WAEC SSCE - Biology - 2019 (Essay)
Question 1 Report
A homozygous tall, red flower plant was crossed with a homozygous dwarf, white flower plant. The F1 were selfed and 160 F2 plants were obtained.
a.(i) Name the type of genetic cross. [1 marks]
(ii) State one reason for the answer in 4(a)(i). [1 marks]
b.(i) What is evolution? [2 marks]
(ii) With the aid of a genetic diagram, state the phenotypic ratio of F2 generation. [13 marks]
(iii) Calculate the number of tall and white flowers that would be obtained in the F2 generation. [3 marks]
(a)(i) The cross is a dihybrid cross.
(a)(ii) It involves two pairs of contrasting characters, namely plant height (tall/dwarf) and flower colour (red/white).
(b)(i) Evolution is the gradual change in the inherited characteristics of a population over many generations, which may result in new forms or species.
(b)(ii) Genetic diagram
Let T represent tallness and t dwarfness; R represent red flowers and r white flowers. Tallness and red flower colour are dominant.
P generation: Tall, red plant (TTRR) × Dwarf, white plant (ttrr)
Gametes: TR × tr
F1 generation: all TtRr, tall and red.
Selfing of F1: TtRr × TtRr
Each F1 plant produces the four types of gametes: TR, Tr, tR and tr.
From the genetic diagram, the F2 phenotypic ratio is:
9 tall red : 3 tall white : 3 dwarf red : 1 dwarf white.
(b)(iii) Tall, white plants are represented by 3 out of 16 plants.
\[\text{Number of tall, white plants}=\frac{3}{16}\times160=30\]
Therefore, 30 tall, white-flowered plants would be obtained in the F2 generation.
Answer Details
(a)(i) The cross is a dihybrid cross.
(a)(ii) It involves two pairs of contrasting characters, namely plant height (tall/dwarf) and flower colour (red/white).
(b)(i) Evolution is the gradual change in the inherited characteristics of a population over many generations, which may result in new forms or species.
(b)(ii) Genetic diagram
Let T represent tallness and t dwarfness; R represent red flowers and r white flowers. Tallness and red flower colour are dominant.
P generation: Tall, red plant (TTRR) × Dwarf, white plant (ttrr)
Gametes: TR × tr
F1 generation: all TtRr, tall and red.
Selfing of F1: TtRr × TtRr
Each F1 plant produces the four types of gametes: TR, Tr, tR and tr.
From the genetic diagram, the F2 phenotypic ratio is:
9 tall red : 3 tall white : 3 dwarf red : 1 dwarf white.
(b)(iii) Tall, white plants are represented by 3 out of 16 plants.
\[\text{Number of tall, white plants}=\frac{3}{16}\times160=30\]
Therefore, 30 tall, white-flowered plants would be obtained in the F2 generation.
Question 2 Report
b. List five parts of the alimentary canal of an earthworm. [5 marks]
c. State six ways by which water could be polluted by agricultural. [6 marks]
d. Describe briefly the life cycle of a housefly. [5 marks]
e. Make a diagram, 6 cm – 8 cm long of the hypogeal germination in a mature maize seedling and label fully. [8 marks]
b. The five parts of the alimentary canal of an earthworm are the mouth, pharynx, esophagus, crop, and gizzard.
c. Six ways by which water could be polluted by agricultural activities are:
- Runoff from fields carrying pesticides and fertilizers into nearby water bodies.
- Livestock waste and manure runoff into rivers and streams.
- Leaching of nutrients and chemicals from fields into groundwater.
- Erosion from tilled fields, which can carry sediment and nutrients into waterways.
- Irrigation runoff carrying chemicals and salts.
- Discharge of untreated or poorly treated agricultural wastewater into surface waters.
d. The life cycle of a housefly involves four stages: egg, larva, pupa, and adult. Adult female houseflies lay eggs on organic matter such as garbage, animal waste, or decaying food. The eggs hatch into larvae, commonly known as maggots, which feed on the organic matter for several days. The larvae then move to a dry location to pupate, forming a hard outer casing known as a puparium. Within the puparium, the larvae transform into an adult fly. The adult emerges from the puparium, mate, and begin the cycle anew.
e. Hypogeal germination is a type of seed germination where the cotyledons (seed leaves) remain below the soil surface. In a mature maize seedling, the embryo is located at the base of the kernel, and the radicle (embryonic root) emerges first during germination. The radicle grows downward, elongates, and forms lateral roots. The shoot also emerges from the embryo and grows upward, carrying the plumule (embryonic shoot) and the first true leaves above the soil surface. The diagram of hypogeal germination in a mature maize seedling should show the radicle, the embryonic shoot, the first true leaves, the kernel, and the soil.
Answer Details
b. The five parts of the alimentary canal of an earthworm are the mouth, pharynx, esophagus, crop, and gizzard.
c. Six ways by which water could be polluted by agricultural activities are:
- Runoff from fields carrying pesticides and fertilizers into nearby water bodies.
- Livestock waste and manure runoff into rivers and streams.
- Leaching of nutrients and chemicals from fields into groundwater.
- Erosion from tilled fields, which can carry sediment and nutrients into waterways.
- Irrigation runoff carrying chemicals and salts.
