Mark Scheme Bio 2 Trial Selangor 2007

1

Mark scheme bio 2 trial selangor 2007

MARK SCHEME Section A Question 1 Item Number 1 (a)

Marking Criteria

(Any two)

(c) (ii)

2

1 1

2

Able to name one substance that is able to move across part X, and explain its characteristic. Sample answer:  Fatty acids / fats / lipids / glycerol /vitamins A, D, E,K  Soluble in fats / lipid or  Water / oxygen / carbon dioxide  Very small (molecule)

(c) (i)

1 1 1

Able to explain the structure of X. Sample answer:  (Each molecule is) made up of head and tail.  The head is hydrophilic / polar / soluble in water, and the tails are hydrophobic / hydrocarbon chains / insoluble in water.

(b) (ii)

Total Mark

Remark

Able to name the labelled parts on the plasma membrane. Answer: X – Phospholipid(s) Y – Pore protein Z – Carrier protein / protein

(b) (i)

Marks

1 1 or 1 1

2

Able to explain the transport of oxygen gas into the cell. Sample answer:  By (simple) diffusion.  From higher concentration to lower concentration (of oxygen)./ down the concentration gradient  Through the protein pore / (phospho)lipid layer / X. (Any two) Able to state why the carrier proteins are specific to certain substance. Sample answer:  Glucose and amino acid are big molecules  These molecules enter the cell through the carrier proteins  Carrier protein have specific shape / site.

1 1 1

2

1 1 1

2

3 √ - 2marks 2 √ - 1 mark 1 √ - 0 mark

Mark scheme bio 2 trial selangor 2007

2 (Any two)

(d)

Terangkan perkaitan pernyataan ini dengan kesihatan tulang wanita it Able to explain the effect of the herbicide to the transport of potassium ions in the weed plants. Sample answer:  The (active sites on the) carrier proteins are denatured / destroyed.  Potassium ions are not transported into (the cells of) the weed plants or No active transport of potassium ions.  No energy / ATP produced / available

1 1 1

TOTAL

3

13

Question 2 Item Number (a) (i)

(ii)

(b)

(c)

(d)

Marking Criteria Able to state what L and M are. Answer L : glucose M: carbon dioxide Able to explain what happen to L after it has been produce. Sample answer • Stored as starch in cells/leaf • Used in cellular respiration//complete explanation of all respiration Able to explain how does M get into cell P Sample answer • By diffussion from atmosphere through stoma into air space between the cells • diffuses through cell wall and cell membrane into cell P / cytoplasm Able to explain how the treatments affects photosynthesis Sample answer • (The wax)blocks/ closes the somatal pore • Hinders/blocks diffusion of carbon dioxide • Rate of photosynthesis decreases (Any two) Able to give suggestion how to increase the crop

Marks

Total Mark

1 1

2

1 1

2

1 1

2

1 1 1 2

Remark

Mark scheme bio 2 trial selangor 2007

3

yield of greenhouse plants. (i)

(ii)

Sample answer • Carbon dioxide is raw material/ used to produce sugar • (Increasing the carbon dioxide concentration) will stamulate/encourage sugar/glucose to be formed Sample answer • (Continuous light results) in continuous light reaction/ photosynthesis • More carbon dioxide can be reduced (by H ion) to form glucose

1 1

2

1 1

TOTAL

2

12

Question 3 Item Number 1 (a)

Marking Criteria

Total Mark

1 1

2

Able to draw and label the respiratory structure of organism Z

•

The respiratory surfaces consist of a single layer of cells which allow the gas to move across the surface • Moist respiratory surface allow the gases to dissolve in water before diffusion takes place. • Many folds on the respiratory surfaces to increase the total surface area for the

3 √ -2 marks 2 √ - 1 mark 1 √ - 0 mark Drawing – 1 mark (include spiracle)

2 (b) (ii)

Remark

Able to name respiratory structures of organism X and Y X – lungs Y – gills

(b) (i)

Marks

1 1 1

Labeling – 1 mark (at least 2 correct labels of spiracle trachea and tracheole with chitin rings)

Mark scheme bio 2 trial selangor 2007

•

(c)

exchange of gases Covered with a dense network of blood capillaries to increase the surface area for the diffusion of gas. (Any two)

•

A lot of respiratory surface increase surface area ratio Thin layer of cells of respiratory surface for diffusion process

2

1 1

2

1

1

Sample answer: Because each cell in the insect’s body has a tracheole system to supply oxygen directly from the surrounding.

