Biology Paper 2 16 Tests m s

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2010Pyramid Projections©

231/2 BIOLOGY PAPER22010 Pyramid Projections 001 Kenya Certificate of Secondary Education MARKING SCHEME

1. (a) X - Vacuole / Sap vacuole;

Y – Tonoplast;

Z – Chloroplast;

(b) Cellulose;

(c) Active transport;

(d) The cell sap is hypertonic to the solution / distilled water; hence water

molecules move into the cell; by osmosis; making it to swell and eventually burst;

( Total 4 marks , max. 3mks)

2. (a) Pneumatophores / Aerial Breathing roots;

- Stomata; - 2

(b) (i) A – Gill rakers;

B – Gill bar /arch;

C - Gill filaments;

(ii) Trap food / solid particles hence prevent them from clogging the gill

filaments;

(iii) Highly vasculorised to transport away oxygen that has diffused in;

- Then epithelium to reduced the distance gases diffuse across;

- Numerous to increase surface area for maximum absorption oxygen;

3. (a) Procreation;

- Improves the quality of species; - 2mks

(b) (i) Prophase - 1mk

(ii) Mitosis Meiosis

- occurs in somatic cells - sex cells

( body cells)

- No chiasma formation - There is chiasma formation

- Provides 2 daughter cells - Produce 4 daughter cells

- Daughter cells are diploid - daughter cells haploid

- Takes place in I phase of 4 - 2 phases of 4 stages

stages

- Homologous chromosome don’t - homologous chromosomes

Associate associate;

Any three correct = 3mks

(c ) Protandry;

Protogony;

Self incompatibility; Any 2 = 2mks

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4 (a) Aa; Aa; 2mks

(b) Aa X Aa; 1mk

( c) ¼ or 25% ; 1mk

(d) Crossing a homozygous recessive parent / organism with an offspring /

organism which shows dominant character;

(1mk)

5. (a) Region of elongation / region of rapid growth in a root / radicle; ( 1mk)

(b)

( c) Provide moisture / water for growth; (1mk)

(d) Oxygen for respiration / oxidation of stored food; to provide energy for

germination;

(ii) Cotyledons – Store food necessary for germination; ( 1mks)

(e) Small seeds store very little amount of food; (1mk)

6.

;

;

;

Normal pigmented children Albino child

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(b) (i) 229 + 5;

(ii) 102 – 97;

= 5KJ;

(c) Bigger body size, small surface area volume ratio; hence loss less heat to

the surroundings; require less energy to compensate for less energy lost.

(3mks)

(d) Extra polation;

2.5 kg;

(2mks)

(e) Lower;

(ii) Reptiles are Poikibilotherms; doesn’t require energy to maintain body

P = 3mks All 8 points correctly plotted = 3mks 7 correctly plotted= 2mks 6 correctly plotted= 1mk 5 correctly = 0 mks C = 1mk Rej. If line is broken If origin O s missing Penalize one scale½ mk Round of ½ mkto 1mk at graphlevel

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temperature;

(2mks)

(f) – Activity / occupation

- sex;

- age;

- State of health; Any 2 correct = 2mks

(g) Proteins; (1mk)

(h) Provide grip hence prevent constipation; add bulk to the food;

Any 1 correct = 1mk

7. The heart is muscular; to pump blood over along distance;

- It has myogenic cardiac muscles; which contract and relax rhythmically

without fatigue hence continuous pumping.

- cardiac musles fibres are interconnected to form network of fibres; to ensure

rapid and uniform spread of excitation throughout the walls of the heart;

- The heart is divided into four chambers which are hollow, to accommodate a

lot of blood.

- Ventricles wall are thicker than auricle walls; to generate higher pressure to

pump blood over longer distance;

- There is a longitudinal septum, which separate it into two halves to prevent

mixing of the more oxygerated blood and the less oxygenated blood;

- It has valves; which prevents backflow of blood;

- It has valves are connected with tough strands of connective tissue (chordae

tendinae); to prevent then from being pushed inside out when ventricles

contracts;

- The heart is joined by blood vessels ( aorta, vena cava, pulmonary vein,

pulmonary artery); which channel blood to and from are body parts;

- It has the coronary artery and coronary vein which supply the myocardium with

oxygen and nutrients; and remove waste products respectively;

- The fibrous layers of pericardium surrounds the heart; which keeps the heart in

position and prevent over –dilating;

- The inner layer of pericardium secretes pericardiac fluid; which reduce friction

between the two layers during systole and diastole;

- Outer layer pericardium, is surrounded by a layer of fats; to acts as shock

absorbers protecting it from mechanical damage;

- It has Sino artricole node (SAN); to acts as pacemaker by regulating rate of

beating and excitation of the heart; Marks 28

Max. 20

8. Palaentology/fossil records;

- Fossils are past materials remains of ancestral for of organisms accidentally

preserved in some naturally occurring materials e.g sedimentally rocks. They

provide a direct evidence of gradual change from one type of organism to

another;

- Their age is determined by radioactive carbon 14 dating; Used to construct

evolutionary history of development of a certain – organism; example human

skull, horse limbs etc.

- Comparative anatomy;

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Members of a phylum or group shows similarities in structures or organs

performing similar functions; Homologous structures/divergent evolution;

same embryonic origin but performing different; arose by divergent evolution

e.g pentadactyl limb of vertebrates.

- Analagous structure / convergent evolution; structures with different embryonic

origins but evolved to perfom similar functions due to exploitation of different

ecological niches; They evolved by convergent evolution; e.g wings of birds

and those of insects.

- Comparative embryology;

Study of embryos in their early stages of development e.g those of vertebrates

look similar; the more the similarity the more the evolutionary relationship (

recapitulation theory);

- Cell biology;

Structure and functioning of cells from almost all kinds of organisms are

basically similar. All Eukaryotic cells contain same cell organelles; such as

mitochondria, lysosome, E.R etc indicating a common ancestry;

- Geographically distribution;

Initially there was one land mass ( pangae); split and parts driffed away

forming the present continents;

After the split members of some species occupying originally similar regions

were isolated by barriers and evolved along their own line for instance ilamas

in S. America, Leopards and cheetahs in Africa;

Marks 21

Max. 20

2010 Pyramid Projections 002 Kenya Certificate of Secondary Education (K.C.S.E) 231/2 BIOLOGY PAPER 2 MARKING SCHEME CONFIDENTIAL 1. a) A – Sapvacuole; B – Chloroplast; C – Plasmodesma; ( 3mks)

b) Stoma contains enzmes which speed up process of photosynthesis; - Lamellae // granum provides a large surface area to accommodate more chlorophyll molecules; ( 2mks)

c)

Check cell of a frog One above i) No cell wall Cell wall present ii) Sap vacuole absent Sap vacuole present iii) Centrioles present Centrioles absent iv) Chloroplast absent Has chloroplast

Any 2 = 2mks

d) Packing and transport of glycoproteins //secretion of synthesized proteins and carbohydrates; (1mk)

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(Total 8mks) 2. i) Q – Piliferous layer; R – Endodermis; S –Pericycle; ( 3mks)

ii) - Casparian strip, water proof, thus water can only pass via cytoplasm; - numerous starch granules acts as a source of energy; (2mks)

b) see diagram (1mk) c) evaporation; (1mk)

c) - Thickening to prevent collapsing; - Impregnated with ligmin for support; any 1 (1mk) Total 8mks

3. a) I – ovulation; (1mk)

b) T – Grafianfollicle; R – Corpus luteum; ( 2mks)

c) T – follicle stimulating hormone;

R – Lutenizing hormone; ( 2mks)

d) ( When sperm comes into contact with egg) A crosome bursts open and releases lytie enzmes; which dissolve egg membrane; - Acrosome then turns inside – out forming a fine filament that penetrates the egg; - Head of sperm enters ovum, tail is left outside the ovum; - Vitelline membrane charge to fertilization membrane which stops any other sperm from entering ovum; - once in cytoplasm head burst releasing male nucleus which then fuses with female nucleus to form a diploid zygote; (max x 3, 3) (Total 8mks)

4. a) Parent Bb � X bb�

Genotypic ratio Bb : 1 bb� (6mks)

b) phenotypic ratio of children 1 Broadlipped : 1 thin lipped� (1mk)

c) - identifying positions occupied by specific genes

- Analying DNA to reveal order of bases in all chromosomes�. (1mk)

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on a chromosome ( Total 8)

5. a) K – ilium; L – pubis; M- ischchium; (3mks) b)i Sacrum; rej sacral vertebrae (1mk)

ii) Well developed traverse process which are fixed to pelvic girdle; - Vertebrae are fused for strength // Transmits weight of a

stationary animal to the rest of the body; - Sacrum has large base // short neural spine for attachment of

(back) muscles; Any 1 (1mk)

c) Pubis symphis, in human female has flexible cartilage which allow widering of pelvic girdle during birth; ( 1mk)

d) Acetabulum; ( 1mk) e) Anticular cartilage, soft // smooth // slippery // absorbing to reduce

shock; - Capsular ligament, strong// fibrous holding bones together; - Synovial membrane secreates synovial fluid, reduces friction//

absorb shock between 2 bones; - Rounded head fits into socket and moves freely // 3600;

any 1 (1mk) (Total 8marks)

b)i 8 min ± 1 min; (1mk) ii) 41 seed ± 1 seed; (1mk)

c) pea seeds have soft testa; heat easily penetrates and denatures enzmes // hormones responsible for germination; (2mks)

d) Increase temperature softens testa; thus making testa more permeable; (2mks)

e) There was a decrease in number that germinated; since after 20th min testa was now permeable thus heat penetrated and denatured emzmes // hormones// damaged embryo;

(2mks)

f)i Decrease in O2 cone; and increase CO2 cone; (2mks) ii) Seeds respired and used O2 & released CO2 ; (1mk) f) - Absence of light;

- Impermeable seed coat to water; - Undeveloped embryo; - Lack of growth stimulator // enzmes // hormones // giberrelins; - Pressences of inhibitors; Any 3 = 3 (Total 20mks)

7. a) Process Products

Exudation//guttation; - Resins, gums, latex, rubber, calcium oxalate and pectate for exulation

- Salts and water for guttation; Any 2 products ( 2mks)

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Traspiration//Diffusion; - water, CO2, O2

Any 2 products (2mk)

Deposition// leaf fall - tanmn, caffeine pyrethrines, morphine, nicotine, quinine, cocaine etc

Any 2 products – (2mks)

N.B – If process is wrong - no marks for products.

b)i Liver

- Amino acids can’t be store in humanbody; Excess amino acids are deaminated; to form ammonia; which combines with Co2 to form urea; and carbohydrate residue; for use in the body; Urea is transported to kidneys through blood steam;

ii) Kidneys

- In kidneys urea is ultra// pressure filtered; from the blood in the glomeruli; as capsular // glomeruli filtrate; into Bowmas capsule; filtrate passas through the ( proximal convoluted) tubule; where urea is concentrated by removal of water; and finally into collecting tubule as part of urine;

8. - Describe the structure and function of various parts of mammalian skin.