- Discharge of untreated or poorly treated agricultural wastewater into surface waters.
d. The life cycle of a housefly involves four stages: egg, larva, pupa, and adult. Adult female houseflies lay eggs on organic matter such as garbage, animal waste, or decaying food. The eggs hatch into larvae, commonly known as maggots, which feed on the organic matter for several days. The larvae then move to a dry location to pupate, forming a hard outer casing known as a puparium. Within the puparium, the larvae transform into an adult fly. The adult emerges from the puparium, mate, and begin the cycle anew.
e. Hypogeal germination is a type of seed germination where the cotyledons (seed leaves) remain below the soil surface. In a mature maize seedling, the embryo is located at the base of the kernel, and the radicle (embryonic root) emerges first during germination. The radicle grows downward, elongates, and forms lateral roots. The shoot also emerges from the embryo and grows upward, carrying the plumule (embryonic shoot) and the first true leaves above the soil surface. The diagram of hypogeal germination in a mature maize seedling should show the radicle, the embryonic shoot, the first true leaves, the kernel, and the soil.
Question 3 Report
The table below shows the percentage composition of fat and protein in six different meat types. Study it and answer questions 2(a) to 2(d).
| Meat Type | Fat (%) | Protein (%) |
| A | 07.2 | 21.3 |
| B | 25.3 | 10.6 |
| C | 20.0 | 22.5 |
| D | 03.1 | 28.2 |
| E | 12.6 | 17.3 |
| F | 13.2 | 14.3 |
a. (i) Which two of the meat types may be recommended for an obese patient? [2 marks]
(ii) State one reason for the answer in 2(a)(i). [1 mark]
b. (i) Which two of the meat types would provide the most energy? [2 marks]
(ii) State two reasons for the answer in 2(b)(i). [2 marks]
c. (i) Which three of the meat types could be recommended for a child suffering from kwashiorkor? [3 marks]
(ii) State one reason for the answer in 2(c)(i) [1 mark]
d. Which of the meat types would most likely be suitable for:
(i) an active teenager; [1 mark]
(ii) a 70-year old human? [1 mark]
e. Which other class of food provides energy? [1 mark]
f. State three uses of fat in the human body. [3 marks]
g. Describe briefly the procedure for testing for fat in a meat sample using a piece of white paper. [3 marks]
The data to reason from is the composition table. Fat yields about \(37\ \text{kJ g}^{-1}\) of energy while protein yields about \(17\ \text{kJ g}^{-1}\), so fat is the richer energy source and protein is the tissue builder. Approximate energy per 100 g is \(E = (\text{fat}\times 9) + (\text{protein}\times 4)\) kcal.
| Meat type | Fat (%) | Protein (%) | Approx. energy (kcal/100 g) |
|---|---|---|---|
| A | 7.2 | 21.3 | 150.0 |
| B | 25.3 | 10.6 | 270.1 |
| C | 20.0 | 22.5 | 270.0 |
| D | 3.1 | 28.2 | 140.7 |
| E | 12.6 | 17.3 | 182.6 |
| F | 13.2 | 14.3 | 176.0 |
(a)(i) Meat types D and A.
(a)(ii) They have the lowest fat content (3.1% and 7.2%), so they add the least fat and energy to the body and will not worsen the obesity.
(b)(i) Meat types B and C.
(b)(ii) (1) They carry the highest fat contents (25.3% and 20.0%), and fat provides more than twice the energy per gram of protein or carbohydrate. (2) Their calculated energy values (about 270 kcal/100 g) are the highest in the table.
(c)(i) Meat types D, C and A.
(c)(ii) Kwashiorkor is a protein-deficiency disease, and these three have the highest protein contents (28.2%, 22.5%, 21.3%), so they best replace the missing protein.
(d)(i) An active teenager: C (it combines high energy with high protein, 22.5%, needed for activity and growth).
(d)(ii) A 70-year-old person: D (very low fat, 3.1%, to avoid cardiovascular strain, yet high protein for tissue repair).
(e) Carbohydrates also provide energy.
(f) Three uses of fat in the human body:
- It is a concentrated reserve store of energy.
- It insulates the body against heat loss (fat layer beneath the skin).
- It cushions and protects delicate internal organs such as the kidneys.
(g) Testing for fat with white paper: rub a small piece of the meat firmly on the white paper so that any oil is transferred. Allow it to stand (or gently warm it) so that water and moisture evaporate. Hold the paper up to the light. A permanent translucent (greasy) spot that does not dry off shows that fat is present.
Answer Details
The data to reason from is the composition table. Fat yields about \(37\ \text{kJ g}^{-1}\) of energy while protein yields about \(17\ \text{kJ g}^{-1}\), so fat is the richer energy source and protein is the tissue builder. Approximate energy per 100 g is \(E = (\text{fat}\times 9) + (\text{protein}\times 4)\) kcal.
| Meat type | Fat (%) | Protein (%) | Approx. energy (kcal/100 g) |
|---|---|---|---|
| A | 7.2 | 21.3 | 150.0 |
| B | 25.3 | 10.6 | 270.1 |
| C | 20.0 | 22.5 | 270.0 |
| D | 3.1 | 28.2 | 140.7 |
| E | 12.6 | 17.3 | 182.6 |
| F | 13.2 | 14.3 | 176.0 |
(a)(i) Meat types D and A.
(a)(ii) They have the lowest fat content (3.1% and 7.2%), so they add the least fat and energy to the body and will not worsen the obesity.
(b)(i) Meat types B and C.