(e)

1

Able to explain two similar characteristics of the respiratory structure of animal X,Y and Z which increase their efficiency in gaseous exchange

•

(d)

4

Sample answer: F1- X/human has a double circulatory system, whereas Y/fish has a single circulatory system

1

P1- In X/human has blood flows directly to the body cells, whereas in Y/fish the blood from the heart flows to the lungs and then to the body cells/not directly //the blood pressure flow to the body cells in human is higher compared to that in fish.

1

TOTAL

2

11

Question 4 Item Number 4 (a)

(b)

Marking Criteria Able to name the types of blood cell Answer P : Red Blood cell / erythrocyte Q: White blood /cells/leucocytes/eusinophyll/neutrophill Able to explain the function of R in body defense. Criteria • Explanation of blood clotting mechanism Sample answer:

Marks

1 1

Total Mark

2

Remark

Mark scheme bio 2 trial selangor 2007

5

1. (When the blood vessel is damaged),

(c)

(blood flows) platelets produce thrombokinase enzyme. 2. Thrombokinase converts prothrombin to thrombin 3. Thrombin converts fribrinogen to fibrin 4. Fibrins traps erythrocytes ,scab formed (Scab dried , wound is healed ) Able to explain one difference between P and Q Criteria • Difference in shape and explanation// • Presence of nucleus and explanation

1 1 1 1

F1: P has a biconcave disc shape but Q has irregular shape//has no nucleus but Q has large nucleus. P1: Biconcave disc shape/absents of nucleus gives large surface area to volume ratio to speed up exchange of respiratory gases.

1

F2 : P has haemoglobin, but Q has no haemoglobin P2 : P can transport oxygen (in the form of oxyhemoglobin) but Q can not.

1

3

Any three

1

1

2

(Any pair of F and P) (d)

Able to explain how the abnormal mitosis of cells Q in the bone marrow affects the person’s health based on the following criteria: • The ways cell divide • Type of cell produced • Defect of cell • Effect on health Sample answer: 1. cells multiply rapidly ( within the bone marrow) 2. producing immature / abnormal cells 3. Cells not able to protect the body against disease / act as body defense. 4. causes cancer // Leukemia

(e)

1 1 1 1

3

Any three

Able to state two importance of blood cells Sample answer: 1. Acts as a medium/ to transport dissolved substances / oxygen /carbon dioxide / digested food /amino acid/glucose/ hormones /antibodies/urea 2. Acts as body defense / protect the body against disease/phagocytosis

1

1

2

Mark scheme bio 2 trial selangor 2007

6 TOTAL

12

Question 5 Item Number 5 (a)

Marking Criteria

(i)

1

1 1 1 1

2

Able to explain how the concentration of the substances present in Q changes: after eating meat and eggs Sample answer • meat and eggs contains high protein/ main source of amino acid • (Excess) amino acids are deaminated / converted into ammonia / urea in the liver • The urea is transported to the kidneys and removed as urine • The concentration of urea in the urine would increase

(ii)

1

Remark

Able state one features and explain why. Sample answer: • They have many/abundant mitochondria • Produce a lot of energy needed for active transport or • Numerous/many microvilli • Increase total surface area for reabsorption

(c)

Total Mark

Able to state one difference the chemical componen between blood vessel X and Y Sample answer: Urea concentration is lowest in Y but higher in X.

