WAEC SSCE - Biology - 2015
Tambaya 1 Rahoto
A person had an accident that affected the skull but not the nose and later lost the sense of smell. The accident must have a affected the
Bayanin Amsa
The sense of smell is facilitated by olfactory receptors located in the nose. When odors are detected, these receptors send signals to the olfactory lobes of the brain, which interprets and processes the information as the sense of smell. Therefore, if a person loses their sense of smell after a head injury that affected the skull but not the nose, it is likely that the accident caused damage to the olfactory lobes of the brain. So, the correct option is (b) olfactory lobes of the brain.
Tambaya 2 Rahoto
The diagram above is illustration of a crocodile. Use it to answer this question.
Two characteristic features of the class to which the crocodile belongs are labelled
Tambaya 3 Rahoto
According to Mendel's first law of inheritance, segregation of genes occur when
Tambaya 4 Rahoto
Which of the following natural resources is most readily available to all organisms?
Tambaya 5 Rahoto
A plant that commonly undergoes vegetative propagation by means of leaves is
Bayanin Amsa
The plant that commonly undergoes vegetative propagation by means of leaves is Bryophyllum. Bryophyllum is a succulent plant that produces plantlets at the edges of its leaves. These plantlets fall off the leaf, take root in the soil, and grow into new plants. This method of asexual reproduction through leaves is known as leaf-budding. The new plants produced through this method are genetically identical to the parent plant.
Tambaya 6 Rahoto
Use the diagram above to answer this question.
What are the genotypes of the parents if the gene for skin colour is H?
Bayanin Amsa
The given diagram shows a cross between a Spotted Leopard and a Black Panther. The gene for skin color is represented by the letter H. From the diagram, we can see that all the offspring have the genotype Hh for the skin color gene. This means that one parent must have the genotype Hh and the other parent must have the genotype hh. Since the Black Panther parent has the homozygous recessive genotype (hh), it must have inherited one h allele from each of its parents. Therefore, the Spotted Leopard parent must have the genotype Hh, with one H allele inherited from one parent and one h allele from the other parent. So, the answer is: Spotted Leopard Hh Black Panther hh.
Tambaya 8 Rahoto
How many chromosomes will be in a gamete if the normal cell has four chromosomes?
Bayanin Amsa
If a normal cell has four chromosomes, it means that it has two pairs of chromosomes. During gamete formation, meiosis occurs and the chromosome pairs separate, producing haploid gametes that have only one chromosome from each pair. This means that each gamete will have only two chromosomes, one from each pair. Therefore, the answer is 2.
Tambaya 9 Rahoto
Which of the following resources is non-renewable?
Bayanin Amsa
A non-renewable resource is a natural resource that cannot be replaced once it is used up. Among the given options, the mineral deposit is a non-renewable resource because once it is extracted from the earth and used up, it cannot be replenished. Forest reserves and wildlife can be conserved and replenished with proper management and conservation efforts, while water is a renewable resource that can be replenished through the water cycle.
Tambaya 11 Rahoto
A community reaches a climax when
Bayanin Amsa
A community reaches a climax when the composition and size of the community remains relatively constant over a long period of time. At this stage, the community has reached a stable state and is in balance with the surrounding environment. The climax community is characterized by a complex and diverse mix of plant and animal species that have evolved to live in harmony with each other and their environment. The community has reached a state of equilibrium, where the growth of each species is balanced by the growth of other species and the availability of resources in the ecosystem. This means that there are no longer new habitats being formed, and there is little to no introduction of new plant and animal species within the community. The pioneer organisms that were present during the early stages of the community's development have been replaced by more complex and specialized organisms. In summary, a climax community is a stable and diverse community that has reached a state of balance with its environment, and it is characterized by a constant composition and size over a long period of time.
Tambaya 12 Rahoto
Which of the following components make up an ecosystem?
Bayanin Amsa
The component that make up an ecosystem are living and non-living factors. An ecosystem is a community of living organisms (biotic factors) interacting with the non-living (abiotic) components of their environment. Biotic factors are the living organisms such as plants, animals, and decomposers that interact with one another within an ecosystem. Abiotic factors, on the other hand, are the non-living components of the ecosystem such as air, water, sunlight, soil, and temperature. Plants and animals depend on abiotic factors for survival. For example, plants use sunlight to produce energy through photosynthesis, and they also need water and nutrients from the soil to grow. Similarly, animals require water, air, and food to survive. Decomposers such as bacteria and fungi are also an important part of an ecosystem, as they break down dead plant and animal matter into nutrients that can be reused by other living organisms. In summary, an ecosystem is made up of living (biotic) and non-living (abiotic) components that interact with one another to form a complex web of relationships.
