Paper 2 Answer

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SULIT 4551/2 Biologi 2007

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PEPERIKSAAN PERTENGAHAN TAHUN TAHUN 2007

BIOLOGY PERATURAN PERMARKAHAN KERTAS 2

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Kertas soalan ini mengandungi 13 halaman bercetak.

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1 Diagram 1 shows the structure of a pancreas cell as seen under an electron microscope.

P

Q: Rough endoplasmic reticulum

R: Golgi apparatus

S : Secretory vesicle

DIAGRAM 1 (a )(i) Name the type of the cell. Animal cell [1 mark] (ii) State one reason for your answer in a (i). Because it does not has a cell wall // has centriole [1 mark] (b) (i) State the function of P Carry / store genetic material / genetic information

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[1 mark]

(ii)

State the material component of P ? DNA / deoxyribonucleic acid [1mark]

(iii)

Diagram 2 shows part of the structure of a component P

X

Y Z

DIAGRAM 2 Name the parts labelled X, Y and Z. X: Y:

Phosphate group Pentose (five carbon ) sugar Z:

(c)

Nitrogenous base

[3 marks]

On Diagram 1, label the structures Q, R and S. [3 marks]

(d)

Q and R are involved in the synthesis of external enzymes. Explain the functional relationship between Q and R. - Q contains ribosome which synthesis protein. - Proteins are transported to the R by transport vesicles which buds off from Q. - In R, the proteins are modified ( to form an enzymes ) and repacked into secretory vesicle - buds off from R to fuse with plasma membrane to be excreted. [ 3 marks ]

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Able to label the structure a) P: carrier protein Q: pore R: cholestrol S: phospolipid b) hypertonic c)(i) plasmolysis (ii) P1: there is a different concentration between inside and outside cell P2: water molecules in the cell diffuse out by osmosis P3: the cytoplasm contract and the plasma membrane moves away from the cell wall d)(i)

(ii) deplasmolysis e)

F : erythrocytes undergo haemolysis P1: water molecules diffuse into the erythrocytes P2: cell will expend and burst because it does not have cell wall

f)

F: vegetables will die P1: excess fertiliser cause water to diffuse out from the hair root cell by osmosis P2: plant cells undergo plasmolysis // become flaccid and plant will die

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3 a (i) Answer : Cell Y

a (ii) Able to give the reasons to the answer in a(i) Suggested answer P1 : There is synapsis between the homologous chromosomes P2 : Each pair of homologous chromosomes (bivalent) consists of (tetrad)

four chromatids

b (i) Able to draw the anaphase stage for cell X Answer :

b (ii) Able to state one difference between anaphase in cell X and anaphase I in cell Y. Anaphase in cell X separation of the sister chromatids

Anaphase I in cell Y separation of the homologous chromosomes (bivalent) c (i) Able to name a process that occurs at the part labeled T in cell Y. Answer : Crossing over c (ii) Able to explain the role of crossing over process in producing variation in organisms Suggested answer: P1 : Crossing over involves the exchange of DNA segments between non-sister chromatids P2 : Crossing over results in a new combination of genes on a chromosome d (i) Able to state the number of chromosomes in the daughter cells for each cell X and cell Y. Answer : Cell X – 4 chromosomes

Cell Y – 2 chromosomes Both must correct

d (ii) Able to explain the answer in d(i) Answer : Cell X – duplication once, dividing once Cell Y – duplication once, dividing twice 4 A series of experiment in Diagrams 4.1 and 4.2 were conducted to study the effect of the 4551 /2 [Lihat sebelah SULIT

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tip on the growth of corn coleoptile. In the dark The tip is removed

coleoptile

After 7 days

DIAGRAM 4.1

In the dark The tip is removed and replaced coleoptile

After 7 days

DIAGRAM 4.2 Notes : Diagram 4.1 – The coleoptile / tip should not exceed the dotted line @ shows no elongation. Diagram 4.2 – The coleoptile / tip must exceed the dotted line @ elongation occurs / straight upward.

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(a) (i) On the Diagram 4.1 and 4.2, draw your observation in the space given. [ 2 marks ] (ii) Give the reason for the answer in (a) (i).

o

4.1 = The tip produce / contains plant hormone / auxin been cut

o 4.2 = Auxin diffuses / moves downward and Auxin stimulates telongation of cells (in zone of elongation) [ 2 marks ] Black box

light

The tip is removed and replaced After a few days Coleoptile

DIAGRAM 4.3 (b)The result in Diagram 4.3 shows that the coleoptile bends towards light. Explain the result. o

Auxin moves away from the light side // auxin accumulates on the shaded side

o

Cells on the shaded side elongate more compare to light side.

