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Biology
WAEC / WASSCE 2014
Nutrition in Plants & Animals (Photosynthesis & Food Substances)
Solution
Study specimens L, M and N carefully and use them to answer questions 4(a) to 4(c).
(a)(i) State the habitats of specimens L, M and N.
(ii) State the modes of nutrition of specimens L. M and N.
(iii) Explain how each of specimens Land M are adapted to their modes of feeding.
(b)(i) Describe briefly the relationship that exists among specimens L, M and N.
(ii) State two ways in which specimen M is of economic importance.
(iii) State one way in which specimen M can be controlled.
(c) Make a drawing 8cm - 10cm long of specimen M and label fully.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Habitats:
Specimen L- Grassland/farmland/garden.
Specimen M-Farmland/garden/grassland/field.
Specimen N-Moist/damp soil/environment/decaying tree/decaying wood/pastures/meadows.
(ii) Mode of nutrition: Specimen L autotrophic/photosynthetic/holophytic.
Specimen M heterotrophic/holozoic Specimen
N- Saprophytic.
(iii) Adaptation of Specimen to their mode of feeding:
Specimen L- has green leaves/chlorophyll; for absorption of sunlight tor photosynthesis; root for absorption of water/nutrient/ minerals.
Specimen M - has mandible; for Cutting/chewing; labrum/maxilla; for cutting/chewing.
(b)(i) Relationship among Specimen L, M and N:
Specimen L/grass produces food for Specimen M grasshopper; also serves as habitat for specimen M .
Decay/decomposition of Specimen L and M serve as food substrate for Specimen N.
-The decomposition activities of Specimen N provides nutrients for Specimen L.
- Specimen L releases oxygen during photosynthesis; which is used up by Specimen M and N respiration.
- Specimen M releases carbon-dioxide during respiration, for the use of Specimen L for photosynthesis.
- The excreta of Specimen M serves as nutrient for Specimen L and N.
(ii) Economic Importance of Specimen M/grasshopper: - Serves as source of income generations
- It destroys crops/plant/reduces crop yield/reduces income.
- It is edible/- Serves as food/source of food.
(iii) Control of Specimen M/grasshopper: - Use or insecticide/pesticide.
- Introduction of biological enemies e.a birds/predators.
(e)Drawing of Specimen M:
Source: Official Exam / Past Question Archive
Biology
WAEC / WASSCE 2014
Variation & Genetics (Heredity)
Solution
(a)(i) Explain the term co-ordinance.
(ii) Calculate the number of individuals with co-dominant blood group.
(iii) What is the total number of individuals in the table that are able to donate blood to an accident victim with blood group B?
(b) A man whose blood group is heterozygous A is married to a
woman whose blood group is AB. With the aid of a genetic diagram, suggest the possible blood groups of their children.
A survey to determine blood groups was carried out on 250 people living in a community. The results are represented in the table below.
Blood group
Percentage
A
8.0
B
14.0
AB
32.8
O
45.2
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Co-dominance: Co-dominance occurs when both the alleles in an heterozygous individual are fully expressed; that is both alleles exert their effects so that the hybrid has a phenotype; that is different from what the parents exhibit.
(ii) Calculation of individuals with co-dominant blood group: Mention of co-dominant blood group is AB. Evidence of calculation 32.8% of 250 OR 32.8/100 x 250 = 82 individuals/ people.
(iii) Total number of individuals that can donate blood to blood group B: The total number of individuals will be from groups B and 0.
—14% of the population is Blood group B 14% of 250 OR 14/100 x 250 = 35 individuals/people.
— 45.2% of the population is Blood group 0 45.2% of 250 OR 45.2/100 x 250 = 113 individuals/people.
— Total number of people = 113 of group 0+ 35 of group B = 148 Individuals/people can donate blood to the accident victim in B. OR The total number of individuals from (B and 0) blood group B and 0
— 14% + 45.2 = 59.2%.
— 59.2% of 250. 592/100 x 250 = 148 people/individuals.
(b)
Source: Myschool Past Questions Archive (Record #52725)
Biology
WAEC / WASSCE 2014
Population Ecology, Habitats & Conservation
Solution
(a) State five reasons why animals move from place to place.
(b) State one function each of the following structures found in plants:
(i) epidermis
(ii) phloem
(iii) sclerenchyma.
(c) Give two examples each of the following types of organisms:
(i) parasitic plants
(ii) saprophytes.
(d) State two adaptive features of plants which inhabit salt water swamp.
(e) State two factors each which result in
(i) increase
(ii) decrease, in population density.