Tambaya 14 Rahoto
In explaining the term camouflage, to a class, the diagram above show four varieties, P, Q, R, and S of the same species of fish living amongst water plants in a river were used
Which of the varieties is likely to decrease most in number if a predatory fish is introduced into the river?
Bayanin Amsa
Tambaya 15 Rahoto
Use the diagram above to answer this question.
What of the following statements about the cross is correct?
Bayanin Amsa
The correct statement about the cross in the diagram is "The gene for the spotted coat is dominant over the gene for the black coat." This can be deduced from the fact that all the offspring in the F1 generation have the spotted coat, despite having one parent with a black coat (which is a recessive trait). If the gene for the black coat were dominant over the gene for the spotted coat, then all the offspring in the F1 generation would have black coats. Similarly, if the two genes were co-dominant, then the offspring in the F1 generation would have coats that show both black and spotted patterns, which is not the case in the diagram. Therefore, the correct answer is that the gene for the spotted coat is dominant over the gene for the black coat.
Tambaya 16 Rahoto
Which of the following statements about sex-linked characters is not true?
Bayanin Amsa
Tambaya 17 Rahoto
A student used the following steps in testing for a non-reducing sugar.
I. added Benedict's solution to the sugar solution
ll. added dilute hydrochloric acid to the sugar and boiled it;
lll. added sodium hydroxide solution to the solution in ll. and boiled;
IV. added Benedict's solution to the cooled solution in lll.
Use this information above to answer the question.
Why did the student add the dilute hydrochloric acid to the sugar solution in step ll? To
Bayanin Amsa
The student added dilute hydrochloric acid to the sugar solution in step ll to hydrolyze the non-reducing sugar into its reducing sugar components. This step breaks down the non-reducing sugar into smaller reducing sugar units that can then react with Benedict's solution in the subsequent steps. Without hydrolyzing the non-reducing sugar, it would not be able to react with Benedict's solution and give a positive result. Therefore, the addition of hydrochloric acid is an important step in testing for non-reducing sugars.
Tambaya 19 Rahoto
The following organisms have structures for movement except
Bayanin Amsa
Among the given options, Spirogyra is the organism that does not have structures for movement. Spirogyra is a type of algae that has a filamentous structure and does not possess any specialized structures such as cilia or flagella for movement. Amoeba moves by the extension and contraction of its pseudopodia, Paramecium moves by the coordinated action of cilia, and Volvox moves by the synchronized beating of flagella on its surface. Hence, the correct answer is Spirogyra.
Tambaya 20 Rahoto
Use the diagram above to answer this question.
What are the genotypes of the parents if the gene for skin colour is H?
Bayanin Amsa
The diagram above shows a Punnett square used to predict the possible genotypes of the offspring of a cross between a spotted leopard (Hh) and a black panther (hh) with the gene for skin colour represented by the letter H. The possible genotypes of the offspring are shown in the boxes formed by the Punnett square. Looking at the Punnett square, it can be observed that the genotypes of the parents are Spotted Leopard Hh and Black Panther hh, since each parent contributes one allele to each offspring. Therefore, the correct answer is "Spotted Leopard Hh Black Panther hh".
Tambaya 21 Rahoto
Blood pressure is higher in the arteries as a result of
Bayanin Amsa
Blood pressure is higher in the arteries as a result of contraction and relaxation. Arteries are the blood vessels that carry oxygenated blood away from the heart to other parts of the body. The heart pumps blood into the arteries, and the contraction of the heart muscles forces blood into the arteries, creating pressure in them. This pressure is known as systolic blood pressure. When the heart relaxes, the pressure in the arteries decreases, which is known as diastolic blood pressure. The alternating contraction and relaxation of the heart muscles along with the elasticity of the arterial walls maintain the blood pressure in the arteries. This pressure is necessary for proper circulation of blood and delivery of oxygen and nutrients to the body tissues.
Tambaya 22 Rahoto
Which of the following features could be used to determine the growth of a seedling?
Bayanin Amsa
The feature that could be used to determine the growth of a seedling is the length of the radicle. The radicle is the first part of the seedling to emerge during germination, and it is the embryonic root of the plant. As the radicle grows, it anchors the seedling in the soil and begins to absorb water and nutrients from the surrounding environment. Therefore, the length of the radicle is a good indicator of how well the seedling is growing and developing. The other options - number of flowers, number of leaves, and length of flowers - are not as good indicators of seedling growth. These features are more relevant to the reproductive phase of the plant's life cycle, and they may not be present or fully developed in a young seedling. Therefore, the length of the radicle is the best feature to use when determining the growth of a seedling.
Tambaya 23 Rahoto
Ability of the human eye to focus images accurately on the retina is called
Bayanin Amsa
The ability of the human eye to focus images accurately on the retina is called accommodation. This is the process by which the lens of the eye changes shape to adjust the focal length of the eye and ensure that objects at different distances are focused correctly on the retina. When we look at a distant object, the ciliary muscles in the eye relax, making the lens thinner and flatter. When we look at a nearby object, the ciliary muscles contract, making the lens thicker and more curved. This change in the shape of the lens is what allows us to see objects at different distances clearly.