(b)(ii) (1) They carry the highest fat contents (25.3% and 20.0%), and fat provides more than twice the energy per gram of protein or carbohydrate. (2) Their calculated energy values (about 270 kcal/100 g) are the highest in the table.
(c)(i) Meat types D, C and A.
(c)(ii) Kwashiorkor is a protein-deficiency disease, and these three have the highest protein contents (28.2%, 22.5%, 21.3%), so they best replace the missing protein.
(d)(i) An active teenager: C (it combines high energy with high protein, 22.5%, needed for activity and growth).
(d)(ii) A 70-year-old person: D (very low fat, 3.1%, to avoid cardiovascular strain, yet high protein for tissue repair).
(e) Carbohydrates also provide energy.
(f) Three uses of fat in the human body:
- It is a concentrated reserve store of energy.
- It insulates the body against heat loss (fat layer beneath the skin).
- It cushions and protects delicate internal organs such as the kidneys.
(g) Testing for fat with white paper: rub a small piece of the meat firmly on the white paper so that any oil is transferred. Allow it to stand (or gently warm it) so that water and moisture evaporate. Hold the paper up to the light. A permanent translucent (greasy) spot that does not dry off shows that fat is present.
Question 4 Report
a. (i) Complete the table below with the respiratory surface of each of the listed organisms.
| organism | respiratory surface |
| Tadpole (2days old) | |
| Cockroach | |
| Domestic fowl | |
| Talinum | |
| Earth Worm | |
| Amoeba |
(ii) State five characteristic features of respiratory surfaces in organisms. [5 marks]
b. Make a drawing, 4 cm – 6 cm long of the respiratory organ in Tilapia and label fully. [9 marks]
(a)(i) Respiratory surfaces of the organisms
| Organism | Respiratory surface |
|---|---|
| Tadpole (2 days old) | External gills |
| Cockroach | Tracheae and tracheoles |
| Domestic fowl | Lungs |
| Talinum | Stomata |
| Earthworm | Moist skin or body surface |
| Amoeba | Cell membrane or body surface |
(a)(ii) Characteristics of respiratory surfaces
- They have a large surface area for rapid gaseous exchange.
- They are thin, usually one cell thick, giving a short diffusion distance.
- They are moist so that oxygen and carbon dioxide can dissolve before diffusion.
- They are permeable to respiratory gases.
- They are well ventilated to maintain a steep concentration gradient.
- In organisms with circulatory systems, they are richly supplied with blood capillaries.
(b) Respiratory organ of Tilapia: gill
The labelled drawing below should be made about 4 cm to 6 cm long.
The numerous gill filaments provide a large, thin, moist and richly supplied surface for gaseous exchange.
Answer Details
(a)(i) Respiratory surfaces of the organisms
| Organism | Respiratory surface |
|---|---|
| Tadpole (2 days old) | External gills |
| Cockroach | Tracheae and tracheoles |
| Domestic fowl | Lungs |
| Talinum | Stomata |
| Earthworm | Moist skin or body surface |
| Amoeba | Cell membrane or body surface |
(a)(ii) Characteristics of respiratory surfaces
- They have a large surface area for rapid gaseous exchange.
- They are thin, usually one cell thick, giving a short diffusion distance.
- They are moist so that oxygen and carbon dioxide can dissolve before diffusion.
- They are permeable to respiratory gases.
- They are well ventilated to maintain a steep concentration gradient.
- In organisms with circulatory systems, they are richly supplied with blood capillaries.
(b) Respiratory organ of Tilapia: gill
The labelled drawing below should be made about 4 cm to 6 cm long.
The numerous gill filaments provide a large, thin, moist and richly supplied surface for gaseous exchange.
Question 5 Report
a. State three features of the small intestine that increase the rate of absorption of digested food. [3 marks]
b. Explain briefly what happens to the glucose formed in a plant during photosynthesis. [3 marks]
c. Explain briefly the mode of feeding in each of the following organisms:
(i) Euglena; [4 marks]
(ii) Spirogyra. [4 marks]
d. State three characteristics of an Estuarine habitat. [3 marks]
e. State three differences between an aquatic habitat and a terrestrial habitat. [3 marks]
(a) Three features of the small intestine that increase the rate of absorption
- It is very long and coiled, providing a large surface area for absorption.
- Its inner wall is folded and carries numerous finger-like villi (and microvilli) which greatly increase the absorbing surface.
- The villi have thin (one-cell-thick) walls and a rich supply of blood capillaries and lacteals, so absorbed food passes quickly into the blood and lymph.
(b) Fate of the glucose formed during photosynthesis
Some of the glucose is used at once in respiration to release energy. The rest is converted to starch for temporary storage in the leaf, or changed to sucrose and translocated through the phloem to other parts of the plant, where it is used for growth or stored (as starch, oils or proteins) in fruits, seeds, roots and stems. Glucose is also used to build cellulose for cell walls and, with nitrogen, to make proteins.
(c) Mode of feeding
(i) Euglena: Euglena is mainly holophytic (autotrophic). It contains chloroplasts and manufactures its own food by photosynthesis in the presence of sunlight. In darkness it can feed saprophytically (saprozoically) by absorbing dissolved organic matter through its body surface. This double mode of nutrition makes it mixotrophic.
(ii) Spirogyra: Spirogyra is holophytic (autotrophic). It has a spiral chloroplast containing chlorophyll and manufactures its own food by photosynthesis, using water, carbon dioxide and sunlight.