(f) Mention four ways in which modern agricultural activities may threaten the survival of species.
(g) State three reasons why mitosis is important to living organisms
(h) State three causes of food spoilage.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a) Reasons why animals move from place to place:
(i) To search for food.
(ii) Escape predators.
(iii) Find mates for reproduction.
(iv) To disperse/reduce competition/overcrowding by moving to new locations.
(v) Avoid dangers. (vi) Find new and favourable habitats.
(b) Function of structures found in plant:
(i) Epidermis: For protection of inner cells from injury, infection, prevent excessive loss of water.
(ii) Phloem: Responsible for the transportation of enzymes/hormones; manufactured food.
(iii) Sclerenchyma: Strengthens the stem/provide mechanical support; makes plants flexible.
(c) Examples of organism:
(i) Parasitic plants: Viscum/mistletoe; dodder plant/Cassytha; Cuscuta; phytophisor; peronospora.
(ii) Saprophyta: Bacteria; fungi/Rhizopus; Mucor, Aspergillus; Mushroom/Agaricus.
(d) adaptive features of plants in salt water swamps: -
Succulent stem/root/store water; to dilute the salt.
Presence of pneumatophores/breathing roots; for breathing atmosphere air/oxygen.
Presence of stilt roots; for anchorage.
Viviparous seedling; to ensure their survival/continuity.
Some a secrete excess salt into swamps; to maintain balance of body salt.
(e) Factors that: (i) Increase population density:
Increased birth rate/natality
immigration, increase a food availability
low mortality rate.
(ii) Decrease population density:
Emigration
diseases
drought/famine/natural disaster
high mortality rate
low natality rate.
(f) Ways modern agricultural activities may threaten the survival of species: -
Deforestation/commercial farming destroys habitats of species.
Destruction/loss of natural vegetation/habitat.
Over hunting of games reduces population of species.
Improper/Indiscriminate application of chemical/salinizationteutrophication, pesticides destroys species.
Lack of afforestation practices.
Inappropriate fishing methods threaten the population of fish.
(g) Reasons why mitosis is important to living organisms: -
It ensures growth.
Involved asexual ,reproduction e.g. fission, budding, spore formation.
Ensures retention of diploid condition. Repair of tissues within organisms.
Transfer of exact genetic code from parents to offspring.
(h) Causes of food spoilage: -
Microbial attacks by bacteria or fungi/or saprophytic action.
Pest attack by beetles, weevils that destroy stored grains.
Improper storage which results in deterioration and spoilage of stored foods.
Destruction of stored fruits and meat by some worms, usually larva of flies.
Delay in transportation due to bad roads.
Lack of market facilities for the commodity.
Poor harvesting method/damage to fruits/ vegetables/tubers during harvesting.
Poor unhygienic handling/processing/exposure to dust.
Storage in contaminated containers.
Use of contaminated water to wash food.
Source: Myschool Past Questions Archive (Record #52726)
Biology
WAEC / WASSCE 2014
Nutrition in Plants & Animals (Photosynthesis & Food Substances)
Solution
(a) (i) List three characteristics of living organisms.
(ii) Explain briefly the characteristics listed in (a)(i) above.
(b) In a tabular form, state four differences between plants and animals.
(c)(i) Name one organism that exhibits both plant and animal features.
(ii) State four animal features and two plant features possessed by the organism named in (c)(i) above.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Characteristic of Living organisms:
- Movement/locomotion
- Excretion
- Respiration
- Reproduction
- Irritability/sensitivity/response
- Growth/development
- Nutrition/feeding
- Competition
- Adaptation
- Life span/death
(ii) Movement: Living things change the position of the whole body/parts of the body; from one place to another; in search of food/shelter/mates, etc.
Excretion: This involves the removal/getting rid of waste products of cell metabolism; membole wastes are poisonous if allowed to accumulate.
Respiration This is the breakdown/burning of food substances; by aerobic/anaerobic respiration; to release energy needed for all processes of life.
Reproduction: This is the ability of living organisms to produce new individuals of their own kind; this ensures the continuity of the species of an organism.
Irritability: Response/ reaction of organisms; towards changes in the environment; to ensure survival. Nutrition: the process involved in obtaining/manufacturing food; can be autotrophic/heterotrophic; and utilizing it for growth and maintenance/life processes.
Growth: This is the irreversible increase; in size and weight of an organism; which leads to complexity and maturation.
Competition: Ability of organisms to struggle; for all necessity of life; to survive in their environment.
Adaptation: Ability of organisms to adjust; to changes in environment; for survival.