- The skin is made up of epidermis and dermis the epidermis is made up of three layers. The outer most layer is known as cornfield layer; made of dead cells that protect against mechanical damage// dessication // microbes;

- The granular layer; is made up of living cells that give rise to cormfied layer;

- The malphigian layer; with actively dividing cells that give rise to a new epidermal cells// granular layer that contain melanim; that protects the skin against U.V rays;

- Has sweat glands// sudorific glands; that produce sweat which evaporates thus reducing body temperature;

- Under cold conditions little// no sweat is produced thus heat is conserved;

- The sweat contains water, sodium chloride// uric acid //urea and hence the skin acts as an excretory organ;

- Has hair; the hair stands erect to trap air to insulate// reduce heat loss when the temperature is low// lies flat to allow heat loss when the temperature is high;

- Has nerve endings; which are sensitive to stimuli//heat //cold pain// pressure//touch;

- Has subcutaneous fats//Adipose time;that unsulates the body against heat loss;

- Has arteries// capillaries// blood vessels; that supply food//oxygen//remove excretory products;

- Arterioles vasodilate when temperatures are high loss of heat by radiation//convection//; vasoconstricts when temperature is low to conerve heat// minimize heat loss;

- Has sebaceous gland; which secretes sebum, an antiseptic// water repellant that prevents skin drying// craking of skin// keeping skin sapo; (max 20)

Total 23

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231/2 BIOLOGY PAPER TWO 2010 Pyramid Projections 003 Kenya Certificate of Secondary Education MARKING SCHEME 1. (a) S - White blood cell / leucocyte; T - Red blood cell / Erythrocyte; (2 marks) (b) Protection against infection/destroys pathogens; Rj: for defense alone unless it is qualified. (1 mark) (c) Platelets/thrombocytes release thrombokinase/thromboplastin; (when exposed to air) which converts

inactive prothrombin into active thrombin; the enzyme Thromin converts the soluble fibrinogen into insoluble fibrin; (3 marks)

(d) (i) Biconcave shape to provide large surface area for diffusion of gases; (ii) Has Haemoglobin that readily combines with oxygen/ has high affinity for

oxygen (iii) Absence of nucleus to give more room/space for packing/accommodating

more haemoglobin; (iv) Has enzyme carbonic anhydrase to speed up combination of carbon (iv) oxide with water;

(Mark 1st two) (2 marks) 2. (a)(i)573;

(ii) Tall; (b) Parental genotypes

(c) To prevent self pollination/fertilization; To increase chances of cross pollination; (2 marks) (d) A condition in flowers where stamens are shorter than pistils/pistils are longer than stamens (to prevent self pollination/fertilization); (1 mark) 3. (a) A- Lenticel;

B - Cork cambium; (Accept cambium alone) (2 marks) (b) X - Carbon (iv) oxide; Y - Oxygen; (2 marks) (c) Secondary thickening of the bark/form secondary cortex on the inner side and cork cells o the outer side. (1 mark) (d) (i) Respiration;

(ii) Glucose + Oxygen. Carbon (iv) oxide + water + Energy/ATP

OR Glucose Alcohol + carbon (iv) oxide + energy (1 mark)

(e) Less respiration / ATP produced hence reduced active transport; (1 mark) 4. (a) A - eye piece ; Rj; eye piece lens

B - Coarse adjustment knob/screw; (b) E - Concentrate/focuses light; (2 marks)

F-Magnification of image; Rj; magnification of object. (2 marks) (c) Eye piece magnification X objective magnification (1 mark) (d) (i) make them transparent; OWTTE (1 mark)

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(ii) make different parts/organelles more distinct; Rej: Visible for distinct, (1 mark) (iii) make/keep cells turgid; 1mark 5. (a) Hypogeal; (b) P - coleoptile/plumule sheath; (2marks) R - Root hair; Rj; plural/hair root (2marks) (c) - absorption of water; - Absorption of mineral salts (2marks) (d) -Thin cell walls; - Dense cytoplasm - Lacks cell vacuoles AWARD 1st two (2marks) (e) - storage of food/starch/carbohydrates; - Storage of enzymes; (ANY 1 ( 1mark) SECTION B (40 marks) 6. (a) see graph paper - labeling of axes (2marks) - scale (2marks) -plot (1mark) - curve (1mark) (b) 76; ±2 (1 mark)

(i) Rapid decrease in number of ticks; acaricide poisoned/killed ticks/ticks susceptible to acaricide; {2 marks)

(ii) (ii) Increase in number of ticks; mutation occurred; that was inherited. The new variety / breed of ticks developed resistance / ticks produced chemicals/enzymes which made them non susceptible to acaricide; (3 marks)

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(d) - Pollution/acaricides may affect/poison non targeted beneficial organism;

- Acaricides remain for along time in the ecosystem; - Acaricides accumulate in food chains/ecosystem;

1ST TWO (2 marks)

(e) - biological control/use predators to prey on ticks; - clearing of bushes - burning of vegetation; - Rotational grazing paddocking/picking picking ticks by hand and destroying them; (accept any other valid answer)

AWARD 1ST THREE (3 marks)

(f) Total counts/counting; (1 mark) (g) (i) Arachnida; (ii) - Four pairs of legs; - 2 body parts/cephalothorax and abdomen; Mark 1st one

TOTAL20 Marks

- - Reduced leaf sizes/leaves reduced to scales; to reduce surface area over which transportation occurs/to reduce transpiration.

- shedding of leaves; (during drought) to reduce transportation; - thick waxy cuticle on leaves; (that is waterproof) to reduce transpiration; folded

leaves; to reduce transpiration; - sunken stomata; where moisture/water vapour accumulates in substomatal air

spaces; hence low vapour diffusion gradient/reducing transpiration; - reversed stomatal rhythm (OWTTE); to reduce transpiration; - have roots that grow deep; to absorb water from low water table; - superficial roots; that grow horizontally close to the soil surface to absorb water

after light showers of rain; - succulent leaves/stems; to store water; - short life cycle; to evade droughts - hairy leaves; to trap moist air reducing vapour diffusion gradient between inside

of the leaf and outside; hence reducing transpiration. (MAX- 20 Marks)

8. blood reaches the kidney through the afferent arteriole; which branches highly to form the glomerulus; the afferent arteriole has a wider lumen/diameter; than the afferent arteriole(leaving the browmans capsule) ; this causes higher pressure in the glomerulus; that results into ultrafiltration; the ultrafiltered materials include glucose, amino acids(vitamins hormones) salts, urea and water; forms the glomerular filtrate; large materials of white blood cells, red blood cells, blood proteins are not ultra filtered ( hence remain in the blood); the filtrate flows to the proximal convoluted tubule; where all glucose, amino acids are selectively reabsorbed/absorbed back (into the blood stream); Active transport is involved in the reabsorption of these materials The filtrate flows to the loop of henle; where water is reabsorbed by osmosis into the blood stream; and sodium chloride ions are actively pumped from the ascending loop of henle into the blood; the filtrateflows into the distal convoluted tubule; where more water is reabsorbed; and mineral salts too into the blood; The filtrate flows into collecting ducts; from where more water is reabsorbed; the remaining filtrate is called urine; and is emptied into the (urinary) bladder via pelvis and ureter; and passed out of the body through urethra;

MAX 20 Marks

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2010 Pyramid Projections 004 Kenya Certificate of Secondary Education (KCSE) 231 /2 BIOLOGY PAPER 2 PAPER 2 MAKING SCHEME SECTION A: 40 MARKS 1 a

4 mks b)i) Presence of long ( conical) curved canines for holding / killing and tearing the prey

Presence of carnassial teeth that have sharp edges to slice/ shearing the flesh/ crush bones 2 mks ii) Absence of incisors on the upper jaw creating a hard pad for pressing and

cutting of grass by incisors on the lower jaw presence of diastema/ gap in the lower jaw to allow the tongue to cut grass / turning of food during grinding 2 mks

2 a) i) D.N.A ii) Has base thymine iii) U-U-C-U-C b) They are rare;

They are inherited; Some are beneficial and others lethal; They arise suddenly / randomly / unpredictable ;

Mark 1ST one only C) Both Aa/ Aa and Aa; Parental Genotypes Gametes Fusion F 1 off springs Genotype of albino child aa; 3 a) Anaerobic (Respiration) b)i) Presence of yeast to produce enzyme / presence of oil layer to prevent entry of air / oxygen

animal Reasons 1 a) Dog ; ii) Sheep ;

Presence of canines on both lower and upper jaw ; Absence of incisor / canines on upper jaw ;

A a ; RJ if X missing Aa

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ii) Flask A- reading on the thermometer was higher / reading higher than the initial reading Flask B- No change in the thermometer reading/ thermometer reading remained at the initial level c) Flask A- Glucose solution was acted upon by the Enzyme (Zymase)

produced by yeast �; thus broken down to form carbon dioxide, alcohol and heat/ energy ; � causing a rise in temperature �flask B – yeast cells are killed by boiling thus no reaction occurs ; � Acc: enzymes were denatured for yeast cells being killed

08 4 a) i) Predators feed on other animals hindering population growth of the prey�;

Distribution of the predators depends on availability of prey�; the more the prey the more the predators/ the more the predator the fewer the preys (Any two)

ii) Light – plant need light for photosynthesis / flowering / germination/ closing

and opening of stomata habitat with light of high intensity has more plants that will support more animals

iii) Competition- organism usually struggle for resources light/ water // air /

nutrients): This results in members better adapted to survive the competition (hence increase in number) 6 mks

b i) Succulent stems and leaves :- for storage of water and food ( for use by the

plants when conditions are harsh) rejects water alone or

food alone ii) Short life cycle: - to flower and form seeds within a short period when water is available 2 mks 5 a) 1- Sensory neurone / Afferent neurone

2- Relay neurone / intermediate neurone 3- Motor neurone/ efferent neurone

3 mks b) Check on the diagram arrows show points towards neurone 3 from 2 and 1 : 1 mk c) Grey matter 1 mks d) Impulse reaching the dentrite end of relay / Neurone 2 causes the synaptic

vesicles, releases acetylcholine / transmitter substances�; into the synaptic cleft�;the acetylcholine / transsmitter chemical diffuses across the cleft�; and causes the depolarization of the motor neuron/ neurone 3

any 3

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6 a) Graph on the graph paper

Axes – 1 mk Scale – 1 mk Curves- 1 mk Identification – 2 1 mk Plotting – 2 mks

8 mks B i) 1010mg ii) Amount of faeces produced decreased rapidly ; because the locusts were

laying eggs hence food eaten was used in egg formation c i) Food is assimilated ; to form new protoplasm / growth; as food

consumption increases body weight increase/ as food consumption decrease , body weight deceased

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ii) As the body weight increase production of faeces increase except during egg production where both decrease

d} Body weight/ faeces production would decreases; and fewer/ smaller eggs would be produced; e) Proteins- for building / eggs production; carbohydrates- for energy production / egg formation 7

� High light intensity: increases evaporation of water from the mesophyll cells of the leaf (rj light intensity alone) increasing the diffusion gradient between the intercellular spaces and the atmosphere ; hence high rate of transpiration

� Low humidity: (reject humidity alone) provides great saturation deficit; hence high rate of transpiration

� Wind/ air currents ; wind blows away saturated air/ water vapour around the leaf; increasing diffusion gradient between the leaf ; increasing diffusion gradient between the inside and out side of the leaf; thus increasing the rate of transpiration .

� Low atmospheric pressure; (reject atmosphere pressure alone) lowers the force that molecule require less energy to escape; thus leads to increase in the rate of evaporation hence high rate of transpiration .

� Water availability; large amount of water in the soil increase absorption into the root hair cell; thus more water is lost out of leaf through transpiration;

b) Absorption of water from the soil, movement of water up to the xylem due to

transpiration pull; absorption of mineral salts from the soil; cooling of the plant temperature regulation; Brings about turgidity in plants

20 mks 8. Nature selects these individual / organisms which are (sufficient) well adapted ;

and allows them to survive; and reject those that are poorly adapted ;. � individual of the same species show variations; that are caused by genes; � these variations can be passed from parents to offspring’s; through genetic

inheritance; some of the variations become more suited to the prevailing envital conditions:

� most organisms produce more offspring than the environ. can support ; hence there is always a struggle for existence ; due to competition among individual;

� individual posses traits / characters- that enables them to have ( competitive ) advantage to survive / stand better chances to survive in the struggle ; in the end , well adapted individuals survive; and reach reproductive age; and pass over their favourable traits to their offspring;

� poorly adapted individual / those without favourable traits perish/ die; and fail to reach sexual maturity ; hence do not pass their traits to their offspring; the fittest individual only survive ;

Total 20 mks

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2010 Pyramid Projections 005

Kenya Certificate of Secondary Education (K.C.S.E)

231/2

BIOLOGY

PAPER 2

Marking scheme:

1. a)i)

(2mks)

ii) Auxins/ produced by shoot and root tip accumulate on lower side of shoot

and root due to gravity;

higher concentration of auxins on lower side of shoot causes rapid growth; than

upper side of the shoot. Therefore, the shoot bends upwards;

higher concentration of auxins on lower side of the root inhibits rapid growth

therefore the root bends downwards;

(4mks)

b) Use of a klinostat;

(1mk)

c) Phototropism that exposed leaves to maximum light for photosynthesis;

Geotropism enables roots to extend deeply in the soil for anchorage;

(any one 1mk)

2 a)

i) Optimum temperature for maximum enzyme activity;

Temperature below optimum level inactivate the enzymes/ temperature above

optimum level denatures enzymes;

ii) Optimum PH for maximum enzyme activity; enzymes are denatured by pH

below optimum/ strong acid or alkalis;

(2mks)

b) When blood sugar level rises above normal. Insulin hormone is produced;

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insulin stimulate liver cells to convert glucose into glycogen. Also stimulate

liver cells to convert glycogen into fats/ stimulates liver cells to increase

oxidation of glucose;

when blood sugar level falls below normal level, glucagon hormone is

produced; glucagou stimulate the liver cells to convert glycogen to glucose;

(4mks)