(d) Three characteristics of an estuarine habitat
- The water is brackish (a mixture of fresh and salt water), so salinity varies.
- Salinity, water level and temperature change with the tides.
- The water is often muddy/turbid and rich in nutrients and silt.
(e) Three differences between an aquatic habitat and a terrestrial habitat
| Aquatic habitat | Terrestrial habitat |
|---|---|
| Medium of life is water | Medium of life is air/land |
| Organisms obtain oxygen dissolved in water | Organisms obtain oxygen from the air |
| Temperature is fairly constant | Temperature fluctuates widely |
| Water is always available | Water may be scarce |
Answer Details
(a) Three features of the small intestine that increase the rate of absorption
- It is very long and coiled, providing a large surface area for absorption.
- Its inner wall is folded and carries numerous finger-like villi (and microvilli) which greatly increase the absorbing surface.
- The villi have thin (one-cell-thick) walls and a rich supply of blood capillaries and lacteals, so absorbed food passes quickly into the blood and lymph.
(b) Fate of the glucose formed during photosynthesis
Some of the glucose is used at once in respiration to release energy. The rest is converted to starch for temporary storage in the leaf, or changed to sucrose and translocated through the phloem to other parts of the plant, where it is used for growth or stored (as starch, oils or proteins) in fruits, seeds, roots and stems. Glucose is also used to build cellulose for cell walls and, with nitrogen, to make proteins.
(c) Mode of feeding
(i) Euglena: Euglena is mainly holophytic (autotrophic). It contains chloroplasts and manufactures its own food by photosynthesis in the presence of sunlight. In darkness it can feed saprophytically (saprozoically) by absorbing dissolved organic matter through its body surface. This double mode of nutrition makes it mixotrophic.
(ii) Spirogyra: Spirogyra is holophytic (autotrophic). It has a spiral chloroplast containing chlorophyll and manufactures its own food by photosynthesis, using water, carbon dioxide and sunlight.
(d) Three characteristics of an estuarine habitat
- The water is brackish (a mixture of fresh and salt water), so salinity varies.
- Salinity, water level and temperature change with the tides.
- The water is often muddy/turbid and rich in nutrients and silt.
(e) Three differences between an aquatic habitat and a terrestrial habitat
| Aquatic habitat | Terrestrial habitat |
|---|---|
| Medium of life is water | Medium of life is air/land |
| Organisms obtain oxygen dissolved in water | Organisms obtain oxygen from the air |
| Temperature is fairly constant | Temperature fluctuates widely |
| Water is always available | Water may be scarce |
Question 6 Report
a. State five:
(i) challenges that make conservation of resources difficult; [5 marks]
(ii) ways by which the Government of a country ensures conservation of natural resources. [5 marks]
b. What is an endangered species? [3 marks]
c. A child with blood group O was born to a mother of blood group A.
(i) What is the blood group genotype of the child? [1 marks]
(ii) What are the possible blood group genotypes of the father? [3 marks]
d. State three reasons why genetic variation is important to plants. [3 marks]
(a)(i) Five challenges that make conservation of resources difficult
- Rapid population growth that increases demand for land, food and fuel.
- Poverty, which forces people to over-exploit resources for immediate survival.
- Ignorance and lack of awareness about the value of conservation.
- Weak enforcement of conservation laws and corruption.
- Inadequate funds and trained personnel to manage protected areas.
Others: industrialisation and urbanisation, and high demand for timber and bushmeat.
(a)(ii) Five ways a government ensures conservation of natural resources
- Enacting and enforcing laws that protect wildlife and forests.
- Establishing national parks, game reserves and forest reserves.
- Educating the public on the need for conservation.
- Encouraging afforestation and reforestation programmes.
- Controlling hunting, logging and fishing through licences and closed seasons.
(b) What is an endangered species?
An endangered species is a species whose population has fallen so low that it is in danger of becoming extinct (dying out completely) if the conditions threatening it are not controlled. Examples include the gorilla and certain species of elephant.
(c) Blood group genetics
Blood group O has the genotype OO. A child of blood group O must receive one O allele from each parent.
(c)(i) Genotype of the child: OO.
(c)(ii) Possible genotypes of the father: Since the child is OO, the father must supply an O allele. His possible genotypes are therefore any group carrying an O allele:
- IᴬO (blood group A carrying O)
- IᴮO (blood group B carrying O)
- OO (blood group O)
(d) Three reasons why genetic variation is important to plants
- It enables some plants to survive changes in the environment (better adaptation).
- It provides the raw material for natural selection and evolution.
- It allows breeders to select and produce improved varieties (higher yield, disease resistance).
Answer Details
(a)(i) Five challenges that make conservation of resources difficult
- Rapid population growth that increases demand for land, food and fuel.
- Poverty, which forces people to over-exploit resources for immediate survival.
- Ignorance and lack of awareness about the value of conservation.
- Weak enforcement of conservation laws and corruption.
- Inadequate funds and trained personnel to manage protected areas.
Others: industrialisation and urbanisation, and high demand for timber and bushmeat.
(a)(ii) Five ways a government ensures conservation of natural resources
- Enacting and enforcing laws that protect wildlife and forests.
- Establishing national parks, game reserves and forest reserves.
- Educating the public on the need for conservation.
- Encouraging afforestation and reforestation programmes.
- Controlling hunting, logging and fishing through licences and closed seasons.
(b) What is an endangered species?
