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Biology Notes, Revision Questions and Answers Form 1-4

 BIOLOGY FORM ONE NOTES

INTRODUCTION TO BIOLOGY

What is Biology?

Biology is the branch of science that deals with the study of living things. In Greek, Bios means life while Logos means knowledge.

Branches of biology

There are two main branches:

  1. Botany: Study of plants
  2. Zoology: Study of animals

The others include:

  1. Ecology: Study of living things in their surroundings.
  2. Genetics: The study of inheritance and variation.
  3. Entomology: Study of insects
  4. Parasitology: Study of parasites
  5. Taxonomy: Study of classification of organisms
  6. Microbiology: Study of microscopic organisms
  7. Anatomy: Study of structure of cells
  8. Cytology: Study of cells
  9. Biochemistry: Study of chemical changes inside living organisms

Name at least six other smaller branches of biology  (6 marks).

Importance of Biology

  1. Solving environmental problemsg. Food shortage, poor health services, pollution, misuse of environmental resources etc.
  2. Choice of careers e.g. Medicine, Agriculture, public health, Veterinary, Animal husbandry, Horticulture, Dentistry etc.
  3. Acquiring scientific skills g. observing, identifying, recording, classification, measuring, analyzing, evaluating etc.
  4. International co-operation g. Development of HIV\AIDS vaccine, fight against severe Acute respiratory Syndrome (SARS), fight to save ozone layer from depletion, management of resources through international depletion.

 

Others

List five professional occupations that require the study of biology. (5 marks)

Characteristics of living things;

  1. Nutrition:Process by which living things acquire and utilize nutrients: plants photosynthesize; animals feed on already manufactured foods.
  2. Respiration:energy-producing process occurring in all the cells of living things.
  3. Gaseous Exchange: where living things take in air (oxygen) and give out air(carbon iv oxide) across respiratory surfaces.
  4. Excretion:Process by which waste or harmful materials resulting from chemical reactions within cells of living things are eliminated. Excess of such materials poison living things.
  5. Growth and Development: Growth –is the irreversible increase in size and Mass.—Essential for body function. Development –Irreversible change in complexity of the structure of living things.
  6. Reproduction: Process by which living things give rise to new individuals of the same kind.
  7. Irritability: Is the ability of living things to perceive changes in their surroundings and respond to them appropriately. E.g. reaction to changes in temperature, humidity, light, pressure and to the presence of certain chemicals.
  8. Movement: Change in position by either a part or the whole living thing. Locomotion – Progressive change in position by the whole living thing. In animals, movement include; swimming, walking, running, flying. In plants, closing of leaves, folding of leaves, closing of flowers, growing of shoots towards light etc.

Question

  1. List four uses of energy obtained from the process of respiration. (4 marks).
  2. List six characteristics of living things (6 marks).

Collection of specimens

Apparatus used

  1. Sweep net: for catching flying insects.
  2. Fish net: For trapping small fish and other small water animals.
  3. Pooter:For sucking small animals from rock surfaces and tree barks.
  4. Bait trap: For attracting and trapping small animals e.g. rats.
  5. Pit fall trap: For catching crawling animals.
  6. Pair of forceps: picking up small crawling animals e.g. stinging insects.
  7. Specimen bottles: keeping collected specimen. Larger specimens require large bottles.
  8. The magnifying lens: Instrument used to enlarge objects. Lenses are found in microscope and the hand lens (magnifier). Its frame is marked e.g. x8 or x10—indicating how much larger will be the image compared to object.

Precautions during Collection and Observation of specimens

Practical activity 2

Practical activity 3

Comparison between plants and animals

Plants Animals
1.   Green in colour( have chlorophyll) 1.   Lack chlorophyll thus feed on readymade food.
2.   Their cells have cellulose cell walls. 2.   Cells lack cellulose cell walls.
3.   Respond slowly to changes in the environment. 3. Respond quickly.
4.   Lack specialized excretory organs. 4. Have complex excretory organs.
5.   Do not move about. 5. Move about in search of food and water.
6.   Growth occurs in shoot and root tips.(apical growth) 6.Growth occurs in all body parts9intercalary growth).

Revision questions

 

CLASSIFICATION I

 

INTRODUCTION

Living things are also known as living organisms.

Organisms (forms of life) have distinguishing characteristics and therefore are grouped.

The Magnifying lens

-Is used for enlarging small objects.

(Diagram)

 

Procedure of its use

Drawing magnification = Length of the drawing/ drawing Length

Length of the object/Actual Length

 (Diagram)

 

External features of plants and animals

External features of plants

  1. Rhizoids as in moss plant.
  2. Fronds in ferns.

External  features of animals

  1. Tentacles in hydra
  2. Feathers in birds
  1. Wings in birds
  2. Fur and hair in mammals
  3. Scales and fins in fish
  1. Locomotory Structures e.g. limbs in insects
  2. Body pigmentation

 

Practical activity 1

To collect and observe animal specimens

To collect and observe plant specimens

 

What is classification?

-Is an area of biology that deals with the grouping of living organisms according to their structure. Organisms with similar structures are put under one group referred to as a taxon—taxa (plural).

The groupings also consider evolutionary relationships (phylogeny)—since all living organisms had a common origin at one time.

Taxonomy—Science of classification.

Taxonomist—Biologist who studies taxonomy.

Need for classification.

Reasons

  1. To identify living organisms into their correct groups for reference and study
  2. To bring together living organisms with similar characteristics but separate those with different features.
  3. To arrange information of living organisms in an orderly manner. This avoids chaos and confusion.
  4. To understand the evolutionary relationship between different organisms

Taxonomic Units

Are groups (taxa) into which organisms are placed as a matter of convenience.

Groups are based on observable characteristics common in the group.

In a classification scheme (taxonomic units or groups, a hierarchy of groups are recognized starting with the first largest and highest group; the Kingdom to the smallest and lowest unit; the species.

There are 7 major taxonomic units.

 

KINGDOM

 

  PHYLUM/ DIVISION  
          CLASS  
         ORDER  
  FAMILY  
  GENUS  

 

 SPECIES

 

The Kingdom

There are five Kingdoms of living organisms, namely:

  1. Kingdom Monera: bacteria
  2. Kingdom protoctista: algae, protozoa, amoeba, paramecium
  3. Kingdom Fungi: Moulds, Yeast, Mushrooms
  4. Kingdom Plantae: Moss plants, ferns, maize, garden pea, pine, meru oak, bean etc.
  5. Kingdom Animalia: hydra, tapeworms, bees, human beings etc.

A kingdom is divided into Phyla in animals or divisions in plants and sorts out organisms based on body plan and form.

Plan is the adaptation to a special way of life.

The Class is further divided into small groups; Orders using structural features.

Orders are divided into families using structural features, then Families into Genera (singular genus) –based on recent common ancestral features that are less adaptive.

Genus is divided into speciesi.e. kind of plant, or animal.

Down the hierarchy, the number of organisms in each group decreases but their similarities increases.

The Species group members naturally interbreed to produce fertile off springs.

Minor differences are exhibited in the species groups e.g. on colour of the skin in human beings and varieties of plants.

The groups of the species are termed to as varieties, races or strains.

Classification of A human being and a maize plant

Taxonomic unit Human being maize bean
kingdom Animalia plantae plantae
Phylum or division Chordata Angiospermaphyta Angiospermae
class Mammalia monocotyledonae Dicotyledonae
order Primates Graminales Rosales
family Hominidae Graminaceae Leguminosae
genus homo zea Phaseolus
species  sapiens mays Vulgaris

 

Scientific name                  Homo sapiens                        Zea mays                               phaseolus vulgaris

Scientific Naming Of Living Organisms

Present naming was developed by carolus Linnaeus 18th c, where organisms were given 2 names in Latin language.

Living organisms have their scientific names and common names i.e. local or vernacular names.

Scientific naming uses the double naming system—Binomial system.

In binomial system, an organism is given both the genus and species name.

Binomial nomenclature (Double –naming system)-Is the assigning of scientific names to living organisms governed by a definite set of rules recognized internationally.

Principles of binomial nomenclature

  1. The first, genus name, should begin with a capital letter and the second name, species, should begin or written in small letters e.g.

Lion—- Panthera leo

Leopard—– Panthera pardus

Domestic dog—– Canis farmiliaris

Human being— Homo sapiens

Maize plant—Zea mays

Lion and Leopard are closely related —Same genus but distantly related—different species.

  1. The scientific names must be printed in italics in textbooks and where hand written to be underlined e.g. Panthera leo.
  2. The specific name (species) is frequently written with the name of the scientist who first adequately described and named the organism e.g.Phaseolus vulgaris i.e. Vulgaris is the scientist who described and named the bean plant.
  3. Biologists should give a Latinized name for a newly described animal or plant species where Latin name is missing e.g.

Meladogyne kikuyuensis – Is a scientific name of a nematode from kikuyu.

Aloe kilifiensis — A member of Aloeceae family from Kilifi discovery.

Garinsoga parviflora waweruensis — a member of Macdonald eye family discovered by Waweru.

Study Question 1

Complete the table below

Taxon Lion Domestic dog Garden pea Napier grass
kingdom        
Phylum/division        
class        
order        
family        
genus        
species        

Scientific name         ———————     ————————    ———————–   ————————

Revision  Questions:

 

CLASSIFICATION 1

Diversity of Living Organisms

Need For Classification

  1. Help in identifying living organisms into their correct groups for reference.
  2. It brings together organisms with similar characteristics and separates those with different features.
  3. Help to organize information about living organisms in an orderly manner avoiding any confusion.
  4. Help to understand the evolutionary relationship between different living organisms.

Historical Background of Classification

Taxonomic Units of Classification

 

  1. Kingdom

Carolus Linnaeus (1707-1778) initially introduced the two kingdom system of classification. However many new life forms have been discovered which are neither animals nor plants. This has led to a more accepted classification system that adopts five kingdoms. These are;

Kingdom is further divided into several phyla in animals or divisions in plants.

  1. Phylum (phyla) or Division in plants.

It is the second largest and further divided into classes.

  1. Class

Each class is divided into several orders.

  1. Order

Orders are divided into smaller groups called families.

  1. Family

Family is divided into several Genera.

  1. Genus

Here members are closely related. It is further divided into the species.

  1. Species

This is the smallest unit of classification.

Species is defined as a group of organisms whose members naturally interbreed to produce fertile offspring’s.

Members of a given species have small differences such as skin colour, height etc.

Classification of Man and Maize plant. ( Table 2.1 Page 15 KLB Bk 1)

Scientific Naming of Living Organisms.

Binomial Nomenclature.

This is the double naming system of organisms where organisms are assigned two names i.e. the generic name and the specific name.

 

In binomial nomenclature the following rules are observed.

Examples

 

Revision Questions

 

CELL PHYSIOLOGY

Membrane Structure and Properties

The Cell Membrane

Diagram

 

Properties of Cell Membrane

  1. Semi-permeability. – It has small pores allowing for the passage of molecules of small size into and out of the cell. Cell Wall however allows all materials to pass through it hence it is referred to as being Permeable.
  2. Sensitivity to Changes in Temperature and pH – Extreme temperature and pH affects the cell membrane since it has some protein layers. Such changes alter the structure of the membrane affecting its normal functioning.
  3. Possession of Electric Charges – it has both the negative and positive charges helping the cell to detect changes in the environment. These charges also affect the manner in which substances move in and out of the cell

Physiological Processes

Diffusion

Practical Activity 1

To demonstrate diffusion using potassium permanganate (VII)

 

Role of Diffusion in Living Organisms

  1. Absorption of Materials
  1. Gaseous Exchange in Plants and Animals
  1. Excretion of Nitrogenous Wastes
  2. Transport of Manufactured Food form Leaves to other Plant Parts.

Factors Affecting Diffusion

  1. Diffusion Gradient
  1. Surface Area to Volume Ratio

Diagrams

  1. Thickness of Membranes and Tissues
  1. Temperature

Osmosis

Diagram fig 4.6

Practical activity 2

Practical activity 3

Osmotic Pressure 

Osmotic Potential

 

Water Relations in Animals

Diagrams

Water Relations in Plants

Diagrams

Study Question 5

 

Practical Activity 4

Wilting

Study Question 6

Role of Osmosis in Organisms

  1. Absorption of water from the soil
  1. Support
  1. Opening and closing of the stomata
  1. Feeding in insectivorous plants
  1. Osmoregulation

 

 

Factors Affecting Osmosis

Active Transport

Role of Active Transport

Factors Affecting Active Transport.

NB/ Any factor affecting energy production affect the rate of active transport.

Revision Questions.

Cell Specialization, Tissues, Organs and Organ Systems

  1. Cell specialization

 

Animal tissues include;

 

Plant tissues include:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Organs
  1. Organ systems.

Revision Questions

MICROSCOPE

Microscope Parts & Function

Parts of the Microscope

1. Eyepiece Contains a magnifying lens that focuses the image from the objective into your eye.
2. Course Adjust For focusing under low magnification
3. Fine Adjust For focusing under high magnification or low
4. Low Power Objective For large specimens or overview
5. High Power Objective For detailed viewing or small specimens
6. Specimen on glass slide What you want to look at
7. Stage Supports specimen in correct location to lens
8. Condenser Focuses the light on specimen
9. Diaphragm (iris or disc) Regulates amount of light and contrast
10. Light Source Illuminates the specimen for viewing

 

Handling and Care of the Microscope

The following rule should be observed:

  1. Use both hand when carrying the microscope. One hand should hold the base and the other holds the limb.
  2. Never place the microscope too close to the edge of the bench.
  3. Do not touch the mirror and the lenses with the fingers.
  4. Clean dirty lenses using soft tissue.
  5. Clean other parts using a soft cloth.
  6. Do not wet any part of the microscope.
  7. Make sure the low power clicks into position in line with the eye piece before and after use.
  8. Always store the microscope in a safe place free from dust and moisture.

Using the Microscope

  1. Place microscope on the bench with the stage facing away from you.
  2. Turn the low power objective lens until it clicks into position.
  3. Ensure the diaphragm is fully open.
  4. Look through the eyepiece with one eye. Adjust the mirror to ensure maximum light can pass through.
  5. Place the slide containing the specimen on the stage and clip it into position. Make sure the slide is at the centre of the field of view.
  6. Again look through the eyepiece while adjusting the mirror to ensure maximum light reach the specimen.
  7. Use the coarse adjustment knob to bring the low power objective lens to the lowest point. While viewing through the eyepiece, turn the coarse adjustment knob gently until the specimen comes into focus.
  8. Use the fine adjustment knob to bring the image into sharp focus.
  9. Make a drawing of what you see.
  10. For higher magnification, turn the medium power into position and adjust the focus using the coarse knob. Use the fine adjustment knob for sharper focus.
  11. For even large magnifications, turn the high power objective lens into position. In this case use only the fine adjustment knob to bring details into sharper focus.

Magnification

Magnification   = Eye Piece Lens Magnification X Objective Lens Magnification.

Study Question 1

Fill the table below.

Eye piece lens

maginification

Objective lens

magnification

Total magnification
X5 X4  
X10 X5  
X10   X100
  X40 X600
X10 X100  

 

Practical Activity 1

Cell Structures as Seen Under the Light Microscope

Diagrams- plant and animal cells

The Electron Microscope.

Study Question 2

Practical Activity 2

 

Cell Structures as Seen Under the Electron Microscope

Diagrams – Plant and Animal Cells

The Cell Organelles

 

  1. Cell membrane (Plasma Membrane).
  1. Encloses all the cell contents.
  2. It allows selective movement of substances into and out of the cell since it is semi-permeable.

Diagram

  1. Cytoplasm
  1. Mitochondria(Mitochondrion)

Diagram

 

  1. Endoplasmic Reticulum (ER)

Diagrams

  1. Ribosomes

Diagram

  1. Centrioles
  1. Chloroplasts
  1. Vacuoles

Study Question 3

 

Differences between Plant and Animal Cells

Preparation of Temporary Slides

Practical Activity 3

 

Estimation of Cell Sizes.

 

NUTRITION IN PLANTS AND ANIMALS

 

Nutrition

Autotrophism

Heterotrophism

Autotrophism

External Structure of a Leaf

A leaf is a flattened organ which is attached to the stem or a branch of a plant.

Diagrams

 

Parts of a leaf

Lamina:  This is the flat surface. It is green in colour and contain the photosynthetic tissue.

Midrib: This is a thick structure running through the middle of the leaf

Veins: They arise from the midrib to forming an extensive network of veins.

Leaf Apex: This is the tip of the leaf and usually it is pointed.

Petiole: It attaches the leaf to the stem or branch.

In some monocotyledonous plants the leaves are attached to the stem by the leaf sheath.

Practical Activity 1: To examine the External Features of a Dicotyledonous and Monocotyledonous leaf

Study Question 1

InternalStructure of a Leaf

 

 

 

 

 

 

 

 

Study Question 2

Adaptations of Leaves to Photosynthesis.

  1. Broad and flat lamina to increase surface area of Carbon (IV) oxide and sunlight absorption.
  2. Thin transparent cuticle and upper epidermis; to allow easier penetration of light to photosynthetic cells;
  3. Thin; for faster diffusion of gases;
  4. Palisade cells placed next to the upper surface; to trap maximum light for photosynthesis;
  5. Palisade cells with numerous chloroplasts; to trap maximum amount of light for photosynthesis;
  6. Large/ intercellular air spaces in the spongy mesophyll layer; for storage of  Carbon (IV) oxide for easier gaseous exchange;
  7. Waxy water proof cuticle; to reduce water loss sand reflect excess light;
  8. Leaf mosaic/ non-overlapping leaves; for maximum exposure to light;
  9. Guard cells, modified cells to open and close stomata; to control amount of water loss from the leaf and allows gaseous exchange;
  10. Leaves have leaf veins; xylem to conduct water to photosynthetic cells, Phloem to translocate products of photosynthesis to other parts of plant;

The Chloroplast

Practical Activity 2: To Observe Distribution of Stomata

Study Question 3.