3 a) – Increase the number of aquatic animal;

- Reduce the number of water weeds;

- Cover the jar with alluminium foil to block light from reacting the water

weeds; (any two 2mks)

b) i) Carbon IV oxide was absorbed from water; and used for

photosynthesis; without replacement. The PH of water rises; (3mks)

ii) Carbon iv oxide was produced; by animals during respiration; this reduced

the ph of water; (3mks)

4 a) Gene for red fur/colour and gene for white fur or colour are co-dominance;

Acc: partial dominance /incomplete dominance

(1mk)

b) Parental genotype RW X RR ;

Parental Gamete

F. Genotype

F. phonotype Red Red Roan Roan

Phenotypic Ratio Red Red : Roan

1 : 1

(4mks)

c) Plant breeding / Animal breeding

Improvement of yields;

(any 1)

d)

– Mutation;

- Crossing over;

- Fertilization;

- Independent assortment of genes

x

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5 a)

S-Pancrease;

U- Duodenum;

b) Emulsification of fats; to increase the surface area for reaction of lipase; provide

alkaline PH; for action for pancreatic amylase; lipase; Trypsin; / Neutralize acidic

chyme from stomach; to provide alkaline medium for efficient enzymatic activity;

(Any four) (4mks)

c) i) Pancreatic trypsin;

pancreatic amylase;

pancreatic lipase; (Any two) (2mks)

ii)

- Trypsin converts proteins to peptides;

- Pancreatic amylase converts starch to maltose;

- Pancreatic lipase converts lipids to fatty;

- Acids and glycerol;

(any two) (2mks)

6a) A-Intermittent /Discountinous growth;

B- Continous growth;

b) Rigid/hard exoskeleton; prevents growth; the exoskeleton is shed during

moulting; this allows rapid growth ;to occur before the new exoskeleton

hardens up; (4mks)

c) Ecdysone/ moulting hormone; (1mk)

(d) i) P&Q

Slow rate of growth, due to small number of cells dividing; the young

organism is still adapting to new environment;

iii) Q & R

- Rapid/ fast rate of growth; due to large number of dividing cells;

- The organisms has adapted to the environment; (4mks)

- No limiting factors like space, wastes e.t.c.;

e) i) Internal fertilization;

ii) Provide protection to the developing embryo from harsh environment;

- Less number of gametes are involved./ No waste of gametes;

- Developing embryo experiences little environmental changes;

(2mks)

(Any two)

f) i) Meiosis; (1mk)

ii) Crossing over; Synapsis/ formation of bivalents; (2mks)

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7. A balanced diet refers to a meal (food) that contains all the nutrients

(carbohydrates, proteins, lipids, vitamins, mineral salts and water) in their right

proportion;

A balanced diet is important to children because it contains:-

(i) Proteins that play the following roles in children:-

- Forms body structures;

- Used in formation of enzymes;

- Forms haemoglobin found in red blood cells;

- Forms hormones used to regulate life processes;

- Forms antibodies that provides immunity against diseases;

- Is oxidized during starvation to release energy,

ii) Carbohydrates that plays the following roles in children:-

- Provide energy required by children during oxidation;

iii) Roughage that plays the following roles:-

-Makes food to be bulk hence a child gets satisfied easily;

- It promotes peristalysis which prevents constipation in children;

iv) Lipids

- are oxidized to release energy;

- Forms part of the structure of the cell membrane;

- Used as an insulator against excess heat loss in children/

- Used to form layers around organs which protect them from mechanical

injury;

- Act as a source of metabolic water

- Stores various types of vitamins ie. Fat soluble

v) Water

- used to dissolve food;

- It cools the body;

- Provides a media for transport;

Facilitates hydrolysis process;

Maintains the shape of the cells;

vi) Mineral salts includes:-

- Calcium for teeth and bone formation to prevent rockets

- Phophorus for teeth and bone formation

- Iodine used in formation of thyroid hormone which prevents goitre;

- Iron used for haemoglobin formation that prevent anaemia;

vii) Vitamins that play at the following roles:-

- Vitamins A prevent poor night vision in children;

- Vitamin B1 prevents beriberi

- Vitamin B2 prevents pellegra

- Vitamin B12 used in formation of blood cells to prevent pernacious

anaemia;

21


- Vitamin C protects against infection that prevent scurvy

- Vitamin D prevents rickets;

- Vitamin E prevents sterelity;

- Vitamin K prevents excessive bleedings;

(Any 20)

8 The gaseous system in mammals includes the nostrils, trachea, bronchus,

bronchioles, the lungs, the rib cage, the intercostals muscles, the diaphragm, the

vertebral column;

The Nostrils:

Is lined with hairs that trap foregn particles in the hair; is lined with goblet cells that

secret mucus which trap solid particles and pathogens;

Tranchea / wind pipe

- Made up of incomplete rings of cartilage a long its length which resist

collapse when pressure is low in chest cavity;

- The incomplete cartilage rings have gaps on the side facing the oesophagus

which allow smooth swallowing;

- It is lined with goblet cells that secrete mucus that trap foreign particles and

pathogens;

- It is lined with cilia which move mucus upwards into the pharynx;

- Has epiglottis that prevents food entering into it during swallowing;

Lungs

- They are covered with pleural membrane;

- The pleural membrane secrets pleural fluid that lubricates the lungs (reduce

friction) during movement;

- The lungs are elastic hence expand and contract to hold and expel large

volume of air;

- The lungs are made up of very many alveoli that gives a large surface area

for gaseous exchange (70m2)

- The alveoli are:-

- Moist which dissolve the diffusing gases;

- Covered by a thin layer of wall for easy diffusion of gases a cross it;

- Highly vascularised which enhances faster transportation of gases;

Diaphragm

- a muscular layer that contract and relax to facilitate inhalation and

exhalation respectively;

The ribcage

-they protect the lungs from mechanical damage;

Provide surface for the attachment of the intercostals muscles;

The ricage and intercostals muscles forms an air-tight thoracic cavity that

maintains constant air pressure;

22


Intercostals muscles

They contract and relax which changes the thoracic volume to enhance

inhalation and exhalation in mammals;

231/2

BIOLOGY

PAPER 2

THEORY

TIME: 2 HOURS


Kenya Certificate of Secondary Education

MARKING SCHEME 1. A student set up the following apparatus for an experiment.

a) The student boiled the yeast solution before the experiment. Explain. (1mk)

To remove any dissolved oxygen, to provide suitable conditions for

anaerobic respiration;

b) The mixture of yeast suspension and glucose solution was cooled to about 370

C. Explain.

(1mk)

To provide optimal temperature for enzyme/yeast activity. Higher

temperature denatures enzymes and prevents yeast action;

c) Why was a layer of oil added to the mixture of glucose and yeast? (1mk)

To exclude atmospheric oxygen from the mixture;

d) State the observations expected in

(i) Tube A (1mk)

Bubbling, warmth/heat is produced;

(ii) Tube B (1mk)

Lime water turns cloudy/a white precipitate formed;

e) Account for the observations in d(i) and (ii) above. (1mk)

(i) Bubbling is due to production of carbon(IV) Oxide from anaerobic

respiration of yeast;

(ii) Carbon(IV) Oxide produced in tube A from anaerobic respiration reacts with

calcium hydroxide to form calcium carbonate, which is insoluble in water;

f) (i) Apart from carbon (IV) Oxide and energy, what other products can be found

in tube A?

(1mk) Ethanol;

(ii) Explain how you can confirm this. (1mk)

-By smelling it/ performing iodoform test;

2. (a) Explain the role of enzymes in living cells. (1mk)

-Enzymes speedup (Slow down) the rate of chemical

reactions/metabolism in the body; without being used up.

23


(b) The graph below shows the effects of temperature on the rate of reaction

of the enzyme

salivary amylase

(i) Account for the change in the curve between C and D. (1mk)

The rate of reaction decreases/drops as temperature increases

(beyond 400C); due to denaturation of enzymes;

(ii) What does the dotted line represent? (1mk)

The normal human body temperature/ the optimum temperature for

the enzyme/salivary amylase;

(c) Explain how the following factors affect the rate of enzyme activity:

(i) Temperature (2mks)

Enzymes being proteins are sensitive to temperature changes/work

best within a narrow range of temperature/have maximum activity at

optimum temperature; Temperatures above optimum destroy enzyme

structure/denature enzyme while those below optimum lower

enzyme activity due to inactivation of enzymes;

(ii) Substrate concentration. (2mks)

Increase in substrate concentration increases enzymes activity up to a

certain level; This is due to an increase in the number of active

sites; until all active sites of the enzyme are occupied;

3. In cats, sex is determined by X and Y chromosomes in the same way as in

humans. One gene for coat color in cats is present on the X chromosome but

not on the Y chromosome. This gene has two alleles Orange (B) and black

(b) and X chromosome bearing the B allele is represented by XB and one

bearing the b allele by Xb.

Female cats that are homozygous for the Xb allele have black coats;

female cats that are heterozygous have tortoiseshell coats. (Orange

with dark patches).

Give the genotype of a female cat with tortoiseshell coat (1mk)

XB Xb;

A male cat with an orange coat. (1mk)

XBY;

A male cat with a black coat. (1mk)

XbY;

A black coated male cat is mated with a tortoiseshell coated female

cat. Use a genetic diagram to explain what would be the expected

24


ratios of the genotypes and the phenotypes of the kittens that could

be produced by this cross. (5mks)

♂ ♀

Phenotype black coated male Tortoiseshell

Male female

Genotype XBY X XBXb

Orange Tortoiseshell Orange

Black

2 orange : 1 Tortoiseshell :

1Black

4. (a) What is meant by the term homeostasis? (1mk)

Homeostasis is the process of maintaining a constant internal

environment; eg. Body

temperature, osmotic pressure;

B

The figure below shows the body temperature of a person before, during

and after taking a cold bath. The temperature of the bath water is 220C.

(b) For how long was the person in the bath? (1mk)

-20 minutes;

(c) Explain why the person’s body temperature fell. (1mk)

-In a cold bath, the body tends to lose much more heat through

convection and conduction, causing the body temperature to fall;

(d) Explain the role played by the following in helping to return the body

temperature to normal.

(i) The liver (2mks)

In the liver, heat is produced as a result of numerous chemical

activities occurring there; The heat is distributed by blood to

other parts of the body;

(ii) Blood vessels in the skin. (2mks

The low external temperature brings about a reflex

constriction of the skin arterioles; less heat flows through the

skin and less heat is lost;

(iii) Muscle of the body. (2mks)

Tissue respiration is very active in muscles setting free large

amounts of heat; which is distributed to different parts of the

body;

5. The diagram shows the internal structure of a leaf

25


(a) Name the pats labeled A, B, D, & F

A Cuticle

B Upper epidermis

D Vascular bundle/xylem

F Stoma

(b) State the functions of the parts labeled A, C, D and F

A Water proof – hence prevent water loss;

C Contain chlorophyll, that traps light for photosynthesis;

D Xylem transport water and mineral salts, phloem transports

synthesized food;

F Site for gaseous exchange;

(c) State two structural differences between guard cells and other epidermal

cells. (2mks)

Guard cells have chloroplast while other epidermal cells don’t;

Guard cells have walls of unequal thickness. (Thin inner walls and

thick outer walls) while other epidermal cells have walls of equal

thickness;

6. The figure below shows the changes in volume of air whilst breathing in and

out during a

breathing exercise. The letters A to H represents successive breaths.

From the information given in the above figure,

(d) (i )How much air leaves the lungs each time you breathe out during

normal quiet breathing? (1mk)

5 – 6dm3;

(ii) What has the person done to achieve peak X on the graph? (1mk)

At peak X, the person has inhaled deeply;

(iii) What is the greatest volume of air that is expelled in a single,

outward breath? (1mk) 5dm3;

(e) State one difference in the flow of the respiratory medium (air or water

in mammals and fish. (1mk)

During inhalation, the air is taken into the lungs of the mammal but

in fish, dissolved air in water flows over the respiratory organs i. e

gills;

(f) State one reason why exercise cause an increase in breathing rate. (1mk)

-Respiration increases and thus the amount of CO2 that needs to be

removed increases the rate of breathing increases to enable the

exchange of gases to occur at a faster rate;

(g) (i) What happens to breathing rate when a person breathes in 100%

oxygen? (1mk)

The breathing rate decreases gradually;

(ii) Suggest what happens in the lungs and blood of the person.