An endangered species is a species whose population has fallen so low that it is in danger of becoming extinct (dying out completely) if the conditions threatening it are not controlled. Examples include the gorilla and certain species of elephant.
(c) Blood group genetics
Blood group O has the genotype OO. A child of blood group O must receive one O allele from each parent.
(c)(i) Genotype of the child: OO.
(c)(ii) Possible genotypes of the father: Since the child is OO, the father must supply an O allele. His possible genotypes are therefore any group carrying an O allele:
- IᴬO (blood group A carrying O)
- IᴮO (blood group B carrying O)
- OO (blood group O)
(d) Three reasons why genetic variation is important to plants
- It enables some plants to survive changes in the environment (better adaptation).
- It provides the raw material for natural selection and evolution.
- It allows breeders to select and produce improved varieties (higher yield, disease resistance).
Question 7 Report
a. Name five vertebrae in mammals with their corresponding locations. [10 marks]
b. State one function each of the following parts in a dicotyledonous leaf:
(i) palisade;
(ii) vascular bundle;
(iii) epidermis. [3 marks]
c. State one function each of the following parts in the stem of a flowering plant:
(i) sieve tube;
(ii) cortex;
(iii) pith.
d. In a tabular form, state:
(i) two external structural differences between the stem and root of a maize seeding. [2 marks]
(ii) two internal structural differences between the stem and root of a maize seeding. [2 marks]
(a) Five vertebrae in mammals with their locations
| Vertebra | Location in the body |
|---|---|
| Cervical vertebrae | Neck region |
| Thoracic vertebrae | Chest (thorax) region, articulating with the ribs |
| Lumbar vertebrae | Lower back (abdominal) region |
| Sacral vertebrae | Hip (pelvic) region, fused to form the sacrum |
| Caudal vertebrae | Tail region |
(b) One function each in a dicotyledonous leaf
- Palisade: contains many chloroplasts and is the main site of photosynthesis.
- Vascular bundle (vein): conducts water and mineral salts to the leaf (xylem) and carries manufactured food away (phloem); it also supports the leaf.
- Epidermis: protects the inner tissues and, being covered by a cuticle, reduces water loss.
(c) One function each in the stem of a flowering plant
- Sieve tube: conducts (translocates) manufactured food from the leaves to other parts of the plant.
- Cortex: stores food and water and gives support to the stem.
- Pith: stores food and water in the centre of the stem.
(d)(i) Two external structural differences between the stem and root of a maize seedling
| Stem | Root |
|---|---|
| Bears leaves and buds | Bears no leaves or buds |
| Green in colour (photosynthetic) | Not green (whitish/brown) |
Also: stem grows upward, root grows downward; root has root hairs and a root cap, the stem does not.
(d)(ii) Two internal structural differences between the stem and root of a maize seedling
| Stem | Root |
|---|---|
| Vascular bundles scattered throughout the ground tissue | Vascular tissue arranged in a central ring (central cylinder) |
| Xylem and phloem lie side by side in each bundle (conjoint) | Xylem and phloem lie on separate radii (radial arrangement) |
Answer Details
(a) Five vertebrae in mammals with their locations
| Vertebra | Location in the body |
|---|---|
| Cervical vertebrae | Neck region |
| Thoracic vertebrae | Chest (thorax) region, articulating with the ribs |
| Lumbar vertebrae | Lower back (abdominal) region |
| Sacral vertebrae | Hip (pelvic) region, fused to form the sacrum |
| Caudal vertebrae | Tail region |
(b) One function each in a dicotyledonous leaf
- Palisade: contains many chloroplasts and is the main site of photosynthesis.
- Vascular bundle (vein): conducts water and mineral salts to the leaf (xylem) and carries manufactured food away (phloem); it also supports the leaf.
- Epidermis: protects the inner tissues and, being covered by a cuticle, reduces water loss.
(c) One function each in the stem of a flowering plant
- Sieve tube: conducts (translocates) manufactured food from the leaves to other parts of the plant.
- Cortex: stores food and water and gives support to the stem.
- Pith: stores food and water in the centre of the stem.
(d)(i) Two external structural differences between the stem and root of a maize seedling
| Stem | Root |
|---|---|
| Bears leaves and buds | Bears no leaves or buds |
| Green in colour (photosynthetic) | Not green (whitish/brown) |
Also: stem grows upward, root grows downward; root has root hairs and a root cap, the stem does not.
(d)(ii) Two internal structural differences between the stem and root of a maize seedling
| Stem | Root |
|---|---|
| Vascular bundles scattered throughout the ground tissue | Vascular tissue arranged in a central ring (central cylinder) |
| Xylem and phloem lie side by side in each bundle (conjoint) | Xylem and phloem lie on separate radii (radial arrangement) |
Question 8 Report
a. Classify the following biological associations under the headings in the table below:
(i) Remora and shark;
(ii) Lichen;
(iii) Cattle and white Egret;
(iv) Tapeworm in the gut of humans;
(v) Flowers and Honeybees;
(vi) Mistletoe and Cacao plant.
| Parasitism | Mutualism | Commensalism |
[6 marks]
b. (i) State three adaptive features of parasites. [6 marks]
(ii) State two effects of parasites on their hosts. [2 marks]
c. (i) What are saprophytes? [2 marks]
(ii) Give four examples of saprophytes. [4 marks]
(a) Classification of the biological associations:
| Parasitism | Mutualism | Commensalism |
|---|---|---|
| Tapeworm in the gut of humans | Lichen (fungus and alga) | Remora and shark |
| Mistletoe and Cacao plant | Flowers and Honeybees | Cattle and white Egret |
In parasitism one partner (the parasite) benefits while the host is harmed; in mutualism both partners benefit; in commensalism one partner benefits while the other is neither helped nor harmed.