The Process of Photosynthesis

6H2O + 6CO2 ———-> C6H12O6+ 6O2

Water      +   Carbon (IV) oxide                   Glucose            +   Oxygen.

The above chemical equation translates as:

Six molecules of water plus six molecules of carbon (IV) Oxide produce one molecule of sugar plus six molecules of oxygen

Light stage (Light Dependent Stage)                                               

–      Occurs in the grana containing chlorophyll which traps / absorbs sun light energy.

–      This Energy is used to split water molecules into hydrogen ion and oxygen gas.

–      This process is called photolysis of water and is shown below.

 

2H2O                             4H          +             O2

(Water)                           Hydrogen atom            Oxygen

–      Hydrogen atoms produced here enter into the dark stage.

–      Oxygen gas removed through stomata or is used for respiration within the plant;

–      Some Light energy is used in Adenosine Triphosphate (ATP) formation; ATP an energy rich compound.

–      ATP is later used in the dark stage.

Dark stage. (Light Independent Stage)

–      Carbon (IV) oxide combines with hydrogen atoms to form glucose/simple carbohydrate.

–      This is called Carbon (IV) Oxide fixation.

 

Carbon (IV) oxide         +     Hydrogen Atom                    Simple Carbohydrate

CO2                       +     4H                                         C6H12O6

–      This stage takes place in the stroma and proceeds whether light is present or not.

–      ATP Energy from light stage is used to provide the required energy in this reaction;

–      Simple sugars formed are used for respiration to provide energy or are converted to storable forms e.g lipids, proteins, starch, cellulose, etc.

Study Question 4

Practical Activity 3: To Investigate the Presence of Starch in a Leaf.

Study Question 5

Factors Affecting the Rate of Photosynthesis

 

 

 

 

 

 

 

 

 

 

 

 

Range of optimum light intensity

 

 

 

 

Light intensity

 

 

 

 

 

 

 

 

 

 

Range of optimum CO2 concentration

 

 

 

Carbon (IV) oxide concentration

Study Question 6

Practical Activity 4: To Investigate Factors Necessary for Photosynthesis.

  1. Light

Study Question 7

  1. Carbon (IV) oxide.

Study Question 8

Study Question 9

Study Question 10

Practical Activity 5: To Investigate the Gas Produced During Photosynthesis.

Study Question 11

Chemical Compounds Which Constitute Living Organisms

  1. Carbohydrates
  1. Monosaccharides

Properties of Monosaccharides

  1. They are soluble in water to form sweet tasting solutions.
  2. They are crystalissable.

Functions

  1. They are oxidized to release energy during respiration.
  2. When condensed together, they form polysaccharides such as starch, cellulose or glycogen.
  3. ii) Disaccharides

 

        Condensation

Monosaccharide    +     Monosaccharide                                   Disaccharide + Water.

C6H12O6              +     C6H12O6                                       C6H22O11 + H2O

Examples

Glucose           +    Glucose                                         Maltose   + Water.

Glucose          +    Fructose                                       Sucrose   + Water

Glucose           +    Galactose                                             Lactose   + Water.

Properties of Disaccharides

  1. i) Soluble in water to form sweet tasting solutions
  2. ii) They are non reducing sugars. Some such as the maltose can reduce copper sulphate in Benedict’s solution when heated together and are therefore referred to as complex reducing sugars.

iii)       They are readily broken into their constituent monosaccharide molecules in a process known as Hydrolysis in the presence of water.

Hydrolysis

Disaccharide      +     Water                                    Monosaccharide + Monosaccharide

 C6H22O11                 +     H2O                Hydrolysis                      C6H12O6             +    C6H12O6

Sucrose              +     Water             Hydrolysis                       Glucose       +        Fructose

Lactose              +     Water             Hydrolysis                       Glucose       + Galactose

Maltose              +     Water             Hydrolysis.                     Glucose       + Glucose.

Functions

iii)       Polysaccharides.They are made of many monosaccharide molecules hence are long and more complex.

Examples of polysaccharides

  1. Starch
  1. Cellulose

Properties of Polysaccharides

  1. All are insoluble in water.
  2. Do not have a sweet taste hence are referred to as non-sugars.

Study Question 12

Practical Activity 6:      To Carry out Food Tests for Carbohydrates

  1. i) Starch
  2. ii) Reducing sugars

iii)   Non Reducing Sugars

  1. Lipids

Diagram

Properties of Lipids

  1. When fats are heated they change into liquid while oils solidify under low temperature.
  2. Both fats and oils are insoluble in water. They however dissolve in organic solvents such as alcohol to form emulsions and suspensions.
  3. Lipids are inert hence can be stored in the tissues of organisms.

Functions of Lipids

  1. Source of energy
  1. Source of metabolic water
  1. Heat insulation
  1. Protection

Study Question 13

Practical Activity 7: testing for the Presence of Lipids

  1. i) The Grease Spot
  2. ii) The Emulsion Test
  3. Proteins

Properties of Proteins

Functions of Proteins

  1. a) Enzymes
  1. b) Hormones

Study question 14

Practical Activity 8

To Test for Proteins

Enzymes

  1. a) Extracellular Enzymes
  1. b) Intracellular Enzymes

 

Naming of the Enzyme

  1. i) Trivial Naming
  1. ii) Use of suffix –ase

Example 1

Substrate Enzyme
Carbohydrate Carbohydrase
Starch e.g. amylose Amylase
Sucrose Sucrase
Maltose Maltase
Protein Protease
Lipid Lipase

Example 2

Reaction Enzyme
Hydrolysis Hydrolase
Oxidation Oxidase
Reduction Reductase

Properties of Enzymes

  1. They are protein in nature hence are affected by changes in temperature and pH.
  2. They are substrate specific.
  3. They are efficient in small amounts as they are not affected by the reactions they catalyze. They can be used again and again.
  4. They are catalysts that speed up the rate cellular reactions and are not used up in the reactions they catalyses.
  5. Most of the enzyme controlled reactions are reversible.

Factors Affecting the Rate of Enzyme Controlled Reactions

Diagrams

Diagrams

Diagrams

  1. a) Competitive Inhibitors
  1. b) Non Competitive Inhibitors

Importance of Enzymes

Study Question 15

Practical Activity 9

Study Question 16

Study Question 17

Practical Activity 10

 

 

 

 

 

 

 

FORM TWO BIOLOGY NOTES

 

EXCRETION AND HOMEOSTASIS

Excretion-Process by which living organisms separate and eliminate waste products formed during metabolic processes from the body. They include; carbon IV oxide, excess water and mineral salts, nitrogenous wastes etc. accumulation of these substances may become toxic to cells.

Homeostasis-This is the maintenance of internal environment of cells under constant conditions E.g. temperature, osmotic pressure, blood sugar and chemical constituents.

Egestion. –      This is the removal of undigested and indigestible materials from the alimentary Canal of animals.

Secretion.       –      This is the release of certain useful substances produced by cells e.g. hormones, Enzymes, sebum, saliva and mucus.

 

Excretion in Plants

  1. There is very little accumulation of toxic wastes such as nitrogenous wastes.
  2. Some waste products are re-used in the same plant such as Co2, oxygen and water.
  1. Some plants store wastes in their tissues in non-toxic forms such as nicotine, caffeine, tannins, resins, quinine, morphine etc.

Economic Importance of Plant Excretory Products

  1. Tannins – They are deposited in dead tissues of wood and barks of trees e.g. in acacia and wattle tree. Tannin is used to treat leather.
  2. Caffeine – it is stored in coffee berries and tea leaves. It is used as a stimulant.
  1. Cocaine – it is obtained from the leaves of coca plant and is used as an anesthetic.
  2. Cannabis – found in the leaves and flowers of Cannabis sativa (bhang). It is used to manufacture some drugs.
  3. Nicotine – found in leaves of tobacco plant and is used in the manufacture of insecticides and narcotic drugs. It also manufactures cigarettes.
  1. Pappain– it is obtained from raw paw paw and it is used as a meat tenderizer.
  2. Khat/miraa – it’s chewed and acts as a mild stimulant.

Excretion and Homeostasis in Unicellular Organisms

Diagram

Excretion in Mammals

A     Structure and Function of the Mammalian Skin

  1. Protection of the underlying tissues from entry of micro-organisms, physical damage and ultra violet rays from the sun.
  2. Regulation of body temperature.
  1. Reception of stimuli such as heat, cold, pain, touch and pressure.
  2. Synthesis of vitamin D.
  3. Storage of fats.

Diagram

  1. Epidermis (upper and outer layer)
  2. The dermis (inner layer)
  3. a) Epidermis (upper and outer layer)
  1. Cornfield layer.
  2. Granular layer.
  1. Cornifield layer
  1. Granular layer
  1. The Dermis (inner layer)

Sebum also prevents bacterial attack due to its antiseptic property;

In hot conditions, they dilate; increasing blood flow near the skin surface enhancing blood flow near the skin surface; minimizing heat loss;

Blood vessels supply nutrients and oxygen to skin tissues and also remove waste products and carbon IV oxide.

Sweat also excretes excess water, mineral salts, urea and lactic acid.

B     The Lungs

C     Structure and Function of the Kidney

Diagram fig. 4.3; generalized urinary system of a mammal (page 88 KLB)

Diagram fig. 4.4(a) and 4.4(b) (page 89 KLB)

Cortex

Medulla

Pelvis

Nephron

Diagram fig. 4.6. The structure of the kidney nephron

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The renal tubule has 5 main parts.

  1. Bowman’s capsule
  2. Proximal convoluted tubule
  3. Loop of Henle
  4. Distal convoluted tubule
  5. Collecting tubule
  6. Bowman’s capsule

Diagram: structure of the nephron

Diagram of ultra-filtration at the Glomerulus

  1. Proximal convoluted tubule
  1. Presence of mitochondria in the cells lining to provide with energy required
  2. Cells of the tubule have micro-cilli (infoldings) which increase surface area for re-absorption.
  1. Coiling of the tubule reduces the speed of flow of filtrate giving more time for efficient re-absorption.
  2. Tubule is well supplied with blood capillaries.
  3. Loop of Henle
  1. Distal convoluted tubule

–   Active intake of sodium salt into the blood at the loop of Henle increases the osmotic potential of the blood.

–   The antidiuretic hormone (ADH) secreted by the pituitary gland. ADH increases the permeability of the tubule and blood capillaries to water

  1. Collecting tubule

Urine Composition

Substance % Composition.
Water  95%
Urea 2%
Uric acid 0.03%
Creatine 0.1%
Salts 1.4%
Ammonia 0.04%
Proteins 0%
Glucose 0%

 

  1. Physiological adaptations.
  2. Habitat of an organism e.g. terrestrial, desert or aquatic.

Study Questions. Page 93.

Comparison Between Aquatic and Desert Animals

Fresh Water Animals Desert Animals.
i)               Have many glomeruli to increase ultrafiltration. Few glomeruli to reduce ultrafiltration.
ii)            Short loop oh Henle to reduce water reabsorption. Long kidney tubules to increase water reabsorption.
iii)          Produce large quantity of dilute urine. Produce small quantity of concentrated urine.

 

Comparison of Composition of urine with that of Glomerular Filtrate and Blood Plasma.

Substance % Composition of;
Plasma Glomerular Filtrate. urine
Urea 0.03 0.03 2.0
Uric acid 0.005 0.005 0.03
Ammonia 0.001 0.001 0.004
Glucose 0.1 0.1 0
Amino acids 0.05 0.05 0
Mineral salts 0.70 0.70 1.4
Blood proteins. 8.00 0 0

Functions of the kidney include:

  1. Regulation of PH

Kidney Diseases

  1. Nephritis

This is the inflammation of the glomerulus of the kidney. It is caused by bacteria or infections such as small pox and measles.

Symptoms

Treatment

  1. Use of just adequate amounts of salts and proteins in diets Kidney stones

Causes

Symptoms

Control & Treatment

  1. Albiminuria (Proteins in Urine).

D     The Liver and its Structure

 

Homeostatic Functions of the Liver

Regulation of blood sugar level

  1. Deamination

 

2NH3   +     CO2                CO(NH2)2       +     H20

Ammonia            Carbon iv                              Urea                    Water

(Toxic)        Oxide                                    (less toxic)

 

  1. Detoxification

 

 

Hydrogen Peroxide         Water      +     Oxygen

(H2O2)                                                  (H2O)          (O2)

  1. Regulation of plasma proteins
  1. Storage of vitamins A, B,D,E and K and mineral salts
  1. Heat production (Thermoregulation)
  1. Inactivation of hormones and drugs
  1. Storage of blood
  1. Regulation of cholesterol and fat metabolism
  1. Manufacture of red blood cells in foetus.

Liver Diseases and Disorders

  1. Liver Cirrhosis

Control

  1. Hepatitis
  1. Jaundice

Homeostasis

Principles of Homeostasis

  1. Negative Feedback Mechanism
  1. An increase in the level if it is dropping
  2. A decrease in the level if it is increasing

Further Excess

 

 Positive feed back

 

 

 

 

 

(Negative Feedback)

 

 

 

 

 

 

(Negative feedback)

 

 

 

 

 

                     Positive feedback

 

                      

Further deficiency

  1. Positive Feedback Mechanism

The following are some of the factors regulated through homeostasis.

  1. Temperature Regulation. (Thermoregulation)

Skin and Thermoregulation

The skin is adapted in the following ways to effect thermoregulation;

  1. It has Hair shaft;
  1. Blood vessels

Diagrams

3)       Sweat gland

4)           Has subcutaneous layer; contains fat which acts as a heat-insulating layer. Organisms in cold areas have thick subcutaneous layer for heat insulation.

Homoiotherms and Poikilotherms

Homoiotherms (Endotherms)

Poikilotherms (Ectotherms)

Methods of Regulating Body Temperature in Animals.

  1. Metabolic activities of the Body, such as shivering to raise body temperature.
  2. Insulatory mechanisms such as dilation or constriction of blood vessels, hair movement etc.
  1. Presence of adaptive features such as hair/fur, subcutaneous tissue etc.

Hibernation is where animals go into deep sleep for long period of time due to cold.

Aestivation is where animals go into deep sleep due to dry and harsh conditions.

Differences BetweenHomoiotherms and Poikilotherms.

Poikilotherms Homoiotherms
i)               They are sluggish under cold conditions. i)               They remain active even under cold conditions.
ii)            They hibernate to avoid death by freezing under very cold conditions. ii)            Only the small animals hibernate because they have large surface area to volume ratio hence lose a lot of heat.
iii)          They aestivate under very hot conditions. iii)          They do not aestivate because they can maintain constant body temperature.
iv)          They are easy prey to predators due to their hibernation and aestivation. iv)          Not easy to prey because they active always.
v)            Require less food because they get heat from the environment to warm their bodies. v)            Require more food because they use it to generate heat for maintaining the temperature constant.
  1. Osmoregulation (Water and Salt Balance).
  1. When the osmotic pressure of the blood rises due to dehydration, the hypothalamus is stimulated and sends an impulse to the pituitary gland which secretes the Antidiuretic Hormone (ADH) or Vasopressin into the blood. ADH increases permeability of the kidney tubules to water. More water is reabsorbed, osmotic pressure of the blood falls hence production of little but concentrated urine.
  2. When osmotic pressure of the blood falls due to excess water in the body there is less production of ADH and less water is reabsorbed hence production of large amounts of dilute urine.

Diabetes Insipidus

  1. Regulation of Ionic Content
  1. Regulation of Blood Sugar Level.
  1. Converts excess glucose into glycogen for storage.
  2. Inhibits conversion of glycogen to glucose.
  1. Increases breakdown of glucose to release energy.
  1. Increases the breakdown of glycogen into glucose.
  2. Increases the conversion of fats and proteins into glucose.

NB/. The hormone adrenaline produced by the adrenal glands also has homeostatic effect on glucose.

It is released during emergencies to avail glucose for fight or flight.

Diabetes Mellitus (Sugar Disease)

Symptoms

Control

Question

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Revision questions

Gaseous Exchange

 

Diagram

Importance of Gaseous Exchange

  1. Promote oxygen intake for respiration.
  2. Facilitate carbon IV oxide removal from the body as a metabolic waste product.

Gaseous Exchange in Plants

Examples of respiratory Surfaces in Plants

Structure and Function of the Stomata

Adaptations of Guard Cells

  1. They are bean shaped/sausage shaped.
  2. Contain chloroplast hence can photosynthesize.

Diagram

Mechanism of Opening and Closing of Stomata

  1. Photosynthetic theory
  2. Starch Sugar inter-conversion Theory. (effect of changes in pH of guard cells)
  1. Photosynthetic theory
  1. Starch Sugar inter-conversion Theory. (effect of changes in pH of guard cells)

Mechanism of Gaseous Exchange in Plants

  1. Gaseous Exchange through the Stomata

Xerophytes: These are plants adapted to life in dry areas.

Hydrophytes: These are the aquatic plants (water Plants)

Diagrams

 

Mesophytes: They are plants growing in areas with adequate amounts of water.

  1. Gaseous Exchange through the Lenticels

Diagram

 

  1. Gaseous Exchange through the Roots

 

Gaseous Exchange in Animals

 

Types and Characteristics of Respiratory Surface

 

 

Type of Respiratory Surface Environment/Medium of Operation Example of Organism
1.   Cell Membrane. Water Amoeba
2.   Gill Filaments Water Fish
3.   Tracheoles Air Insects
4.   Alveoli/Lungs Air Mammals
Birds
Frogs
Reptiles
5.   Skin Water Frog
Air Earthworm
6.   Buccal Cavity Air Frog

 

  1. Must have a large surface area.
  2. Must be moist to allow gases to diffuse in solution form.
  1. Have a thin membrane to reduce the diffusion distance.

Gaseous Exchange in Insects

Insects have their gaseous exchange system made of many air tubes forming the tracheal system.