o Hence, the coleoptile grows (and bends) toward light. [ 3 marks ] (c) (i) Name a plant hormone that can be found in the shoot tip? Auxin / IAA (ii) What is the effect of plant hormone in c (i) on the growth of plant? Stimulate / promote the cells elongation. [ 2 marks ]

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(d) Plant hormones are used extensively in agriculture to modify plant growth and development. (i)

What is the function of the hormone in culture tissue? To stimulate cells division / mitosis / cell differentiation in callus [ 1 mark ]

(ii)

Explain the use of hormone in parthenocarpic fruit development.

o Auxin is applied / sprayed to the unfertilized flowers o

Ovary develops to become fruit without fertilisation

o The ovary wall develops into a seedless fruit. [ 2 marks ] 5

Two individuals P and Q were injected to acquire immunity. The level of antibody in the blood of individual P and Q is shown in Diagram 5.1 and 5.2. (a) What is the substances injected into the blood of individual P and individual Q ? P : Dead or weakened bacteria / viruses / antigens// vaccine Q : Serum containing antibodies // antiserum [ 2 marks ] (b) Explain the type of immunity obtained by individual P and individual Q. P : Artificial active immunity The body produces its own antibodies to against infections by pathogens. Q : Artificial passive immunity The body receives antibodies to against infections by pathogens. [ 4 marks ]

A boy was bitten by a snake.

He was

unconscious and now he is in a hospital. (c) Using biology knowledge, describe how you could save that boy.

o Injection of serum which contain instant / ready antibodies / antiserum / anti-toxin must be given to the patient. 4551 /2

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11 o

Quantity of antibody higher than immunity level to ensure the body get the artificial passive immunity

o

Antitoxin reacts with toxin / snake venom and neutralize it / he saved. [ 2 marks ]

Table 5 shows a schedule of immunisation given for every new born until 2 year old in Malaysia. Age New born

Types of Immunity Tuberculosis (B.C.G) Hepatitis B

( First dose )

1 month

Hepatitis B

( Second dose )

3 month

Triple Antigen Polio

( First dose )

4 month

Triple Antigen Polio

( Second dose )

5 month

Triple Antigen Polio

( Second dose )

Hepatitis B

( Third dose )

9 – 24 month

Germans measles

1 ½ - 2 year

Triple Antigen Polio

( Third dose ) TABLE 5

(d) Based on Table 5, explain why every parent must be obey that schedule to ensure that their baby are safe from certain diseases.

o Immunisation is given to avoid the diseases like Tuberculosis, Hepaptitis B, Polio, diphtheria, whooping cough, tetanus. German measles ( state at least 3 example ) o

New born get Artificial Active Immunity

o

First dose are given to promote baby lymphocyte to produce antibody which specific to antigens / bacteria / virus

o 2nd and 3rd dose are booster dose to produce more and fast production of antibodies 4551 /2

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12 o

To achieve the immunity level // antibody remained in the blood for a long time and protect them for the next infection. Any 3 [ 3 marks ]

Section B [ 40 marks ] Answer any two question. The time suggested to complete this section is 30 minutes 6

Able to list the general characteristics of enzymes Sample answer P1 – Enzymes are proteins which are synthesized by living organisms. P2 – Enzymes bind to their substrates and convert them to product in the enzymatic reaction P3 – Enzymes have specific sites called active sites to bind to specific substrates // enzymes are highly specific in their reaction // each enzyme can only catalyse one kind of substrate / specific substrate (lipase can hydrolyse/react the lipid --- f .a n gly.) P4 – Enzymes speed up the rates of chemical reactions but remain unchanged (at the end of the reaction ) // They are not destroyed by the reactions they catalyse. P5 – Enzymes are needed in small quantities because they are not used up (but released at the end of a reaction) P6 – Most enzyme-catalysed reactions are reversible // enzymes can catalyse the reaction in either direction. P7 – The activity of an enzyme can be slowed down or completely stopped by inhibitors // In order to function well , many enzymes require helper molecules, called cofactors. Any 4 = 4 marks

Able to discuss the uses of enzymes in industrial processes and our daily life, using suitable examples Sample answer

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Type of industry/ Application (T) 1. Food processing industry

Enzymes used (E) Rennin Lipase

• •

Solidifies milk proteins Ripening of cheese

(a)Dairy products

Lactase

•

Hydrolyses lactose to glucose in the making of ice-cream

(b)Bread and other bakery products (baking industry)

Amylase

•

Uses (U)

Zymase

amylase convert starch flour into sugar in the making of the bread • protease convert protein in the making of biscuit • amylase convert starch in malt into glucose for the fermentation of yeast (in wine and beer production.) • Converts sugars into ethanol during fermentation of yeast (in wine and beer production.)