Account for your answer in (i) above? (2mks)

• The concentration of oxygen in the lungs and blood increases

while the carbon dioxide concentration decreases beyond the

normal level;

26


• The rate of diffusion of oxygen into blood and carbon dioxide out

of blood increases;

• This decreases the rate of breathing;

(h) Distinguish between breathing and respiration. (2mks)

Breathing is any process which speeds up the rate of gaseous

exchange between an animal and its surrounding; Respiration is

oxidation/breakdown of food within cells to release energy;

(i) Describe the breathing mechanism in man during inhalation. (4mks)

• Internal intercostals muscles contract while external intercostals

muscles relax;

• The ribcage moves upwards and outwards;

• The diaphragm contracts and flattens; the volume of the thoracic

cavity increases and air pressure within it decreases relative to

atmospheric pressure;

• The lungs inflate and atmospheric air moves in;

(j) Explain the disadvantages of anaerobic respiration over aerobic

respiration. (6mks)

• Less energy is produced in anaerobic respiration since food is

partially oxidized; while more energy is produced in aerobic

respiration where food is completely oxidized;

• Some metabolic wastes (intermediate products) accumulate in the

cells; affecting cellular functions (Ethanol produced in plants poison

the tissues; while lactic acid causes muscle fatigue/cramp in animals

that may stop muscle contractions); such intermediate wastes are not

produced in aerobic respiration;

7. Discuss the various ways in which seeds and fruits are adapted for

dispersal. (20mks)

seeds and fruits are dispersed through various modes – self – explosive

mechanism;

• Wind;

• Water;

• Animals;

Adaptations include :

• Dry pericarp / pod along lines of weakness sutures splits forcefully

dispersing seeds;

• Small and light weight; to reduce the density ;hence float in air

• Hair – like structure /floss /pappus/parachute; to increase the surface

area; hence buoyancy easily blown by wind

• Winged/ extensions; Increase surface area . resulting in buoyancy;

• Air – spaces; trap air in mesocarp reducing density /weight hence

buoyancy /float;

• Tough and water proof epicarp; protect the seed from absorbing

water; hence float in water and easily transported to far off places.

• Impermiable seed coat / testa ; allow the seed to remain viable;

27


• Hooks and spines; on pericarp and persistant calyx to cling to the

body / hair /fur; on the animal body.

• Sweet scented; to attract animals; to feed on them.

• Succulent/fleshy; to attract animals to feed on them.

• Bright coloured; attract animals to feed on them

(iii)Describe how the mammalian ear is adapted to perform its functions.

(20mks)

• The pinna is flap made of skin and cartilage; for collection and

concentration of sound waves;

• Auditory canal/meatus is a tube lined with hairs which trap solid

particles like dust; It has wax secreting cells; that secrete wax for

trapping solids and insects entering the ear;

• The eardrum/tympanic membrane is thin with double layer of

epidermis; It vibrates translating sound waves into sound vibrations;

Sound vibrations are transmitted to ear ossicles;

• Ear ossicles are malleus, incus and stapes; they amplify and

transmit vibrations to the oral window;

• The oval window is a thin membrane which transmits sound

vibrations to the fluid of the inner ear; perilymph and Endolymph;

• Eustachian tube connects middle ear with pharynx equalizing air

pressure in the ear with atmospheric pressure; to prevent distortion

of the eardrum;

• The cochlea is highly coiled tube with system of canals (and

sensory cells) to occupy a small space/increase the surface area for

accommodating many sensory cells to detect sounds vibrations; and

generate impulses transmitted to the brain;

• Auditory nerve transmits nerve impulses to the brain for

interpretation;

• Semicircular canals are tubular cavities containing fluid;

• The canals are arranged at right angles to each other in the three

planes of spaces; to detect changes in position of the body; the

canals have ampulla’s: utriculus and sacculus; to detect position of

body in relation to gravity;

• Utriculus has otoliths attached to sensory cells which generate

impulses which are then transmitted to the brain through the

auditory nerve;

• The per lymph and endolymph fluid in the inner ear absorb

mechanical shock/transmit sound vibrations/protect delicate parts;

Total 25 max 20 marks


MARKING SCHEME BIOLOGY

28


Paper 2 Q. 1

(a) Kingdom protoctistab (1mk) (b) B – Vacuole b (1/2 mk) Y – Pyrenoid b (1/2 mk) ( c) A – for movement b ( 1mk) X – for photosynthesis b ( 1mk) (d) because the nucleus is surrounded by a nuclear membrane b ( 1mk)

Q2. (a) x – incisor b ( 1mk) (b) crush the bones b (1mk) Slice flesh from bones ( c) (i) Flesh b ( O. W. T. T. E. ) (ii) – sharp canines for piercing flesh b ( 1mk) - strong jaws for holding the prey firmly b (1mk) - chisel shaped incisors for and tearing fleshb ( 1mk) ( Any 2 ) Q3. (a) (i) Juvenile hormone b ( 1mk) (ii) Ecdysone / moulting hormone b (1mk) (b) Prothoracic gland b ( 1mk) ( c) Has a flattened body to fit in crevicesb (1mk)

- Has wings and legs to enable it escape predatorsb ( 1mk)

- Has long antennae for sensing - Has a brown colour to camouflage against predators ( Any 2)

4. (a) Individual 7 - XH Y ( 1mk) Individual 9 - XH XH or XH Xh ( 1mk)

♀

♂

XH Xh

Xh XHXh

XhXh

Y XHY

XhY

(c) The gene for hemophiliac disease is linked to the x – Chromosome and male have only one x – chromosome.b

5 (a) Living beetles used oxygen for respiration giving out carbon (iv) oxide that was absorbed by sodium hydroxide

pellets; hence a decrease in pressure.bb ( 1mk) (b) Control experiment b (1mk) ( c) Y – coloured liquid didn’t move; becauseb the carbon (iv) oxide given out by beetles was used up by the green plant; the oxygen given out is utilize in respiration by both organisms. Z – Coloured liquid rises, carbon (IV)b oxide is utilized by

plants creating a low partial pressure in the flask. 6. (a) Thoracic vertebrate b ( 1mk) (b) (i) Neural spine b ( 1mk) (ii) For attachment of musclesb ( 1mk) ( c) The arrow should point to the L. H. S of the neural spine.b (1mk) (d) part Y – for articulation with capitulum facet of the rib.b 7. (a) 30 – 10 = 20 min. b ( 1mk) (b) Lost heat to the cold water throughb conduction ( 1mk) (c ) (i) The liver – increased metabolic rateb to generate more heat (1mk)

(ii) The blood vessels – vasoconstricted to prevent loss of excess heat.b (iii) The body muscles – contracted and relaxed ( shivering) to generate more heat. ( 1mk)

8. (a) Divergent (1mk)

(b) As insects explored different ecological niches, a different food was available for consumption (1mk) the mouth parts changed to adopt themselves for the consumptionb of the food available in their ecological niches.b ( 1mk)

29


(c ) – Beaks of birds ( 1mk) - Dental structures of mammals (1mk) - Pentadactyl limbs - Dispersal of seeds ( Any 2)

(d) – Reduced in size / rudimental ( structures) and became functionless through evolution.b ( 1mk)

9. (a) A natural unit where and abiotic factors interactb (1mk), to form a stable self sustaining system. b (1mk) (b) (i) Photosynthesisb (1mk) (ii) Respiration b ( 1mk) (c ) decomposers / Detrivores / decomposing bacteria (d) (i) A = ( 1500 + 500 + 1000) KJ = 3000KJ (1mk) B = 500 – ( 320 + 100) KJ = 500 – 420 KJ = 80KJ b ( 1mk) (ii) Total energy = 2000 KJ + 1500 KJ + 3000KJ = 6500KJ b (1mk)

(iii) Amount lost as heat = (2000 + 1500 + 320) KJ = 3820KJ 3820 x 100 = 58.77%b 6500 ( 1mk)

(iv) 80 x 75 = 60KJ lost as heat 100 C = ( 80 – 60) KJ = 20KJ.b (e) Because during transfer of energy at each trophic level some

amount of energy is lost as heat; hence most of the energy will have been lost before man benefits from the bag of maize. b( 1mk)

(f)

1/2 mk for every correct arrows

(g) (i) There would be an increase in the number of squirrels and gazelles due to reduced preparation.

(ii) Fire would clear all the grass ( producers) causing food shortage for insect larvae, gazelles and squirrels that would die leading to reduced food for lizards and wild dogs that would die or migrate.

10. (a) Role of Gibberellins - Stimulate rapid cell division and cell elongation in dwarf varieties. - Induce flowering - Induce parthenocarpy - Break bud dormancy, promotes side branching /lateral growth - Inhibit sprouting of adventitious roots from stem cutting

- Reduce leaf fall /retards leaf abscission ( any 5; 1 x 5) (b) Meristematic cells - Lack vacuoles - they have a capacity to divide continuously - have a dense cytoplasm - have thin cell walls - tightly packed together - are small and cuboid in shape. ( 1 x 5)

(c ) Production of auxinsb ( 1mk) from shootb apex; auxins migrate baway from light; stimulate cell, growth and cell elongation on the side opposite direction of light; more growth, occurs leading to curving of the shoot towards the light.

11. Scrotum/ Scrotal sac : Located outside the body to provide

optimum temperature for sperm production Testes ; Produce sperms and hormones (androgens/testosterone) Penis; Introduce sperms into the female reproductive system

30


Epididymis; Storage of sperms Sperm duct; contract to expel sperms Prepuce; protects the glans Glans; Sensitive to touch Urethra; Conveys sperms into the female reproductive systems Prostate gland; produce semen to nourish the sperms Cowpers gland; produce alkaline fluid to neutralize acidity caused by urine in the

urethra. ( Structure must be tied to function (s) ).

2010 Pyramid Projections-012 Kenya Certificate of Secondary Education (K.C.S.E) 231/2 BIOLOGY (THEORY) MARKING SCHEME: 1. (a) The buffalo consume less oxygen(47gm/hr)per unit body weight than mouse(1510gm/hr); (b)i) Buffalo;

Has a smaller surface area to volume ratio of the body hence it retains more heat leading to low respiration rate hence less oxygen consumed;

(ii) The mouse;

Has a large surface area to volume ratio of the body hence it loses more heat to the surrounding leading to high rate of respiration, hence more oxygen consumed;

(c) The volume/amount of oxygen consumed increase with a decrease in body weight or vice versa;

2. (a) To show that germination require an optimum amount of water; (b) (i) A-most seeds germinated; D-the seeds got rotten;

(ii) A-Germination takes place in adequate water i.e.; D- Excess water cuts off oxygen supply to the seeds which end up

rotting instead of germinating; (c)

� breakdown of insoluble (stored) food substances into soluble form by hydrolysis;

� oxidation of food substances to release energy for cell division ; � conversion of hydrolysed food into new plant tissues;

3 (a) A-Hepatic artery; B-Hepatic portal vein; C-Hepatic vein;

(b) i) B - hepatic portal vein; ii) B - hepatic portal vein; iii) C- hepatic vein;

(c) No glucose comes from alimentary canal making glucose concentration in

31


Vessel B low. Vessel C obtain glucose derived from hydrolysis of glycogen in the liver;

4 (a) YY and yy; b) (i) yellow; (ii) Gene for yellow colour is dominant over that of white colour; (c) If ¾=880 ¼=? 4/3x880x1/4 =294 plants; (d)

� Down syndrome(mongolism); • Klinefelters syndrome; � Turners syndrome ;

5. (a) i) produce spindle fibre; ii) Contain lytic enzymes which digest and destroy worn out organelles cells

(b) Ability/capacity to distinguish between two structures that look very similar;

(c)

� transportation of manufactured food in the phloem ; � absorption of mineral salts by root hairs;

(d)

� water is hypotonic to contents of the red blood cell; water molecules/heamolyse entered into R.B.C by, osmosis, hence swell and burst due to weak cell membrane.