(b)(i) Three adaptive features of parasites:
- Presence of hooks and suckers for firm attachment to the host so they are not dislodged.
- A thick, resistant cuticle (and anti-enzymes) that protects them from the host's digestive juices.
- High reproductive capacity (production of very many eggs) to ensure that some offspring reach new hosts.
Other acceptable features: reduced or absent locomotory organs, reduced sense organs, and a reduced digestive system since food is absorbed already digested.
(b)(ii) Two effects of parasites on their hosts:
- They deprive the host of nutrients, causing malnutrition and weakness.
- They cause diseases and damage host tissues or organs (for example anaemia from blood loss).
(c)(i) Saprophytes are organisms that feed on dead and decaying organic matter, secreting digestive enzymes onto the material externally and absorbing the soluble digested products.
(c)(ii) Four examples of saprophytes: Mushroom, Rhizopus (bread mould), Mucor, Yeast (also Penicillium and many bacteria of decay).
Answer Details
(a) Classification of the biological associations:
| Parasitism | Mutualism | Commensalism |
|---|---|---|
| Tapeworm in the gut of humans | Lichen (fungus and alga) | Remora and shark |
| Mistletoe and Cacao plant | Flowers and Honeybees | Cattle and white Egret |
In parasitism one partner (the parasite) benefits while the host is harmed; in mutualism both partners benefit; in commensalism one partner benefits while the other is neither helped nor harmed.
(b)(i) Three adaptive features of parasites:
- Presence of hooks and suckers for firm attachment to the host so they are not dislodged.
- A thick, resistant cuticle (and anti-enzymes) that protects them from the host's digestive juices.
- High reproductive capacity (production of very many eggs) to ensure that some offspring reach new hosts.
Other acceptable features: reduced or absent locomotory organs, reduced sense organs, and a reduced digestive system since food is absorbed already digested.
(b)(ii) Two effects of parasites on their hosts:
- They deprive the host of nutrients, causing malnutrition and weakness.
- They cause diseases and damage host tissues or organs (for example anaemia from blood loss).
(c)(i) Saprophytes are organisms that feed on dead and decaying organic matter, secreting digestive enzymes onto the material externally and absorbing the soluble digested products.
(c)(ii) Four examples of saprophytes: Mushroom, Rhizopus (bread mould), Mucor, Yeast (also Penicillium and many bacteria of decay).
Question 9 Report
a. Complete the table below.
| Organism | Mode of feeding | Feature of mouthpart that adapts organism to mode of feeding |
[6 marks]
b. List five parts of the alimentary canal of an earthworm. [5 marks]
c. State six ways by which water could be polluted by agricultural practices. [6 marks]
d. Describe briefly the life cycle of a housefly. [5 marks]
e. Make a diagram, 6 cm – 8 cm long of the hypogeal germination in a mature maize seedling and label fully. [8 marks]
(a) Completed table of organisms, mode of feeding and the adapting mouthpart feature:
| Organism | Mode of feeding | Feature of mouthpart that adapts organism to mode of feeding |
|---|---|---|
| Mosquito | Piercing and sucking | Long needle-like proboscis (stylets) for piercing skin and sucking blood |
| Housefly | Sponging and lapping | Soft, spongy proboscis (labellum) for mopping up liquid or dissolved food |
| Grasshopper | Biting and chewing | Strong toothed mandibles for biting and grinding plant material |
(Butterfly, with a long coiled proboscis for sucking nectar, and honeybee, with mandibles plus a long tongue for chewing and lapping, are also acceptable entries.)
(b) Five parts of the alimentary canal of an earthworm: pharynx, oesophagus, crop, gizzard, intestine (the full canal runs mouth, buccal cavity, pharynx, oesophagus, crop, gizzard, intestine, anus).
(c) Six ways water can be polluted by agricultural practices:
- Chemical fertilizers washed (leached) into rivers and lakes, causing eutrophication.
- Insecticides and pesticides running off farmland into water bodies.
- Herbicides washed from sprayed crops into streams.
- Animal droppings and livestock waste washed into water.
- Soil erosion from cultivated land silting up water bodies.
- Improper disposal of decaying farm and organic waste into water.
(d) Life cycle of a housefly (complete metamorphosis): The adult female lays batches of eggs on decaying organic matter such as refuse or faeces. Each egg hatches into a larva (maggot) that feeds actively, grows and moults. The larva then changes into an inactive pupa enclosed in a puparium, within which the body is reorganised. Finally an adult fly (imago) emerges, matures and mates, and the cycle repeats: egg to larva to pupa to adult.
(e) Hypogeal germination in maize: the diagram should show the grain remaining below the soil (the cotyledon and food-storing endosperm stay underground), the coleoptile pushing vertically up through the soil, the first foliage leaf emerging from the tip of the coleoptile above ground, the primary (radicle) root growing downward, and the fibrous seminal/adventitious roots. The soil line should be marked, with the grain and roots below it and the shoot above it.