(d)Fish products (e)Meat products (f)Cereal grain products (g)Seaweed products

Protease Protease Cellulase

• • •

(h)Starch products

Amylase

(c)Alchoholic drinks (beer / wine making industry)

Protease Amylase

Cellulase

Glucose isomerase 2. Leather products 3. Medical / phartimaceutical product 4 .Biological washing powder or detergent

Trypsin / Protease (Pancreatic) tripsin (Microbial) trypsin Protease and amylase

Protease removes the skin of fish Tenderises meat Breaks down cellulose and removes seed coats from cereal grain • Digest cell wall(cellulose) and extracts agar from seaweed • Change starch to sugar in the making of syrup • Convert glucose into fructose // Production of high fructose syrup •

Removal of hair from animal hides

•

Treats inflammation

•

Dissolves blood clots

•

Dissolve protein and starch stains in clothes

Grant marks : All the three corresponding ( T + E + U ) should be correct 4551 /2

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Any 6 ( T + E + U ) = 6 marks

Able to explain how extracellular enzyme is produced by emphasising on the role of P, Q, R and S Sample answer P1– P : nucleus, store genetic imformation / gene (for the synthesis of enzymes) in chromosome / DNA / is carried by the DNA P2– The messenger RNA/ mRNA is synthesised according to the instruction on the DNA // The genetic information to synthesis the enzyme in DNA is transfered to RNA in code form // mentioning of the transcription process briefly. P3– Q : mitochondrion, produce energy by cellular respiration (used in the production of extracellular enzyme P4 –The messenger RNA / mRNA / RNA then leaves the nucleus and moves to the ribosome (which is the site of protein synthesis) P5– The messenger RNA / mRNA /RNA attaches itself to the ribosome P6– Protein that are synthesised at the ribosome are transported through the spaces within the rough endoplasmic reticulum P7– Proteins depart from the rough endoplasmic reticulum wrapped in vesicle that bud off from the sides of the rough endoplasmic reticulum / from the membranes of the rough endoplasmic reticulum P8 – These transport vesicles fuse with the membrane of the R, Golgi apparatus / body and empty their contents into the membranous space P9– These proteins are modified during their transport in the Golgi apparatus, R . P10– For example, sugar to make glycoproteins/ carbohydrate are added to protein P11– S, secretory vesicle containing these modified proteins bud off from the Golgi membrane and travel to plasma membrane P12– These vesicle will then fuse with the plasma membrane before releasing the proteins outside the cell as extacellular enzymes. Grant marks: If student mention the names of P,Q, R and S before or after explaining the process.

7

Able to describe the transmission of nerve impulses across synapse • •

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When an impulses arrives in the axon terminal Stimulates (synaptic) vesicles to move towards and bind with the presynaptic membrane [Lihat sebelah SULIT

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•

The vesicles fuse / release the neurotransmitter into the synapse • The neurotransmitter molecules across the synapse to the dendrite of another neurone • Stimulated to trigger a new impulses which travels along the neurone Able to explain the situation given. F1 - P is afferent neurone which transmits nerve receptors to the interneurone. E1 - If P damaged, impulse from receptor cannot interneurone. E2 - (As a result), individual A cannot feel any pain

impulse from the be transfered to the

F2 - R is efferent neurone which transmits nerve impulse from interneurone to the effector E1 - If R damaged, impulse from interneurone cannot be transfered to the effector E2 - (As a result), individual A cannot withdraw the finger // pull the hand away from the pointed needle Able to explain the differences between reflex arc in two Diagram

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Diagram 7(b)(i)

F1 E1 F2 E1 E2 E3

E4 E5

F3 E1

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Diagram 7(b)(ii) Withdraw finger from a Knee jerk reflex sharp needle reflex Leg swings forward // knee jerk while the hand is withdrawn from the sharp needle Involved afferent Involved afferent, and efferent interneurone and neuron/two types efferent neuron/three of neurone types of neurone - Receptor detect the stimulus and triggers a nerve impulse - Afferent neurone carries impulse from receptor to the spinal cord - Synapes with the efferent neurone (in grey matter) / synapes with interneurone and sinapse with the efferent neurone (in grey matter) - Efferent neurone carries impulse from spinal cord to effector - stimulates the quadriceps/biseps muscle to contract Produce response Produce response fast faster Involved one synapse in diagram 7(b)(i) while two synapse in diagram 7(b)(ii)

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Able to explain how endocrine system is functioned in the negative feedback mechanism o

The endocrine system consist of various ductless glands, (synthesised and) secreted hormones into t he bloodstream.

1

o

The hormones regulates the activity of the target organs

1

o

The hormones stimulates and coordinates body activities/

1

Max; 1

regulates the chemical coordination o

The stimulus is detected by the receptor, the impulse is sent to the control centre (CC)/ brain

o

The impulse from CC is sent to effector/ endocrine glands

o

Endocrine

glands

secretes

hormones

for

the

corrective

mechanism

1 1 1 5 Max: 4

a(ii)

Able to explain the corrective mechanism occurs in the regulation of the blood osmotic pressure 1.