6 (a) see graph

NB: � all points plotted correctly-1 mark; � smooth curve-1mark; � axes labelling-2marks; � scale-2marks;

(b)(i) Gas evolved increased with increase in light intensity; an increase in

light intensity leads to more splitting of water molecules into hydrogen ions and oxygen; hence increase in the rate of photosynthesis;

(ii) The amount of gas evolved remained constant as the rate of

photosynthesis is constant due to limiting factors; Glucose formed is converted into start for storage in the tissue

(c) Glucose formed is converted into starch is osmotically inactivate

hence stored;

(d) By heating the leaves in an oven at 1100 C and weighing them; the heating and weighing is repeated several times till constant weight is obtained;

(e) Has the stroma where carbon (iv) oxide fixation takes place; has the

grana containing chlorophyll that traps light energy; for photosynthesis;

32


-Has intergrana lamellae that hold the grana in position;

7 (a) Homologous structures;

Structures of same embryonic origin; that become modified in the course of evolution to perform different functions in different ecological niches;

(b) Analogous structures; Structures of different embryonic origin; that become modified in the course of evolution to perform similar functions in the same ecological niches;

(c) � Organisms in the same environment are always competing for

resources such as food,mates,shelter etc; � There follow struggle for existence and those organism that are

well/best adapted to survive and breed in the prevailing environmental condition survive to the reproductive maturity and give rise to off spring of the next generation;

� The less/poorly adapted die young hence leading to survival for the fittest ;

� The favourable characteristic possessed by the fittest organism are genetic hence are passed on to the offspring ;

� This lead to a natural occurrence of variation within a species; � The variations that are genetic are passed on/transmitted to

successive generations of offspring; � Consequently there is a gradual change in the characteristics of

the species making it better adapted to its environment; � Accumulation of small variation over along period of time lead

to the emergence of new forms of life i.e. species; � If suited to and well adapted to the new environment these new

form reproduce successfully and pass on their characteristics; � If not suited these new form are eliminated by nature, leaving

mutant forms which are better adapted to the environment; � Through this process nature selects those organism with better

adaptations; � While ensuring the elimination of those not able to adapt to the

changing environment ; � Thus the changing environment forces an organism /species to

adapt/be eliminated; 8.

� Anterior lobe of pitutary gland secretes follicle stimulating hormone; FSH causes Graafian follicle to develop in the ovary; and also stimulates ovarian tissues to secrete oestrogren; oestrogen cause the repair of uterine wall;oestrogen also stimulates anterior lobe of pituary gland to secrete luteinizing hormone; LH cause ovulation also causes graafian follicle to change into corpus inteum LH stimulates corpus liteum to secrete progesterone which causes proliferation of uterine; wall in preparation for implantation; progesterone inhibit the production of FSH; thus no more follicles develop and oestrogen production drops/reduces;

� In the next two weeks progesterone level rises; and inhibits production of LH; from the pituitary gland; corpus luteum stops secreting progesterone and menstruation occurs; when the level of progesterone drops; anterior lobe of pituitary gland starts secreting FSH again; and cycle repeats;

2010 Pyramid Projections 013 Kenya Certificate of Secondary Education (K.C.S.E)

33


231/2 BIOLOGY PAPER 2 MARKING SCHEME 1 a) M- Chitinous rings / Rings of chitin; N- Tracheole; 2 marks

b) Keep the trachea lumen open/ prevent trachea from collapsing; 1 mark

c) Tracheole ; trachea; spiracle; 3 marks

d) M- Trachea; N- Bronchioles; 2 marks 2 i) FSH - Cause development of graafian follicle; - Stimulate tissues of ovary to secrete oestrogen; max. 1 mark

ii)Progesterone - Cause proliferation/ thickening of the uterine wall (in preparation for implantation); - Inhibits production of FSH; - Increase blood supply into endometrium ; max 2 marks

iii) Luternising hormone (LH) - Stimulates corpus luteum to secrete progesterone; 1 mark - Cause ovulation;

b) Anthers; Ovary; 2 marks

c) (i) Prophase I- ; (RJ prophase alone) (ii) Formation of chiasmata / chiasma/ crossing over occurs / synapsis occurs / homologous chromosomes intertwine/Bivalent is formed; 2 marks

3 a) (i) Thoracic region;

� Presence of long neural spine for attachment of muscles; � Presence of neural can al for the passage of spinal cord; � Transverse process for attachment of muscles; � Facets for articulation with other vertebral bones / vertebrae; � Neural arch and Centrum to protect spinal cord; � Demifacets for attachment of ribs;

4 marks

b) Exoskeleton; Hydroskeleton; Endoskeleton; 3 marks 4 a) ¼ x 7324; = 1831;

34


b) Gene for smooth seed coat is recessive/gene for wrinkled seed coat is dominant over that for smooth seed coat; 1 mark c) X- insertion;

Y- inversion; 2 marks

d) Deletion; Translocation; Non- disjunction; 3 marks

5 a) 100

15x 100; = 15%;

2 marks

b) (i) The allele O appears in many blood groups/ blood groups A,B and O/blood group A and B are heterozygotes; there fore higher chances of being inherited in a population;

(ii) Allele A and B are co- dorminant hence express themselves only in blood group AB;

3 marks

c) (i) To increase the surface area for increased supply/ efficient

transport of oxygen ( to tissues );

(ii) At high altitude air is less dense / partial pressure of oxygen is low / there is low concentration of oxygen hence the number of RBC s / structure C increases to increase oxygen carrying capacity of blood

(iii) Phagocytosis

3 marks

35


6 a) Axes x 2 Scale x 2 Plotting x 2 Curve x 1 Curve identity x 1

Total 8 marks

b) (i) Population increases rapidly because of fewer predators / less predation;

(ii) Population decreases / declines/ reduces due to shortage of food / inadequate supply of food / few number of prey;

c) (i) 100 ±1;

(ii) 65 ± 1;

d) (i) Population reduces; due to many predators;

(ii) Population reduces; due to competition over the same food source;

e) Using a sweep net the houseflies were caught, marked using ink that cannot be erased easily counted and recorded; The flies were then released to the population. After 48 hours the procedure was repeated the flies in the second capture counted and recorded; In the second capture, the marked re captured were also counted and recorded; Population of flies = first marked x second captures; Marked recaptured

7 a) Describe how insect pollinated flowers are adapted to pollination 14 marks

� Large/ broad / wide flowers /conspicuous flowers; � Brightly coloured / corolla/ in florescence/florets; to attract insects;

(Rj coloured flowers alone) � Scented ; to attract insects; � Have nectary guides/ nectaries ; that secrets nectars ; to direct insects

into the flower; � Pollen grain have rough / spiky/ sticky surface ; to stick to insect

body; � Special shaped corolla tube; to enable insects to land; � Anthers situated inside the flowers; to ensure contact with insects; � Sticky stigma ; for pollen grain to stick/adhere; � (award to attract insect once )

Total 15 pts max 14 marks)

b) Explain how a seed is formed in a flowering plant

(6 marks) One male nuclei; fertilized two polar nuclei; to form a zygote (which undergoes mitosis) to form embryo ; which have radical and plumule; cotyledon/endosperm develop; testa develops from intergaments;

36


8 Discuss the various evidences which show that organic evolution has taken place (20 marks)

TAXONOMY/ COMPARATIVE ANATOMY; Members of a phylum show similarities; the organisms have similar structure / organs performing similar functions / e.g. presence of digestive, urinary, nervous systems;

� Vestigial organs/ coccyl, appendix; � Pentadactyl / limbs in different animals eg monkey and rats, have

bones arranged in a similar way/Theses are examples of homologous structures / structure of similar origin but perform different functions;

� Convergent / analogous structures like wings of birds and insects all these point towards common origin;

COMPARATIVE EMBRYOLOGY; Vertebrate embryos are (morphologically) similar; suggesting the organisms have a common origin / Acc when two embryos of mammals, reptiles and amphibians are compared they show similarities; FOSSILS RECORDS; These are remains of organisms preserved naturally occurring materials for many years; They show morphological changes of organisms over a long period of time; e.g. skull of man; GEOGRAPHICAL DISTRIBUTION; Present continents are thought to have been a large land mass joined together ; continental drift lead to isolation that lead to patterns of evolution; e.g. camels in S America resemble those in Africa / tiger in Asia resemble Jaguars in S. America; CELL BIOLOGY; Occurrence of organelles/ eg mitochondria, cytoplasm, nucleus; point towards a common origin; 20 marks

2010 Pyramid Projections 014 BIOLOGY PAPER 2

MARKING SCHEME CONFIDENTIAL

1. a) Endotherm/ Homeoiotherm (1mk)

b) It ensures that metabolism occurs without being affected by external factor; c) Sweating; excretion; breathing; defecation; (any 2mks) c) - Erector filli muscles relax; thus body hair lies flat decreasing insulation and

increasing heat loss; - Vasodilation occurs; - it increases blood flow in superficial blood vessels,

hence increasing heat loss by convection and radiation; - Sweat glands secrete more sweat; which evaporates increasing loss of heat

through latent heat of vaporization. (3mks)

e) To prevent overheating in the body / to prevent heat stroke (1mk) 2. a) A – Gill filament B – Gill bar (2mks)

37


b) Filter solid particles/ food particles from inflowing water (1mk)

c) Water in the gill filament flow in opposite direction to the flow of the blood in the filaments; creating a steep diffusion gradient for faster/ maximum

gaseous exchange;

d) Concentration of O2 in water is higher than its concentration in the protozoa; O2 diffusises a cross the cell membrane into the protozoa ; concentration of CO2 in the protozoa is higher than in the water surrounding due to respiration ; CO2 diffuses out into the water across the cell membrane; (Any 3 mks for any 3 points)

3. a) Curve Y; - it starts with a higher population than X, the predator; (2mk)

b) The number of predators dropped due to increase competition for food; as a result of a decrease in the number of prey; (2mks)

c.i) Prey – predator relationship or predation (1mk) ii) It maintains the numbers of prey and predators at the carrying capacity of the

habitat; It removes weak organism from the population by natural selection;

d) There would be an increase in the herbivorous population due to reduced

predation; Herbivores would feed on a lot of vegetation which would reduce primary productivity – hence causing soil erosion; ( 2mks)

4. a) Aa; Aa; (2mks)

b) Parental phenotype

Genotype of normal pigmented children :- AA; Aa; (2mks) Genotype of albino child – aa; (1mk) c) 1/5 (1mk) 5. a) Oestrogen - Heals and repairs uterine wall

- together with progesterone causes prolifenation of the uterine lining in preparation for implantation; - Makes uterine muscle more sensitive to oxytocine; (any 1 pint, 1mk) Progesterone: - Causes prolifenation of the uterine wall; - Maintains thickening of the uterine wall; - Prevents miscarriage / maintains pregnancy; ( any 1 pt 1mk)

b)i Umbilical artery; (1mk) ii) Umbilical vein; (1mk) c) -Umbilical artery – Co2; Nitrogenous wastes

- Umbilical vein – Oxygen; food nutrients; - Antibodies; (any 1 pt, 1mk)

d) Oestrogen inhibits the production of FSH, no follide grows; - Progesterone inhibits the production of FSH and LH. No ovulation can occur; (2mks)

6. a) Nitric acid; carbonic acid; (2mk) b) on the graph;

Title - ½ mk Scale - ½ mk Correct curves - ( 2mks) Y axis – mean length in mm (2mks) X axis – acid concentration (mol dm x 10) (2mks)

P. genotype Aa Aa�1 ½

Gametes

Fusion

Normal Normal

38


c)i Growth of shoot – as the acid conc. increases, growth of the shoot decreases; due to the low pH which is toxic to the cells; (2mks)

ii) The length of the root increases slightly at the beginning; but as the acid concentration increase, the growth of the root decreases; low pH is not suitable for the growth of root; (2mks)

d) At 5 x 10-3 - Mean shoot length 2.0mm ± 0.1; - Mean root length 2.0 mm ± o.1; (2mks) e) - Kills organisms in water and soil;

- Corrodes walls and roofs of buildings; - Causes leaching of aluminum from soil; (any 2 points 2mks)

f) - Use of substances that extract sulphur from sulphur containing substances; - Fitting chimneys with scrubbers that dissolve gases like sulphur dioxide and

nitrogendioxide; - More use of electricity instead of fossil fuels; - Fitting automobiles with filters and catalytic converters in their exhaust pipes

to reduce emission of sulpur oxides; ( any 3 points,

3mks)

SECTION B 7. In the mouth;

There is mechanical breakdown of food using teeth; to increase the surface arae for enzyme action; food is moistened; and lubricated by saliva; for easy swallowing; The enzyme ptyalin; converts starch; to maltose; food is then rolled into bolluses; and pushed down the gullet by the tongue; There is no chemical digestion of protein in the mouth; since no protein digesting enzymes are there; In the stomach HCL acid provides a suitable PH; to activate pepsinogen; to pepsin; protein in the egg is converted by pepsin; to peptides; and peptones; No digestion of starch; because the pH is not suitable for ptyalin to continue acting on starch; Food is churned; by rhythmical contraction of the muscular wall of the stomach; to form chyme; In the duodenum; Bile; provides a suitable PH and neutralizes the acid chime; Pancreatic amylase; converts starch to maltose; Trypsin; converts proteins into peptides and peptones; Bile salts; emulsify the fats in the egg; pancreatic lipase; digests fat to fatty acids and glycerol’s; In the ileum; Maltase; converts maltose into glucose; Peptidase; convert peptides into amino acids; Lipase converts fats to fatty acids and glycerol; ( Any 20 points 20mks)