(b)

Marks

if a person suffering from diabetes had not taken enough insulin. Sample answer: • Lack of insulin result in glucose not converted into glycogen • The glucose concentration rises in blood • Glucose is lost in the urine/glucosuria occurs • No storage of energy rich molecules

1 1 1 1

3

Any three

3

Any three

1 1 1 1

Mark scheme bio 2 trial selangor 2007 (d)

7

Able to state what is reabsorption base on diagram Sample answer: (i) • Reabsorption is the absorption of glucose water and salt from the kidney tubule to the capillaries. (ii) • Reabsorption is a process which requires energy

1 1

TOTAL

2

11

Question 6 Num 6(a)

6(b)

Criteria Able to explain the role of X in digestion based on the following criteria • Function as a gland • Name of any one enzyme secreted • Site of enzyme • What happens in the duodenum • Example specific enzyme reaction

Marks

Sample answer 1. ( organ x is pancreas ) secretes pancreatic juice 2. containing lipase / amylase / trypsin enzyme 3. ( Secretes ) into the duodenum 4. complex food of lipids // starch // proteins are digested 5. Lipase hydrolyses lipids to fatty acids and glycerol / Amylase hydrolyses starch to maltose /Trypsin hydrolyses peptones to peptides Any 4 Able to explain the assimilation of digested food of glucose , amino acids and lipids

1 1 1 1 1 4

Sample answer In the liver Glucose •

Excess of glucose in the blood is converted to glycogen and stored • When glucose level in blood is low glycogen is converted to glucose • Excess of glycogen is converted to lipids Amino acid • Synthesis of plasma protein from amino acids • Excess amino acids are deaminated into urea In the body cell Glucose • glucose is oxidized to release energy (in cellular respiration) Amino acid

1 1 1 1 1 1 1 1

Mark scheme bio 2 trial selangor 2007 • •

6(c)

8

Amino acids is used to synthesise new protoplasm/ repair of damaged tissues Amino acids is used to synthesise enzymes/antibodies/hormones

Lipids • Excess lipids is stored in adipose tissues • Phospholipids and cholesterol make up the plasma membrane Any 10 Able to calculate the energy value taken daily Answer Total energy value is 5250+1500+2400+2400+750+2500+400+600=15800KJ

1 1 1 10

1

1

Able to explain the consequences of the eating habit when taking the menu daily for a long time Sample answer

• • •

1

The menu is not a balanced diet /does not contain the 7 classes of food in appropriate ratio Menu is highly rich in carbohydrates and fats// no vegetables and lack of vitamins Higher energy intake compare to energy requirement for a girl aged 15

Consequences

•

Constipation /description – lack of fiber , faeces moving to slowly through the colon • Scurvy - lack of vitamin C //any other vitamins deficiency with explanation • Obesity – increase in body weight drastically

1 1

1 1 1 1 1 1 1 1

•

Diabetes mellitus/glucosuria – excess of glucose contain in blood

1 1

• Arteriosclerosis – fats deposited in the lumen of blood vessel

•

Heart attack/angine - blockage in the coronary artery // Any other cardiovascular diseases with explanation Any 5

1 1 5

6

9

Mark scheme bio 2 trial selangor 2007

TOTAL

20

Question 7 Ques . No 7(a)

1

Criteria Problems faced by mangrove plants (Fact) Soft muddy soil/substrate

Adaptive characteristics of mangrove plants (Explanation)

• •

2

3

4

Marks

Waterlogged conditions of the soil/ Very little oxygen for root respiration

•

The high content of salt/salinity makes the water in the soil hypertonic compared to the cell sap of the root cells/ Water diffuse out from plant/ the root cells by osmosis// dehydration

•

Excessive exposure to sunlight/ intense heat// High rate of transpiration.

•

•

Highly branched root system to support themselves./ Eg. Avicennia have long/underground/horizonta ls cable/ roots (Avicennia) have breathing roots /pneumatophores /Gaseous exchange occurs through pores/ lenticels. Cell sap of (the root) cells are hypertonic compared to the soil water/. The root does not lose water but seawater enters the root cells instead/ Excess salt in the plant is eliminated by the salt glands The leaves (of mangrove trees) have a thick cuticle/ sunken stomata to reduce transpiration/ The leaves are thick /succulent to store water.