 

Diagram

Mechanism of Gaseous Exchange in the Tracheal System of an Insect

 

Gaseous Exchange in Fish

Diagram

 

Mechanism of Gaseous Exchange in the Gills of a Bony Fish

Diagram

 

Diagram

Mechanism of Gaseous Exchange in Amphibians

  1. Gaseous exchange through the lining of the buccal cavity
  2. Gaseous exchange through the lungs
  3. Gaseous exchange through the skin
  4. Gaseous exchange through the mouth (buccal) cavity

 

  1. Gaseous exchange through the skin

Mechanism of Gaseous Exchange in Mammals

  1. Nose
  1. Larynx
  1. Trachea
  1. Lungs

Diagrams 

 

 

The mechanism of breathing

  1. Inspiration (Inhalation) i.e. breathing in.
  2. Expiration (Exhalation) i.e. breathing out.

Inspiration (Inhalation) i.e. breathing

Diagrams page 62

Expiration (Exhalation) i.e. breathing out

 

Gaseous exchange in the alveolus

Diagram page 64 KLB

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Percentage composition of gases in inhaled and exhaled air

Gas %  in inhaled air. % in exhaled air
Oxygen 20 16.9
Carbon (iv) oxide 0.03 4.0
Nitrogen and other gases 79.97 79.97

 

Regulation of Breathing

This is controlled by a part of the brain called Medulla oblongata.

 

Factors affecting the rate of breathing in humans

  1. Exercise

Breathing rate increases during vigorous activity.

  1. Age

Younger people have a faster breathing because their bodies have more energy demand.

  1. Emotions

Things like anxiety, fear and fright increases the breathing rate.

  1. Temperature

Relatively high temperatures increase the rate of breathing.  However, very high temperatures reduce the breathing rate.

  1. Health

If there is fever (high body temperature), the breathing rate increases. Some respiratory diseases however, make breathing difficult.

Lung Volumes

  1. Lung capacity

This is the total amount of air the lungs can hold when completely filled. The lungs of an adult have a capacity of about 5,500cm3

  1. Tidal volume

This is the amount of air taken in and out of the lungs during normal breathing. Tidal volume is about 500cm3

This is an additional volume attained after having a forced inhalation in addition to the tidal volume. It is about 2000cm3

  1. Inspiratory capacity

This is the tidal volume +Inspiratory reserve volume.

  1. Expiratory reserve volume

This is air removed after a forced exhalation. It can be up to 1,300cm3

  1. Vital capacity

This is the deepest possible exhalation. This air can only be forcibly pushed out of the lungs.

This is the air that normally remains in the lungs after the deepest exhalation. It is normally about 1,500cm3

Diagram

 

Diseases of the Respiratory System

  1. Asthma

It is caused by:

Symptoms

 

Treatment and Control

  1. Bronchitis

There are two types; Acute and Chronic

Symptoms

Treatment

It is caused by a bacterium called Bordetella pertussis.

Symptoms

Treatment

Control

  1. Pneumonia

It is caused by a bacterial called Streptococcus pneumoniae

Symptoms

Treatment

Control

  1. Pulmonary Tuberculosis

It is caused by a bacterium calledMycobacterium tuberculosis.

Symptoms

Treatment

Control

  1. Lung cancer

Cancer is uncontrolled cell growth in the body causing tumours.

Some general causes

Treatment and control

Revision Questions

 

RESPIRATION

Diagram

Practical Activity 1

To investigate the gas given off when food is burnt.

Types of Respiration

Aerobic Respiration

C6H12O6 + 6O2   6CO2       +     6H2O      +     Energy (ATP)

First Phase (Glycolysis)

Second phase (Krebs Cycle)

  1. Cells must be provided with glucose/food.
  2. Oxygen must be present.
  1. Temperature must be favourable for efficient functioning of enzymes.
  2. End products of the reaction (energy, water and carbon (iv) oxide) must be constantly removed from the mitochondrion.

Practical Activity 2

To investigate heat production in germinating seeds.

Anaerobic Respiration in Plants and Animals

Glucose                     Ethanol   +     Energy.   +    Carbon (iv) oxide

(C6H12O6)          (2C2H5OH)            (ATP)                            (CO2)

Glucose                     Lactic acid     +     Energy.

(C6H12O6)          (2C3H6O3)             (ATP)

Oxygen Debt

Application of Anaerobic Respiration

  1. Baking industry
  2. Beer brewing and distillery industry.
  1. Production of vinegar citric acid, oxalic acid, butyric acid and some drugs.
  2. Production of power alcohol which is used as a substitute for petrol.
  3. Silage making.

Practical Activity 3

To investigate gas produced during fermentation.

Comparison between Aerobic and Anaerobic Respiration

Aerobic Respiration Anaerobic Respiration.
     i.          Oxygen is required Oxygen not required
   ii.          High amount of energy is released as one molecule of glucose yields 38 ATP molecules (2880 KJ) Low amount of energy is released as one molecule of glucose yields 2 ATP molecules (210 KJ)
 iii.          There is complete breakdown of the substrate into carbon (iv) oxide and water. There is incomplete breakdown of substrate hence lactic acid or alcohols are produced.
 iv.          End products are energy, water and carbon (iv) oxide End products are energy, alcohol in plants and lactic acid in animals.
   v.          Water molecules are produced. Water molecules are not produced.
 vi.          Over a short period of time, energy is not released faster Over short period of time, energy is released faster.
vii.          Occurs in the cytoplasm and in the mitochondrion. Occurs only in the cell cytoplasm.

 

Respiratory Substrates

  1. Carbohydrates
  1. Fats

Assignment

Respiratory Quotient (RQ) and its Significance

RQ              =

Factors Affecting the Rate of Respiration

  1. Oxygen concentration. When the amount of oxygen increases, the respiration rate also increases. Decrease in oxygen concentration will lead to decreased respiration rate.
  2. Substrate concentration. Increase in sugar concentration increases respiration and vice versa.
  1. Surface area to volume ratio (Body size). If the SA/volume ratio is high, the organism would lose more heat energy. As more heat is lost to the surrounding more is required to replace the lost energy hence more respiration.
  2. Age. Young people require more energy because their cells are actively dividing hence respiration rate is higher in them than in older people.
  3. Occupation. People engaged in heavier tasks have higher rate of respiration.

Revision Questions

 

 

FORM THREE BIOLOGY NOTES

ECOLOGY

Introduction

Ecology is the study of the interrelationships of organisms to each other and to their environment (biotic and Abiotic factors).

Autecology; study of single species within a community and how it relates with both the biotic and Abiotic factors.

Synecology. This is the study of many different species of organisms’ interacting among themselves within an ecosystem.

Ecology helps to address the following issues.

Concepts of ecology

Study Question 1

Factors in an Ecosystem

They are divided into two:

  1. Abiotic factors or the non living factors
  2. Biotic or the living factors

Abiotic Factors

This is the measure of acidity or alkalinity of water in aquatic habitats or soil solution. This influences the distribution of plants and animals in soil and aquatic habitats. Different organisms have different pH requirements. pH is determined using the pH meter.

Study Question 2

Practical Activity 1

Study Question 3

Biotic Inter-Relationships

 Competition

Living organisms compete for resources such as nutrients, space, light and mates. There are two types of competition.

 

 

However, closely related species can live together without competition. For example, when Paramecium caudatum and Paramecium bursoria are grown in the same culture, there is no competition because each species occupies a different part of the culture. Similarly, browsers and grazers can occupy same habitat without competition because they feed at different levels of the same plants. For example, the zebras eat the softer shoots, followed by the wild beasts, and the gazelles which eat the fibrous left over of the same grass.

Study Question 4

When there is competition the best adapted organisms survive while the less adapted ones may die or be forced to migrate.

This is the relationship where one organism kills another for food and feed on it either as a whole or a part of it. The predator is the one which kills while the prey is the one being killed for food.

Predators have various adaptations to enable them to be efficient in capturing the prey. These include;

 

NB/. When the number of the prey increases that of the predators also increases. An increase in the number of predators leads to a decrease in the population of the prey. This decrease in prey population leads to a fall in predator population which in turn gives space for the increase in the population of the prey. This is the basis of biological control. See the graph below.

 

Study Question  5

This is an association between two of different species in which both benefit. For example the association of colon bacteria with humans and other animals, especially plant-eating animals, the ox-pecker bird and the ox etc.

The Rhizobium bacteria help the leguminous plants to fix nitrogen while the bacteria obtain shelter and carbohydrates from the plants.

Diagram

This is where organisms obtain nutrients from dead organisms causing decomposition hence releasing nutrients into the ecosystem. Saprophytes include the bacteria and fungi.

The Nitrogen Cycle

This refers to the cycling of nitrogen and its compounds in the natural environment.

Diagram

Practical activity 2

Study question 6

 

Energy Flow in an Ecosystem

The sun is the natural source of energy. This energy is transferred to the following feeding levels;

Decomposers

They break down organic materials into simple substances which are made available for re-use by other organisms. Decomposers are mainly fungi and bacteria.

Food Chains

This is the representation of energy flow from a producer to other organisms linearly. Green plants are eaten by herbivores which are eaten by carnivores.

Producers’          Primary consumers        Secondary consumers        Tertiary consumers Quaternary consumers

Some energy is lost as it is moved from one trophic level to the next. This is through respiration, defecation, excretion and in form of heat.

Fig. 2.7

Examples

When the decomposers are included in a food chain, they are placed at the end.

Study Question 7

Food Webs

These are several interconnected food chains. Simple food chains rarely exist since in any ecosystem, many populations interact.

Examples

Study Question 8

Ecological Pyramids

These give a simplified representation of feeding relationships and energy flow in an ecosystem. They are of three types.

Pyramid of Numbers

There is a progress decrease in the number of organisms as one move from the producers all the way to the quaternary consumers. Producers have the greatest number followed in a decreasing order by primary, secondary, tertiary and quaternary consumers.

 

Construction of Pyramid of Numbers

Study Question 9

Interpretation of Pyramid of Numbers

 

Pyramid of Biomass

Biomass of an organism is its constant dry weight. In an ecosystem, the producers have the highest biomass followed in decreasing order by primary, secondary, tertiary and quaternary consumers.

 

Study Question 10

Practical activity 3

Study Question 11

Population

Populations change in size, structure and organisation.

Characteristics of a population.

Population Estimation Methods

Usually a representative sample is used to estimate the population of organism in a big habitat. A sample is a small number of individuals taken from the habitat that is a representative of the whole population. The following methods are used when sampling.

Adaptations of plants to various Habitats

An adaptation is a change to suit environment: the development of physical, physiological or behavioural characteristics that allow organisms to survive and reproduce in their habitats. There are four main groups of plants namely;

Xerophytes

These are plants adapted to survive in the dry habitats. These habitats have the following characteristics.

Adaptations of Xerophytes

Mesophytes

These are plants growing in well watered areas. Such habitats have the following general characteristics.

Adaptations of Mesophytes

They show various adaptations depending on where they grow. Some of these adaptations are for reduction of water loss, others for increased water, loss and some are also adapted to light conditions.

Forest Ecosystem

 

Hydrophytes

These are plants growing in fresh water either partially or wholly. Such habitats have the following general characteristics.

Adaptations of hydrophytes

Diagrams

Halophytes

These are plants which are able to tolerate very salty conditions in soil and marine water. Such habitats have the following general characteristics.

Adaptations of Halophytes

 

CLASSIFICATION II

Classification, in biology is the identification, naming, and grouping of organisms into a formal system based on similarities such as internal and external anatomy, physiological functions, genetic makeup, or evolutionary history.

Study Question 1

General Principles of Classification

Organisms that have similar and common features are grouped together while those that have different features are grouped separately.

Taxonomy is the study of grouping of organisms according to their relationship. There are seven major taxonomic units (taxa).

As you move from the kingdom to the species the differences decrease as the similarities increases.

Species is a group of organisms that can freely interbreed to give rise to viable/fertile offsprings.

Sometime members of different species may interbreed to give an offspring which is sterile. E.g. a donkey and a horse can interbreed to give rise to a mule which is infertile.

Binomial Nomenclature

This is the double naming system of organisms where organisms are assigned two names i.e. the generic name and the specific name.

Examples

In binomial nomenclature the following rules are observed.

Study Question 2

The Five Kingdoms of Classification

Carolus Linnaeus initially introduced the two kingdom system of classification. However many new life forms have been discovered which are neither animals nor plants. This has led to a more accepted classification system that adopts five kingdoms. These are;

Fig. 1.2

  1. Kingdom Monera

The kingdom is made up of mainly the bacteria e.g. nitrobacter, azotobacter. Vibrio cholerae etc.

General characteristics

Fig. 1.4

Diagrams

Study question 3

  1. Kingdom Protoctista

Examples include paramecium, amoeba, plasmodium, chlamydomonas, euglena, spirogyra, and trypanosome.

General characteristics

Diagrams

Practical Activities 1 and 2

 

  1. Kingdom Fungi

Examples

Saprophytic ones include mushrooms, toadstools, bread moulds, penicilia, yeast etc.

Parasitic ones cause plant diseases such as wheat rust, potato and tomato blight and animal diseases such as athlete’s foot and ringworm.

Practical Activities 3

General characteristics

Examples

Study Question 4

 

4. Kingdom Plantae

Study question 5

General Characteristics

 

The kingdom Plantae is divided into three main divisions.

 

  1. Division Bryophyta

These are the mosses and the liverworts.

General Characteristics

Diagrams.

 

  1. Division Pteridophyta

This includes ferns and horsetails.

They are more advanced compared to the bryophytes.

General Characteristics

Study Question 6

Practical Activity 4

Study Question 7

Practical Activity 5

  1. Division Spermatophyta

This comprises of all the seed bearing plants.

General Characteristics

The division Spermatophyta is made up of two main subdivisions i.e.

Gymnospermaphyta

General Characteristics

 

This subdivision has three main classes.

  1. i) Class Coniferales

Diagrams.

  1. ii) Class Cycadales

iii) Class Ginkgoales

Angiospermaphyta

General characteristics

This subdivision is divided into two classes.

 

Class Monocotyledonae Class Dicotyledonae.

 

·      They have seeds with one cotyledon. ·      Have two cotyledons.
·      They have narrow-long leaves with parallel venation. ·      Broad leaves with reticulate venation.
·      Most of their leaves have a modified petiole to form a leaf sheath. ·      Leaves have distinct petioles.
·      Their stems have scattered vascular bundle. ·      Vascular bundles are arranged to form a concentric ring.
·      Pith is usually absent. ·      Pith is present.
·      Vascular cambium is usually absent hence no secondary growth. ·      Vascular cambium is present hence there is secondary growth.
·      They have a fibrous root system ·      They have a tap root system
·      Floral parts are in threes or in multiples of three. ·      Floral parts are in fours, fives or their multiples.
·      In the root vascular bundles are arranged in a ring with phloem and xylem alternating. ·      In roots, the xylem is centrally placed and star shaped with the phloem alternating with the arms of the xylem.

 

Study question 8

Practical activity 6

Study Question 9

General characteristics

Kingdom Animalia has nine phyla but only two will be discussed i.e. Arthropoda and chordata.

Phylum Arthropoda

Practical Activity 7

General Characteristics

The phylum arthropoda is divided into five classes.

Different members of the phylum are placed to their respective classes based on;

  1. Class Crustacea

Examples.  Daphnia, crayfish. Crab and prawn.

 

 

General Characteristics

 

  1. Class Chilopoda

These are the centipedes.

Diagram

General Characteristics

  1. Class Diplopoda

These are the millipedes.

Diagram

 

 

 

 

General Characteristics

  1. Class Arachnida

These include the scorpions, spiders, ticks and mites.

Diagrams

 

General Characteristics

  1. Class Insecta

They include grasshoppers, bees, houseflies, butterflies, termites, beetles etc.

Insects form half the population of animals on earth. They occupy all habitats i.e. air, water, and land. Their food is varied such as plant tissues, animal fluids, dead animals and excretions of animals making them to be found almost everywhere on earth.

General Characteristics

Assignment

Discuss the economic importance of arthropods.

Study Question 10

Practical Activity

 

Phylum Chordata

Chordate, common name for animals of the phylum Chordata, which includes vertebrates as well as some invertebrates that possess, at least for some time in their lives, a stiff rod called a notochord lying above the gut. About 43,700 living species are known, making the chordates the third largest animal phylum.

In animals such as the Amphioxus the notochord persists but in others it is replaced at later stages of development by the vertebral column.

Members in this phylum inhabit both aquatic (marine and fresh water) and terrestrial (burrowers and arboreal) environments.

General Characteristics

The main classes of the phylum chordata are;

 

Pisces

Diagram

These are the fishes. They include those with a skeleton made of cartilage e.g. shark and those with a bony skeleton such as the tilapia, Nile perch, lung fish, dog fish, and cat fish etc.

General Characteristics

They include the toads, newts, salamanders and frogs.

The toad is the most advanced amphibian. Its skin is less moist and therefore uses the lungs more for gaseous exchange. They therefore stay mostly on land and only return to the ponds during reproduction.

Diagrams

General Characteristics

Examples include tortoise, turtles, snakes, crocodiles, lizards and chameleons.

General Characteristics

Examples include doves, chicken, hawks, eagles and turkeys.

They are terrestrial and arboreal while some have been adapted for aquatic life.

General Characteristics

Study Question 11

General Characteristics

 

The Dichotomous Key

The word dichotomous means separating into two. I.e. Separation of different or contradictory things: a separation into two divisions that differ widely from or contradict each other. As you move down the key you progress from general characteristics to more specific characteristics. The last single choice reveals the identity of the unknown organism.

Rules Used in Constructing a Dichotomous Key

  1. a) leaf simple.
  2. b) Leaf compound
  3. a) Leaf net veined.
  4. b) Leaf parallel veined.

Be very specific in your description such as

Some common Features Used For Identification.

Part of plant Some characteristics.

 

Leaf ·      Phylotaxy

·      Leaf type

·      Leaf venation

·      Margin

·      Lamina

·      Colour

 Flower ·      Inflorescence

·      Flower shape

·      Number of floral

Stem ·      Type of stem( woody, herbaceous or fleshy)

·      Shape (rectangular or cylindrical)

·      Texture of the stem (smooth or spiny/thorny)

Roots ·      Root system (taproot or fibrous)

·      Storage roots.