8. Evidences of evolution Comparative anatomy/ Taxonomy;

Members of a phylum/ group show similarities; organisms have similar structures/ similar organs performing the same functions; eg digestive system, urinary system, nervous system/ any correct example; vestigial organs e.g appendix, coccyx tail; vertebrate heart; The pendactyl limb/ any correct example; Analogous structures/ different structures performing the same function; e.g wings of insects bats and birds – show convergent evolution;

39


Homologous structures/ structures with same origin but performing different functions; show divergent evolution; Fossilrecord / palaentology; These are remains of organisms preserved in naturally occurring materials for many years; show morphological changes of organisms over a long period of time; e.g skull of man, leg of horse;

(4 mks) Comparative embryology;

Vertebrate embryos are morphologically similar; suggesting the organisms have a common origin / ancestry; ( 3mks) Geographical distribution; Present continents are thought to have been a large land mass joined together; and as a result of continental drift isolation occurred; bringing about different patterns of evolution; eg kangaroos in Australia, Jaquar in South America, Liama in Amazon; (any correct example) Coparative serology/ physiology; Antigen – antibody reactions/ Rh. Factors blood group/ haemoglobin structure; reveal some relationship among organisms/ common ancestry; (2mks) (Total 23 max, 20mks)

2010 Pyramid Projections 015 Kenya Certificate of Secondary Education (K.C.S.E) 231/2 BIOLOGY PAPER 2 MAKING SCHEME 1 A) Photosynthesis; 1 mk b) i) Starch test ; 1 mk ii) plant B - Does not change the colour of iodine solution;/ colour of

iodine solution persists,/ colour remains brown ; 1 mk – plant C- Stain blue black with iodine solution; 1 mk c) Sodium hydroxide absorbs carbon (iv) oxide which is a requirement for

photosynthesis, hence no photosynthesis; thus absence of starch 2 mks d) Plant D withered . The high amount of concentrated sodium chloride made

the soil water/ solution hypertonic; leading to no absorption of water by the root of the plant; while the rate of transpiration is high; 2 mks

2 a) (i) 8.0; 1 mk ii) It is the PH value at which the film took the shortest time to clear; 1 mk b) Use the same volume of trypsin but boiled ; to ensure it is denatured ;1 mk c) It is the action of the enzyme that is under test;/ to show that the change

was due to trypsin acting enzymatically and that the PH was therefore affecting trypsin and not the film directly;

1 mk d) To maintain a constant temperature ; 1 mk e) To bring them up to the experimental temperature (35 0 c) , otherwise

temperature would be changing during the experiment; 1 mk f) The film would not clear ;/ infinite ; 1 mk

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Reason- The high temperature (650c) would denature trypsin, and it would fail to catalyze the breakdown of gelatin; 1 mk

3 a) Mango tree; sap sacking ants; Insect eating Bird carnivorous insect Herbivorous moths ; b) C i) Mango tree; 1 mk ii) Sap sacki sap- sucking ants and herbivorous moths ; 1 mk

(Both must be correct to score full mark)

4 (a) increase in somatic cell/ growth and development / increase in size of organisms

For repair of worn out tissues / regeneration; For a sexual production / sporulation/ budding/ fission; any one = 1 mks

b) i) E – ureter / Ureter tube; F- Cervix; ii) A- Excretion of urea/ excess water/ excess salts ; 1 mk

B- Site for fertilization of ovum by spermatozoons; 1 mk

C- Formulation of Ovum/ Ova; - Production of hormone s/ Oestrogen/ Progestorone ; any one

= 1 mk

D- Site of implantation of Embryo/ Zygote / Fertilized ovum;

G- Storage of urine / Temporary storage of urine; 1 mk 5 (a) Hairy ears; hairy nose; baldness; porcupine man (or spiny skin);

Carnivorous insect

Sap- sucking ants Herbivorous insects

Mango tree

Insect eating bird

Correct labeling = 1 mk Proportionality = 2 mks

41


Any one point = 1 mk

b) i) XH Y; XHXh; = 2 mks II) Parental phenotypes normal x carrier female

Parental genotypes XHY X XHXh

The genotypes are XHXH, XHXh , XHY and XhY;

II) 2 completely normal: 1 haemophiliac 2: 1;

(Don’t award mark for ratio alone without indicating what the numbers represent) iv) Haemophilia is sex linked and they chromosome of male is empty , hence

once the recessive gene is present on the X- chromosome , it shows itself , unlike females where they are carriers ,

6 (a) Graph

Gametes

F1 offspring :

F1 offspring X h

42


6 (i) B(i) Bamboo; 1 mk c) i) The growth cells have maximumly divided hence no further growth; there is

further development resulting into the reproductive parts ; hence an increase in the dry weight

ii) Dry weight provide more accurate results than fresh weights. This is because

the fresh weight includes the weight of water in the plant which fluctuate according to the environmental conditions;

iii) Average height: Involves measuring all the heights of plants, then dividing

by the total number of the respective plants; Average dry weight: cutting the plants and drying them; until a constant weight is obtained; divide the total dry weight by the total number of the respective plants;

d) When both height and dry weight are used more accurate results is obtained 1 mk e) They lack cambium / cambium cells; 1 mk 7(a) In natural setting , the tree trunks are covered by light coloured lichens

which grow on them; light coloured moths blend in well/ camouflaged against the light coloured lichens ; As a result they are not easily seen and eaten by predatory birds; The light colour of the moth is a favourable variation/ beneficial variation to the moth; . However , the dark coloured moths/ the melanic forms do not blend or camouflage against the light coloured lichen; As a result the dark coloured moths / melanic forms were readily seen and eaten by the predatory birds; The dark colour of the moths is an unfavourable variation / a disadvantage to their survival ; As a result of this selection, the light coloured moth are many and the dark coloured moths are few; (OWTTE). This is because the light coloured moths survive and reproduce more light coloured moths; and are therefore better adapted to this environment; compared to the melanic form;/ dark coloured. As a result of industrialization, a lot of smoke and soot is released to the natural environment.This kills the light coloured lichen growing on the trunks and blackens the trunk; and blankness ; In this new environment the dark form/ melanic form is better adapted; Its colour is a favourable variation / beneficial variation because it blends well against the black background; predatory birds do not see it easily and it is not quickly eaten; The light coloured moth is disadvantaged by its colour which is unfavourable variation because it does not blend well against the blackened tree trunks; It is easily seen and eaten by predatory birds; due to this form of selection caused by change in the environment; more dark coloured moths survive and reproduce more than the light coloured moths; Total 18 mks , max = 16 mks

b) Lamark’s Theory The theory is based on ;

43


� Influence of environment on development of structure / organs; It stipulates that when environment demands / requires the use of a particular structure . Development of structure/ organism the organism develops it in response to the demands;

� Use and disuse of structures: It postulates that the frequent use/ exercise of an organ leads to proper development or enlargement of such organs; disuse of an organ (over an extensive period of time) results in its degeneration;

� Inheritance of acquired characteristics: It suggest that the traits acquired by an individual due to the environment all pressure are passed on to the offspring. Total = 4 mks

8 a) Has villi and micro villi, to increase surface area for absorption / digestion of food;

� Long; to increase surface area for absorption / digestion of food; � Highly/ extensively folded; to slow down movement of food for

effective digestion/ absorption of food for effective digestion/ absorption of food.

� Has lacteal; for increased surface area for absorption of fats � Narrow lumen: to allow food to be in close contacts with the walls

for affective digestion/ absorption � Walls have secretory cells: which secrete mucus which prevents the

digestion of the walls and brings smooth flow of food; � Secretory cells produce enzymes; for digestion of food; (Acc.

Specific enzymes with their function) � Thin wall / Epithelium; to increase the rate of diffusion of digested

food � Highly vascularised ; for affective transport of digested food;

(Total 18 mks , max 12 mks )

� The bile salts from bile emulsify fats to small droplets ; for the action of

pancreatic lipase to change the fats into fatty acids and glycerols ; � Bile contains sodium bicarbonate; which makes the medium alkaline for the

actions of the pancreatic enzymes ; � The pancreatic amylase ; breaks down starch to maltose; � Typsin; breaks down proteins to polypeptides ; Total 8

mks 2010 Pyramid Projections 016 Kenya certificate of Secondary Education 231/2 BIOLOGY PAPER 2 MARKING SCHEME 1. a) A – ilium (1mk) B- Acetabulum (1mk)

44


b) Allows passage of blood vessels /and nerves to the leg. (1mk) c) Suture / immovable /fixed joint. (Any one 1 mark) d) - The end of the cells are marked with thickened regions /intercalated disc;

that form bridges between fibres hence transmit impulses rapidly through out the heart.

- Are myogenic: - contract and relax without nervous stimulation. - Have more mitochondria to sustain their energy demands;

(Any 2 = 4mks) 2. a) To find out the production of the heat by germinating seeds. (1mk) b) Antiseptic killed micro-organism which could have caused decay; during

which heat could be produced. (2mks) c) Acts as a control for Y. (1mk) d) To prevent heat loss or gain. (1mk) e) Absence of certain wavelength of light; Freezing of seed during winter; (2mks) f) Respiration (1mk) 3. a) Sweep nets - Buckets /container - Gloves. - Water proof marking pain + brush. (½ mk each) b) Method: capture – recapture (1mk) Reason: crabs are highly mobile (1mk)

c) Wear gloves to avoid stinging by crabs. (1mk)

Wear protective footwear to prevent infection. (1mk) d) - Direct observation. - Examine gut contents. - Examine droppings. - Infer from feeding apparatus/ mouth parts. (Any 2 points = 2mks) 4. a) i) Pupil becomes smaller (1mk) ii)Pupil widens (1mk) b) Fovea centralis A cc Retina (1mk) c) (i) A condition in which light rays from a distance object are brought to focus

in front of the retina while those near are clearly focused. (1mk)

d) Cones lack retinal convergence (Each one has its own Bipolar neurone

which links it to optical nerve); Rods show retinal/ convergence (several rods connected to one bipolar neurone.) (2mks)

45


e (i) Cochlea (1mk) (ii) Vestibular apparatus. Acc. Semi circular canals and vestibule. RJ. Vestibule alone. 5. a) G U A C U C A (1mk) b) A (sudden) change in structure of DNA/gene /chromosome at a particular locus. (1mk) c) Radiations / X-rays / gamma /befa /U-V-rays chemical e.g. mustard gas, colchicines (Any two, 2mks) d) Inversion; deletion; translocation; duplications; non disjunction; (Any two, 2mk) e) Result in exchange of genetic material; leading to variation; (2mk)

6. a) (i) 1001100

1100000,23x

−; = 1990.9%;

(2mks)

(ii) 1001300

1300400x

−;= 69.23%;

(2mks) b) (i) At rest there is more blood flow through the gut because skeletal muscles

have less physical (metabolic) activity hence less glucose and oxygen required; more blood available at the gut to carry away digested food. (2 mks)

(ii) During streneuous exercise, more energy is required in skeletal muscles,

hence more glucose and more oxygen are required by muscles; there is need for faster removal of metabolic wastes (urea, co2, H2O and heat) hence more blood required. (2mks)

c) By stronger contractions; faster contractions; increased heart beat.; (2mk) d (i) Body temperature normal; hence less blood required at the surface of the skin. (2mk) (ii) Body temperature increases hence more blood flow to the skin

/vasodilatation is necessary to take away excess heat. (2mks)

c) Water; urea/uric acid/ nitrogenous waste/ creatinine/ hormones/ mineral salts/minerals (Any 3) 7. a) Nature selects those individuals which are sufficiently adapted and allow

them to survive; and reject; those that are poorly adopted; (3 mks)

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b) Individuals of the same species show certain variations/differences; the differences are caused by genes/mutations; some of the variation confer an advantage to the individuals while others are disadvantages; Most organisms produce large number of offspring’s than the parental generation, however only a few survive to adulthood and are able to reproduce, This is due to environmental pressures; such as predation / diseases / and competition for food / and breeding sites; this is referred to us the struggle for existence; Individuals with advantageous vanations /with desirable variations; are well adapted for survival; they survive and reach reproductive maturity; and pass over the favourable characteristics to their offspring; Those that do not have advantageous variations/ traits/ genes; die while young /do not reach sexual maturity, and therefore do not pass their genes/ traits to their offsprings; and thus perish /become eliminated; This is called survival of the fittest;

(17 mks)

8. a) Secondary thickening in flowering plants.