Tambaya 24 Rahoto
When a Spirogyra filament is placed in a concentrated salt solution for 30 minutes, the cell would become
Bayanin Amsa
When a Spirogyra filament is placed in a concentrated salt solution for 30 minutes, the cell would become plasmolysed. This is because the salt solution has a higher concentration of solutes than the cell, causing water to move out of the cell and into the surrounding solution through osmosis. As a result, the protoplasm and cell membrane shrink and detach from the cell wall, causing the cell to shrink and the cytoplasm to become concentrated. This process is called plasmolysis, and it is reversible if the cell is placed in a hypotonic solution (one with a lower concentration of solutes) to allow water to move back into the cell through osmosis.
Tambaya 25 Rahoto
The variety P,Q,R and S Outlive the others because the
Bayanin Amsa
The variety that is most likely to outlive the others is the one whose markings are similar to those of the environment (option C). This is because such a variety is better able to blend in with its surroundings and is less likely to be seen by predators. As a result, it has a better chance of survival and passing on its genes to the next generation. This is an example of natural selection, where individuals with advantageous traits are more likely to survive and reproduce.
Tambaya 26 Rahoto
A student used the following steps in testing for a non-reducing sugar.
I. added Benedict's solution to the sugar solution
ll. added dilute hydrochloric acid to the sugar and boiled it;
lll. added sodium hydroxide solution to the solution in ll. and boiled;
IV. added Benedict's solution to the cooled solution in lll.
Use this information above to answer the question.
What is the importance of step ll in the testing process? To
Bayanin Amsa
Tambaya 27 Rahoto
In explaining the term camouflage, to a class, the diagram above show four varieties, P, Q, R, and S of the same species of fish living amongst water plants in a river were used
Which of the varieties is most likely to outlive the others?
Tambaya 28 Rahoto
Which of the following statements about protoplasm of a cell is not correct? It
Bayanin Amsa
The statement that is not correct is "consists the nucleus only". This is because the protoplasm of a cell is the living substance that makes up the entire cell, including the nucleus, cytoplasm, and cell organelles. The nucleus is just one component of the protoplasm, which also includes the cytoplasm and cell organelles such as mitochondria, ribosomes, and the endoplasmic reticulum. Therefore, the correct statement would be "comprises of the nucleus, cytoplasm, and cell organelles".
Tambaya 30 Rahoto
Which of the following organs of the alimentary canal is not correctly matched with its function?
Bayanin Amsa
The organ that is not correctly matched with its function in the given options is the Appendix - releases enzymes. The Appendix is actually a small, finger-shaped pouch that is located near the junction of the small and large intestines. While the function of the appendix is not entirely clear, it is believed to play a role in immune function and may also have a role in maintaining a healthy balance of bacteria in the gut. The other options are correctly matched with their functions. The gallbladder stores bile, which is produced by the liver and helps to digest fats in the small intestine. The liver also plays a role in storing glycogen, which is a form of glucose that can be used by the body for energy when needed. Teeth are responsible for mechanically breaking down food by grinding and crushing it, which makes it easier to swallow and digest.
Tambaya 31 Rahoto
A farmer who wants to keep seeds for three years before planting and wants to prevent them from sprouting uses
Bayanin Amsa
The farmer can prevent the seeds from sprouting by using abscisic acid (ABA). ABA is a plant hormone that is involved in the regulation of seed dormancy and the prevention of premature germination. When seeds are exposed to ABA, it inhibits the action of gibberellins, which are responsible for seed germination. By applying ABA to the seeds, the farmer can prevent them from sprouting for up to three years while keeping them in storage. Therefore, the correct answer is abscisin.
Tambaya 32 Rahoto
Which of the following processes removes carbon (IV) oxide from the atmosphere?
Bayanin Amsa
The process that removes carbon (IV) oxide from the atmosphere is photosynthesis. During photosynthesis, plants absorb carbon (IV) oxide from the atmosphere and use it, along with water and sunlight, to produce glucose and oxygen. The oxygen is released into the atmosphere as a byproduct, while the glucose is used as an energy source by the plant. This process helps to reduce the amount of carbon (IV) oxide in the atmosphere, which is important for maintaining the balance of gases and preventing global warming.
