S Bio K2 Tspm09

MARKING SCHEME BIOLOGY 2 (4551/2) SPM TRIAL EXAMINATION 2009

No. 1(a)

Marking Criteria Able to name the parts labelled Q and R.

Mark

Sample answer : Q : Carrier protein R : Channel protein / pore protein 1(b)(i)

1 1

2

Able to state the component of structure P. Sample answer : It is composed of two layers of phospholipids

(ii)

1

Able to explain the main function of P. Sample answer : Acts as a barrier between the internal and external environment of the cell // Allows only specific molecules to pass through it // provide the structural basis for all cell membrane.

1(c)

1

2

1

1

Able to give the meaning of ‘semi-permeable’. Sample answer : A semi-permeable plasma membrane is a membrane that allows only certain substances to move freely across it.

1 (d)(i)

Able to state the concentration which is isotonic to blood plasma. Sample answer : 0.45 g/100 cm3

1(d)(ii)

Able to explain the answer in (d)(i).

1

SampleAnswer : Both percentage of haemolysis of red blood cells and percentage of crenation of red blood cells are zero (0%). 1(d)(iii)

1

Able to comment on the osmotic pressure at Q. SampleAnswer : F

: The osmotic pressure inside the red blood cells is equivalent to its environment. P2 : Amount of water moving in and out of the cells are the same, P3 : therefore the size and structure of the red blood cells does not change. ( F + Any P2/P3 ) 1(e)(i)

1

1

4

Able to define active transport. Sample answer : Active transport is a movement of substances / molecules / ions against the concentration gradient / from low to high concentration across the plasma membrane with the help of carrier protein and energy / ATP.

1

(ii) Able to explain what will happen to the uptake of the ions by root cells if the roots are immersed in a solution containing metabolic poisons such as cyanide. Sample answer : P1 – there is no uptake of ions by root cells P2 – metabolic poisons kill/ damaged the (root) cells P3 – no energy/ ATP is produced P4 – active transport does not occur (Any three)

3

4

TOTAL

13 marks

2(a)(i)

Able to name organ X and organ Y. Sample answer : Organ X : Ileum // small intestine Organ Y : Liver

(ii)

Able to name molecule K, molecule M and enzyme L. Sample answer : Molecule K : Starch Molecule M : Glucose Enzyme L : (Pancreatic) Amylase

(b)

1 1

1 1 1

5

Able to state two characteristics of enzyme L based on Diagram 2.1 Sample answer : 1. Enzyme remains unchanged at the end of the reaction (and can be used again). 2. Enzyme is substrate specific / reaction is very specific

(c)

1 1

2

Able to explain why molecule M is needed in muscle cells. Sample answer : Pt. 1 Molecule M / glucose is the substrate for respiration Pt. 2 As the muscle cells contract and relax, energy is needed for activities Pt. 3 therefore, molecule M is needed in muscle cells to provide energy from respiration process.

(d)

1 1 1

Able to explain the importance of forming glycogen. Sample answer : Pt.1 : Glycogen is the main reserve of carbohydrates in animals

1

3

Pt. 2 It can be converted back to glucose when energy is needed from respiration process

3(a(i))

1

2

TOTAL

12 marks

1 1

2

Able to name stage X and Y. Sample answer : X : Prophase I Y : Metaphase I

(ii)

Able to Able to state two differences between chromosomal behaviour at X and Y. Sample Answer:

1 1

1 2 (b)(i)

1 1

2

(ii)

(c)(i)

Prophase I Metaphase I (Paired homologous chromosomes) are arranged randomly. (Paired homologous chromosomes) are arranged on the metaphase plate / equatorial plane. Spindle fibre does not hold on the centromere of the chromosomes . Spindle fibre holds on the centromere of the chromosomes.

(ii)

(The homologous chromosomes paired and) crossing over take place. (The homologous chromosomes paired) crossing over does not take place.