 

 

 

 

 

 

 

·

 

Summary of the Five Kingdoms.

Practical Activities.

Revision Questions.

 

REPRODUCTION IN PLANTS AND ANIMALS

Diagram

Diagram

Importance of Reproduction

This ensures that a species does not become extinct.

Reproduction allows for mixing of genetic materials bringing

about variations.

These variations are important tools in the refinement of quality of offsprings.

Cell Division

Chromosomes

Diagram

  1. Mitosis
  2. Meiosis

Mitosis

  1. Interphase
  2. Prophase
  1. Anaphase
  2. Telophase.

1)Interphase

During this stage the following activities take place within the cell in preparation of the division.

Diagrams

2)Prophase

The following events take place in this stage.

Diagram

3)Metaphase

Diagram

Diagram

In animal cells, division of cytoplasm is by constriction of cell membrane.

In plant cells, a cell plate forms within the cytoplasm and grows to separate the cell into two.

Diagrams

Significance of Mitosis

  1. Forms basis for asexual reproduction e.g. budding and spore formation.
  2. Causes cell growth when the cells formed increase in number and size.
  1. Replaces damaged and dead cells in the body.

Meiosis

First Meiotic Division

 Interphase I

The cell prepares for division by the following.

 Prophase I

NB/. During chiasma formation homologous chromosomes may exchange genetic material during crossing over. These genetic exchanges are important because they bring about variations in offsprings.

Metaphase.I

Diagram

Anaphase I

Diagram

Telophase I

Diagram

Second  Meiotic Division.

In this stage the sister chromatids are separated from each other.

Interphase II

Prophase II

Metaphase.II

Anaphase II

Telophase II

Significance of Meiosis

  1. Gamete formation (sperms and ova) forming basis for sexual reproduction
  2. Provides opportunities for genetic variations during crossing over

Similarities between mitosis and meiosis

  1. Both take place in plants and animals.
  2. Both involve division (multiplication) of cells.

Differences between meiosis and mitosis

Meiosis Mitosis.
1.   Homologous chromosomes associate with each other. No association of homologous chromosomes
2.   Takes place in 2 nuclear divisions. Takes place in one nuclear division.
3.   4 daughter cells are produced each haploid (n) 2 daughter cells are produced each diploid (2n)
4.   Occurs in reproductive organs leading to gamete formation. Occurs in somatic (body) cells leading to growth.
5.   Chiasma formation takes place leading to crossing over hence variation No chiasma formation therefore no crossing over hence no variation.

 

Asexual Reproduction

Types of Asexual Reproduction

  1. Binary fission in amoeba, plasmodium and bacteria
  2. Sporulation in rhizopus
  3. Budding in yeasts
  4. Binary fission in amoeba

Diagrams

  1. Sporulation in Rhizopus

Diagrams

 Budding in Yeast

Under favourable conditions such as plenty of sugar, moisture, oxygen and optimum temperature, the yeast cell reproduces asexually by budding.

Diagrams

 

Sexual Reproduction in Plants

Structure and Function of a Flower

  1. Calyx (sepals)
  2. Corolla/petals
  1. Gynoecium – female parts

 

  1. Calyx (sepals)

 

  1. Corolla/petals

Diagrams

  1. Gynoecium – female parts
  1. Epigynous (inferior) ovary

Diagram

  1. Hypogynous (superior) ovary

Diagram

Diagram

The gynoecia can also be grouped into different types dependi.ng on the number of carpels present i.e. monocarpous or syncarpous.

Monocarpous Gynoecium

Diagram

Polycarpous Gynoecium

  1. Apocarpous gynoecium

Diagrams

  1. Syncarpous gynoecium

Diagrams

Terms Used in Describing a Flower

  1. Complete flower – has all the four floral parts; calyx, corolla, androecium and gynoecium.
  2. Incomplete flower – has one or two floral parts missing.
  1. Staminate flower – male flower.
  2. Pistillate flower – female flower.
  3. Monoecious plant – bears both male and female parts of the flower.
  1. Regular or actinomorphic flower – a flower that can be divided into tow similar halves by any vertical section passing through the center i.e. radial symmetry such as in morning glory.
  2. Irregular or zygomorphic flower – can be divided into two similar halves on one particular plane only i.e. bilateral symmetry e.g. in clotalaria.
  3. Pedicillate flower- flower with a stalk.
  1. Non essential floral parts – are the calyx and corolla.

Pollination

This is the transfer of pollen grains from the anther to the stigma.

Types of Pollination

  1. Self pollination. – Transfer of pollen grains from the anther to the stigma of the same flower.
  2. Cross Pollination – transfer of pollen grains from the anther of one flower to the stigma of another flower but of the same species.

Agents of Pollination

Adaptations of Insect Pollinated Flowers (Entomophilous)

  1. Flowers are large, conspicuous with brightly coloured petals and inflorescence to attract insects.
  2. Flowers are scented and produce nectar to attract insects.
  1. They have small and firmly attached anthers to a firm filament.
  2. Stigmas are small, sticky and contained within the flower. This ensures that pollen grains from the body of an insect stick onto it.
  3. Flowers have a tubular or funnel shaped corolla, landing platforms and honey guides.

Adaptations of Wind Pollinated Flowers (Anemophilous)

e.g. maize and other grasses

  1. Small flowers with inconspicuous petals, bracts or inflorescence.
  2. Flower structure is simple and flowers have no particular shape.
  1. Pollen grains are small, smooth and light to be easily carried by the wind.
  2. Flowers are not scented and lack nectar.
  3. Anthers are large and loosely attached to a flexible filament to be easily released when the wind blows.

Diagram of a grass flower

 

 

 

 

 

 

 

 

 

Features and Mechanisms Hindering Self Pollination and Self Fertilization

  1. Heterostyly – condition whre the stigma na d style have different arrangements e.g. coconut flowers have shorter stamens than pistils hence pollen grains from the anthers cannot reach the stigma.

Diagram

  1. Self sterility or incompatibility – condition where pollen grains of a flower fail to germinate if they land on the stigma of the same flower.

Protandry – stamen mature before the stigma e.g.in sunflower.

Protogyny – stigma matures before the anthers mature e.g. in maize.

  1. Dioecious plants and presence of features that promote cross pollination such as brightly coloured petals which attract insects hence cross pollination.

Fertilization in Flowering Plants

Fertilization in plants is the fusion of the male and female nuclei in the embryo sac.

Diagram

Female gamete (egg cell) is found in ovules contained in the embryo sac.

 

 

Process of Fertilization

 

 

Seed and Fruit Development

Development of the Seed

Development of Fruits

Classification of Fruits

 

Succulent fruits

They are divided into berry and drupe.

Berry– has a succulent pericarp divided into epicarp, mesocarp and endocarp e.g. orange, tomato, passion fruit, melon, paw paw etc.

Diagram

Drupe – they have a thin epicarp, fleshy or fibrous mesocarp and a very hard endocarp enclosing the seeds. In mango the fleshy edible part is the mesocarp while in coconut the mesocarp is a fibrous cover just before the hard endocarp.

Diagram

Dry Fruits

Dry Dehiscent fruit

They dehisce to release their seeds. They are divided into;

  1. Legume e.g beans

 

Diagram

 

  1. Capsule e.g poppy

 

Diagram

 

  1. Schizocarp e.g. castor.

Diagram

Dry indehiscent fruits

  1. Caryopsis – pericarp and seed coat are fused together to form a thin covering round the seed e.g. maize.

Diagram

  1. Cypsela – it’s a one seeded e.g. the blackjack.

Diagram

  1. Nut – the pericarp becomes hard and woody and it is separate from the seed coat e.g. macadamia.

Diagram

Placentation

  1. Marginal Placentation.

Diagram

  1. Basal placentation

Diagram

  1. Axile Placentation

Diagram

  1. Parietal Placentation

Diagram

  1. Free central placentation

Diagram

Adaptations of Fruits to Various Agents of Dispersal

  1. Water dispersal
  1. Wind dispersal
  1. Animal dispersal
  1. d) Self dispersal

 

SEXUAL REPRODUCTION IN ANIMALS

External Fertilisation in Amphibians

Internal Fertilisation

Study Question 8

Reproduction in Mammals

Platypus

The duck-billed platypus, Ornithorhynchus anatinus, found only in eastern Australia, belongs to an unusual group of egg-laying mammals called monotremes. It lives in streams, rivers, and occasionally lakes. The duck-billed platypus feeds on bottom-dwelling aquatic insect larvae, which it finds by probing the streambed with its pliable, sensitive bill.

Mother Kangaroo and Baby

Kangaroos are a type of mammal called a marsupial. Baby marsupials are unable to survive on their own when they are born, so they must live in a pouch on their mother’s belly. A newborn kangaroo, called a joey, stays in its mother’s pouch for about six months, where it feeds on her milk.

Reproduction in Human beings

Structure and Function of The male Reproductive System

Male Reproductive System

The organs of the male reproductive system enable a man to have sexual intercourse and to fertilize female sex cells (eggs) with sperm. The gonads, called testicles, produce sperm. Sperm pass through a long duct called the vas deferens to the seminal vesicles, a pair of sacs that lies behind the bladder. These sacs produce seminal fluid, which mixes with sperm to produce semen. Semen leaves the seminal vesicles and travels through the prostate gland, which produces additional secretions that are added to semen. During male orgasm the penis ejaculates semen.

Internal View of Male Reproductive System

The reproductive anatomy of the male human is largely external. Beginning at puberty, sperm are produced within seminiferous tubules of the testicles, a pair of glands that reside in a pouch called the scrotum. The external location of the scrotum keeps the temperature of sperm slightly below body temperature, which is necessary for their healthy development and survival. From each testicle, sperm migrate to a long, coiled tube known as the epididymis, where they are stored for one to three weeks until they mature. Also located outside the body is the penis, the erectile organ responsible for the excretion of urine and the transfer of sperm to the vagina of the female. Just before ejaculation during sexual arousal, mature sperm travel from the epididymis, a coiled tube behind each testicle, through a long duct called the vas deferens. Sperm leave the body in semen, a fluid produced by the seminal vesicles.

Study question 9 and Practical.

Structure and Function of The Female Reproductive System.

Diagram

Practical

The Human Sperm

Diagram

Formation of The Ova

Diagram

Study Question 11

 

 

Fertilisation

Diagrams

Study Question 12

Implantation

Diagrams

Formation of Placenta

Diagram

The Role of The Placenta

Refer to the table below

Diagram

Study question 13

Major functions of oestrogen and progesterone during pregnancy

Oestrogen Progesterone.
    i.)          Growth of mammary glands

ii.)          Inhibits FSH release.

iii.)          Inhibits prolactin release.

iv.)          Prevent infection in uterus

v.)          Increase size of the uterine muscle cells.

vi.)          Increase ATP and creatine phosphate formation.

vii.)          Increases sensitivity of myometrium to oxytocin.

  i.)          Growth of mammary glands.

ii.)          Inhibits FSH release

iii.)          Inhibits prolactin release.

iv.)          Inhibits contraction of myometrium.

What is allowed to pass through the placenta What is not allowed to pass through the placenta  
From the mother to the foetus.

i.)          Oxygen

ii.)          Vitamins

iii.)          Mineral salts

iv.)          Hormones

v.)          Water

vi.)          Antibodies and antigens.

vii.)          Glucose, amino acids, fatty acids and glycerol.

From the foetus to the mother

i.)          Carbon (iv) oxide.

ii.)          Nitrogenous wastes.

  i.)          All blood cells.

ii.)          Plasma proteins.

iii.)          Most bacteria.

 

Gestation Period

 

 

Birth/Parturition

Diagrams

Caesarean delivery

Parental care

Milk production in various environments

Positive Environment Negative Environment
·      Sucking at the breast, smell of the baby or crying of the baby trigger milk let down.

·      Hypothalamus relays impulses to pituitary gland which releases hormone oxytocin

·      Oxytocin reaches the breasts and causes alveoli to contract forcing milk into the ducts.

·      Ducts conduct milk into the reservoirs behind the areola

·      Baby sucks the milk from this reservoir.

·      Milk let down may be inhibited or blocked if the breastfeeding mother experiences embarrassment, fatigue or anxiety.

Assignment

Child labour

Role of Hormones in Human Reproduction

Secondary sexual characteristics

These are physiological, structural and mental changes associated with masculinity and femininity.  They are controlled by oestrogen in females and androgens in males. They occur at puberty.

Secondary sexual characteristics in males

Secondary sexual characteristics in females

Menstrual Cycle

Menstruation

 

Assignment

Sanitary Health

Advantages of Asexual reproduction

Disadvantages

Advantages of sexual reproduction

Disadvantages

Revision Questions

 

BIOLOGY FORM FOUR NOTES

 

THE EYE

The mammalian eye is spherical; fluid filled and has three layers;

Adaptations of the eye

Part Properties Functions
1.   Sclera Tough, white opaque layer. ·      Protects delicate inner parts of the eye.

·      Maintain shape of the eye.

2.   Cornea Transparent and front layer of the sclera. ·      Allow light into the eye.

·      Refract light entering the eye

3.   Conjunctiva Thin and transparent layer before the cornea. ·      Allow light into the eye.

·      Protects the cornea.

4.   Choroid Have cells that have melanin, arteries & veins ·      Provide nourishment to the eye.

·      Pigment prevents reflection of light within the eyeball.

5.   Ciliary body Have thin rings of thickened tissue arising from choroids. ·      Produce the aqueous humour

·

6.   Ciliary muscle Have circular and radial muscles which are antagonistic. ·      Muscles alter the tension of suspensory ligaments.
7.   Suspensory ligaments. fibrous ·      Alter the shape of the lens.
8.   Pupil Hole at the center of iris. ·      Regulates the amount of light entering the eye.
9.   Iris Contain melanin, circular and radial muscles. ·      Give the eye its colour.

·      Regulates the amount of light entering the eye

10.                  Lens Transparent, biconvex and elastic structure.

Found behind pupil.

·      Refracts light onto retina.

·      Involved in accommodation of the eye.

·      Separates the aqueous humor from vitreous humor.

11.                  Aqueous/

Vitreous humor

Transparent fluids ·      allow light pass / refract light,

·      Hydrostatic pressure – maintain shape of eyeball.

·      Contains sugars / proteins / salts – provide nutrients to eye.

12.                  Retina Contains photoreceptors (Cones & Rods) and blood vessels. ·      Rods are sensitive to low light intensity and detect black and white and more in nocturnal animals. Have photochemical pigment called rhodopsin.

·      Cones are sensitive to high light intensity, detect colour and present in large numbers in diurnal animals. Have photochemical pigment called iodopsin.

13.                  Fovea centralis Contain mainly the cones. ·      It’s the most sensitive part of the retina.

·      Image is formed here.

14.                  Blind-spot Has no photoreceptors ·      Optic nerve leaves the retina.

·      Blood vessels emerge here to nourish the eye.

15.                  Optic nerve Made of nerve fibres ·      Transmits nerve impulses to the brain for interpretation.
16.                  Lachrymal gland Secrete tears ·      Tears moisten the cornea and washout foreign particles.

·      Tears have antiseptic effect.

17.                  Nictitating membrane Transparent membrane in birds, reptiles and fish. ·      Draw across the eye cleaning and protecting it.
18.                  External eye muscles Contractile ·      move eyeball within socket
19.                  others  -externally-   ·      orbit- protective

·      eye lids-protect the eye by closing

·      Eye lashes-prevent entry of small foreign particles.

·      Eye brows-prevent dust & sweat from entering eye.

Image formation and interpretation

Diagram  pages 100 KLB

Binocular vision or Stereoscopic Vision

Advantages

  1. Larger field of view
  2. Provide much accurate assessment of distance, height or depth of objects.
  3. Damage to one eye is compensated by the other
  4. Cancels the effect of blind spot

Accommodation of The Eye

Accommodation of a close object

Diagram

Accommodation of a distant object

 Diagram

 

Control of Light Entering the Eye

  1. Bright Light

 

Diagram

 

  1. Dim Light.

Diagram

Defects of the Eye

  1. Short sightedness(Myopia)

Correction

Diagram

  1. Long sightedness (Hypermetropia)

Correction

Diagram

  1. Astigmatism
  1. Squintedness
  1. Old sight (Presbyopia)
  2. Cataracts
  3. Colour blindness

 

THE EAR

Functions of the ear

  1. Hearing
  2. Maintaining body balance and posture

 

Diagram

Adaptations of the ear to its Functions

The ear is divided into:

  1. Outer ear
  1. Middle ear. Its air filled

 

 

 

 

 

  1. Inner ear. Its fluid filled

Diagram

 

Diagram

 

 

NB/ Auditory nerve – transmits nerve impulses to the brain for interpretations.

 

Deffects of the Ear

GENETICS

 

This is the study of inheritance and variation.

Terms used in genetics

  1. Inheritance; transmission of characteristics from the parents to the offsprings
  2. Variation; possession of characteristics different from those of the parents and other offsprings.
  3. DNA; De-oxyribonucleic acid
  4. RNA; Ribonucleic acid
  5. Monohybrid inheritance; inheritance of one characteristic controlled by one pair of hereditary factors e.g. Tallness
  6. Dihybrid inheritance; inheritance of two characteristics at the same time e.g. colour and Texture/shape in the garden pea plant
  7. Dominance; ability of a trait to only express itself
  8. Recessiveness; a trait that only expresses itself when in homozygous state.
  9. Heterozygosity; presence of two dissimilar members of an allele e.g. Rr, Tt etc.
  10. Homozygosity; presence of two similar of an allele e.g. TT, RR, tt, rr etc.
  11. Allele; one pair of genes which occupy corresponding loci/positions in homologous chromosomes
  12. Phenotype– the physical appearance of an individual or organism. It’s influenced by the genotype and environment.
  13. Genotype– the genetic constitution of an organism. Its purely genetical.
  14. F1 generation (first filial generation) – are the offsprings that represent the first generation of organisms or individuals under study.
  15. F2 generation(second filial generation) – these are offsprings obtained after self crossing the F1 gen.
  16. Incomplete dominance/ co-dominance– a condition where no allele is dominant over the other. The phenotype of the offspring is intermediate between that of the parents.
  17. Multiple allelism– are characteristics determined by more than two variant forms of a single gene e.g. inheritance of the Blood groups in man (ABO)
  18. Test cross / back cross– it’s a crossing involving a homozygous recessive to determine the genotype of an organism.
  19. Mutation- these are spontaneous changes in the individual’s genetic makeup.