Answer Details
(a) Completed table of organisms, mode of feeding and the adapting mouthpart feature:
| Organism | Mode of feeding | Feature of mouthpart that adapts organism to mode of feeding |
|---|---|---|
| Mosquito | Piercing and sucking | Long needle-like proboscis (stylets) for piercing skin and sucking blood |
| Housefly | Sponging and lapping | Soft, spongy proboscis (labellum) for mopping up liquid or dissolved food |
| Grasshopper | Biting and chewing | Strong toothed mandibles for biting and grinding plant material |
(Butterfly, with a long coiled proboscis for sucking nectar, and honeybee, with mandibles plus a long tongue for chewing and lapping, are also acceptable entries.)
(b) Five parts of the alimentary canal of an earthworm: pharynx, oesophagus, crop, gizzard, intestine (the full canal runs mouth, buccal cavity, pharynx, oesophagus, crop, gizzard, intestine, anus).
(c) Six ways water can be polluted by agricultural practices:
- Chemical fertilizers washed (leached) into rivers and lakes, causing eutrophication.
- Insecticides and pesticides running off farmland into water bodies.
- Herbicides washed from sprayed crops into streams.
- Animal droppings and livestock waste washed into water.
- Soil erosion from cultivated land silting up water bodies.
- Improper disposal of decaying farm and organic waste into water.
(d) Life cycle of a housefly (complete metamorphosis): The adult female lays batches of eggs on decaying organic matter such as refuse or faeces. Each egg hatches into a larva (maggot) that feeds actively, grows and moults. The larva then changes into an inactive pupa enclosed in a puparium, within which the body is reorganised. Finally an adult fly (imago) emerges, matures and mates, and the cycle repeats: egg to larva to pupa to adult.
(e) Hypogeal germination in maize: the diagram should show the grain remaining below the soil (the cotyledon and food-storing endosperm stay underground), the coleoptile pushing vertically up through the soil, the first foliage leaf emerging from the tip of the coleoptile above ground, the primary (radicle) root growing downward, and the fibrous seminal/adventitious roots. The soil line should be marked, with the grain and roots below it and the shoot above it.
Question 10 Report
Use the information below to answer questions (a)(i) and (a)(ii).
Hb Are presents normal haemoglobin,
HbS represents sickled haemoglobin.
a. A female heterozygote for sickle cell married a sickler. With the aid of a genetic diagram, determine the:
(i) possible genotypes of their offspring; [8 marks]
(ii) phenotypic ratio of the offspring. [2 marks]
b. Explain briefly the reason why a Rhesus negative woman married to a Rhesus positive man might lose her second pregnancy. [5 marks]
c. Name two examples of features in animals that support the theory of use and disuse of body parts as used by Lamarck. [2 marks]
d. List three structures in mammals that are vestigial. [3 marks]
(a)(i) Genetic cross
Let HbA represent the allele for normal haemoglobin and HbS represent the allele for sickle-cell haemoglobin.
Female heterozygote: HbAHbS
Male sickler: HbSHbS
The female produces gametes HbA and HbS, while the male produces only HbS gametes.
Therefore, the possible genotypes of the offspring are:
- HbAHbS, heterozygous normal carrier or sickle-cell trait, 2 out of 4;
- HbSHbS, sickler, 2 out of 4.
Genotypic ratio: HbAHbS : HbSHbS = 2:2 = 1:1.
(a)(ii) Phenotypic ratio
Normal carriers (non-sicklers) : sicklers = 1:1.
(b) Rhesus incompatibility and loss of the second pregnancy
A Rhesus-negative woman has no Rh antigen on her red blood cells. If her first foetus is Rhesus-positive, some foetal red blood cells may enter her circulation, especially at delivery when the placenta separates. The Rh antigen stimulates the mother to form anti-Rhesus antibodies. In a later pregnancy with a Rhesus-positive foetus, these antibodies cross the placenta and destroy the foetal red blood cells. The resulting severe haemolysis, called erythroblastosis foetalis, may cause death of the foetus and miscarriage.
(c) Features explained by Lamarck's use and disuse theory
- The long neck of the giraffe, attributed to continual stretching to reach high leaves.
- The webbed feet of ducks or geese, attributed to constant use in swimming.
(d) Vestigial structures in mammals
- Vermiform appendix in humans.
- Coccyx, or reduced tail vertebrae, in humans.
- Pelvic and hind-limb bones in whales.
Answer Details
(a)(i) Genetic cross
Let HbA represent the allele for normal haemoglobin and HbS represent the allele for sickle-cell haemoglobin.
Female heterozygote: HbAHbS
Male sickler: HbSHbS
The female produces gametes HbA and HbS, while the male produces only HbS gametes.
Therefore, the possible genotypes of the offspring are:
- HbAHbS, heterozygous normal carrier or sickle-cell trait, 2 out of 4;
- HbSHbS, sickler, 2 out of 4.
Genotypic ratio: HbAHbS : HbSHbS = 2:2 = 1:1.
(a)(ii) Phenotypic ratio
Normal carriers (non-sicklers) : sicklers = 1:1.
(b) Rhesus incompatibility and loss of the second pregnancy
A Rhesus-negative woman has no Rh antigen on her red blood cells. If her first foetus is Rhesus-positive, some foetal red blood cells may enter her circulation, especially at delivery when the placenta separates. The Rh antigen stimulates the mother to form anti-Rhesus antibodies. In a later pregnancy with a Rhesus-positive foetus, these antibodies cross the placenta and destroy the foetal red blood cells. The resulting severe haemolysis, called erythroblastosis foetalis, may cause death of the foetus and miscarriage.
