll

F2 generation Round : Long : Short : LongPhenotype: Green Green Striped Striped

F2 generation 9 : 3 : 3 : 1Phenotypic Ratio:

2

1

1

12 m

Max: 10m

b. Able to explain how the inheritance can be prevented based on the schematic diagram given.

Sample Answer

1. Colour blindness are sex-linked inheritance disease and is carried by recessive gene.

2. Males are homozygous, receiving only X chromosome from their mother.

3. Females are heterozygous, inheriting X chromosomes from both parents.

4. A normal father and a heterozygous carrier mother pass the gene for colour blindness on to possibly one-half of their children.

5. Half the daughters will be carriers

6. Half the sons will be colour blindness.

1

1

1

1

1

1

1


13


Question Answer Marks Remark

7. The rest of the siblings will be normal.

8. Daughters, as long as one parent is genotypically normal, can only be carriers.

9. The normal gene on the second X chromosome counteracts the defect and the daughters do not suffer from trait.

10. When a son receives the defective gene from his mother he will be colour blindness because the Y chromosome cannot counteract the defective gene located on his X chromosome.

11. Thus, colour blindness is more common in males than females.

12. Inheritance of colour blindness can be prevented by avoid marriage of the grandchildren which consists of carriers or colour blindness through a few generation

13. So that the recessive gene will disappear.

1

1

1

1

1

1

113 m

Max: 10m


14


SECTION C

QUESTION 9

ITEM SCORING CRITERIA MARK NOTES

9 (a)

(b)

Able to explain process of ultrafiltration (Y process)

Sample Answer

Process Y = ultrafiltration

A process whereby F1 – water and solutes from glomerulus being forced to filter through the membrane of Bowman’s capsule by the high hydrostatic pressure F2 - forming glomerular filtrate Able to explain the formation of urine F1 - Able to state three processes in urine formation E1 - Ultrafiltration, reabsorption and secretion.

F2 - Able to explain the ultrafiltration process

P1 - Blood is under relatively high pressure when it reaches the nephron.P2 The high blood pressure in the glomerulus, forces fluid to filter through the filtration membrane into the lumen of Bowman’s capsuleP3 - forming glomerular filtrate;P4 - contains water, glucose, amino acids, urea, mineral salts and other small molecules F3 - Able to explain the reabsorption process

P5 - The glomerular filtrate will flow into proximal convoluted tubuleP6- selective reabsorption occurs; all the glucose, amino acids, vitamins and many inorganic ions are reabsorbed back into the blood P7- by active and passive transportP8- forming a relatively high solute concentration in the peritubular capillaries P9 - thus a large volume of water is reabsorbed into the blood by osmosis andP10- increase the concentration of urea in the convoluted tubule

1

1

1

1

1

1

11

1

1

11

1

1

Max 2

Max 3

Any 3

Any 3


15



P10- glomerular filtrate then flow into loop of henle and distal convoluted tubule P11- more water and minerals being reabsorbed back into the blood

F4 - Able to explain the secretion process

P12 -takes place in the distal convoluted tubuleP13 -urea/toxins/certain drugs / hydrogen ions/potassium ions/ammonia being secreted by passive diffusion and active transport from the blood capillary into the distal convoluted tubuleP14- the filtrate reaches the collecting duct ; now called urine P15 -flows down the ureter, the bladder and the urethra and is finally excreted.

1

1

11

1

1

TOTAL

Max 3

Max 2

____10

____

Any 2


16



9 (c) Able to evaluate the suitability of potato rings as the main source of food for a child. Sample answer : F1 : Potato rings contains very little protein compared to its carbohydrate & fat content. E1 : children needs more protein to build new tissues during the period of rapid growth. F2 : do not contain vitamins & minerals E2 : the child will fall sick easily due to lack of vitamins & minerals for optimum health / prevention of diseases. F 3 : contains very high fat E 3 : hence, the child will also be obese F 4 : contains saturated fat E 4 : have high levels of cholesterol in their blood; face a higher risk of developing cardiovascular disease F5 : does not contain fibre E5 : the child will get constipation lead to painful defaecation

1

1

11

1111

11

TOTAL

Max10

10


17


APPENDIX A

QUESTION 7 (b)(ii)

Criteria Nervous System Endocrine System Mark

F1 : Organ / tissue involve

E1 : Sensory organ E1 : Gland F1 -1E1 -1

F2 : Transmission of information

E2 : Electrical signals/ Nerve impulse

E2 : Chemical signals/ Hormone

F2 -1E2 -1

F3: Medium of electrochemical transmission

E3 : Neurone E3 :Blood F3 -1E3 -1

F4: Response E4 : Muscle contraction

E4 : Production of hormone

F4 -1E4 -1

F5: Period of message conveyed

E5 : Rapid/fast E5 : Slowly F5 -1E5 -1

F6: Destination of Message

E6 : Specific location E6 : Various destination

F6 -1E6 -1


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Documents

Mark Scheme Biology Paper 2 Ujian Pra 2007