Life span/death: All living organisms have definite period of existence; all organisms die.
b Differences between plants and animals;
Animals
plants
move from place to place, locomotion, move with the whole body
Are fixed/do not move from place to place, and move only parts of the body, root/stem/leaf
Heterotrophic mode of nutrition that depend on already-made food
Autotrophic mode of nutrition/ manufacture their own food using the energy from the sun
No chlorophyll/chloroplast
chlorophyll/chloroplast present
No cell wall
Has a cell wall made of cellulose
sense organs present
sense organs absent
Growth is limited/definite, growth in all parts of the body
growth is unlimited/indefinite .growth is apical or terminal
stores carbohydrates as glycogen
stores carbohydrates as starch
Has special excretory body
No special excretory body
Response is fast, short termed
response very slow/short termed
Respiratory organs present
No respiratory organs
(c)(i) Name of organism:
Euglena/Chlamydomonas.
(ii) Animal features: For Euglena
— Gullet for feeding.
— Has flagellum for locomotion in water.
— Myonemes for movement.
— Has pellicle that gives it a definite shape.
— Has eye spot to detect light intensity.
— Has contractile vacuole for osmoregulation and excretion.
Plant features:
- Definite shape.
— Has chioroplasts containing chlorophyll used to absorb light.
— Paramylum granules.
— Pyrenoid.
Source: Myschool Past Questions Archive (Record #52709)
Biology
WAEC / WASSCE 2014
Digestive System & Feeding Mechanisms
Solution
(a) Define the following in
(i) elimination
(ii) excretion
(iii) secretion.
(b) Describe the digestion of boiled fish in the gut of humans
(c) Make a diagram 8cm - 10am long of a longitudinal section of a villus and label fully.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a) Definition of terms:
(i) Elimination: Removal of undigested food; from the body of living organisms.
(ii) Excretion: Removal of waste products of metabolism; from the cells/tissues of an organism; may be toxic if retained in the body.
(iii) Secretion: Production and release of substances; useful to the body; by specialized glands of living organism.
(b) Digestion of boiled fish in the gut of humans: Boiled fish is a source of protein; chewed and swallowed into the stomach; digestion of protein starts from the stomach; stomach secrets gastric juice; containing pepsin, rennin and hydrochloric acid; pepsin in acidic medium acts on the boiled fish and convert it to peptides; peptides move into the duodenum/small intestine; into which pancreatic juice is secreted; undigested protein is acted upon by trypsin in alkaline medium; into peptides/peptones, erepsin finally acts on peptides/peptones in alkaline medium; and converts it into amino acids.
(c) Diagram of villus
Title TL: - Longitudinal section of villus. Size: (8 — 10cm long).
Quality (Q): Clarity of lines. Neatness of labels.
Details: Epithelium layer shown (EL) Conical shape of villus (one cell thick) (CV).
Labels (L): Epithelium layer blood capillary, lacteal, lymphatic vessels, goblet cells, venule, arteriols, cryps of lieberkuhn, cells of paneth, muscularis mucosa.
Source: Myschool Past Questions Archive (Record #52714)
Biology
WAEC / WASSCE 2014
Nutrition in Plants & Animals (Photosynthesis & Food Substances)
Solution
(a) Explain briefly the following ecological terms:
(i) biosphere
(ii) habitat
(b) Describe the mode of nutrition in a named carnivorous plant.
(c) State three reasons why the dispersal of fruits and seeds are important.
(d) State six methods of conserving soil.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Biosphere: Total/whole area, part of the earth/lithosphere/ hydrosphere/water, and atmosphere/air, inhabitable by living things/where life exists.
(ii) Habitat: A place/home; within an environment, where an organism naturally lives/dwells.
(b) Mode of nutrition of named carnivorous plants:
- Carnivorous plants have green leaves and are able to manufacture their own food carry out photosynthesis.
They have roots to absorb water and mineral salts from the soil.
Usually grow in places with little nitrogenous salts.
Use insects/other small animals as source of Nitrogen.
They have devices/modified leaves with bright colour/sents/sugary liquids/stcky wet surfaces for attracting and trapping insects.
Once the insects is trapped, the plant secretes enzymes to kill and digest it.
The soluble nitrogenous compounds are absorbed/assimilated by the plant.
Name of carnivourous plants: -
Butterwort/Pinquicula
Bladderwort/ Utricularia
Sundew/Drosera
Venus fly-trap/Dionaea
Pitcher plants/Nepenthes/Sarracenta.
(c) Importance of fruits and seed dispersal: -
Prevents overcrowding of plants.