Tambaya 33 Rahoto
A person suffering from obstruction of the bile duct is advised not to eat fats and oil because
Bayanin Amsa
The correct answer is: "bile emulsifies fat and oil." Bile is a digestive fluid produced by the liver and stored in the gallbladder. It helps in the digestion and absorption of fats and oils in the small intestine. Bile contains bile salts which act as emulsifying agents. Emulsification is the process by which large fat globules are broken down into smaller droplets, making it easier for enzymes to break down the fats and for them to be absorbed into the bloodstream. When the bile duct is obstructed, bile cannot flow into the small intestine, leading to a decrease in the amount of bile available for digestion. As a result, the digestion and absorption of fats and oils become compromised, leading to symptoms such as abdominal pain, bloating, and diarrhea. Therefore, people with bile duct obstruction are advised to avoid consuming fats and oils to prevent worsening of their symptoms.
Tambaya 34 Rahoto
Which of the following organisms is not a protozoan?
Bayanin Amsa
Ascaris is not a protozoan. Ascaris is a roundworm, which belongs to the phylum Nematoda. The other three options, Amoeba, Plasmodium, and Paramecium are all protozoans, which are unicellular eukaryotic organisms.
Tambaya 35 Rahoto
Which of the following cells is a specialized cell?
Bayanin Amsa
A specialized cell is a cell that has a specific structure and function to perform a particular task within an organism. Of the options given, the cell that is a specialized cell is the guard cell. The guard cell is a specialized cell found in the epidermis of leaves, which regulates the opening and closing of stomata, tiny pores on the surface of leaves that allow for gas exchange. The guard cell has a kidney-shaped structure with a thickened inner wall and a thinner outer wall. When water enters the guard cell, it swells, and the thin outer wall expands, causing the guard cell to bend and create an opening between the two cells. This opening allows for gas exchange between the plant and its environment. The amoeba and Plasmodium are unicellular organisms that do not have specialized cells as they perform all the necessary functions within a single cell. The meristematic cell is a type of plant cell that has the ability to divide and differentiate into various types of plant cells, but it is not considered a specialized cell as it does not have a specific structure and function to perform a particular task. In summary, the guard cell is a specialized cell because of its specific structure and function to regulate the opening and closing of stomata in plant leaves.
Tambaya 36 Rahoto
The diagram above is illustration of a crocodile. Use it to answer this question.
Which of the labelled parts enable the animal to stay under water most of the time?
Bayanin Amsa
The labelled parts of the crocodile that enable it to stay underwater most of the time are the following: 1. The eyes (part I): Crocodile's eyes are equipped with a special membrane called the nictitating membrane that allows them to see underwater. This helps the crocodile to locate its prey, predators, and also avoid obstacles while underwater. 2. The lungs (part II): Crocodile's lungs are specially adapted to hold more oxygen than most reptiles. This enables them to stay underwater for long periods without the need to come to the surface to breathe. 3. The tail (part IV): The tail of the crocodile is a powerful swimming tool that propels it through the water. The crocodile uses its tail to move quickly and efficiently underwater, allowing it to stay submerged for extended periods. Therefore, the correct answer is option (II) I, II and V.
Tambaya 37 Rahoto
Which of the following characteristic features is not associated with monocotyledonous plants?
Bayanin Amsa
Tambaya 39 Rahoto
The diagram above illustrates a part of the mammalian skeleton . use it to answer this this question
The part of the skeleton illustrated in the diagram is the
Bayanin Amsa
The part of the mammalian skeleton illustrated in the diagram is the cervical vertebrae. This is because the cervical vertebrae are the seven vertebrae that make up the neck of the mammal, and the diagram shows seven vertebrae with the characteristic features of cervical vertebrae, such as a small body, a large vertebral foramen, and transverse foramina for the passage of blood vessels.
Tambaya 40 Rahoto
The following diagram above show the sequence of events in early cell division. Study the diagram and answer the question.
In which of the following cells is this divisin likely to take plce?
Bayanin Amsa
Tambaya 41 Rahoto
The colour change to be observed in step IV is
Tambaya 42 Rahoto
Study the diagram above and answer the question
The structure labelled I is
Bayanin Amsa
Tambaya 43 Rahoto
Oil applied to the surface of water kills the larvae of mosquitoes through
Bayanin Amsa
Oil applied to the surface of water kills the larvae of mosquitoes through suffocation. Mosquito larvae breathe air through a tube called a siphon, which they use to reach the surface of the water to breathe. When a layer of oil is applied to the surface of the water, it creates a film that covers the siphon and prevents the larvae from accessing air. As a result, the larvae suffocate and die. This method is often used as a non-toxic and environmentally friendly way to control mosquito populations.
Tambaya 44 Rahoto
Which of the following statements about human blood groups is not true?
Bayanin Amsa
Tambaya 46 Rahoto
The diagram above is illustration of a crocodile. Use it to answer this question.