Concepts of Variation

Variation refers to observable differences among living organisms.

Types of variation

  1. Discontinuous variation – in this type of variation, there are distinct and definite groups of individuals with no intermediate forms. E.g.
    • sex either male or female,
    • blood groups- one can only belong to one of the four blood groups A,B,AB,O
    • ability to role the tongue
    • Presence of long hair in the nose and on the ear pinna.
    • Presence of a free or attached ear lobe.

All these traits are controlled by one or two major genes. These traits are not influenced by the environment

  1. Continuous variation – this variation has a wide range of differences for the same characteristic from one extreme to the other e.g.
    • Height
    • Skin colour/pigmentation
    • Weight
    • Length of internodes
    • Number of leaves, fruits on a tree etc
    • Finger prints

When these traits are plotted on a graph a normal distribution curve is obtained.

 

 

 

 

 

 

 

 

 

A normal distribution curve of heights

This type of variation is brought about by the interaction of both the genetic environmental factors.

E.g. a plant with genes for tallness may fail to grow tall due to climate and poor soils.

Practical Activity 1

Causes of variation

  1. Gamete formation – during gamete formation two processes contribute to variation. These are
    • Independent assortment– during metaphase I of meiotic division, homologous chromosomes come together in pairs and segregate into daughter cells independently of each other. This independent assortment produces a variety of different gametes. The total number of combinations is given by 2n where ‘n’ is the haploid number of chromosomes. In man n = 23 hence 2n = 223 which is equal to 8,388,608.
    • Crossing over during the prophase I of meiotic cell division. I.e. when homologous chromosomes break and rejoin at certain points called
  2. Fertilization – during fertilization parental genes ca come together in different combinations. Therefore desirable and undesirable qualities of parents can be combined in the offsprings.
  3. Mutation– these are spontaneous changes in the genetic makeup of an organism. Mutation brings about changes in the living organisms.

The Chromosome

Each chromosome is made up of two parallel strands called chromatids. Each pair of chromatids is connected at a point by the centromere. Chromosomes contain the hereditary material called the genes. All cells including the sperms and ova have chromosomes.  Chromosomes are present in the nucleus and are only visible under the microscope during cell division – mitosis and meiosis. There is a definite number of chromosomes in each cell for every species of animal or plant.

E.g. in man somatic cells (body cells) have 46 chromosomes while the sex cells/gametes have 23 chromosomes. During fertilization fusion of the sperm- 23 chromosomes and the ova -23 chromosomes restores the 46 chromosomes to form a diploid zygote.

Chromosomal numbers in different organisms

Organism Number of chromosomes
Somatic cells – 2n Gametes – n
Sheep (Ovis auries) 56 28
Cow (Bos Taurus) 60 30
Fruit-fly (Drosophila melanogaster) 8 4
Maize (Zea mays) 20 10
Wheat(Triticum vulgare) 14 7
Tobacco (Nicotiana tabacum) 12 6
Man ( Homo sapiens) 46 23

 

Practical Activity 2

GENES AND DNA

Genes occupy specific positions on the chromosomes called the gene loci( gene locus)

The gene is a chemical in nature. The genes are in the form of a nucleic acid molecule called De-oxyribonucleic acid (DNA). In 1953 two Biologists Francis Crick and James Watson worked out the structure of the DNA. DNA was found to be composed of three different components;

  1. A five carbon sugar-pentose
  2. phosphate molecule
  3. nitrogenous base

There are four types of the nitrogenous base;

A combination the pentose sugar, a phosphate molecule and a nitrogenous base forms a nucleotide.

I.e.

 

 

 

DNA structure contains several nucleotides fused together to form long chains called DNA strands. Two parallel strands twist on one another forming a double helix structure. Adenine always combines with Thymine and Cytosine with Guanine.

Diagram

Role of DNA

DNA REPLICATION

Diagram

Role of the DNA in protein synthesis

The sequence of bases along the DNA strand acts as the alphabet and determines the sequence of amino acids when they join to form a polypeptide chain. Protein synthesis takes place in ribosome’s found in the cytoplasm. Since the DNA molecules are confined in the nucleus, there has to be a way of communicating the DNA information to the ribosome’s where actual protein synthesis occurs.

The cell therefore has a special molecule called the Ribonucleic acid (RNA). Its role is to carry genetic information from the DNA to the site of protein synthesis in the cytoplasm. It’s referred to as messenger RNA (mRNA). RNA is formed from the DNA strands.

During formation of the mRNA a section of the DNA strands acts as the template strand. The double helix of the DNA unzips and free nucleotides align themselves opposite the template. The base sequence of the template strand is copied onto a new strand.

In RNA, Thymine is replaced by another base called Uracil (U)

The transfer of DNA sequence on the mRNA strand is referred to as Transcription.

Diagram

After its formation the mRNA leaves the nucleus with instructions from the DNA about the kind of protein to be synthesised by the cell. This information is in the form of base triplets known as Codons which code for a particular amino acid of a protein molecule e.g.

Differences between DNA and RNA
DNA RNA
1.   Has De-ox ribose sugar Has ribose sugar
2.   Double stranded Single stranded
3.   Confined in the nucleus Found in nucleus and cytoplasm
4.   Have organic bases as cytosine, guanine, adenine and thymine. Has organic bases as cytosine, guanine, adenine and uracil

 

THE FIRST LAW OF HEREDITY

MENDEL’S EXPERIMENTS

An Austrian monk known as Mendel is considered to be the father of genetics. He carried out various breeding experiments and observed variations in different characteristics of the garden pea. The characteristics include:

He selected a group of dwarf plants and self pollinated them by dusting mature pollen grains onto the stigmas of the same plant. He collected the resulting seeds and planted them. He noted that these seeds grew into dwarf plants only. He repeated the experiment for several generations and obtained the same results.

In another experiment, Mendel selected tall plants and self-pollinated them. He planted the resulting seeds and observed that they grew into a mixture of tall and dwarf plants. He took seeds from the tall offsprings only and repeated the experiment for many generations until he obtained only tall plants.

This way he was able to obtain a pure line of tall garden peas and a pure line of dwarf garden peas.

He then cross-pollinated pure bred tall garden pea s with the pure bred dwarf variety. He planted the resulting seeds and he observed that all the offsprings were tall plants.

Diagram

He further crossed two of these tall offsprings and planted the resulting seeds. Mendel observed that this second generation consisted of a mixture of tall and dwarf plants. After counting these plants he noted that the ratio of tall to dwarf plants was approximately 3:1 respectively. He observed that this ratio was always obtained when crosses were made between the non-pure breeds of tall plants.

Diagrams

Mendel concluded that the traits of an organism are determined by hereditary factors which occur in pairs. Only one of pair of such factors can be represented in a single gamete. This later became Mendel’s First Law, The Law of Segregation

At this time Mendel had no idea of genes and so he called them factors. He postulated that these factors are found on the chromosomes and are passed from the parents to the offsprings via gametes.

Reasons behind Mendel’s success

  1. He used favourable materials i.e. the garden pea plant which is normally self pollinated. This made it easy for him to employ cross pollination at will.
  2. the pea plant he used had several contrasting traits
  3. His study was focused on particular traits while those before him had been attempting to determine wholesome heredity of each organism.
  4. He kept accurate data on all his experiments and fro the analysis of this data he was able to formulate definite hypothesis.

Mendel chose the garden pea plant because of the following reasons

Monohybrid Inheritance

This is the inheritance of one trait like height in the garden pea plant that is controlled by a single pair of hereditary factors (genes) contributed by both parents. Genes occur in pairs on chromosomes and such gene pairs are known as alleles.

The genetic constitution of an organism is called the Genotype while the physical appearance is known as the Phenotype.

The genotype of an organism is represented using paired letter symbols. Capital letters represents the dominant gene while small letters represent the recessive gene.

Genetic Cross

Components of a genetic cross
  1. The conventional symbol for male is and that of female is

Example 1

During gamete formation in the dwarf plat, each gene in the pair segregates into different gametes. When the female and male fuse during fertilization, the offspring produced contain the same number of genes as in each parent. The inheritance of dwarf ness in the pea plant can be illustrated diagrammatically by the following genetic cross

Diagram

Example 2

Similarly the pair of genes in the pure breed tall plants will segregate into different gametes during gametogenesis. When self fertilised the resulting seeds will have half the number of genes from each parent i.e.

Diagram

Example 3

When the purebred tall plant is crossed with dwarf plants, the resulting seeds grow into tall plants only. These offsprings represent the first generation (F1 gen)

Diagram.

In the genetic cross above, the male plant is tall and the female plant is dwarf. If the cross is reversed so that the female is tall and the male a dwarf, this is referred to as a reciprocal cross. The F1 results will be the same for either cross.

Diagram

Example 4

When the F1 offsprings are self pollinated, they produce offsprings which that grow into a mixture of tall and dwarf plants. These offsprings are known as the F2 gen.

Diagram

A Punnet Square can also be used to work out genetic crosses e.g.

Parental phenotype   tall          tall

Parental genotype     Tt     X   Tt

Gametes                T t            T  t

 

             ♀

T t
T TT Tt
t Tt Tt

When the allelic genes are identical, as in TT and tt, the condition is known as homozygous. An individual

with such a condition is known as a homozygote.

When the allelic genes are not identical as in Tt, the condition is referred to as heterozygous. An individual with such a genotype is referred to as a heterozygote. An individual with genotype Tt, will be physically tall because the gene T is dominant over t.  The allele t is recessive.

A dominant gene expresses itself in both the homozygous (TT) and heterozygous (Tt) states while a recessive gene only expresses itself I its homozygous state (tt).TT is therefore referred to as homozygous dominant and tt is homozygous recessive.

The ratio 3 tall: 1 dwarf, in the F2 gen is characteristic of monohybrid inheritance where one gene is completely dominant over the other. This is referred to as complete dominance.

The monohybrid crosses are based on Mendel’s first law, The law of Segregation which states the characteristics of an organism are determined by internal factors which occurs in pairs. Only one of a pair of such factors can be represented in a single gamete.

Diagram – the process of segregation

Ratios and Probability

The 3:1 ratio in monohybrid inheritance can be represented in the form of probability. When a large number of heterozygous garden pea plants are selfed, the probability of getting tall plants is ¾ or 75% and that of dwarf will be ¼ or 25%.

Diagram

  1. The inheritance of characteristics involves probability. The chance that a particular gamete will fuse with another is a random occurrence, in genetics this done by showing all possible fusions.

Practical Activity 3 and 4

Similar monohybrid inheritance results as those of Mendel have been obtained by using the fruit fly (Drosophila melanogaster) the insect has many observable characteristics that are contrasting such as,

Assignment

Using appropriate letters work out the following crosses with respect to the fruit fly

The fruit fly is suitable for genetic study because of the following reasons.

  1. The female lays very many eggs hence increasing the sample size for study.
  2. Have many observable characteristics that are distinct and contrasting.
  3. It is easily bred in the laboratory with minimum requirements.
  4. It has a short generation time 10-14 days. Therefore many generations can be studied in a short period of time.
  5. Offsprings can be crossed with their parents at will (backcrossing)
  6. Flies are safe to handle because they do not transmit any known human diseases.
Study Question 10

Practical Activity 5

Incomplete Dominance (Co-dominance)

In Mendel’s experiments with garden pea p[plants, the genes determining  the various traits were clearly dominant or recessive. However in some species, alleles determining several contrasting traits do not have a clear cut dominant-recessive relationship. This implies that neither of the alleles is completely dominant over the other.

Heterozygous individuals are phenotypically different from either of the parents. Mostly the phenotype of the heterozygous offspring is intermediate between that of the parents. This phenomenon is called Incomplete Dominance.  Examples of incomplete dominance.

  1. Inheritance of flower colour in the 4 o’clock plant (Mirabilis Jalapa). If a true breeding plant producing red flowers is crossed with a true breeding plant producing white flowers, all the F1 offsprings will have pink flowers. When the F1 plants are self pollinated, they yield red flowered, pink flowered and white flowered offspring at a ratio of 1:2:1 respectively.

Diagrams

  1. Incomplete dominance in short horn cattle. Mating red and white shorthorn cattle yields Roan light Red) calves due to presence of both red and white hairs. A mating between two roan coloured shorthorns yields a mixture of red, roan and white coloured calves at a ratio of 1:2:1 respectively.
Diagrams
Study Question 12

Inheritance of the Blood groups (Multiple allelism)

In all the kinds of inheritance discussed so far, each phenotypic characteristic is determined by 2 variant forms of a single gene located at a specific locus on the homologous chromosome. However some characteristics are determined by more than two variant forms of a single gene. This phenomenon is referred to as multiple allelism and the genes involved are called multiple alleles. E.g. in the ABO blood groups in humans, there are three genes involved and they are responsible for the presence of antigen types on the red blood cells.

These are gene A responsible for the presence of antigen A, gene B for antigen B and gene O responsible for absence of antigens on the red blood cells.

Genes A and B have equal degree of dominance i.e. are co-dominant. They both express themselves when present together as in the blood group AB.

Genes A and B are dominant over gene O. Gene O is recessive and only expresses itself in the homozygous condition. The genotypes for the four blood groups in the ABO system are therefore,

Blood group

(Phenotype)

Genotype Antigens
A AA, AO A
B BB,BO B
AB AB A and B
O OO O- Zero

i.e.

 

A marriage between a man of blood group A and a woman of blood group B will produce children of all the four blood groups if both parents are heterozygous.

Diagram

Marriage between a man of genotype AA (blood group A) and Woman of genotype BB (blood group B) results in all the offsprings having blood AB.

Diagram.

Assignment

Work out the following crosses
Study Question 13

Inheritance of the Rhesus factor

In man the possession of Rhesus antigens makes one Rh+ and this is dominant over Rh–ve. If blood from a Rhesus positive person is transfused into a rhesus negative person, this induces antibodies against the Rhesus factor of the donor. This causes agglutination of red blood cells of the recipient.

If a Rh-ve woman is married to a Rhe+ve, when she becomes pregnant, the child will be Rh+ve. Rhesus antigens cross the placenta into the mother’s blood stream. This stimulates the mother’s immune system to produce Rhesus antibodies. When these antibodies get into the foetal circulation, an antigen-antibody reaction takes place and the red blood cells of the foetus are destroyed (Haemolysed).

During the second pregnancy, the amount of Rhesus antibodies are more and cause a lot of damage to the foetus’s red blood cells resulting to death. This is called Haemolytic Disease of the Newborn or Erythroblastosis foetalis.

Determining Unknown Genotypes

This can be done in two ways.

  1. Carrying out a Test Cross

A test cross is a cross between an individual of unknown genotype with an individual of a recessive genotype. A test cross where an offspring is crossed with one of its parents is called a Back Cross.

In garden pea plants the gene that determines red flowers is dominant over that which determines white flowers. A plant with red flowers may either be homozygous (RR) or heterozygous (Rr) for this characteristic. To establish its correct genotype it is crossed with a homozygous recessive plant i.e. a white flowered one (rr)

If all their offsprings bear red flowers then this indicates that the red flowered plant is homozygous or it’s from a pure line.

Diagram

If the offsprings bear a mixture of red and white flowers in the ratio of 1:1, this indicates that the red flowered plant was heterozygous.

Diagram

Selfing

Unknown genotypes can also be determined by carrying out selfing experiments. For example, a phenotypically tall plant is either homozygous (TT) or heterozygous (Tt) for this trait.

If selfed and all its offsprings are tall, the parental genotype is TT that is homozygous dominant.

Diagram

But if after selfing both tall and dwarf offsprings are produced in the ratio 3:1 respectively, then the parental genotype is heterozygous (Tt).

Diagram

Sex Determination

The sex of an organism is a genetically determined characteristic. Cells of most organisms contain a pair of chromosomes called sex chromosomes in addition to the ordinary chromosomes. In man there are 46 chromosomes (23 pairs of homologous chromosomes in everybody cell). The genes determining whether a child becomes a female or a male are located on the specific pair of sex chromosomes called the X and the Y named after their shapes.

The remaining 22 pairs of chromosomes are called Autosomes.  Autosomes are responsible for other inheritable traits.

A male human being carries the XY chromosome i.e. he is Heterogametic

The female carries the XX chromosomes i.e. Homogametic.

After meiosis in a male the spermatozoon can either carry the X or Y chromosome while the female ova contain only the X chromosome. The sex of a child is a matter of chance and depends only on whether a spermatozoon that fertilizes the ovum carries X or Y chromosome.

There is therefore a 50% chance that fertilization can result in either XY (Boy) or XX (Girl) i.e.

       Female

(XX)

Male (XY)

X X
X XX XX
Y XY XY

I.e. 2 girls: 2 Boys

In terms of probability, the chance that a boy or a girl is produced in a family is ½.

NB/ in birds the female is XY – heterogametic and the male is XX – homogametic.

In some insects, the female is XX and the male is XO with the Y chromosome absent.

In the fruit fly (Drosophila melanogaster) sex determination is as exactly as in man, i.e. male XY and Female XX.

Linkage

An organism has a large number of traits controlled by many different genes. Because the number of chromosomes is limited, each gene cannot be located on its own chromosome. Therefore chromosomes must accommodate many genes each controlling particular characteristics. Those genes located on the same chromosome are called linked Genes. All the linked genes constitute a linkage group. Linked gene are inherited together and do not segregate/separate during meiosis. They are therefore transmitted into the same gamete.

Diagram

If genes Q, R and T are linked, then all the three pairs of genes are accommodated on a homologous pair of chromosome.

In Drosophila sp, it has been found that the genes for wing length, abdomen size and body colour are located on the same chromosome. Therefore these characteristics are usually inherited together.