(c) Features explained by Lamarck's use and disuse theory
- The long neck of the giraffe, attributed to continual stretching to reach high leaves.
- The webbed feet of ducks or geese, attributed to constant use in swimming.
(d) Vestigial structures in mammals
- Vermiform appendix in humans.
- Coccyx, or reduced tail vertebrae, in humans.
- Pelvic and hind-limb bones in whales.
Question 11 Report
a. Explain briefly the importance of the following factors and organisms in the nitrogen cycle
(i) lightning; [3 marks]
(ii) Nitrosomonas; [2 marks]
(iii) Azotobacter. [2 marks]
b. Name the excretory organ in:
(i) insects: [1 mark]
(ii) earthworm.
c. Define
(i) hepatitis; [4 marks]
(ii) kidney stones [4marks]
d. (i) State three effects of lack of sense receptors in the skin to humans. [3 marks]
(ii) List three layers of the epidermis in the skin of humans. [3 marks]
e. (i) What is metamorphosis? [2 marks]
(ii) State the type of metamorphosis exhibited by each of the following insects listed in the table below.
| Insect | Type of metamorphosis |
| Grasshopper | |
| Cockroach | |
| Butterfly | |
| Mosquito | |
| Housefly |
(a)(i) Lightning: The high electrical energy of lightning supplies enough energy to make atmospheric nitrogen combine directly with oxygen to form oxides of nitrogen (nitric oxide and nitrogen dioxide). These oxides dissolve in rain water to form dilute nitric acid, which reaches the soil and forms nitrates that plants absorb. Lightning therefore fixes free nitrogen into a usable form.
(a)(ii) Nitrosomonas: It is a nitrifying bacterium that oxidises (converts) ammonia or ammonium compounds in the soil to nitrites.
(a)(iii) Azotobacter: It is a free-living, nitrogen-fixing soil bacterium that fixes atmospheric nitrogen into nitrogenous compounds, enriching the soil with usable nitrogen.
(b) Excretory organs:
- (i) Insects: Malpighian tubules.
- (ii) Earthworm: Nephridia.
(c) Definitions:
- (i) Hepatitis is the inflammation of the liver, usually caused by a viral infection, which damages liver cells and may lead to jaundice.
- (ii) Kidney stones are hard, solid crystalline deposits of mineral salts (such as calcium oxalate and urates) that form inside the kidney from substances concentrated in the urine.
(d)(i) Three effects of lack of sense receptors in the skin:
- Inability to feel pain, so injuries occur without the person being aware.
- Inability to detect temperature changes, leading to burns or excessive cold exposure.
- Inability to detect touch and pressure, so harmful contact is not sensed.
(d)(ii) Three layers of the epidermis of human skin: the cornified (horny) layer, the granular layer, and the Malpighian (germinative) layer.
(e)(i) Metamorphosis is the series of distinct changes in body form and structure that an insect undergoes during its development from the egg to the adult stage.
(e)(ii) Type of metamorphosis in each insect:
| Insect | Type of metamorphosis |
|---|---|
| Grasshopper | Incomplete metamorphosis |
| Cockroach | Incomplete metamorphosis |
| Butterfly | Complete metamorphosis |
| Mosquito | Complete metamorphosis |
| Housefly | Complete metamorphosis |
Answer Details
(a)(i) Lightning: The high electrical energy of lightning supplies enough energy to make atmospheric nitrogen combine directly with oxygen to form oxides of nitrogen (nitric oxide and nitrogen dioxide). These oxides dissolve in rain water to form dilute nitric acid, which reaches the soil and forms nitrates that plants absorb. Lightning therefore fixes free nitrogen into a usable form.
(a)(ii) Nitrosomonas: It is a nitrifying bacterium that oxidises (converts) ammonia or ammonium compounds in the soil to nitrites.
(a)(iii) Azotobacter: It is a free-living, nitrogen-fixing soil bacterium that fixes atmospheric nitrogen into nitrogenous compounds, enriching the soil with usable nitrogen.
(b) Excretory organs:
- (i) Insects: Malpighian tubules.
- (ii) Earthworm: Nephridia.
(c) Definitions:
- (i) Hepatitis is the inflammation of the liver, usually caused by a viral infection, which damages liver cells and may lead to jaundice.
- (ii) Kidney stones are hard, solid crystalline deposits of mineral salts (such as calcium oxalate and urates) that form inside the kidney from substances concentrated in the urine.
(d)(i) Three effects of lack of sense receptors in the skin:
- Inability to feel pain, so injuries occur without the person being aware.
- Inability to detect temperature changes, leading to burns or excessive cold exposure.
- Inability to detect touch and pressure, so harmful contact is not sensed.
(d)(ii) Three layers of the epidermis of human skin: the cornified (horny) layer, the granular layer, and the Malpighian (germinative) layer.
(e)(i) Metamorphosis is the series of distinct changes in body form and structure that an insect undergoes during its development from the egg to the adult stage.
(e)(ii) Type of metamorphosis in each insect:
| Insect | Type of metamorphosis |
|---|---|
| Grasshopper | Incomplete metamorphosis |
| Cockroach | Incomplete metamorphosis |
| Butterfly | Complete metamorphosis |
| Mosquito | Complete metamorphosis |
| Housefly | Complete metamorphosis |