2

2

2

2

Remarks

10

Mark scheme bio 2 trial selangor 2007

5

High mortality rate//low survival rate of seedings

•

Have viviparous seedling // the seeds are able to germinate while still attached to the mother plant.

2

One fact and one explanation = 2 marks TOTAL

10

B

Marks Impact

1

Exposed to storm and tsunami

• •

2

Coastal erosion

•

• 3

Greenhouse effect

• • • • •

4

Destruction of natural habitat

remarks

Explanation

• • •

Mangrove plants (protect the coastline)by acting as wave breakers Strong waves are not stopped on reaching coastline Mangrove root, e.g prop/buttress/plank roots hold soil/mud particles and stabilize Substrates/soil are easily washed away by waves Removal of mangrove plants results in exposure to direct sunlight/heat No plants to absorb carbon dioxide produced High evaporation rate of sea water Air /weather has very low humidity/dry/hot/heat waves Resulting in sea water very saline Mangrove plants are breeding places for birds/fish /prawns Animal migrate to other habitats Loss of gene pool

1 1

1

1 1 1 1 1 1 1 1 1 10

Mark scheme bio 2 trial selangor 2007

11 (Any 10) TOTAL

20

QUESTION 8 Ques. No 8(a)

Criteria Able to give main points of the flow chart • Pollen parental / mother cell • Meiosis • Microspore • pollen

Sample answer Pollen parental cell C1 Meiosis C2 Microspores C3 Mitosis C4 Pollens

Able to describe how pollen grains are formed in plants

score 1 1 1 1

notes

4

Mark scheme bio 2 trial selangor 2007

12 1

Sample answer

• • • • • •

•

(b)

Pollen grains are male gametes (produced in the anther/pollen sac (pollen sac) contains diploid pollen parental cell/ pollen mother cell/ microspore mother cell (each pollen mother cell) undergoes meiosis Producing 4 haploid microspores(n) nucleus of (every) microspore undergoes mitosis forming two nuclei The generative nucleus and the tube nucleus Microspore with surfaces adapted for dispersion and is now called pollen

1 1 1 1 1 1

Able to describe how infertility is overcom in human using the following methods. Sperm bank • Used when the man/spouse/male suffers from low 1 sperm count/production of weak/unhealthy sperms • Sperms are donated and donor’s identity is 1 recorded/secret • Genetic biodata of donor and the woman recipient 1 should match for successful fertilization Artificial insemination • Practiced by a couple when the man is sterile • Donor’s sperm are injected into the womans uterus during ovulation • No sexual intercourse is involved • The genetic background and health of donor is screenned before acceptance In vitro fertilization • Practiced when the woman’s ovum cannot be fertilized due to blockage of fallopian tube • Ovum is extracted (from the woman) and fertilized outside the body/in Petri dish/test tube • Fertilized ovum/zygote grows in culture medium • Than it is implanted into the woman’s uterus Surrogate mother • Method practiced when a woman cannot be

1 1 1 1

1 1 1 1

1

6

Mark scheme bio 2 trial selangor 2007

• • •

13

pregnant in normal way The zygote is obtained by invitro fertilization Another woman is implanted with the couple zygote (Surrogate mother) carries the baby only until it is born

1 1 1

10 [any 10 points] TOTAL

QUESTION 9

20

Item Number 9 (a)

Scoring Criteria

Marks

Able to explain the crossbreed based on the following 14 C1 – The traits of the parent plants in the crossbreed. C2 – The gametes formed. C3 – The random fertilisation of the gametes. C4 – The products of the crossbreed / offsprings / F1.