The increases of blood osmotic pressure is detected by the osmoreceptor cells in the hypothalamus

1

Corrective mechanism 2.

Hypothalamus stimulates the pituitary gland to release more

1

ADH into the blood 4551 /2

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3.

The permeability of the distal convoluted tubule/collecting duct to water is increases.

1

4.

More water is reabsorbed back into the blood

1

5.

Adrenal glands not stimulated to release aldosteron

6.

Less amount of salts reabsorbed of distal convoluted tubule.

7.

So that the osmotic pressure falls to normal range, (Urine

1 1

Max: 5

1

produced is concentrated and small in amount)

6 8b

Max 6

Able to explain the process of urine formation.

•

The formation of urine involves three main processes – ultrafiltration, reabsorption and secretion.

Ultrafiltration: • Blood enters the kidney from the renal artery and flows into the glomerulus through the afferent arteriole. • The hydrostatic pressure of blood in the glomerulus forces almost all the blood plasma contents to be filtered into the Bowman’s capsule. • The glomerular filtrate collected in the Bowman’s capsule contains all the substances that present in the plasma such as water, glucose, salt, amino acid and urea except the plasma proteins. Reabsorption: • The glomerular filtrate moves into proximal convoluted tubule. Here, the water and nutrients such as glucose and amino acids are reabsorbed into the blood. • Wastes such as urea, ammonia and excessive salts remaining in the blood after filtration go into the tubule. • The filtrate then flows into the Loop of Henle. The Loop of Henle concentrates the filtrate by removing water from it. • The filtrate then passes into the and collecting tubule – only water, and the distal convoluted tubule - water and salts are reabsorbed here. • The remaining filtrate in the collecting tubule is called urine, which is the channeled into the urinary bladder to be excreted out of the body to external environment. 4551 /2

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1

1 1

1

1

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19 Secretion: • Secretion is the process by which substances move into the distal convoluted tubule and collecting tubule from blood in the capillaries around these tubules. • Secretion moves substances such as urea, uric acid, drugs and ammonia out of the blood into the tubules where they mix with the water and urine is produced. These substances are secreted through either active transport or as a result of diffusion across the membrane.

1 1 1

1

o The functional unit of urine formation is nephron o The formation of urine involves three processes namely ultrafiltration, reabsorption and secretion o

The blood enters the glomerulus through

1

afferent arteriole

which has a bigger diameter than the efferent arteriole // As a result, there is a high hydrostatic pressure in the glomerulus o

Hydrostatic pressure causes many constituents of the blood to be filtered out from the glomerulus into the Bowman’s capsule. This called ultrafiltration

o

The blood constituents like glucose, amino acid, urea and salts

13

Max: 10

such as Na+ are filtered out of the glomerulus to form glomerular filtrate.// -

the concentration of glucose, amino acid, urea and salts/Na+ in the filtrate is the same as the concentration (of glucose, amino acid, urea and Na+) in the plasma

o

No RBC, platlet and plasma proteins in the glomerular filtrate because these substances are too large and cannot be filtered out of the glomerulus.

o

As the filtrate flows through the proximal convoluted tubule, reabsorption occur to the useful substance.

o

Glucose and amino acid are reabsorbed by active transport ( into blood capillaries)

o

(Meanwhile 65% / most of water is reabsorbed by osmosis)

o

Urea is not reabsorbed as this substances to be eliminated

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As the filtrate passes along the loop of Henle, (water) and salts/ Na+ are reabsorbed

o

At the distal convoluted tubule and collecting duct, water and salts/ Na+ are still reabsorbed by active transport, and (water is reabsorbed by osmosis depends on the content water and salts in the blood)

o

The absorption of water is regulated by ADH from pituitary gland.

o

Along the distal convoluted tubule (DCT) , left waste product in the blood capillaries such as urea, (uric acid and ammonia) are pumped out into DCT by active transport. This process is called secretion.

o

-(Some drugs and other toxic substances are also secreted out of blood capillaries by simple diffusion.)

o

Eventually, the filtrate is now referred as urine. It contain more urea Na+ and water. -

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ITEM NO

9(a)

ACCEPTED POINTS / DESCRIPTION / EXPLANATION Able to state the type of skeleton in organism P and Q

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Able to explain each type of skeleton 21 correctly

MARKS

REMARKS

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Sample answer

(b)

Organism P - Hydrostatic skeleton - The support is derived from body fluid contained within the bodys cavity Organisma Q - endoskeleton - The support is derived from hard skeleton of bones inside the body.

8c Able to state one similarity Explanation

20 marks

Sample answer 4551 /2

-

The skeleton supports important body organ It protects the organ from damage.

Able to state two differences Sample answer

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END OF MARKING SCHEME

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