(iii)

1 1

1 1

Able to state the occurrence at Z. Sample Answer: P1 : Four daughter cells formed P2 : Each daughter cell has two chromosomes / haploid / n

Able to state the chromosome number in each of the daughter cell in Z and able to give reason. Sample answer :

2

1 1

( Any 2 )

2

2

1 TOTAL

12 marks

P1 : 6 (chromosomes). P2 :(During meiosis) the daughter cell receives half the number of chromosome from the parent cell / 2n // Daughter cell haploid / n, parent cell diploid / 2n Able to state either cell A, cell B and cell C are genetically identical and explain. Sample answer : F : Cell A is similar to cell B but is different from cell C. P : Cell A and cell B are products of mitosis whereas cell C is a product of meiosis. Able to state the number of chromosome in Cell if Cell B undergoes an improper cell division. Sample answer : 24 (chromosomes) Able to state the syndrome of the individual. Sample answer : Down’s syndrome // Klinefelter’s syndrome 4(a)

Able to state the function of the eosin solution. Sample answer : To stain the xylem (vessels) (with red dye)

4(b)

1

1

1 1

2

Able to name the parts labelled K and M. Sample answer : K : Xylem M : Phloem

4(c)

Able to name the tissue which is responsible for transporting water and mineral ions from the roots to the upper parts of the plant. Sample answer :

Xylem 4(d)

1

1

1 2

3

Able to draw and label the observation of the root cut across. Sample answer :

Phloem

Xylem

Pericycle Cortex // ground tissue Drawing – 1 m Any 2 labels – 2 m

(e)(i)

Able to state the type of transport involved in Diagram 4.3. Sample answer : Translocation

(ii)

1

Able to explain why does the part above the ring become swollen after two weeks. Sample answer : F : The products of photosynthesis cannot be transported to the parts below the ring P : as tissue M / phloem is removed

1 1

(iii) Able to explain why have the leaves not wilted after two weeks. Sample answer : F : Water can still be transported to the leaves P : as tissue K / xylem is not removed from the stem

1 1

5

TOTAL

12 marks

5(a)

Able to complete the drawing the appropriate neurons involved in the reflex action. Sample answer :

3 neurones – 2 m 2 neurones - 1 m 5(b)

2

2

Able to explain the transmission of impulse from one neurone to another neurone. Sample answer : Pt..1 When an impulses arrives in the axon terminal Pt. 2 it stimulates (synaptic) vesicles to move towards and bind with the presynaptic membrane Pt. 3 The vesicles fuse / release the neurotransmitter into the synapse Pt. 4 The neurotransmitter molecules across the synapse to the dendrite of another neurone Pt. 5 Stimulated to trigger a new impulses which travels along the neurone ( Max 4 )

5(c)

1 1 1 1 1 4

Able to name the structure M and N. Sample answer : M : Sensory reseptor // finger tip N : Effector // muscles tissues

1 1

2

5(d)

Able to differentiate the reflex action with the voluntary action. Sample answer : The reflex action is governed by the spinal chord whereas the voluntary action is governed by the cerebrum.

5(e)

1

1

1

1

Able to state the importance of reflex action to us. Sample answer : To protect the body against injuries

5 (f)

Able to predict the effect on O if it is injured or damaged. Sample answer: 1. The nerve impuls will be sent from afferent neurone to the effector 2. The effector / muscles will not contract 3. The hand will not be removed immediately from the needle. (Any one )

1 1 1

1

TOTAL

11 marks

6(a)

Able to relate the tissues involved in producing the running movement Sample Answer: Pl- Tendons, ligaments, bones, muscles and joints are important features in a movement, P2- Tendons connect muscles to bones P3- Tendons are strong and non elastic P4- Force is transferred to bones through tendons. P5- Movement at the joint is possible with the aid of ligaments. P6- Ligaments connect two bones together P7-to give support and strength to the joint. P8- Ligaments are strong and elastic. P9- The quadriceps / extensor muscles contract while the biceps femoris muscles relax and the leg is straightened. P10- The biceps femoris muscles contract while the quadriceps / extensor muscles relax and the leg is bent. P11- Calf muscles contract to lift up the heels. P12-Feet push downward and backward P13-Repeated contraction and relaxation of muscles result in the running movement. MAXIMUM: 10 marks

1 1 1 1 1 1 1 1 1 1 1 1 10

(b) Able to give example and explain how the support system in woody plants differs from that of non-woody plants. Examples – 2 marks , Facts – 8 marks Sample answers: Non-woody plants (herbaceous plants) Example: Balsam plant/ any suitable answer P1: (Support in herbaceous plants is) provided by the turgidity of the parenchyma / collenchyma cells

1 1 1

P2: (When there is enough warm in the ground). the cells take in water by osmosis and become turgid. P3: The turgor pressure of the fluids in the vacuoles pushes the cell contents / plasma membrane against the cell wall P4: creating support for it stem/ roots /leaves P5: The thin thickening die cell walls with cellulose / collenchyma cells gives support to herbacous plants Woody plants : Example : Rambutan tree/ hibiscus/ any suitable example

