Mark Scheme Biology Paper 2 Ujian Pra 2007
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Transcript of 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
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4551/2 PRA TRIAL BIOLOGY Mark Scheme CONFEDENTIAL
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
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(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
© 2007 PRA TRIAL BIOLOGY SPM
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or
Plat cell
4551/2 PRA TRIAL BIOLOGY Mark Scheme CONFEDENTIAL
(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
© 2007 PRA TRIAL BIOLOGY SPM
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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
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(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
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( 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
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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
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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
© 2007 PRA TRIAL BIOLOGY SPM
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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
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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
© 2007 PRA TRIAL BIOLOGY SPM
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4551/2 PRA TRIAL BIOLOGY Mark Scheme CONFEDENTIAL
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
© 2007 PRA TRIAL BIOLOGY SPM
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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
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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
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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
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ITEM SCORING CRITERIA MARK NOTES
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
© 2007 PRA TRIAL BIOLOGY SPM
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4551/2 PRA TRIAL BIOLOGY Mark Scheme CONFEDENTIAL
ITEM SCORING CRITERIA MARK NOTES
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
© 2007 PRA TRIAL BIOLOGY SPM
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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
© 2007 PRA TRIAL BIOLOGY SPM
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