Sex-linked Genes

All the genes located on the sex chromosomes are said to be sex-linked. They are therefore transmitted together with those that determine the sex. In Drosophila melanogaster, the gene, which determine eye colour, is located on the X chromosome. However the corresponding allele on the Y chromosome is absent. This is because most sex-lined genes are carried on the X chromosome whereas the Y chromosome carries very few genes and is almost empty.

In humans there are few genes located on the Y chromosome, which control traits that are exclusively found in males. These are, Premature baldness and tufts of hair in the in the inner pinna and in the nose.

Diagrams.

The characteristics controlled by genes located on the X chromosome include Colour blindness and Haemophilia. These characteristics can arise in either male or females.

Colour blindness

This is the inability to distinguish Red and Green colours by some people. This trait is linked to the X chromosome. The gene that determines normal colour vision is dominant over that for colour blindness.  A marriage between a colour-blind man and a woman homozygous for normal colour vision results in their daughters being carriers but with normal colour vision. The daughters are said to be carriers because they are heterozygous and colour blindness is suppressed/masked by the dominant gene for colour vision.

All the sons are of the two parents are however normal. This is illustrated below. Let N represent the gene for normal colour vision and n represent gene for colour blindness. Since the gene is linked to X chromosome, its alleles are represented as XN and Xn.

        Colour blind male

( XnY)

Normal

Woman

(XNXN)

 

 

Xn

 

 

Y

 

XN

 

XNXn

 

XNY

 

XN

 

XNXn

 

XNY

 

All the daughters are carriers- XNXn

All the sons have normal colour vision-XNY

If a carrier daughter from the above parents married a normal man, some of their sons will suffer from colour blindness while the daughters will either be carriers or homozygous for normal colour vision as shown below.

           Carrier female

(XNXn)

Normal male

(XNY)

 

XN

 

Xn

 

XN

 

XNXN

 

XNXn

 

Y

 

XNY

 

XnY

Offsprings are; XNXN              -Daughter with normal colour vision

XNXn                 -Carrier Daughter

XNY              -Son with normal colour vision

XnY               -Colour blind son.

The above examples show that the gene for colour blindness is passed from mother to sons.

This is because the only X chromosome a man inherits is from the mother. If the X chromosome carries the gene for the trait, then this gene will be expressed since allele on the Y is absent. Therefore there are more male sufferers in a population compared to females.

Females only suffer when in homozygous condition of the recessive gene. Inheritance of colour blindness through several generations can be clearly illustrated using a pedigree. A pedigree is a record in table form showing the distribution of one or more traits in different generations of related individuals. Fig. 1.24

Haemophilia

This is another sex-linked trait where the blood of the sufferer takes abnormally long time to clot. There is prolonged breeding in the event of a cut hence the term Bleeder’s Disease. A recessive gene on the X chromosome causes haemophilia.

If a normal man is married to a carrier woman for haemophilia, there is a probability of ½ that if their child is a boy will be a haemophiliac and if a daughter, she will be a carrier. None of the daughters of the couple will be haemophiliacs.

Let H represent the gene for normal blood clotting and h to represent gene for haemophilia i.e.

       Carrier woman

(XHXh)

Normal man

(XHY)

 

XH

 

Xh

 

XH

 

XHXH

 

XHXh

 

Y

 

XHY

 

XhY

Their offsprings will be;      -XHXH -Normal daughter

XHXh   -Carrier daughter

XHY    -Normal son

XhY     -Haemophiliac son.

Study question 14

Apart from carrying the sex-linked traits, the X chromosome in the females and the Y in males bring about the development of both the primary and secondary sexual characteristics. At puberty, secondary sexual characteristics in females include breast enlargement, widening of the hips, and growth of pubic hair and onset of menstrual cycle. The X chromosome controls these.

In males, they include growth of pubic hair and beard, deepening of the voice, widening of the shoulders etc.

Effects of Crossing Over on Linked Genes

Some of the linked genes separate and are transmitted on different chromosomes. This happens during crossing over (prophase I of meiosis) when sections of chromatids of a bivalent intertwine and may break off. Some of these sections get rejoined to different chromatids thus separating genes that were previously linked. The fusion of such gametes containing chromatids whose genes have changed places produces new combinations (recombinants). Crossing over results in chromosomal mutations, which in turn cause variations.

Diagrams

Mutations

Mutation is brought about by spontaneous changes in the individual’s genetic makeup. Mutations are normally due to recessive genes most of which are transmitted in the usual Mendelian fashion. Therefore they are quite rare. Individuals with mutations are referred to as mutants. Mutation can be induced by certain factors. Such factors are called Mutagens. They include,

NB: Mutations occurring in gametes are more important than those in somatic cells. Mutational changes are the basis of discontinuous variation in population.

Types of Mutations

  1. Chromosomal mutation
  2. Gene mutation
Chromosomal Mutations

This involves the change in the structure or the number of chromosomes. During crossing over in meiosis homologous chromosomes intertwine at points called chiasmata. These points are later broken creating various opportunities for changes on the chromatids. There are five types of chromosome mutations (chromosome aberrations).

Deletion

This occurs when some sections of chromatids break off and fail to recombine. They are therefore completely lost and the genetic material they contain is said to be deleted out. Most deletions are lethal since the offspring may lose genes responsible for the synthesis of some vital protein molecules.

Diagrams

Duplication

In this case a section of chromatids replicates and adds an extra length to itself. Duplication can produce serious effects depending on the chromosome sections involved.

Diagrams

Inversion

In this case a chromatid breaks at two points. When rejoining, the middle piece rotates and joins in an inverted position. This reverses the gene sequence along the chromatid. This might bring together genes whose combined effects are advantageous or dis-advantageous.

Diagrams

Translocation

This occurs when a section of one chromatid breaks off and becomes attached to another chromatid but of a non-homologous pair.

Translocation therefore involves the movement of genes from one non-homologous chromosome to another.

Diagrams

Non-disjunction

This leads to addition or loss of one or more whole chromosomes. If it occurs at anaphase of the first meiotic division, two homologous chromosomes fail to segregate and they move into the same gamete cell. If it happens at anaphase of the second meiotic division, sister chromatids fail to segregate. This results in half the gametes containing two of the same chromosome while the others have none.

Diagrams
Non-Disjunction causes the following
  1. Downs’s Syndrome: this is where there is an extra somatic chromosome number 21. such individuals have;

NB/ these conditions are common among children born of mothers above 40 years old.

  1. Klinefelter’s Syndrome: in this case individuals have an extra sex Such individuals have a total of 47 chromosomes in their cells i.e. XXY (male) and XXX (female). This occurs as a result of non-disjunction during spermatogenesis or oogenesis. The symptoms of Klinefelter’s syndrome are

 

Diagrams
  1. Turner’s syndrome: This is where an individual lacks one sex chromosome hence there are 45 chromosomes (XO or YO).
  2. Polyploidy: sometimes during meiosis chromosomes might undergo non-disjunction. This results in half the number of gametes having two of each type of chromosome i.e. diploid the rest having none. If the resulting diploid gamete fuses with a normal haploid gamete a triploid zygote is formed. If two diploid gametes fuse, a tetraploid individual is obtained. This is what is called polyploidy.

Polyploidy is rare in animals but common in plants where it’s considered to be advantageous. Polyploidy increases yields, early maturity and resistance to pests and diseases. It can be artificially induced using a chemical called colchicine, which prevents spindle formation during mitosis leading to a cell with double the number of chromosomes (4n).

Gene Mutation

This involves a change in the structure of a gene. Gene mutations are also referred to as point mutations. A gene mutation arises as a result of a change in the chemical nature of the gene. The change may involve some alterations in the DNA molecule. A change in the DNA molecule is passed onto the m-RNA. This alters the sequence of amino acids during protein synthesis. This may result in unintended protein molecules being synthesised, which may be lethal. Types of gene mutations;

Insertion

This is the addition of an extra base onto the existing DNA strand.

Diagram

By this insertion no polypeptide chain is formed as it were intended.

Deletion

This is the removal of a gene portion. If the base Thymine is deleted from its position as indicated below, the base sequence becomes altered at this point.

This results in the wrong proteins being synthesised.

 

Diagrams

Substitution

This is the replacement of a portion of the gene with a new portion. If Adenine is substituted by Guanine on a DNA strand, the base sequence is altered at this particular portion.

Diagrams

Inversion

If a portion of the DNA strand is rotated through 1800 that portion is said to be inverted as shown below. This alters the base sequence at this point.

Diagrams

Study Questions

Disorders Due to Gene Mutations

Such disorders include albinism, sickle cell anaemia, haemophilia, colour blindness and chondrodystrophic dwarfism.

  1. Albinism

This a condition where the synthesis of skin pigment called melanin fails. The victim has a light skin, white hair and pink eyes. Such a person is referred to as an Albino. Melanin is derived from two amino acids – Phenylalanine and Tyrosine. Melanin is synthesised through a series of reactions controlled by a specific gene.

Gene ‘A’ is responsible for presence of melanin and ‘a’ is responsible for its absence. Gene ‘aa’ in homozygous state blocks in one or two places in the synthesis of melanin hence no melanin is formed. This occurs as a result of one enzyme (Tyronase) failing to be formed in the presence of the recessive gene.

A person with genotype AA has normal skin pigmentation.

One with genotype Aa is a carrier and has normal skin pigmentation. In a family an albino can be born under three conditions only.

Assignment

  1. Sickle Cell Anaemia

This is a gene mutation as a result of substitution. Normal haemoglobin Hb A consists of two polypeptide chains. In the sickle cell condition, one amino acid called glutamic acid is substituted by another amino acid called valine in each of the two-polypeptide chains of the haemoglobin molecule. The resulting haemoglobin is known as Haemoglobin S – Hb S and is different from the Hb A in several ways.

Comparison between Hb A and Hb S

Normal haemoglobin (Hb A) Defective Haemoglobin (Hb S)
1. A position in each polypeptide chain is occupied

by glutamic acid

The same position is occupied by valine in each polypeptide chain
2. Does not easily crystallise in low oxygen

concentration

Easily crystallises in low oxygen concentration
3. The haemoglobin is efficient in oxygen loading

and transportation

Not efficient in oxygen loading and transportation
4. The red blood cells are biconcave in shape Red blood cells are sickle shaped (crescent shape)

Sickle cell anaemia is therefore the condition where the victim is homozygous for the defective gene that directs the synthesis of haemoglobin S. Most of the victims’ red blood cells are sickle shaped and the person frequently experiences oxygen shortage to the body tissues. Such a person cannot carry out strenuous physical exercises.

Many sickle cell victims die young due to insufficient oxygen supply to body tissues.

In the heterozygous condition, less than half the number of the red blood cells is sickle shaped. The rest are normal and efficient in oxygen transport. This is referred to as sickle cell trait. An individual with the sickle cell trait experiences a mild case of anaemia but leads a normal life.

Inheritance of Sickle Cell Anaemia

If a man with sickle cell trait marries a normal woman, the probability that any of the offspring will carry the sickle cell trait is ½.

If both parents are carriers the probability of getting an offspring with sickle cell anaemia is ¼.

  1. Haemophilia: This condition where the blood takes abnormally long time to clot. A haemophilic gene that prevents the production of the clotting factors causes the condition.
  2. Colour blindness: There are different forms of colour blindness. The most common one is the red-green colour blindness. In this case an individual is unable to distinguish between red and green colours.
  3. Most disadvantageous genes are recessive. Very few are dominant e.g. the gene for chondrodystrophic dwarfism

Study Question 16

Effect of Environment on Heredity

The genotype and the environment influence the development of an individual. In animals genetically identical individuals reared under different environments will appear different than those reared under very different conditions. Consider identical twins.

Practical Applications of Genetics

Man chooses those plants and animals with the desirable qualities. This is referred to as artificial selection. Inbreeding or crossbreeding does this. Inbreeding however increases the chances of undesirable genes whereas crossbreeding increases heterozygosity with the offspring’s having better performance than both parents. This is referred to as hybrid vigour e.g. a cross between Boran and Hereford.

Polyploidy has also been used in planting. The original wheat had a diploid number of 14 chromosomes but the commercial wheat has either 28 or 42 (tetraploid-4n or hexaploid-6n).

Examples of characteristics, which have been selected in agriculture.

  1. Resistance to diseases e.g. cassava resistant to cassava mosaic, coffee variety resistant to CBD.
  2. Early maturity in animals and plants.
  3. Adaptations to various conditions e.g. rainfall, temperature etc.
  4. Ease of harvesting e.g. in coffee and bananas where dwarf varieties have been developed
  5. Increased productive season e.g. in chicken
  6. Higher productivity
  7. Production of flowers such as roses for their colour and aroma.

 

2) Blood Transfusion

Before blood is given to a recipient, blood typing is first done. This is done to ensure compatibility between the donor and the recipient.

Blood typing also can be used to solve disputed parentage. However the most recent technique in establishing parentage is the DNA matching.

3)    Genetic Counselling

This is the provision of information and advice on genetically inherited disorders to individuals. The individual is given such advice to enable him or her make the best choice.

Examples of disorders for which genetic counselling may be done include
In order to confirm the disorder the doctors can do the following

4)    Genetic Engineering

This deals with identification of a desirable gene, altering, isolating and transferring it from one living organism to another.

 

STIMULUS AND RESPONSE REVISION QUESTI-ONS

  1. The diagrams below repents a nerve cell
  2. Identify the nerve cell.            (1mark)

(b)  (i)   Give a reason for your answer in (a) above                                    (1mark)

(ii) Show by use of an arrow the direction of flow of the nerve impulses.  (1mark)

  1. Below is a diagram showing parts of a synapse observe and other the questions that follow.
    • Name the parts labeled: A, B.        (2mks)
    • What is the role of part labeled C. (1mk)
  1. A student set up an experiment as shown in the diagram below.

The set up was left for 4 days.

  1. a) What was the aim of the experiment. ( 1mk)
  2. b) i) State the expected results after 4 days.                                                                   ( 1mk)
  3. ii) Account for the results you have stated in ( b) (i) above.                               ( 4mks)
  4. c) In another experiment, a student placed a seedling horizontally on moist cotton wool. Later the shoot grew upwards while the Radicle grew downwards. Explain why the radicle showed a downward curvature.                                ( 2mks )
  5. Describe how different types of tropisms adapt plants for survival in their habitats. (20mks)
  6. Diagram below shows the structure of motor neuron.

(a)   Name the parts labelled.     A, B                                                                                     (2mks)

(b)   State three adaptations that enable the neurone to carry out its functions efficiently.    (3mks)

(c)  State two features that would distinguish sensory neurone from the above neurone.       (2mks)

  1. The diagram below shows the structure of its human ear.
  2. a) State the functions of the ear.                                                 (2mks)
  3. b) Give the names of the structure labelled C,G and F.                        (3mks)
  4. c) (i) What is the function of the structure labeled H?                         (1mk)

(ii)  Name the structure in the ear that detects sound waves.                            (1mk)

  1. d) In which structure of the ear is the velocity of the sound waves fastest? (1mk)
  2. The diagram below represents a nerve cell. Study it and answer the questions that follow.
  3. a) (i)  Identify the cell                                                         (1mk)

(ii)  Give a reason for your answer in a (i) above                                 (1mk)

  1. b) Name the parts labelled N, P, Q and R.                        (4mks)
  2. c) State the functions of the parts labelled N and Q.                               (2mks)
  3. Use the diagram below to answer the questions that follow
  4. a) (i) Name the eye defect represented above                        (1 mk)

(ii) What is the cause of this defect                                              (1 mk)

(iii)How can the defect you have named (a) (i) be corrected?                      (1 mk)

9      The diagram below shows three different types of neurones along a reflex   arc

 

  1. a) Identify the neuron labelled 1 ,2 and 3 (3 mks)
  2. b) Using arrows show the direction of impulse transmission on the diagram (1 mk)
  3. c) Name the part of the spinal cord where the cell bodies of neurone 2 and 3 are located (1mk)
  4. d) Describe the transmission impulse across the part labelled P (3 mks)

10    A response exhibited by a certain plant tendril is illustrated below

11    Describe how the mammalian ear is adapted to perform its functions.               (20mks)

  1. a) Describe how the structure of the eye is adapted to its function.             (16mks)
  2. b) Identify two defects of the eye and how they can be rectified.             (4mks)

13    Differentiate between nervous and endocrine communication in animals              (3mks)

  1. The diagram below represents a section through the mammalian ear. Study it and answer the questions that follow.

(a)   Name the structures labeled H and J                                            (2mks)

(b)   State how the structures labeled H, M and N are adapted to their functions       (3mks)

(c)   State what would happen if the structure labeled K was completely damage (1mk)

(d)   Name the fluid contained in structure N                               (1mk)

(e)   Apart from hearing, state the other role performed by the human ear    (1mk)

15    The following experiment was set up in a chamber made from two connected Petri dishes. Housefly maggots were introduced at the centre of the chamber, so the maggots could move to either Petri dish A or B as shown below.

(a) Name the type of response being investigated in the set up.                            (1mk)

(b) State the survival value of the response named in (a) above.                           (1mk)

(c) Give the role of calcium chloride in the experiment above.                               (1mk)

16    (a) What is accommodation?                                                                                   (1mk)

(b) Describe the sequence of events that occur in the eye for one to be able to see clearly

(i) a distant object                                                                                                   (4mks)

(ii) if one moved from a dim lit room to bright light.                                       (3mks)

 

MARKING SCHEME STIMULUS AND RESPONSE REVISION QUESTIONS

1 (a) Motor neuron/Motor nerve cell;

(b) (i) cell body is terminally situated/located at the end of the axon;

(ii) Arrow should point away from the cell body

  1. (a) A-Synaptic cleft

B- Mitochondria

  1. a) Show the effect of unilateral light on growth of seedling / plants;             (1mk)
  2. b) (i)    Curvature of the tip of the shoot  toward the source light;

(ii)   Auxins / IAA/ Growth harmones; produced by the apical bud; move away from light / move to the dark side; causing faster elongation; hence curvature;

  1. In the roots / radicle higher  concentration of auxins / IAA inhibits growth; hence the upper side with less auxins grows faster than the lower side; (hence curvature downwards)                                                             (2mks)
  2. Phototropisms; -enables plant shoots to grow and get light for maximum photosynthesis; Allows for leaf mosaic;

Chemotropism;   -Growth curvature in response to contact/ hard surface;

-Make plants with weak stems to get support on large plants /trees; this makes then to

reach and get light for maximum photosynthesis;

Geotropism;       -Growth curvature in response to gravity; enables plant roots to grow deep into the soil

to Maximum support/ anchorage;

Hydrotropism; – Growth curvature in response to moisture / water;

Chemotropism; -Growth curvature in response to chemical concentration gradient;

Thermo tropism; -Growth curvature in response to temperature changes;

-Enables some plants to grow to where they can acquire optimum temperature for  effective plant processes; e.g. (Sunflower orientates towards the directions of the sun.