1 1 1 1 1 1

P6: Woody plants have specialised tissues/ sclerenchyma tissues/ xylem vessels / tracheids. to give them support; P7: These tissues have cellulose walls which have deposits of lignin for added strength. P8: Sclerenchyma cells have very thick walls (which do not allow water to pass through). P9: (These cells are dead cells and) their function is to provide support for the plant. P10: Xylem vessels have thick walls of lignin which are deposited during the plant's secondary growth. P11: The lignified xylem vessels form the woody tissues of the stem. P12: This makes the plant stronger and also provides support for the plant. P13: Tracheids are also dead cells with thick walls and very small diameters. P14: They are found with the xylem vessels and together they support the plants.

1 1 1 1 1 1 1

10

MAXIMUM: 10 marks

20 marks

TOTAL 7(a)

(i) Able to explain how the transport of oxygen and carbon dioxide takes place in the body cells Sample answers: P1: The blood circulatory system transport oxygen from the alveoli to the body cells. P2: Oxygen combines with the haemoglobin in the red blood cells P3: to form oxyhaemoglobin (which is unstable.) P4: Oxygen is carried (in form of oxyhaemoglobin) to the tissues (which have a low partial pressure of oxygen.) P5: The (unstable) oxyhaemoglobin breaks down into oxygen and haemoglobin again. P6: Oxygen (molecules are) transferred to the body cells P7: Carbon dioxide binds (itself) to the haemoglobin P8: (and is) transported in the form of carbaminohaemoglobin. P9: Carbon dioxide is (also) transported as dissolved carbon dioxide (in the blood plasma.) P10: Most of carbon dioxide is carried as bicarbonate ions (dissolved in the blood plasma.) P11: When the blood carrying carbon dioxide reaches the body cells, the carbon dioxide diffuses into the blood plasma and combines with the red blood cells. P12:Carbon dioxide reacts with water to form carbonic acid. P13:Carbonic anhydrase in the red blood cells catalyse the formation of carbonic acid. P14: The carbonic acid then dissociates into a hydrogen ions and bicarbonate ions. MAXIMUM: 6 marks (ii) Able to describe the adaptations of the alveolus for gaseous exchange Sample answer: F1: The millions of alveoli P1: provide a large surface area for gaseous exchange. F2: The walls of the alveoli are moist P2: and this allows respiratory gases to dissolve easily to them. F3: The walls of the alveoli are very thin (one-cell thick) P3: forquick / easy diffusion of gases. F4: The alveoli are richly supplied with blood capillaries

1 1 1 1 1 1 1 1 1 1 1 1 1 1 6

P4: to increase the rate of diffusion / the rate of the transportation of gases MAXIMUM: 4 marks (b)

4

Able to explain how an oxygen debt is built up when an athlete is running and how it is settled after he stops running. Sample answer: P1: During a vigorous exercise /running, the breathing rate is increased. P2: This is to supply more oxygen (quickly to the muscles) P3:for rapid muscular contraction). P4: However, the supply of oxygen to muscles is still insufficient P5: and the muscles have to carry out anaerobic respiration (to release energy). P6: The glucose is converted into lactic acid, P7: with only a limited amount of energy being produced P8: An oxygen debt builds up in the body as shown in the graph P9: High levels of lactic acid in the muscles P10: cause them to ache. P11: After running, the athlete breathes more rapidly / deeply than normal for 20 minutes (shown in the graph) P12: There is a recovery period (from the 10th minute until the 20th minute) P13:when oxygen is paid back (during aerobic respiration) P14: About 1/6 lactic acid is oxidised to carbon dioxide, water and energy.

10

MAXIMUM: 10 marks

20 marks

TOTAL

8(a)

Able to explain how the inheritance happen Answer : P1: The situation involved is monohybrid inheritance. P2: The genotype of blood group A can be IAIA /1A10 P3: while the genotype of blood group B can be I BI B or IBIO. P4: Blood group 0 has a genotype, I OI O (while the genotype of blood group AB is I AI B. P5: Alleles 1A and IB are codominant P6: IO allele is recessive. P7: Mr. Nick is heterozygous dominant/IAIO (for his blood group A) P8: while his wife is heterozygous dominant/ IBI0 (for blood group B) P9: Mr. Nick and his wife produce haploid gametes/sperm/ovum (as a r e s u l t o f m e i o s i s ) P10: Mr. Nick produces (gametes with) genotypes IA /IO P11: (while) his wife (will) produce (gametes with) genotypes 1A/ lO P12: The gamete (IO) of Mr. Nick fuses with his wife's gamete (10) P13: to produce a zygote with genotype I°Io. P14: (Thus, they will) produce an offspring with blood group 0.