Mark schemecriteria. bio 2 trial selangor 2007

Sample answer: C1: P1 – The crossbreed involves 2 traits of the pea plants // It is a dihybrid cross. P2 – The first traits is the shape of the pea seed // The characteristics are smooth and wrinkle. P3 – The second traits is the colour of the pea seed // The characteristics are green and yellow. P4 – Smooth and green are the dominant characteristics. P5 – The smooth and green pea seed is heterozygous dominant in both traits. P6 – Wrinkle and yellow are the recessive characteristics. P7 – The wrinkled and yellow pea seed is homozygous dominant in both traits. (Any 5 of P1 to P7) C2: P8 – The smooth and green pea seed plant produced 4 types of gametes / shown in the crossbreed diagram or in the Punnette Square. P9 – The wrinkled and yellow pea seed plant produced 2 types of gametes / shown in the crossbreed diagram or in the Punnette Square. (Any 1 of P8 and P9) C3: P10 – Fertilisation of the gametes occurs randomly / any gametes of one pea plant can fertilise any gametes of the other pea plant / shown in the crossbreed diagram or in the Punnette Square. C4: P11 – 4 Phenotypes of offsprings (F1) are produced / Smooth and green pea seed, smooth and yellow pea seed, wrinkled and green pea seed, and wrinkled and yellow pea seed / shown in the crossbreed diagram or in the Punnette Square. P12 – 4 genotypes of offsprings (F1) / shown in the crossbreed diagram or in the Punnette Square. P13 – Phenotype / genotype ratio is 1:1:1:1 / shown in the crossbreed diagram or in the Punnette Square. (Any 3 of P10 to P13)

1 1 1 1 1 1 1 5 1 1

1

1 1 1 3

Crossbreed diagram: Smooth and green SsGg

Parent:

X

Wrinkled and yellow ssgg P8 / P9

Gamete:

SG

sG

Sg

sg

sg P10

Offspring:

SsGg

ssGg

Ssgg

ssgg

P11 P12

Total Marks

Mark scheme bio 2 trial selangor 2007

15

Mark scheme bio 2 trial selangor 2007 (b) (i)

16

Able to complete the pedigree chart of Mr. Jay’s family and explain the inheritance of albinism. Sample answer: Pedigree chart:

Albino daughter

Mr. Jay

Mdm. Kay

Albino son

Normal twin girl

PC (Correct pedigree chart) L (Correct labels)

1 1 2

Explanation: P1 – Albinism is controlled by recessive gene / (gene) a. P2 – Phenotype appears if the individual is a homozygous recessive / (alleles) aa. P3 – Mr. Jay and Mdm. Kay / both parents are carriers / (alleles) Aa. P4 – Albinism is not sex-linked. P5 – Albino daughter / albino son receives one recessive allel / gene a from Mr. Jay and one recessive allel / gene a from Mdm. Kay. P6 – The twin girls are homozygous dominant / heterozygous (dominant) / AA / Aa. (Any 3 of P1 to P6)

1 1 1 1 1 1 3

5

17

Mark scheme bio 2 trial selangor 2007

(b) (ii)

Able to explain how to control the inheritance of albinism to the next generation of Mr. Jay’s family, based on the following criteria. C1 – The inheritance of albinism if the albino daughter / son marry an albino person. C2 – The inheritance of albinism if the albino daughter / son marry an albino carrier person. Sample answer: C1: P1 – Albino daughter / son must not marry an albino person. P2 – Both partners will produce gametes with recessive gene / a. P3 – If the (recessive) gametes undergo fertilization, albino offspring will be produced // The chance to get albino offspring is 100%. Parent: Gamete:

Albino aa

X

a aa 100% albino

Albino aa

X

Gamete:

a

Offspring:

Aa 50% albino 1

1

P2 P3

C1: P4 – Albino daughter / son must not marry an albino carrier person. P5 – Both partners are able to produce gametes with recessive gene / a. P6 – If the recessive gametes undergo fertilization, albino offspring will be produced // The chance to get albino offspring is 50% / 1:1. Parent:

1

Albino aa

a

Offspring:

1

1 1 1

Carrier Aa A

a P5

aa 50% normal : 1 (Any 4 of P1 to P6)

P6

ES (Evaluating skill): Any P from C1 + Any P from C2

4 1

TOTAL

5 20

20