Two characteristic features of the class to which the crocodile belongs are labelled
Tambaya 47 Rahoto
Soil erosion could be prevented by
Bayanin Amsa
Soil erosion can be prevented by cover-cropping. Cover-cropping involves planting a crop on the land during the fallow period to cover the soil and protect it from erosion caused by wind or rain. The roots of the cover crop hold the soil in place and help to build soil health by adding organic matter. Cover crops also improve soil fertility, reduce soil compaction, and suppress weeds. This method is a sustainable and effective way to protect the soil from erosion and maintain the health of the land for long-term productivity. Flooding a farmland can cause more harm than good as it can wash away topsoil and nutrients. Deforestation and leaving the soil bare also contribute to soil erosion as they leave the soil exposed to wind and rain.
Tambaya 48 Rahoto
The structure of the cell membrane is a
Bayanin Amsa
The structure of the cell membrane is a middle bi-layer of lipid with a protein layer on either surface. The cell membrane is composed of two layers of phospholipid molecules arranged tail-to-tail, which forms a barrier that separates the inside of the cell from the outside environment. The proteins are embedded within the lipid bilayer, with some extending through the entire membrane and others located only on one surface or the other. This structure allows the cell membrane to control what enters and leaves the cell, as well as to communicate with other cells and the environment.
Tambaya 49 Rahoto
(a) Explain how the leaf of a flowering plant is adapted for photosynthesis.
(b) Describe briefly the mode of feeding in tapeworm.
(c) Name one mineral element each needed for the proper functioning of :
(i) bones (ii) red blood cells
(iii) thyroid gland.
(d)(i) State three benefits of including roughages in the diet of humans.
(ii) State three reasons why proteins are important to humans.
(a) How the leaf is adapted for photosynthesis
- Broad, flat lamina gives a large surface area to trap maximum sunlight.
- Thin lamina so that light and gases diffuse quickly to the inner cells.
- Abundant chloroplasts (rich in chlorophyll), especially in the palisade layer, to absorb light energy.
- Palisade cells are elongated and packed near the upper surface to catch the most light.
- Many stomata allow entry of carbon dioxide and exit of oxygen.
- Air spaces in the spongy mesophyll allow rapid diffusion of gases.
- A network of veins (xylem) supplies water and (phloem) removes manufactured food.
- Leaves are arranged in a mosaic (leaf mosaic) to avoid overshadowing one another.
(b) Mode of feeding in tapeworm: the tapeworm lives in the host's small intestine and feeds by absorbing already-digested food (mainly soluble sugars and amino acids) directly through its flat body wall over its whole large surface area. It has no mouth or gut of its own; it secretes anti-enzymes to protect itself from being digested and simply absorbs the host's nutrients (parasitic nutrition).
(c) One mineral element needed for the proper functioning of
- (i) Bones: calcium (phosphorus also acceptable).
- (ii) Red blood cells: iron.
- (iii) Thyroid gland: iodine.
(d)(i) Three benefits of roughage in the diet
- Adds bulk to food and stimulates peristalsis, aiding movement of food through the gut.
- Prevents constipation and helps regular egestion of faeces.
- Helps prevent bowel disorders (such as colon cancer) and absorbs/removes harmful substances.
(ii) Three reasons why proteins are important to humans
- For growth and formation of new cells and tissues.
- For repair of worn-out and damaged tissues.
- For making enzymes, hormones and antibodies.
Bayanin Amsa
(a) How the leaf is adapted for photosynthesis
- Broad, flat lamina gives a large surface area to trap maximum sunlight.
- Thin lamina so that light and gases diffuse quickly to the inner cells.
- Abundant chloroplasts (rich in chlorophyll), especially in the palisade layer, to absorb light energy.
- Palisade cells are elongated and packed near the upper surface to catch the most light.
- Many stomata allow entry of carbon dioxide and exit of oxygen.
- Air spaces in the spongy mesophyll allow rapid diffusion of gases.
- A network of veins (xylem) supplies water and (phloem) removes manufactured food.
- Leaves are arranged in a mosaic (leaf mosaic) to avoid overshadowing one another.
(b) Mode of feeding in tapeworm: the tapeworm lives in the host's small intestine and feeds by absorbing already-digested food (mainly soluble sugars and amino acids) directly through its flat body wall over its whole large surface area. It has no mouth or gut of its own; it secretes anti-enzymes to protect itself from being digested and simply absorbs the host's nutrients (parasitic nutrition).
(c) One mineral element needed for the proper functioning of
- (i) Bones: calcium (phosphorus also acceptable).
- (ii) Red blood cells: iron.
- (iii) Thyroid gland: iodine.
(d)(i) Three benefits of roughage in the diet
- Adds bulk to food and stimulates peristalsis, aiding movement of food through the gut.
- Prevents constipation and helps regular egestion of faeces.
- Helps prevent bowel disorders (such as colon cancer) and absorbs/removes harmful substances.
(ii) Three reasons why proteins are important to humans
- For growth and formation of new cells and tissues.
- For repair of worn-out and damaged tissues.