Secondary thickening results into an increase in girths /width; due to the activity of the cambium. In secondary growth new tissues are formed by vascular cambium; and cork cambium; the vascular cambium divides radially to produce new carbrium cells between the vascular bundles. This forms a continuous cambium ring; The new cells of the cambium differentiate to become secondary phloem on the outsides; those to the inner side differentiate to form secondary xylem, more secondary xylem; is formed than secondary phloem; the interfascular cambium divides to form secondary parenchyma thereby increasing the growth of medullary rays; As a result of the increase in volume of the secondary tissue; pressure is exerted on the outer cells of the stem resulting in rupturing of epidermal cells; In orders to replace the protective outer cover a new band of cambium cells are formed in the cortex; The cork cambium/ phellogen differentiate into secondary cortex on the inner side; and cork cells on the outside; cork cells are dead with thickened walls which became coated with a water proof substance called suberin; The cork cells increase in a number and become the bark of the stem which prevents loss of water; prevents infection from fungi and is also insulators. Bark is normally impermeable to water and respiratory gases; periodically the cork cells form a loose mass of cells known as lenticels which makes gaseous exchange possible; The rate of secondary growth is stem varies with seasonal changes; e.g during rainy season xylem vessels and tracheids are formed in large number; the cells are large thin walled and with wood with light texture; In dry season xylem and tracheids formed are few in number, are small and their wood have a dark colour. This leads to development of two distinct layers within the secondary xylem formed in a year called annual rings; ( Allow marks on correctly annoted diagrams.) (14 marks)

b) EITHER

Choose /identify a young leaf (just unfolded); use the same leaf throughout; Measure total length of the whole leaf (accept measurements of any part of the leaf). Record; repeat at regular intervals until no more change is length occurs / constant length. OR. Choose /identify a young leaf (just unfolded); use the same leaf throughout; Trace the outline on a graph paper; and workout the area; Record; Repeat at regular intervals until constant area; average rate of growth is equal to total increase in area divided by the period of time taken to achieve final area; (6 mks)

2010 Pyramid Projections 017 Kenya Certificate of Secondary Education (K.C.S.E) 231/2 BIOLOGY

47


PAPER 2 MARKING SCHEME 1 a) A - Oesophagus B - Stomach C - Pancreas D- Illeum (½ x 4 = 2mks)

b) - Destroys bacteria that may come in with food; - Provides a suitable PH medium for action of (enzyme) pepsin,) - Unfolds proteins enabling action by pepsin; (3 mks max 2)

c) - Long to increase surface area for digestion / absorption;

- Highly coiled to increase surface area / slow down food movement allowing more time for complete digestion / absorption;

- Inner lining contain villi (& microvilli) to increase surface area for absorption;

- Lined with cells containing large numbers of mitochondria to generate energy necessary for active transport ;

(accept any other correct adaptation) (4mks max 3)

d) Blood sugar level is regulated by (hormones) Glucagon and Insulin;

which flows directly to blood stream; (Pancreatic juice that contain) digestion ezymes; can not

pass through to the alimentary canal; (4 mks max 3) 2. a) i) Yeast ; (1mk) ii) Budding; (1mk)

iii) Cause (plants & animals) diseases; Causes food spoilage;

Source of food; Manufacture of antibiotics;

(4mks max 3)

b i) A group of organisms that can naturally freely interbreed & give rise to fertile / viable offspring; (1mk)

ii) The first (generic) name starts with a capital letter, the rest are small letters (inclusive of the specific name) ; - The two names should be underlined separately when handwritten or typed and italized when printed.; (2mks)

3. a) A - Scapula; B – Humerus; C - Radius.; (1x3 =3mks) b) Ball and socket (joint) ; (1mk) c) Synovial fluid; reduces friction / lubricates the joint;. (1 x 2 =2mks)

48


d) Ligament; (1mk) e) When one contracts the other relaxes and viceversa; (1mk) 4 a) i) When a person lives in high altitude; OWTTE ( 1mk) ii) During infection; (1 mk)

b) i) It will produce carbon (II) oxide, which is a respiratory poison/ combine with haemoglobin to form a stable compound/ carboxy haemoglobin; ( 1X2= 2 mk)

ii) The potted plant respires and uses oxygen/ competes for oxygen with the occupants of the room; ( 1mk)

c) For pathogens e.g HIV; and compatibility; (1 x2)= 2 mks

d) They can donate blood to all other groups;

(1 mk)

5 a) RNA; presence of uracil nucleotide ; (1x2 = 2mks) Tie function to structure. b) i) 4–Xc Y ; 5–XC Xc; 7 – XC Y; (1x3=3mks) ii) ¼; (1mk)

d) Baldness; Hairy ears/ (pinna ) / nose; (1x2 = 2mks)

6. a) - Plotting (All points correctly plotted) – (1 x2 =2mks) - Smoothness of the curves (1x 2 = 2mks) - Labelling of the curves (½ x 2= 1mk) - Scale (½ x 2 = 1mk) - Labelling of the axes (½ x 2 =1mk) (7mks)

49


b) -

Seedling grown in darkness – fresh weight increases up to day II due to absorption of water; then it begins to drop due to lack of photosynthesis; seedlings grown in light – fresh weight increases steadily even after day II due to photosynthesis; (1x 3 =3mks)

c) - In the dark – the dry weight would drop immediately due to lack of photosynthesis.

- In the light – the dry weight would drop before the leaves develop; then increase as leaves develop to carry out photosynthesis; (1 x 3 = 3mks) d) 65.5 ± 1.0g. (1mk) e) Optimum temperature; water; oxygen (1 x 3 = 3mks) f) Auxins; gibberellins; abscisic acid; ethylene; cytokinins. (5 mks max 3) a) (i) Secretion – production of useful substances from the body cells (e.g enzyme / hormones); ( 1 mk)

(ii) Excretion – elimination of waste products of metabolism from the body cells/ tissues; (1mk)

(iii) Egestion - Removal of indigestible / undigested food materials from the alimentary canal; (1 mk)

b) Adaptation of the mammalian kidney to its functions.

Time in days

Fresh weight (g)

50


The renal artery; supply blood to the kidney. The renal vein; carry away purified blood from the kidney. Afferent arteriole; from the renal artery supply blood to individual nephron. It is narrower than the efferent arteriole; creating high pressure in the glomerulus; That lead to ultra filtration; Glomerular capillaries; are narrow to create high pressure enhancing ultra-filtration of liquid part of the blood and dissolved substances to pass through the narrow pores into the bowman’s capsule/ capsular space; The proximal convoluted tubule; is highly coiled; to slow down the rate of flow of filtrate allowing reabsorption; the walls of PCT contains a large number of mitochondria; to provide sufficient energy for active reabsorption of useful substances/ glucose/ amino acids; back into blood stream; The loop of henle; is long and narrow and surrounded by numerous blood capillaries; for quick uptake of water and salts; Distal convoluted tubule; and collecting ducts; enhance further reabsorption of water; the remaining filtrate enters the pelvis; and into the ureter ; that connects to the bladder; (22mks max 17)

8 Adaptation of xerophytes to their habitats;

- Some have reduced leaf size/ scaly leaves; reduces the surface area; over which transpiration occur;

- Some shed their leaves; during drought; minimizing transpiration; - Some have thick waxy cuticles; minimizing cuticular transpiration; - Some xerophytes fold their leaves; on hot days; reducing the surface

area exposed to environment; hence reduce transpiration rate; - Some leaves have sunken stomata; that accumulate moisture;

reducing the rate of diffusion; - Some xerophytes have a reduced number; and size of the stomata;

lowering the rate of transpiration; - Some xerophytes have reversed stomatal rhythm; lowering

transpiration rate during day time; - Some have deep roots; that reach the water table;/ others have

superficial roots; that grow horizontally; absorbing water close to the ground surface; (soon after rains);

- Some have succulent leaves/ stems / roots; for storage of water; - Some xerophytes are drought evaders; going through the dry season

in form of seeds/ or underground perenating organs; ( 23 mks mxm 20)

2010 Pyramid Projections 018 231/2 BIOLOGY PAPER 2 2 ½ HOURS MARKING SCHEME Key to marking ; end of marking point. Rj Reject Acc Accept answer OWTE - Other words to that effect. / Alternative answer. Words not important.

51


1. a) . (4mks) vultures leopards grasshopper guinea fowl termites gazelles grass b) - Grasshopper, Guinea fowl, termites and gazelle;

(mark as a whole) c) (2mks) - They will feed on gazelles which would reduce; - The leopards will reduce due to starve/migrate; - Grass will increase; d) (1mk) - scavenger; 2. a) (1mk) - Potometer; b) :- i) Cutting shoot under water. (1mk) - To prevent the entry of air bubbles in the xylem; Rj vascular

bundle or vessel. ii) Keeping the system air tight. (1mk)

- Prevent air lock; c) i) In a dark room. (1mk) - Position( of the air bubble )remains constant /no change in position( of air

bubble); ii) In an air current created by a fan. (1mk)

- (air bubble) moves towards the shoot; iii) In the sun with leaves smeared with Vaseline on both sides. (1mk) - No change in position of (air bubble); d) . (2mks) - Stomata remain closed; in the dark, no transpiration; 3.a) (2mks) - Early death is caused by the fact that the homozygous recessive plants do

not have chlorophyll; and cannot make their own food;

52


b) (4mks) NN – Genotype for normal green plant Nn- Genotype for pale green plant. Parents genotype NN X Nn Gametes F1 generation NN NN Nn Nn Normal green pale green Or Parents genotype NN X Nn c) (2mks) Nn – genotype for seeds from heterozygous plant. Parents Nn x Nn (no mark for working) genotype Gametes

NN Nn Nn nn ; F2 generation

Phenotypes 1 normal green 2 pale green 1 white (will not grow to maturity)

Phenotypic ration 1: 2 Green: pale green; Or 1 green: 2 pale green

4. Concentration of sugar solution (%) Diameter of cells (micrometer) 1 5.0 5 4.0 10 3.0 15 2.0

a) . (2mks) conc. Of average values of 3 µ and 2 µ = 10+ 15; = 12.5%; 2 b) (1mk)

N

n

N NN Nn N NN Nn

N N N n

N n N n

53


- Isotonic; c) . (3mks) - 1% sugar solution is hypotonic( to the cell sap); cell gains water by osmosis;

and swelling/ increases in the cells diameter; d)i)

. (1mk)

- After- Sap vacuole will appear smaller in size; tied

ii) (1mk) - Plasmolysis; 5. a) (1mk) - Long sight/hypermetropia; b) (2mks) - The eye ball is too short; - weak eye lens; c) .

d) . i) (1mk) - geotropism; acc. Shoot is negatively geotrophic while Root is positively geotrphic Rj – if root/shoot alone. ii) . (2mks) - High concentration of auxins in the lower part of the root inhibit growth;

while more growth occur on the root upper side; hence the root bend downwards/show positive geotropism.

Max 2mks

54


SECTION B (40MKS) Answer question 6 (compulsory) and either question 7 or 8

a)

Scale 2 Labelling axis 2 plotting 1 Smooth curve 1

b) . (1mk)

55


- 10 – 15 sec; Rj if no range ci) (1mk) - 22mg/100cc of blood; ii) (1mk) - Muscle cramps/cell poisoning; raises osmotic pressure of body

cells/lymph/blood; causing fatigue; (mark 1st correct) iii) . (3mks) - Liver converts lactic acid into glucose; fast /deep breathing; heart beat

increase/blood circulation rate increases; d) i) Occurred (1mk) - 5 – 15( sec); ii) (1mk) - 15th (sec); e)

(4mks)

anaerobic aerobic - Oxygen not used Oxygen used; - Glucose not completely broken down glucose completely broken

down; - Less ATP energy produced. More ATP energy produced; - By-Products – lactic acid By-Product- water, CO2; f)i .