1 1 1 1 1 1 1 1 1 1 1 1 1 1

10

MAXIMUM: 10 marks

(b)

(i) Able to explain how DNA fingerprinting is carried out. Answer:

1 1 1

P1: Tissue samples are taken from the scene of a crime and DNA is extracted. P2: An enzyme breaks down the DNA into fragments. P3: The DNA fragments are classified according to size.

1 1

P4: An alkali is added to separate the double-stranded D N A into single strands. P 5 : Each single strand is laid on a nylon membrane and

1

radioactive matter is added to it. A banding pattern appears. P6: An X-ray film is produced and the positions of black bands are compared with the part of DNA treated with radioactive matter.

4

MAXIMUM: 4 marks (ii) Able to state the advantages and disadvantages of DNA fingerprinting Sample answer:

1

Advantages:

1

P1: DNA fingerprinting is more accurate than common fingerprinting as no two people have the same DNA fingerprints.

1

P2: DNA fingerprinting is more efficient than blood-type identification because many people have the s a m e

1

blood type

1

P3: DNA fingerprinting requires only a small amount of DNA to 1 obtain a highly accurate result P4: DNA samples last longer than fingerprints. P5: Mixed DNA samples can still be used.

1

P6: DNA evidence is harder to clean up compared to 1

fingerprints.

6

Disadvantages: P7: DNA samples may be degraded by adding chemicals, and this will affect the accuracy of the technique. P8: Human errors are possible when different procedures and standards are used in DNA fingerprinting. MAXIMUM: 6 marks TOTAL

20 marks

9(a)

(i) Able to explain why most plants cannot colonise and grow in the swamps. Sample answer: P1: The ground is too soft and unable to support plants, P2: The water-logged / muddy swamps provide very little oxygen for root respiration. P3: The swamp water has a high concentration of salt and is hypertonic. P4: The plants growing in swamp will have the problem of dehydration. P5: Seeds that fall into the muddy swamp will die of dehydration / insufficiency of oxygen. P6: The swamp is exposed to strong sunlight and intense heat. P7: As a result, the plants growing there will lose water very fast by transpiration.

1 1 1 1 1 1 1 5

MAXIMUM: 5 marks (ii) Able to explain how the mangrove trees adapt themselves to the harsh living conditions

1 1

Sample answer: P1: Root system which is highly branched and spreads over a big area to give good support to the plants. P2: Pneumatophores (breathing roots) which grow protruding upwards above the ground. P3: The plant cells have high concentration of cell sap. P4: Hence, the cells are able to withstand the high salt content of the swamp. P5: Excess salt is eliminated through hydatodes found at the lower epidermis of leaves. P6: Viviparous seeds which germinate while still attached to the parent plant. P7: The long radical produced will let the seedling stick into the ground and not submerge or drift away. P8:Thick cuticle and sunken stomata which help to reduce the rate of transpiration.

1 1

5

MAXIMUM: 5 marks (b) Able to describe the effects of unplanned development and improper management of the ecosystem. P1: The leave canopy in the forest protects the soil from excess rain water. P2: When the forest is cleared, the soil is exposed to rain (water) / wind. P3: this will cause soil erosion P4:The soil that is exposed to wind will be blown to another area, P5: while soil that is exposed to rain water will be eroded and deposited at the bottom of the river / pond /lake. P6:The soil at the hill slopes can (also) be washed away by heavy rain water P7: resulting in land slides. P8: (The deposited soil will) cause the water level to increase rapidly when it rains and P9: this will in turn cause flash floods. P10:Wild life species will also be threatened P11: when their habitat is destroyed. P12: Global warming will occur P13:due to an increase in the Earth's temperature, P14:which is caused by excess emissions of carbon dioxide/ methane/ CFC /nitrogen dioxide (into the atmosphere). P15:These gases trap the heat that is reflected by the Earth. P16:The thinning of the ozone layer occurs P17: when the ozone layer (that protects the Earth from ultraviolet radiation) is destroyed by chlorofluorocarbons (CFC). MAXIMUM: 10 marks

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

10

TOTAL

20 marks

20 marks