- For making enzymes, hormones and antibodies.
Tambaya 50 Rahoto
(a) Copy and complete the table below
| Disease | one effect | one remedy |
| kidney stones | ||
| Nephritis | ||
| Diuresis |
(b) Make a labelled diagram of the carbon cycle.
(c)(i) What is Estuarine habitat?
(ii) Construct a food chain typical of an Estuarine habitat.
(d)(i) Name two diseases that can attack plant crops.
(ii) State three ways of controlling plant diseases.
(e) State four functions of a public health authority
(f) State four ways of maintaining food hygiene in the community.
(a) Effects and remedies of kidney diseases
| Disease | One effect | One remedy |
|---|---|---|
| Kidney stones | Severe pain during urination, sometimes with blood in the urine. | Drink plenty of water and seek medical treatment. |
| Nephritis | Oedema, especially swelling of the feet and ankles. | Seek medical treatment; dialysis may be used in severe cases. |
| Diuresis | Excessive loss of water in urine, leading to dehydration and thirst. | Seek medical advice and avoid excessive intake of alcohol or diuretics. |
(b) Labelled diagram of the carbon cycle
(c)(i) An estuarine habitat is the region at the mouth of a river where fresh water mixes with sea water. The water is brackish and is affected by tides.
(c)(ii) A food chain in an estuarine habitat is:
Phytoplankton → barnacle → fish → bird
(d)(i) Two diseases that attack crop plants are:
- Rust disease.
- Mosaic disease.
(d)(ii) Three ways of controlling plant diseases are:
- Plant disease-resistant varieties of crops.
- Spray affected crops with appropriate chemicals such as fungicides or bactericides.
- Remove and destroy diseased plants or affected plant parts.
(e) Functions of a public health authority
- It ensures proper collection and disposal of refuse.
- It provides clean and safe water for the public.
- It controls infectious diseases through immunisation, vaccination and health inspection.
- It provides public health education and ensures sanitation of public places.
(f) Ways of maintaining food hygiene in the community
- Wash food items thoroughly with clean water before preparation.
- Cook food properly at the appropriate temperature.
- Keep cooking areas and utensils clean.
- Keep cooked food covered and preserve perishable food by refrigeration.
Bayanin Amsa
(a) Effects and remedies of kidney diseases
| Disease | One effect | One remedy |
|---|---|---|
| Kidney stones | Severe pain during urination, sometimes with blood in the urine. | Drink plenty of water and seek medical treatment. |
| Nephritis | Oedema, especially swelling of the feet and ankles. | Seek medical treatment; dialysis may be used in severe cases. |
| Diuresis | Excessive loss of water in urine, leading to dehydration and thirst. | Seek medical advice and avoid excessive intake of alcohol or diuretics. |
(b) Labelled diagram of the carbon cycle
(c)(i) An estuarine habitat is the region at the mouth of a river where fresh water mixes with sea water. The water is brackish and is affected by tides.
(c)(ii) A food chain in an estuarine habitat is:
Phytoplankton → barnacle → fish → bird
(d)(i) Two diseases that attack crop plants are:
- Rust disease.
- Mosaic disease.
(d)(ii) Three ways of controlling plant diseases are:
- Plant disease-resistant varieties of crops.
- Spray affected crops with appropriate chemicals such as fungicides or bactericides.
- Remove and destroy diseased plants or affected plant parts.
(e) Functions of a public health authority
- It ensures proper collection and disposal of refuse.
- It provides clean and safe water for the public.
- It controls infectious diseases through immunisation, vaccination and health inspection.
- It provides public health education and ensures sanitation of public places.
(f) Ways of maintaining food hygiene in the community
- Wash food items thoroughly with clean water before preparation.
- Cook food properly at the appropriate temperature.
- Keep cooking areas and utensils clean.
- Keep cooked food covered and preserve perishable food by refrigeration.
Tambaya 51 Rahoto
(a)(i) What is sewage?
(ii) State five effects of releasing untreated sewage into a stagnant water body.
(b) Explain surface terracing as a soil conservation method.
(c) Explain the term adaptation.
(d) Explain two ways each by which the following organisms adapt to their habitats:
(i) hydrophytes
(ii) xerophytes.
(a)(i) Sewage is the liquid and solid waste matter, including human excreta, waste water and refuse, that is carried away from homes, industries and towns in drains and sewers.
(ii) Five effects of releasing untreated sewage into a stagnant water body
- It causes water pollution, making the water unfit for drinking and domestic use.
- It introduces disease-causing organisms (typhoid, cholera, dysentery).
- It promotes rapid growth of algae (eutrophication), forming an algal bloom.
- Decomposition of the sewage and dead algae by bacteria uses up dissolved oxygen, killing fish and other aquatic organisms.
- It produces foul (offensive) odours and breeds mosquitoes and other vectors.