Prevents undersirable competition for space/light/water and nutrients.
Prevents diseases from spreading among plant species.
Results in colonization of new areas with new plants.
Results in adaptation to adverse/harsh/unsuitable environment.
(d) Soil conservation method: -
Contour ploughing/ridges are made along the contour of a slopping land/to prevent water running down the hill/slope/and washing away soil.
Strip cultivation/the land is divided into strips along the contours of the slopes/crops are planted in rows that alternate with the strips planted with cover crops/grasses.
Terracing/constriction of terraces which create barriers to hinder rapid water flow down the slope.
Mulching/materials left after weeding or harvest are left to lie on the soil/ protecting the soil against agents of erosion.
Afforestation/cover cropping/growing grasses, trees and shrubs to cover and protect the soil.
Crop rotation/growing crops on a piece of land in sequence/season after season to maintain its fertility.
Storm drains/construction of barriers/to stop free-flow of surface water through farms.
Application of manure/fertilizer/to stabilize the soil.
Avoid bush burning.
Avoid over grazing.
Source: Myschool Past Questions Archive (Record #52723)
Biology
WAEC / WASSCE 2014
Reproduction in Plants & Animals
Solution
Observe specimens A, B, C and D closely and use them to answer questions 1(a) to 1(e).
(a)(i) Specimen A Into its division and class.
(ii) State three observable features of specimen A which are characteristics of the class mentioned in 1(a)(i) above.
(b) State the relationship between specimen A and: (i) specimen B; (ii) specimen C; (iii) specimen D.
(c)(i) From. your observations of specimens B and C, what type of plant is specimen A? (ii) Suggest the agent of pollination ot specimen B.
(iii) State two reasons for the answer in 1(c)(i) above.
(d) Classify specimen D fully according to origin and structure.
(e) Make a drawing 8 cm 10 cm long of specimen C and label fully.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Classification of Specimen A: Division- Angiospermatophyta/Angiospermophyta/Magnoliophyta/Tracheophyta.
Class Monocotyledoneae/Liliopsida., (ii) Observable features of the Specimen A.
- presence of narrow leaf/leave
-Fibrous roots.
- Parallel venation on leaf,
- Presence of leaf sheath.
(b) Relationship between Specimen A and Specimen B, C, and D:
B is male flower(s), male nflorescence of A .
C-is the female flower(s) female inflorescence of A.
D- is the fruit/seed of A.
(c)(i)Type of plant: A monoecious plant. (ii) Agent of pollination: Wind (not wind pollination or wind pollinated flower).
(iii) Reasons: - The flowers are dull in colour.
- The flowers are small/incospiscous
- Flowers lacks scent.
-The stigmas are borne on long flexible stvle/stigmas hand out to expose themselves to pollen grains.
- The stigmas are feathery.
(d) Full classification of Specimen D according to origin and structure: True; simple, dry indehiscent, caryopsis.
Source: Myschool Past Questions Archive (Record #60302)
Biology
WAEC / WASSCE 2014
Variety of Organisms & Classification of Living Things
Solution
(a) Study specimens E and F carefully and use them to answer questions 2(a)(i) to 2(a)(ii).
(i) State the Class to which each of specimens E and F belong.
(ii) List four observable features each of specimens E and F.
(iii) State how each of the observed features in 2(a)(i) adapt each of specimens E and F to their habitats.
(b) Study specimen G and use it to answer questions 2(b)(i) to 2(b)(ii).
(i) Classify specimen G into its phylum and class.
(ii) State four observable features of specimen C.
(iii) Name one habitat of specimen C.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Class: Specimen E-Mammalia. Specimen F-Aves.
(ii) Observable features of Specimen E and F: Specimen E - Pinna/External ear.
Body covered with fur/hair,
mammary gland.
Heterodont dentition.
Whiskers; (five) clawed digits/pentadactyl limbs.
Specimen F- Body covered with feathers;
mouth modified into beak;
legs have (horny) Scale.
Fore-limb modified into wings/presence of wings; stream-lined body.
(iii) Features which adapt Specimen E and F to their habitats:
Specimen E- enlarged external ear/Pinna for trapping sound waves.
whiskers for detection of vibration of vibration/feeling its way througn burrows.
- Pear shape head/narrow body with folded legs for easy movement in burrows.
Position of eyes on either side of the head allows wide range of view.
- Long (muscular) hind limbs for hopping/ jumping.
Sharp incisors/heterodont dentition for nibbing vegetation/ mxed diet.
-Limbs end in claws used for scratching/digging.
-Body covered with fur/hair for temperature regulation.