5        (a)

S.N M.N
– Cell body has no dendrites -Cell body has dendrites
– Cell body at a point

along nerve fibre;

-Cell  body located at the top of nerve fiber;

 

6      a)     –      Hearing / detection of sound;

Body balance / posture;                                 (2marks.)

  1. b) (i)    C- Ear canal / External auditory meetas;

(ii)   G- Semi – Circular canals;

(iii)  F- Auditory nerve;                                        (3marks)

  1. c) (i)    Equalise air pressure between middle ear and outer ear / prevent damage of ear drum;                         (1mark)

(ii)   Cochlea/ E;                                                   (1mark)

  1. d) –      Ear ossicles / D;                                            (1mark)
  2. a) i)     Sensory neurons / afferent neurone
  3. ii) Cell body located off the axon
  4. b) N-    Axon

P-    Cell body

Q-    Myelin Sheath

R-    Schwann cell

  1. c) N-    Transmission of impulse

Q-    Insulation / speed up impulse transmission.

8 a) Long sightedness / Hypermetropia ;                                                            1 mk)

  1. b) Short eye ball;

Weak lens (any one) ;1 mks

  1. c) Wearing of convex lens / conveying lenses;   1 mk                               rj if a(i) is wrong

9 a) 1- Sensory neurone / Afferent neurone

2- Relay neurone     / intermediate neurone

3- Motor neurone/ efferent neurone                                                            3 mks

  1. b) Check on the diagram arrows show points towards neurone 3 from 2 and 1 : 1 mk
  2. c) Grey matter         1 mks
  3. d) Impulse reaching the dendrites end of relay / Neurone 2 causes the synaptic vesicles, releases acetylcholine  / transmitter substancesü; into the synaptic  cleftü;the acetylcholine  / transmitter  chemical diffuses across the cleftü; and causes the depolarization of the  motor neuron/ neurone

 

10        Thigmotropism/Haptotropism;

Contact with support; causes migration of auxins to the outside; causing faster growth on the side away from contact surface; (Causing dendrils to curl around the support);

NB (a) (ii) is tied to (i)

Escape injurious stimuli/seek favourable habitat

Rej. Seek mate and obtain food

  1. Describe how the mammalian ear is adapted to perform its functions. (20mks)
    • The pinna is flap made of skin and cartilage; for collection and concentration of sound waves;
    • Auditory canal/meatus is a tube lined with hairs which trap solid particles like dust; It has wax secreting cells; that secrete wax for trapping solids and insects entering the ear;
    • The eardrum/tympanic membrane is thin with double layer of epidermis; It vibrates translating sound waves into sound vibrations; Sound vibrations are transmitted to ear ossicles;
    • Ear ossicles are malleus, incus and stapes; they amplify and transmit vibrations to the oral window;
    • The oval window is a thin membrane which transmits sound vibrations to the fluid of the inner ear; perilymph and Endolymph;
    • Eustachian tube connects middle ear with pharynx equalizing air pressure in the ear with atmospheric pressure; to prevent distortion of the eardrum;
    • The cochlea is highly coiled tube with system of canals (and sensory cells) to occupy a small space/increase the surface area for accommodating many sensory cells to detect sounds vibrations; and generate impulses transmitted to the brain;
    • Auditory nerve transmits nerve impulses to the brain for interpretation;
    • Semicircular canals are tubular cavities containing fluid;
    • The canals are arranged at right angles to each other in the three planes of spaces; to detect changes in position of the body; the canals have ampulla: utriculus and sacculus; to detect position of body in relation to gravity;
    • Utriculus has otoliths attached to sensory cells which generate impulses which are then transmitted to the brain through the auditory nerve;
    • The perilymph and endolymph fluid in the inner ear absorb mechanical shock/transmit sound vibrations/protect delicate parts; Total 25 max 20 marks

12             a)       i)     Conjunctiva transparent allow light to enter eye

  1. ii) Cornea transparent / curved allow light / refracts light entering eye;

iii)    Aqueous / vitreous humour – clear / allow light pass / refract light, hydrostatic pressure –

Maintain shape of eyeball. Contains sugars / proteins / salts – provide nutrients to eye.

  1. iv) Iris – contractile – controls light intensity / amount of light entering eye;
  2. v) Ciliary body glandular – secretes humuor
  3. vi) Ciliary muscle contractile – controls curvature of lens;

vii)   Suspensory ligament – fibrous – holds lens in position

viii)  Lens transparent / Biconvex – allow light go through to retina / refract light / focus light.

  1. ix) Retina – rods – rhodopsin – for dim light vision;

– Iodopsin – for bright light vision

  1. x) Fovea centralis – high concentration of cones – for accurate vision
  2. xi) Choroid layer – blood vessels pigmented        – for nutrition

– reduce light reflection / absorb stray light

xii)   Sclera – fibrous – protection / give eye shape;

xiii)  Optical nerves – sensory neurone – transmit impulse from retina to brain.

xiv)  External eye muscle – contractile – move eyeball within socket

  1. xv) Blind spot – cone and rods absent – no image is perceived

Correction

  1. b) i)    Short sightedness;           biconcave / diverging lens
  2. ii) Long sightedness;           converging lens / convex lens

iii)    Astigmatism;                          use of cylindrical lens

  1. iv) Squinting;                              surgery            (any 2 identifyü correctionü 4mks)
  2. Nervous communication Endocrine communication

– Nerve impulse to evoke a response              – Chemical substance/ hormone to evoke

Response

– High speed of transmission- Low speed of transmission;

– Rapid response                            – Response delayed

– Impulse transmitted through neurone   – Hormones transmitted in blood

– Responses specific and localized to one       – Responses affects several parts of the

Target organ                                  body;

Note: Comparison should come out clearly to award.

  1. a) H – Eustachian tube;

J – Semi-circular canals;

  1. b) H – Tube open/ connection to the pharynx and to the middle ear/ opens during swallowing/ yawning and vomiting to equalize the air pressure in the middle ear with the atmospheric air pressure;

M – (pinna) curved/ funnel shaped to receive or collect and direct sound waves into the ear;

N – (cochlea) – long/highly coiled/ spiral in form to increase surface area for sound Perception;

– Has sensory hairs/ cells which convert sound vibrations to impulses/ generate impulses;

– Has endolymph to transmit vibrations;

Mark one for each structure.

Rej. If the adaptation is not tied to function.

  1. c) Total deafness;
  2. d) Endolymph;
  3. e) Balance; acc body balance/ posture.

15    (a) Tactic response;
(b) Move away from a harsh environment/move to favorable environment;

(c) To absorb any moisture from Petri dish A/OWTTE

16 .(a) A reflex mechanism/ability of the eye to adjust to bring an image from near or far object into sharp focus on the retina;
(b) Circular muscles of the iris contract; while the radial muscles relax; Ciliary muscles relax; increasing tension on suspensory ligaments; lens become thinner increasing the focal length; image focused onto the retina;
(ii) Circular muscles of the iris contract; pupil constrict/become smaller; and allows enough light for (sharp) image to be focused onto the retina;

 

SUPPORT AND MOVEMENT IN PLANTS AND ANIMALS

 

Support and Movement in Plants

Importance of Movement in Plants

  1. Enable plants to obtain resources such as sunlight, water and nutrients due to tropic and nastic responses.
  2. Enhances fertilization in bryophytes and Pteridophytes
  1. Helps plants to escape harmful stimuli such as high temperature

Importance of Support in Plants

  1. Hold flowers in position for pollination to occur.
  2. Help plants to withstand forces of the environment such as gravity and air currents.
  1. Increase the efficiency of photosynthesis as the leaves are firm and arranged in mosaic pattern for maximum absorption of light and carbon (iv) oxide.

Arrangement of Tissues in Plants

Diagrams

 

 

Practical Activity 3

 

To Observe Wilting in Plants

Support and Movement in Animals

Importance of Movement in Animals

  1. Enable searching of food, mate and shelter.
  2. Move to avoid predators.
  3. To colonize new areas
  4. Move from areas with unfavourable conditions such as fire, earthquakes, flood etc.

Types and Function of Skeletons

  1. Hydrostatic skeleton
  1. Exoskeleton

Functions of the Exoskeleton

  1. Reduces water loss
  2. Protection against microbial infections and mechanical injury
  1. Provide point for attachment of muscles allowing locomotion to take place.
  2. Enhance flight in insects by means of wings which are the flattened parts of the exoskeleton.
  3. Enhance walking in insects using jointed appendages.

 

NB/.        1. Exoskeleton has a disadvantage as it limits growth. To overcome this limitation it is periodically shed through moulting (ecdysis).

  1. Insects that jump or hop have powerful hind limbs. The femur of the hind limb has powerful antagonistic muscles.

Diagrams

 

 

Functions of the Endoskeleton

  1. Supports the animal’s body
  2. Gives the body its shape
  1. Provide surface for muscle attachment facilitating movement.
  2. Production of blood cells i.e. the long and short bones
  3. Acts as a reservoir of calcium and phosphate ions in the body

Locomotion in Finned Fish (Tilapia)

 

Diagrams

Practical Activity 5

Practical Activity 6

 

How a finned fish is adapted to locomotion in water

  1. Streamlined body/ tapered anteriorly and posteriorly; to minimize water resistance;
  2. Inflexible head; to maintain forward thrust;
  3. Overlapping scales facing posterior end; to bring about less resistance; Overlapping of scales also prevents wetting of the skin;
  4. Slimy/oily substance to moisten scales; hence reduce resistance between water and fish;
  5. Swim bladder; air filled cavity which controls/ brings buoyancy; and depth at which it swims;
  6. Flexible backbone /series of vertebrae with Myotomes/ muscles blocks; which contract and relax alternately bringing about thrust/force; which propels fish forwards;
  7. Pectoral and pelvic fins (paired fins); which bring about balancing effect; braking; and changing direction; they also control pitching i.e. control upward and downward movement;
  8. Dorsal fin, caudal fin and anal fin (unpaired fins); to increase vertical surface area; and therefore prevent rolling from side to side; and yawing;
  9. Tail fins/caudal fins that are long and flexible; for steering/ more force/ thrust;
  10. Lateral line has sensory cells; which enables to perceive vibrations; hence can locate objects so that it escapes / changes direction;

Support and Movement in Mammals

Diagram of a human and rabbit skeleton

 

The skeleton is divided into:

Axial Skeleton

  1. Skull
  1. Ribcage and sternum
  1. Vertebral column
  1. Cervical vertebrae
  2. Thoracic vertebra
  3. Lumbar vertebrae
  4. Sacral vertebrae
  5. Caudal vertebrae

All the vertebrae have a common basic plan.

Structure of a Vertebra

Each vertebra is made up of the following parts.

Diagram

  1. Atlas (First cervical vertebra)

Diagram

  1. Axis (second)

Diagram

  1. The other cervical vertebrae.

Diagram 

Diagram

4)   Sacral vertebrae

  1. All sacral vertebrae fused to form sacrum
  2. Transverse processes of first sacral vertebra large and wing like for articulation with pelvic girdle
  3. Pairs of holes on the lower surface for the spinal nerves to pass through.
  4. Sacrum is broader on the front side and narrow towards the tail.

Functions

  1. Protects alimentary canal on dorsal side.
  2. Provides attachment to hip girdle
  3. Protects the spinal cord
  4. Provides attachment for the muscles

Diagram

 

  1. Caudal vertebrae
  1. Very small in size
  2. No neural canal

 

 

Functions

  1. Provides attachment for tail muscles
  2. Helps in the movement of the tail

 

Diagram

 

Biology Form 1-2 Work Revision

  1. A student set up materials in an experiment as shown below.

 

 

 

 

 

 

 

(b) If the experiment set up was left over-night, state observation in the set up A and B.         (2mks)

(c) Account for the observations in each set up.                                                  (3mks)

(d) If another experiment C was set such that nothing is placed in the potato cup, state and explain the results that would have been obtained.                                                            (2mks)

  1. An experiment was carried out to investigate, haemolysis of human cells. The red blood cells were placed in different concentration of sodium chloride solution. The percentage of haemolysed cells was determined. The results were shown in the table below.
Salt conc. (g/100cm3) 0.33 0.36 0.38 0.39 0.42 0.44 0.48
Red blood cells haemolysed % 100 91 82 69 30 15 0

 

(a) (i) On the grid provided plot a graph of haemolysed red blood cells against salt concentration.  (6mks

(ii) At what concentration of salt solution was the proportion of haemolysed cells equal to non-haemolysed   cells? (1mk)

(iii) State the percentage of red blood cells haemolysed at salt concentration of 0.45.             (1mk)

(b) Account for the results obtained at:

(i)0.33% salt concentration                                                                     (3mks)

(ii) 0.48% salt concentration                                                                   (3mks)

(c) What would happen to the red blood cells if they were placed in 0.50% salt solution.        (3mks)

(d) Explain what would happen to onion cells if they were placed in distilled water.         (3mks)

  1. Explain how various environmental factors affect the rate of transpiration in plants.      (20mks)
  2. (a) State the meaning of the following terms.

(i) Digestion                                                                                      (2mks)

(ii) Ingestion                                                                                      (2mks)

(b) Describe the process through which a piece of ugali undergoes in man from the time of ingestion up to the time of absorption.                                                                    (16mks)

 

  1. The diagram below represents a unit of gaseous exchange in man. Study it carefully and answer thequestions that follow.

 

 

 

 

 

 

 

 

 

 

 

 

  1. a) Name the blood vessel that brings blood to the lungs and the vessel which takes blood away from the lungs.            (2mks)
  2. b) Name the structure above. (1mk)
  3. c) Label A and (2mks)
  4. d) In what form is carbon (IV) oxide transported in structure labeled E. (1mk)
  5. e) Name the gas G. (1mk)
  6. Gastrin is a hormone produced by mammals.

(a) (i) Where is the hormone produced?                                                          (1mk)

(ii) What is the function of gastrin?                                                              (1mk)

(b) What stimulateds the production of gastrin.                                                (1mk)

(c) The diagram below shows part of the human intestine.

 

 

 

 

 

  1. i) Identify the parts labeled A and B (1mk

(ii) To which circulatory system does the part labeled B belong.                                  (1mk)

  1. d) State any two adaptations of the human large intestine to its function.  (2mks)
  2. The diagram below represents part of a xylem tissue.

 

 

 

 

  1. a) (i) Name the parts labeled P and Q (2mks)

(ii) Give the function of the part labeled P.                                       (1mks)

  1. b) State the function of the phloem tissue.   (1mk)
  2. c) (i) State how the functioning of the phloem tissue is affected if the companion cell is destroyed. (1mk)

(ii) Give a reason for your answer.                                                           (1mk

  1. d) State any two structural differences between phloem and xylem tissues.  (2mks)
  2. In an experiment to determine the effect of exercise on the concentration of lactic acid in blood, the following data was obtained. Study the data and use it to answer the questions that follow.

The lactic acid concentration was measured before, during and after the exercise.

Time minutes 0 10 20 25 30 40 50 60 70 80 90 100
Lactic acid conc. (arbituary units) 0.5 0.5 5 13 12 8 6 4 3 2 1 0.9
  1. a) Using a suitable scale, plot a graph of the concentration of lactic acid against time. (6mks)
  2. b) From the graph you have drawn determine

(i) The period of exercise . Explain.                                                            (2mks)

(ii) The time when oxygen debt occurred Explain.                                        (2mks)

(iii)    The duration it took to pay back the oxygen debt.Explain                            (2msk)

  1. c) On the same set of axes plot a hypothetical curve for oxygen intake during the experiment period of 90 minutes.  (2mks)
  2. d) Why does lactic acid level usually continue to rise in the blood after exercise ceases. (2mks)
  3. e) Suggest the two importance of anaerobic respiration to animals. (2mks)
  4. d) What is oxygen debt? (2mks)
  5. What is the role of the liver in the maintance of a constant level of materials in the body. (20mks)

10.The diagram below represents a simple respiratory pathway in cells

 

 

 

 

 

 

 

 

 

  1. a) Name the process marked X and                                 (2mks)
  2. b) State two differences between process X and                               (2mks)
  3. c) State the name of substance B and condition under which it is formed.   (2mks)
  4. d) Explain how body size affects the rate of respiration in animals. (2mks)
  5. The diagram below represent the structure of a nephron. Study it and answer the questions that follow.

 

 

 

 

 

 

 

 

 

 

 

  1. a) (i) State the physiological process by which solutes are selectively re-absorbed back into blood at the part labelled                                                                            (1mk)

(ii) How is the part labeled B adapted to carry out the physiological process named in 3 (a) (i) above.                                                                                       (1mk)

  1. b) In which part of the kidney is the part labelled A abundantly found. (1mk)
  2. c) On the diagram above , indicate the direction of flow of blood using arrows at the part labelled (1mk)
  3. e) State the functions carried out by the following hormones in the functioning of the nephron.

(i)Aldosterone.                                                                                 (1mk)

(ii)Anti diuretic hormone.                                                                                (1mk)

  1. The data below shows the rate of photosynthesis at different temperature in attached leaves of three East African plants. (Crotolaria, Gynandropsis and Amaranthus species) respectively which were grown outside with the same illustration while water and carbon (IV) oxide are not limiting factors in this experiment.