(b) Surface (bench) terracing: this is a soil-conservation method used on steep slopes in which the slope is cut into a series of broad, level, step-like platforms (terraces) supported by ridges or walls along the contour. The flat terraces reduce the speed of run-off water, allow it to soak into the soil, and so check soil erosion while retaining water and nutrients for crops.
(c) Adaptation is the possession of special structural, physiological or behavioural features that enable an organism to survive and reproduce successfully in its particular environment.
(d) Two ways each of adaptation
- (i) Hydrophytes (water plants): possess many large air spaces (aerenchyma) for buoyancy and gaseous exchange; have poorly developed roots and thin/dissected leaves; stomata are on the upper leaf surface (in floating leaves).
- (ii) Xerophytes (dry-habitat plants): have reduced leaves or spines and a thick waxy cuticle to reduce water loss; have sunken stomata and deep or extensive roots; store water in fleshy (succulent) tissues.
Bayanin Amsa
(a)(i) Sewage is the liquid and solid waste matter, including human excreta, waste water and refuse, that is carried away from homes, industries and towns in drains and sewers.
(ii) Five effects of releasing untreated sewage into a stagnant water body
- It causes water pollution, making the water unfit for drinking and domestic use.
- It introduces disease-causing organisms (typhoid, cholera, dysentery).
- It promotes rapid growth of algae (eutrophication), forming an algal bloom.
- Decomposition of the sewage and dead algae by bacteria uses up dissolved oxygen, killing fish and other aquatic organisms.
- It produces foul (offensive) odours and breeds mosquitoes and other vectors.
(b) Surface (bench) terracing: this is a soil-conservation method used on steep slopes in which the slope is cut into a series of broad, level, step-like platforms (terraces) supported by ridges or walls along the contour. The flat terraces reduce the speed of run-off water, allow it to soak into the soil, and so check soil erosion while retaining water and nutrients for crops.
(c) Adaptation is the possession of special structural, physiological or behavioural features that enable an organism to survive and reproduce successfully in its particular environment.
(d) Two ways each of adaptation
- (i) Hydrophytes (water plants): possess many large air spaces (aerenchyma) for buoyancy and gaseous exchange; have poorly developed roots and thin/dissected leaves; stomata are on the upper leaf surface (in floating leaves).
- (ii) Xerophytes (dry-habitat plants): have reduced leaves or spines and a thick waxy cuticle to reduce water loss; have sunken stomata and deep or extensive roots; store water in fleshy (succulent) tissues.
Tambaya 52 Rahoto
(a) Explain briefly sexual reproduction in Rhizopus.
(b) A student suddenly stepped on a big snake, he cried for help and ran away.
(i) Name the hormone produced in the body for the reaction.
(ii) In which part of the human body is the hormone named in (b) (i) produced?
(iii) State five changes that must have taken place as a result of the hormone named in (a) (i).
(c) (i) In tabular form, state three differences between excretion in flowering plants and excretion in humans,
(ii) List two organelles found in plant cell that are responsible for the excretion of metabolic wastes.
(a) Sexual reproduction in Rhizopus: when two hyphae from opposite mating strains (+ and -) grow close together, each puts out a short side branch called a progametangium. The tips swell and are cut off by cross-walls to form two gametangia, each containing many nuclei (gametes). The walls between the touching gametangia dissolve and their contents fuse (plasmogamy then karyogamy). This produces a thick-walled, resistant zygospore. The zygospore undergoes a resting period, then germinates by meiosis to produce a sporangium that releases haploid spores which grow into new mycelia.
(b)(i) Hormone produced: adrenaline (epinephrine).
(ii) Site of production: the adrenal (suprarenal) glands, specifically the adrenal medulla, located above each kidney.
(iii) Five changes caused by adrenaline
- Increased heart rate and force of heartbeat.
- Increased rate and depth of breathing.
- Raised blood sugar level (conversion of glycogen to glucose in the liver).
- Widening (dilation) of the pupils of the eyes.
- Diversion of blood from the gut and skin to the muscles (and erection of body hair; increased sweating).
(c)(i) Three differences between excretion in flowering plants and in humans
| Excretion in flowering plants | Excretion in humans |
|---|---|
| No specialised excretory organs | Has specialised organs (kidneys, lungs, skin) |
| Wastes produced slowly and in small amounts | Wastes produced rapidly and in larger amounts |
| Some wastes are reused (e.g. oxygen, carbon dioxide) or stored in tissues | Wastes are mostly removed from the body, not reused |
| Nitrogenous waste is little; main products are oxygen, carbon dioxide and water | Produces much nitrogenous waste (urea) |
(ii) Two organelles in plant cells responsible for excreting metabolic wastes: the vacuole (stores waste in the cell sap) and the plastids/chloroplasts (release oxygen and other products). (Golgi bodies are also acceptable.)