Specimen F- Forelimbs modified to wings for flying.
Streamined body for easy movement through the air during flight.
- Possession of feathers on the body for temperature regulation.
-Tail feathers used for steering/balancing/brake/courtship display.
- Beak adapted for feeding.
-Clawed digits for scratching for food.
- Position of eyes/laterally placed eyes for wide vision.
(b)(i) Classification: Phylum of G- Arthropoda. Class of G Insecta.
(ii) Observable features of Specimen B: (Upper part of) abdomen constricted into a waist/has a narrow waist.
- Two pairs of (membraneous) wings.
- Fore wings wings larger than hind wings.
-Wings are linked together by hooklets.
-Has piercing and biting mouthparts.
- Metameric segmentation.
- Bilaterally symmetrical.
- Three pairs of (thoracic) legs.
- Three body divisions head, thorax and. abdomen.
- Presence of exoskeleton (of chitin)
.- one pair of antennae.
- One pair of compound eyes.
- Presence of jointed appendages.
(iii) Habitat of Specimen G: Mud nest on ceiling of buildings/wall/furniture.
Source: Myschool Past Questions Archive (Record #60303)
Mathematics
WAEC / WASSCE 2013
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, O is the centre of the circle. OM||XZ and < ZOM = 25º. Calculate ∠XYZ
Attached Figure / Diagram
A
50o
B
55o
C
60o
D
65o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Angle in a semi-circle is 90º when OM is subtended
: (90 - 25)º
= 65º
Source: WAEC / JAMB / NECO Classroom Archive (Record #36922)
Mathematics
WAEC / WASSCE 2013
Fractions, Decimals, Percentages & Approximations
Solution
When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number
A
11
B
\(\frac{15}{2}\)
C
5
D
\(\frac{5}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let the number be y, subtract y from 2 i.e 2 - y
2 - y = 4 < \(\frac{1}{5}\) y,
2 - y = \(\frac{y}{5}\) - 4
2 - y + 4 = \(\frac{y}{5}\)
6 = \(\frac{y}{5}\) + y
6 = \(\frac{y + 5y}{5}\)
6 = \(\frac{6y}{5}\)
multiplying through by 5
6 * 5 = 6y
\(\frac{30}{6}\) = y
= 5
Source: WAEC / JAMB / NECO Classroom Archive (Record #36891)
Mathematics
WAEC / WASSCE 2013
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
Using the histogram, what is the median class?
Attached Figure / Diagram
A
60.5 - 70.5
B
50.5 - 60.5
C
40.5 - 50.5
D
30.5 - 40.5
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: WAEC / JAMB / NECO Classroom Archive (Record #36924)
Physics
WAEC / WASSCE 2013
Newton's Laws, Momentum & Collisions
Diagram
Solution
A block of wood of mass 5kg is pulled on a platform by a force of 40N as illustrated in the diagram above. If the frictional force, F experienced by the block is 12N, calculate the magnitude of the acceleration of the acceleration of the block
Attached Figure / Diagram
A
2.4ms-2
B
5.6ms-2
C
8.0ms-2
D
2.4ms2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
F + Ma = P
a = \(\frac{P - F}{m}\)
= \(\frac{40 - 12}{5} = 5.6ms^{-2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #35729)
Physics
WAEC / WASSCE 2013
Current Electricity & Electric Circuits
Solution
In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [Electrochemical equivalent of copper = 0.00033 g C\(^{-1}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
M = zit
I = \(\frac{M}{Zt}\) = \(\frac{0.66}{30 \times 60 \times 33 \times 10^{-5}}\)
= 1.11A
Error = 1.11 - 1
= 0.11A
Source: WAEC / JAMB / NECO Classroom Archive (Record #53125)
Physics
UTME / JAMB 2013
Atomic, Quantum, Nuclear Physics & Electronics
Solution
In photo-emission, the number of photo-electrons ejected per second depends on the?
A
frequency of the beam
B
work function of the metal
C
threshold frequency of the metal
D
intensity of the beam
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The photoelectric effect is dependent upon various factors including frequency of light, intensity of light, nature of material, energy of light and potential difference.
Source: WAEC / JAMB / NECO Classroom Archive (Record #27370)
Physics
WAEC / WASSCE 2013
Atomic, Quantum, Nuclear Physics & Electronics
Solution
Which of the following phenomena supports the theory that waves have a particle nature?
A
electron diffraction
B
photoelectric effect
C
diffraction of light
D
x-ray interference
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.
Source: WAEC / JAMB / NECO Classroom Archive (Record #35728)