(1mk) - Heart ii) . (1mk)

- Presence of cardiac muscle; that do not get fatigued. 7. (20mks) Mouth Chewed using teeth;to increase surface area for enzyme action; - ptyalin/salivary amylase digest starch; into maltose; - Saliva has water; to moisten/soften food; - Saliva has mucus; for lubrication; - Saliva has slightly alkaline – suitable pH for ptyalin/amylase; Stomach pepsin digests protein; into peptides; No starch digestion due to unsuitable pH/ presence of HCl/Acidic media; HCL activate pepsinogen into pepsin; HCL provide suitable pH for action of pepsin; Duodenum Trypsin digests proteins; into peptides; Pancreatic amylase digest starch into maltose; Sodium bicarbonate : provides suitable pH/neutralize acidic chyle;

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Bile juice: provided suitable pH /neutralise acidic chyle; Ileum Peptidase digests peptides into amino acids; Maltase digest maltose into glucose; 8a). Describe the theory of natural selection by Darwin (15mks)

Organism produce more young ones than can survive; many young ones die due to environmental resistance,( e.g. diseases/ predators/, competition,/ extreme weather, e.t.c.); this keeps population relatively constant.; This is called struggle for existence; variations occur; some are unsuitable; These make organism to die early; without leaving offspring; These unsuitable characteristics tend too reduce in a population; This is called survival of the fittest; Variations occur in a population by chance; Some variations are suitable; These make organism survive to reproductive age; and pass the suitable genes/ variation(s) to their offsprings; These suitable characteristics tend to increase in a population;

b) Explain how resistance to insecticides support the theory of natural

selection. (5mks) e.g housefly (can use another insect|) is resistant to DDT; It has been observed that houseflies are not killed by DDT which used to kill

them previously; However, a few mutant houseflies were resistant to DDT ; After several generations only the resistant houseflies were left; Due to lack of competition these reproduced and their population increased rapidly; NB/ DDT does not cause mutation but only kills susceptible flies.

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2010 PYRAMID PROJECTIONS019

MARKING SCHEME BIOLOGY PAPER 2- 1. (a) Survive malaria attacks.

(1mk) (b)(i)

(ii)HbAHbs (1mk)

HbAHbA HbAHbS HbAHbS HbSHbS (1mk) 1MK (iv) 2/4 = 1/2 (1mk)

2. .

HbA

HbA

HbA

HbA

HbAHbs x HbAHbA

HbAHbA HbAHbA HbAHbs HbAHbS (1mk) Normal Carriers (1mk)

HbA

HbS

HbA

HbS

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b) –In A- The seedling grew upwards coleoptiles were covered with black polythene which does not allow light to pass through thus no migration of auxins. (1mk) - In B – The part where auxins are found was cut off. Thus no effect on the seedlings Acc. OWTTE (1MK) - In C – The seedlings bent towards light due to migration of auxins from the lighted side to the darker side causing more elongation on the darker side thus bending towards the light. (1mk) (c) Phototropism (1mk) (d) Seeks to maximize light energy absorption thus increase the rate of photosynthesis. (1mks) 3. a) To Demonstrate respiration by germinating seeds. (1mk) b) Antiseptic killed micro-organisms which would have caused decay, during which heat could be produced. (2mk) c) Acts as a control for Y. (1mk) d) To prevent heat loss and heat gain. (1mk) e) – Absence of certain wavelength of light. - Freezing of seed during winter/unsuitable temperatures (2mk) f) Respiration (1mk) 4. a) pepsin is produced as pepsinogen by which is the precursor for

pepsin, On inactive form which cannot digest the stomach wall as would the pepsin do. (2mk)

b) The medium of the stomach is Acidic which is not condusive for Amylase activities hence No effect. (1mks) C) It is converted to casein because pepsin can only act on milk protein when it is converted to caseins, solid milk stays in the stomach longer for digestion to take place. d) – To lubricate the food. - To protect the elimentary canal wall from being digested by protein digesting enzymes. (2mk) (e) Allows the tongue to manipulate vegetation material in such a way that material being chewed is kept away from that which is freshly gathered. (1mk)

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5. a) Microscopic algae Mosquito larvae small fish large fish

Crocodiles (2mks) b) Acc. 2500 4000 (1mk) - Should be more than the large fish and fewer than mosquito larvae. (1mk) c) i) Capture – recapture method. (1mks) ii) –Capture a number of fish, mark them and release them back to the pond having noted their number. (1mk) -After sometime/a few days, make another capture and out of them, Note the number marked . i.e 1st captured and marked = x 2nd Captured = Y. (1mk) 2nd Captured with mark = z. The total population = T= XZ Y (1mk) 6. a ) graph b) 41 mg C(i) food stored in endosperm was utilized by the growing/germinating prumule and radicle. (ii) Food from the endosperm was taken to the embryo which was growing thus increased in weight. (iii) This is due to the energy lost through respiration ie. Not all the food from the endosperm is utilized in the growth of the embryo. (iv) The “seedling” has developed foliage leaves thus photosynthesis has started taking place. (d) (i) – Immaturity of the embryo - Absence of enzymes/hormones. - Presence of Inhibitors. - Impermeable seed coat. (ii) –Excess alot/ heat. - lack of water/oxygen. (e) – Thin walls - Dense cytoplasm - Absence of vacuoles. 7. a)

i) Concentrates/ collects and directs sound waves into the outer auditory meatus/ear tube. (1mk)

ii) –Transmits vibrations. (1mks) - Amplifies vibrations (1mk) iii) – Transmits sound impulse to the brain/CNS. (1MKS) B) The pinna collects and concentrates sound waves and directs them into the outer aunditory meatus. The sound waves strikes the tympanic membrane/ear drum which vibrates. These vibrations are picked up by ear ossicles, first the mallens, to the incus and to the stapes. In the ear ossicles, the vibrations are amplified since the ossicles forms a system of levers. The vibrations are taken by the oval window/fenestra ovale and picked up by perilymph of the inner ear vibrations in the inner ear stimulates the sensory cells, triggering a nerve

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impulse which is conducted by the auditory nerve to brain for interpretation. (16mks)

8. The skin is made up of epidermis and dermis, the epidermis is made up of three layers. � The outermost layer known as cornified layer, made of dead

cells that protect against mechanical damage/dessication/microbes

� The granular layer, is made up of living cells that give rise to cornifield layer.

� The malphigian layer, with actively dividing cells that give rise to a new epidermal cells/granular layer that contain melanin, that protects the skin against U.V. rays

� Has sweat glands/sudoritic glands, that produce sweat which evaporates thus reducing body temperature.

� Under cold conditions little/No sweat is produced thus heat is conserved.

� The sweat contains water, sodium chloride/uric acid/Urea and hence the skin acts as an excretory organ

� Has hair, the hair stands erect to trap air to insulate/reduce heat loss when the temperature is low/lies flat to allow heat loss when the temperature is high

� Has nerve endings, which are sensitive to stimuli/heat/cold/pain pressure/touch

� Has subcutaneous fats/Adipose, that insulates the body against heat loss.

� It has arteries/capillaries/blood vessels, that supply food/oxygen/remove excretory products.

� Arterioles vasodilate when Temp. are high/loss of heat by radiation/convection, Vasoconstricts when temperature is low to conserve heat/minimize heat loss.

� Has sebaceous gland, which secretes sebum on antiseptic/water repellant that prevents skin drying/ cracking of skin.

� TOTAL 23 (MAX 20)

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2010 PYRAMID PROJECTIONS- 023 BIOLOGY PAPER 2 (SET 1) 231/1

(MARKING SCHEME) TIME: 2HRS

1(a)Genes controlling other character apart from sex but are located in sex-chromosomes therefore transmitted /inherited together; with sex chromosomes

(b)(i)XcXc XCY; Xc Xc; XCY

Normal Normal Carrier Color

Daughter son daughter r blind

son

(ii)0.25 or 25%;

(c)(i)Inversion - The word mothers inverted to thermos;

(ii)insertion - The letter T inserted in the word say to form word stay ;

2.(a)To destarch the leaves ; (b)C02 is a condition necessary for photosynthesis;

(c)Absorbed / remove carbon IV oxide from the air in flask ;

(d)(i)Leaf B - is control experiment;

(ii)l - Boiling leaf to kill the protoplasm/ to stop photosynthesis from taking place

II - to dissolve chlorophyll / remove chlorophyll from leaf ;

(e)(i)Point when rate of photosynthesis is the same as respiration therefore no exchange of carbon IV

oxide and oxygen in stomata ;

(ii)Photosynthesis is controlled by enzymes therefore optimum temperature for enzyme activity is needed for highest rate of photosynthesis;

3.(a)C - Enamel ; D- Blood capillaries; E - Cement ;

(b)C - Non living and hard therefore provides with a very hard surface for biting and grinding food ;

(c)Premolar - two roof system ;

(d)Trapped sugary food between teeth acted on by bacteria to form organic acids that corrodes the tooth enamel hence exposing the dentine and pulp cavity;

(e)- Regular visit to dentist ;

- Avoid eating of sugary foods ;

- Eating diet rich in calcium and vitamin D ;

4. (a)

Phytoplankton's Mosquito larva Small fish Large fish Fish eagle; Phytoplankton Mosquito larva Small fish Large fish Crocodile;

(b)A lot of energy is lost as the energy flow from one trophic level to another therefore less energy is made available to organisms in quaternary trophic level;

(c)(i)TheSun ;

(ii)Phytoplankton's absorbs sunlight and uses it to make their own foods through photosynthesis

Some of the foods is stored in phytoplankton'sis fed by mosquito larva or small fish which are

then fed by large fish ;

(d)As the depth increases the intensity ofsunlight decreases therefore decreasing the rate

of photosynthesis hence biomass;

(e)- Overfishing;

- Discharge of untreated sewage;

- Discharge of industrial effluents / with poisonous chemicals;

- Oil spillage from garages ;

- Short hunting of birds / fish eagle ;

- Agricultural activities / fertilizers washed into lake;

5. (a)To investigate the rate of transpiration of the shoot under different environmental conditions;

(b)The shoot should be cut under water to prevent the xylem vessels from being blocked by air;

(c)(i)Rate of transpiration higher - high temperature increases rate of evaporation / by increasing kinetic

energy of diffusing water vapor molecules;

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(ii)Low rate of transpiration - polythene bag prevent air currents and provides with humid environment;

(iii)High rate of transpiration - fan provides wi air currents which speeds up transpiration ;

(iv)Low rate of transpiration - plucking off the leaves reduces the surface area for stomata transpiration;

6(a)

6.(b)(i)134mm + showing on graph ; (ii)136mm + Extrapolation on graph ; (c)Shoot A - Removal of apical bud removesource of auxins therefore lateral buds sprouts and grow faster,

Shoot B - Removal of apical bud stimulate the sprouting of lateral buds, the giberrellic acid stimulate growth of lateral branches. This explain why lateral branches in B grew faster.; Shoot C - Presence of apical bud inhibits the sprouting of the lateral buds because it is the source of auxins which inhibits sprouting of lateral buds.;

(d)Shoot C - Control experiment; (e)- Stimulate apical dominance;

- dominance; - Stimulate tropic responses ;

- Stimulate development of adventitious roots;

- Stimulate secondary growth in dlcot;

7(a)

Monosaccharide's Polysaccharides

Sweet in taste - Non-sweet

- -crystallizable-non crystallible - Soluble in water - Insoluble in water

-Reducing property - Does not have reducing property –

-any two

(b)Functions of lipids

- Storage purpose / energy reserve

- Oxidized to release energy - Source of metabolic water - Protection of delicate organs e.g. kidneys, eyeballs

(c)- Place a food substrate in a test tube add a few drops of sodium hydroxide solution and shake well; - Add a few drops of 1% copper - (II) Sulphate solution shake well and observe the colour change; - If proteins are present a purple or pink coagulations is formed; - (d)(i)Pepsin - Breaks down proteins into polypeptides;

(\\)Blle - Contains bile salts whose functions are - Emulsification fats; - Neutralize acid in chyme and provide alkaline medium;

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(iii)Hydrochloric acid - Kills bacteria in the food ;

- Activates the conversion of pepsinogen to active pepsin - Provides with acidic medium for pepsin to function

8.(a)Breathing - taking in and out of air into the

respiratory organ e.g. (lungs)

Respiration - Chemical breakdown of foods to

produce energy needed by the cell

(b)- Age - Low aged individuals higher in young

individuals because they require more energy.

- Body size - Small sized Individuals have large SA to volume ratio therefore higher. - Basal metabolic rate (BMR) - Health - unhealthy individuals respires more faster to produce energy needed for

recovery - Activity - During physical activities more energy needed therefore faster rate of

respiration

(c)Adaptations of red blood cells

- Biconcave in shape - large surface area for exchange of respiratory gases - Small and numerous in number - Large surface area for transportation of oxygen and

C02. - Enucleated / no nucleus - Large surface area for packing hemoglobin. - Presence of hemoglobin in cytoplasm - a pigment with high affinity for 02 - Thin cell membrane - short distant for diffusion of gases - Very short life span 90 days - efficiency in transport of gases.

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2010 Pyramid Projections © 231/2

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Biology Paper 2 16 Tests m s