Bayanin Amsa
(a) Sexual reproduction in Rhizopus: when two hyphae from opposite mating strains (+ and -) grow close together, each puts out a short side branch called a progametangium. The tips swell and are cut off by cross-walls to form two gametangia, each containing many nuclei (gametes). The walls between the touching gametangia dissolve and their contents fuse (plasmogamy then karyogamy). This produces a thick-walled, resistant zygospore. The zygospore undergoes a resting period, then germinates by meiosis to produce a sporangium that releases haploid spores which grow into new mycelia.
(b)(i) Hormone produced: adrenaline (epinephrine).
(ii) Site of production: the adrenal (suprarenal) glands, specifically the adrenal medulla, located above each kidney.
(iii) Five changes caused by adrenaline
- Increased heart rate and force of heartbeat.
- Increased rate and depth of breathing.
- Raised blood sugar level (conversion of glycogen to glucose in the liver).
- Widening (dilation) of the pupils of the eyes.
- Diversion of blood from the gut and skin to the muscles (and erection of body hair; increased sweating).
(c)(i) Three differences between excretion in flowering plants and in humans
| Excretion in flowering plants | Excretion in humans |
|---|---|
| No specialised excretory organs | Has specialised organs (kidneys, lungs, skin) |
| Wastes produced slowly and in small amounts | Wastes produced rapidly and in larger amounts |
| Some wastes are reused (e.g. oxygen, carbon dioxide) or stored in tissues | Wastes are mostly removed from the body, not reused |
| Nitrogenous waste is little; main products are oxygen, carbon dioxide and water | Produces much nitrogenous waste (urea) |
(ii) Two organelles in plant cells responsible for excreting metabolic wastes: the vacuole (stores waste in the cell sap) and the plastids/chloroplasts (release oxygen and other products). (Golgi bodies are also acceptable.)
Tambaya 53 Rahoto
(a) Explain the following terms:
(i) test cross
(ii) monohybrid cross.
(b) In a mango plant the, the allele for bean-shaped seed is r and is recessive to round-shaped seed R. With the aid of a genetic diagram, determine the genotypes of the offspring if:
(i) a homozygous bean-shaped parent with a homozygous round shaped parent;
(ii) a heterozygous bean-shaped parent is crossed with a heterozygous round shaped parent.
(a)(i) Test cross: This is the crossing of an organism showing a dominant character, but whose genotype is unknown, with a homozygous recessive organism in order to determine the unknown genotype.
(a)(ii) Monohybrid cross: This is a genetic cross involving one pair of contrasting alleles, used to study the inheritance of a single character.
(b) Let R represent the allele for round-shaped seed (dominant) and r represent the allele for bean-shaped seed (recessive).
(i) Homozygous bean-shaped parent × homozygous round-shaped parent
Parental genotypes: \(rr \times RR\)
All the offspring have genotype \(Rr\). Therefore, all the offspring are round-shaped.
Genotypic ratio: 100% \(Rr\)
Phenotypic ratio: 100% round-shaped seeds.
(ii) Heterozygous cross
A bean-shaped plant cannot be heterozygous because the heterozygous genotype \(Rr\) produces round-shaped seeds. Thus, the stated heterozygous parents have genotype \(Rr\), and the genetic cross is \(Rr \times Rr\).
Offspring genotypes are \(RR, Rr, Rr\), and \(rr\).
Genotypic ratio: \(1RR : 2Rr : 1rr\).
Phenotypic ratio: \(3\) round-shaped seeds : \(1\) bean-shaped seed.
Bayanin Amsa
(a)(i) Test cross: This is the crossing of an organism showing a dominant character, but whose genotype is unknown, with a homozygous recessive organism in order to determine the unknown genotype.
(a)(ii) Monohybrid cross: This is a genetic cross involving one pair of contrasting alleles, used to study the inheritance of a single character.
(b) Let R represent the allele for round-shaped seed (dominant) and r represent the allele for bean-shaped seed (recessive).
(i) Homozygous bean-shaped parent × homozygous round-shaped parent
Parental genotypes: \(rr \times RR\)
All the offspring have genotype \(Rr\). Therefore, all the offspring are round-shaped.
Genotypic ratio: 100% \(Rr\)
Phenotypic ratio: 100% round-shaped seeds.
(ii) Heterozygous cross
A bean-shaped plant cannot be heterozygous because the heterozygous genotype \(Rr\) produces round-shaped seeds. Thus, the stated heterozygous parents have genotype \(Rr\), and the genetic cross is \(Rr \times Rr\).
Offspring genotypes are \(RR, Rr, Rr\), and \(rr\).
Genotypic ratio: \(1RR : 2Rr : 1rr\).
Phenotypic ratio: \(3\) round-shaped seeds : \(1\) bean-shaped seed.