Rate of photosynthesis was expressed interms of carbon (IV) oxide uptake in mg/mm2/hr at various temperatures as tabulated below.

Temperature oC Rate of photosynthesis (mg/mm2/hr)
  Gynandropsis sp Crotolaris sp Amaranthus sp
5

10

15

20

25

30

35

40

45

50

22

50

60

80

85

80

73

66

2

20

40

49

64

48

45

42

31

15

10

27

42

55

54

50

45

40

11

 

  1. a) Represent the results graphically (rate of photosynthesis against temperature)
  2. b) Using the graph in (a) above indicate optimum temperature for the Gynandropsis and Amaranthus species.       (2mks)

Gynandropsis ………………………………………………

Amaranthus  ………………………………………………

  1. c) Give a reason why Gynandaropsis and Amaranthus could not function photosynthetically at 5o (1mk)
  2. d) What are the possible ecological habitats for the following plants. (2mks)

(i) Amaranthus

(ii) Crotolaria

  1. e) At what temperature was the amount of carbon (IV) oxide around the leaf of Gynandropsis highest?      (1mk)
  2. f) What raw material is required in the light stage of photosynthesis.      (1mk)
  3. g) Name the parts of chloroplasts in which the following stages of photosynthesis take place. (2mks)

(i) Light stage

(ii) Dark stage

  1. h) State one structural similarity and difference between chloroplast and mitochondria. (2mks)

Similarity

Difference

i)What is the compensation point of photosynthesis?                                                  (1mk)

13 (a) Explain why plants lack elaborate excretory organs like those found in animals.                 (3mks)

(b) Name five methods of excretion in plants.                                                         (5mks)

(c) State any six excretory products in plants and give economic uses.                              (12mks)

  1. During a laboratory investigation, a scientist extracted gastric juice from the mammalian stomach. He used it to carry out tests on a food sample B which was suspected to contain proteins. He divided the food sample B into three portions and treated them as below.
  2. On the 1st portion of B, he added Gastric juice and mixed them thoroughly before adding sodium hydroxide followed with copper (II) sulphate drop by drop.
  3. On the 2nd portion of B, he added boiled gastric juice and mixed them thoroughly before adding sodium hydroxide followed with copper (II) sulphate drop by drop.

III. On the 3rd  portion of B, he added Gastric juice, sodium bi-carbonate and mixed them thoroughl before adding sodium hydroxide followed with copper (II) sulphate drop by drop.

  1. a) State the observations he made in each set up.                               (3mks)

– 1st  portion

– 2nd  portion

– 3rd  portion

  1. b) Why was the experiment on the 1st portion included in the tests?                             (1mk)
  2. c) Name the property of the chemical being investigated in these tests.      (1mk)
  3. d) Account for the observations made in 2 (a) above.                                            (3mks)
  4. The diagram below illustrates circulation in certain organs of the mammalian body.

 

 

 

  1. a) Identify the blood vessels represented by A, B and C.      (3mks)
  2. b) Explain why blood from the small intestines goes to the liver before it goes to any other organ of the body.                                                            (2mks)
  3. c) Compare the blood in vessels B and C.                                     (1mk)
  4. d) Outline how a glucose molecule in vessel A finally reaches the heart.       (2mks)
  5. 1 The table below shows how the internal temperature two animals X and Y varied with the external temperature. The temperature was measured regularly and recorded for 12 hours in a day. Study the table and answer the questions that follow.

 

  1. a) Using the same grid, draw graphs of external temperature, and internal temperature of animals X and Y

(Y-axes) against time (X-axes).                                                                      (7mks)

  1. b) Account for the variation of internal and external temperatures for the animals X and Y. (2mks)
  2. c) Identify the classification of organisms whose internal temperature varies as X and Y (2mks)
  3. d) Explain two ways used by organism Y to make its internal temperature vary as shown despite of changes in external temperature.                                                                      (4mks)

18    a) Give the functions of the skin in organisms.                                                         (6mks)

  1. b) How is the mammalian skin modified to enable it perform its functions?      (l4mks)
  2. The diagram below shows how gaseous exchange occurs across the gills in fish.

 

 

 

 

 

 

 

 

 

(a)  According to the diagram water and blood flow in opposite direction across the gills.

(i)  Give the term used to describe this flow.                                        (1 Mark)

(ii)  Explain the advantage of the above flow named in a(i) above.                  (2 Marks)

(b)  What difference would be observed if water and blood flows across the gills in the same direction?                                                                              (2 Marks)

(c)  In which structures in the gills does gaseous exchange take place?                      (1 Mark)

(d)  Name two organs in man which display the flow system named in a(i).             (2 Marks)

  1. An experiment was carried out to investigate the effect of different concentrations of Sodium Chloride on human red blood cells. Equal volumes of blood were added to equal volumes of salt solutions of different concentrations.  The results were as shown below:-
Set up Sodium Chloride concentration Shape of red blood cells at the end of experiment Number of red blood cells at the end of experiment
A

B

0.9%

0.3%

Normal

Swollen

No change in number

Fewer in number

 

  1. If the experiment was repeated with 1.4% Sodium Chloride solution, state the results you would expect with reference to:-

(i)  Number of red blood cells.                                                            (1 Mark)

(ii)  Appearance of red blood cells when viewed under the microscope.                      (1 Mark

Account for the fewer number of red blood cells in 0.3% Sodium Chloride salt solution. (3 Marks)

  1. c) Give the biological term which can be used to describe 0.9% Sodium chloride solution.(1 Mark)
  2. d) Define plasmolysis.                (1 Mark)

 

Marking Scheme

  1. i) Osmosis.
  2. ii) A – solution in potato cup increases.  Level of water in the beaker decrease;

B- Remain the same;

iii)  A – Surrounding the cube is a region with high concentration of water molecules while in the                sugar crystals, there are very few water molecules;

The sugar crystals exert on Osmosis pressure by Osmosis water molecules move across the    potato tissue, which acts as a semi-permeable membrane.  The level rises;

B- No change since boiling denatures the membrane structure of potato cells;

  1. iv) C- No water moves into the potato cup/remains the same; since there is no concentration gradient;

2(a) i)  Graph.

  1. 402 ± 0.01
  1. b) i)  All cells have been haemolysed; cells contains one hypertonic to salt solution; water enters cells by osmosis; cells swell and eventually burst.
  2. ii) No cells were haemolysed; cell contents were isotonic to salt solutions (aments of water entering      the cell was equal to that leaving the cell); no net movement of water into cells;
  3. c) The cells would become crenated; the cell contents would be hypotonic to salt solutions; water would leave cells by osmosis; membranes would shrink.
  4. d) Contents of Onion epidermal cells would be hypertonic to water; water would enter cells by Osmosis; cells would become turgid;
  5. –  Temperature;- High temperature faster rate of transpiration; high temperature increases the capacity of atmosphere to hold water and moisture; also heat increase internal temperature of the leaf       hence water     evaporation;     4 accept converse

        –  Atmospheric pressure; Low atmospheric pressure, high rate of transpiration                 2

        –  Humidity; Low humidity higher rate of transpiration; low humidity increases the saturation      defiant; hence water moves form leaves to drier atmosphere;                      4

        –  Wind; When it is windy the rate of transpiration is higher; wind sweeps away vapour that has accumulated at the surface of leaf; increasing saturation deficit; hence faster rate of transpiration  5

        –  Light intensity;High light intensity faster rate of transpiration high light intensity increase        photosynthesis rate hence stomata opens;                                                              4

        –  Amount of water in soil; More water in the soil increases the rate of transpiration; it wets the xylem (ensure xylem is wet throughout);    3                                                                       Max 20

  1. a)  Define digestion and ingestion.
  2. i) Digestion- It is break down of complex insoluble; √ food substance into simple soluble food substance;√
  3. ii) Ingestion- is introduction of food through the mouth into the digestive system;
  4. b) Describe the digestion of Ugali.
  1. (a) – Pulmonary artery

– Pulmonary vein

(b) Alveolus

(c) A– cavity of alveolus

E – Red blood cell

(d) Hydrogen carbonate ions;

Carbamino haemoglobin;

  1. (a) (i) Walls of stomach;

(ii) Stimulates the secretion / production of gastric juice; √

(b) Presence of  food in the stomach;

(c) A – Blood capillaries; B – Lacteal;

(d) – Produces plenty of mucus to lubricate coarse/indigestible material during peristalsis;

– Wide human accommodates /store indigestible food

– Elongate to increase surface are for absorption of water.

– has muscles to facilitate peristalsis when they contract;

  1. (a) (i) P – Tracheids Q – pits

(ii) P– water conducting elements of xylem

(b) Function of phloem – translocation/ transport of organic substances from the leaves to the of the plant;

(c) (i) Translocation of food will not occur acc. Slow translocation

(ii) Reason – it contains a lot of mitochondria which provide energy for translocation;

(d)

 

Phloem Xylem
1)  Made of living cells

2)  Have companion cells

3)  Have cytoplasmic strands

4)   Lack lignin deposits

–       made of dead cells

–       lack companion cells

–       lack cytoplasmic strands

–       have lignin deposits; (any 2×1=2mks)

 

  1. (a) Photocopy – scale – 1m

Labeling axes – 1

Plotting – 2m

Curves – 2m (curves must be labeled)    rej. Dotted line for curves

A GRAPH OF LACTIC ACID CONCENTRATION AGAINST TIME

 

 

 

 

 

 

 

 

 

 

 

 

(b) (i) 10-15 minutes ; period of rapid increase in lactic acid concentration (2mks)

(ii) 10-20 seconds : period when lactic acid level starts to increase; (2mks)

(iii) 75minutes i.e. from 25th minutes to the 100minutes, this is the time lactic cid took to decrease from the highest level to normal; (2mks)

(c) It would have the same basic shape; but would peak slightly ahead of the lactic acid curve in time;

(e) Because it is still diffusing out of the muscles, where it was made a few minutes earlier;

(e) Allows for energy production even cases of oxygen deficiency; thus enables animals to survive active exercise and to inhabit even in areas with limited oxygen supply;

(f) Oxygen debt is the amount of oxygen to get rid of the lactic acid; that has accumulated due to anaerobic respiration; (2mks)

  1. Regulation of blood sugar level; under the influence of insulin; and glucagons (hormones). When there is excess sugar; the hormone insulin stimulate/causes liver cell to convert it to glycogen; some converted to fats/lipids for storage;
  1. (a) X – glycolysis Y – Kreb’s cycle

(b)

Process X Process Y
–       occurs in cytoplasm

–       independent of oxygen

–       produces less energy

–       raw material is glucose

–       End products are energy, CO2, lactic acid or ethanol

–       occurs in mitochondria

–       Is oxygen dependent

–       produces more energy

–       Raw material is pyruvate

–       End products are energy, CO2 and water

 

(c) lactic acid; under anaerobic conditions

(d) small body size leads to alarge surface area to volume ratio; hence more loss of heat to the environment; leading to increased rate of respiration to replace the lost heat;

  1. (a) (i)Active transport/diffusion

Tied       (ii) Numerous Mitochondria in its wall to generate energy/microvilli/coiling increase surface area/thin epithelium for quick diffusion.

(b) Cortex

(c) on the diagram

(d) plasma proteins; Blood cells; accept specific examples e.g. albumins, red blood cells;

(e)       (i) Regulate re-absorption of Sodium salts;

(ii) Regulate re-absorption of water

  1. (a) Allocation of marks on graph

(b) Gynandropsis – opt To = 30oC

Amaranthus – opt. to = 25oC

(c)At 5oC, the enzymes that catalyse the process of photosynthesis are inactivated.

(d) Amaranthus – Terrestrial;  Crotolaria – terrestrial;

(e) 50oC;

(f) water;

(g)(i) Granum; (ii) Stroma

(h) Similarity: Both have double membrane; 1mk – Both have fluid filled matrix;

Difference : inner membrane of mitochondrion is folded to form cristae while inner  membrane of chloroplast is smooth;

– chloroplast is biconcave shaped while mitochondria is oval/sausage shaped (any 1×1=1mk)

(i) Point at which the rate of photosynthesis equals to the rate of respiration.

  1. (a) – plants wastes accumulate slowly;

– plants produce less toxic wastes;

– some excretory products are recycled by plants e.g. CO2, SO2)

– plant tissues are tolerant to toxic wastes;

–  plant wastes are stored in temporary structures which fall off e.g. leaves (any 3×1=3mks)

(b)   – Diffusion;

– Transpiration;

– Exudation;

– Deposition of wastes/ leaf fall/ flower fall/ storage in bark;

– Recycling;

– Guttation ;                                  (any 5×1=5mks)

(c)

Excretory products Uses
1.   Caffeine;

2.   Popain;

3.   Tannin;

4.   Nicotine;

5.   Latex;

6.   Quinine;

7.   Atropine;

8.   Morphine;

9.   Colchine;

10.                  Cocaine;

11.                  Cannabis;

12.                  Khat/ miraa;

 

Body stimulant;

Meat tenderizer;

Leather tanning;

Stimulant; insecticide;

Manufacture of tyre/rubber products;

Anti-malarial drugs;

Increase heart beat; dilate eye pupil;

Cancer treatment;

Used in genetics to induce polypoidy;

Anesthesia/painkiller/stimulant;

Pain killer;

Stimulant;  (Any 6×2=12mks)

 

  1. (a) 1st portion. – Blue; colour was observed

2nd portion – Purple; colour was observed

3rd portion – Purple; colour was observed

(b) A control experiment;

(c) Proteins are highly sensitive to temperature and pH changes; (award if either temp of pH is stated singly)

(d)      1st portion – Enzyme pepsin broke down proteins into peptones;

2nd portion – Enzyme pepsin works in acidic medium; (not in basic medium)

  1. (a) A – Hepartic portal vein;        B – Hepartic vein;           C – Hepartic artery;

(b) – So that any toxic substances absorbed together with food nutrients from the ileum be detoxified;

– So that food substances e.g. glucose, amino acids can be regulated. Only the required quantity of glucose is left in circulation as excess is either stored as glycogen, fat and excess may be respired.

Excess amino acids are deaminated;

(c) B – Deoxygenated                           C – Oxygenated

(d) From the small intestines, it is transported to the liver through Hepartic portal vein; (It is then transported to the heart through the hepartic vein;

  1. (a) graph

(b) X – Lacks internal mechanisms to regulate its internal temperature

 Y – Has internal means to regulate its internal temperature. hence able to maintain it within narrow range

(c) X – Poikilotherm                      Y – Endotherm

(d)   – Blood vessels vasodilate when temperature is higher than norm to allow for heat loss from blood through radiation, evaporation, etc; when temperature is lower,  blood vessels constrict              to prevent loss of heat from blood through radiation, evaporation etc;

– When temperature is higher, lies flat to allow for heat loss from the body since insulation layer of air is removed; when temperature is lower, hair strands erect to hold air which insulates the body against heat loss through radiation, evaporation etc.

  1. (a) Higher temperature; increases the kinetic energy; of water molecules which makes water turn into  vapour on the leaf surfaces faster and hence increase rate of transpiration
  1. (a) – It protects the underlying tissues against mechanical injury, UV-light rays and entry of pathogens; (Rej. germs)

(b) – Presence of the cornified layer; which tough and has keratin to enable it protect the underlying tissues from mechanical injury; It alsos has sebum; which is antiseptic and enables it to prtect the ody against entry of pathogesn. Presence of melanin; enables it to protect the underlying tissues against damage by t UV-light radiations.

  1. a) i)  Counter current system;
  2. ii) Maintain a diffusion gradient so that there is maximum uptake of oxygen; continue diffusing into blood and Carbon (iv) oxide into water;
  3. Parallel flow lower diffusion gradient; so that less oxygen diffuse into blood/low rate of gaseous exchange
  4. Gill filaments
  5. Placenta

Kidney

20 a)       i)  Remain the same

  1. ii) Crenated
  2. The solution is hypotonic to red blood cells hence the cells grains water; by osmosis; swelling until they burst.
  3. Isotonic solution
  4. Plasmolysis – the process by which plant cells loss water by Osmosis shrink and become flaccid.

 

 

BIOLOGY DIAGRAMS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TOPICAL QUESTIONS FOR BIOLOGY

 

FORM I TOPICS

  1. a) Define biology

-the study of life/living things

  1. b) List the branches of biology

– Zoology (study of animals)

– Botany (study of plants)

– Microbiology (study of microorganisms)

  1. c) Explain the importance of biology

– helps to solve environmental problems

– Helps to learn scientific skills

– For entry into other professions/careers

– To apply knowledge to everyday life situations

– To classify organisms into their right groups

– understanding living organisms

  1. d) State the characteristics of living organisms

– feeding/nutrition

– Growth and development

– respiration (to produce energy)

– sensitivity/irritation/response

– excretion (getting rid of metabolic waste material)

– movement/locomotion

– reproduction

  1. e) State the main differences between plants and animals
Animals

 

Plants
Specialized excretory organs

 

No specialized excretory organs
Respond to stimulus quickly

 

Slow respond to stimulus
All body parts grow equally(intercalary)

 

Grow at shoot tip and root tip only
Move around to look for food

 

Stationery
Heterotrophic

 

Autotrophic
Cells have no cell walls Cells have cell wall made of cellulose

 

 

No chlorophyll

 

Contain chlorophyll

Give parental care to young Plants don’t care for their young
  1. a) i) What is a hand lens?
  1. ii) How is a hand lens used?
  1. ii) When is a hand lens used?
  1. iv) Explain how to calculate drawing magnification

– drawing magnification equals to length of drawing divided by length of object or image length divided by actual length i.e.  length of drawing     or   image length

Length of object actual length

  1. b) i) what is classification?

– Orderly arrangement of living organisms into various groups according to their similarities

  1. ii) List the external features used to classify plants

iii) List the external features used to classify animals

  1. c) Give the reasons why classification is important

Placing/grouping living organisms into correct groups called taxa

– Identification

– arrange information about living organisms into orderly and sequential manner i.e. it is easy to study organisms in groups

-helps in understanding evolutionary relationships

– monitoring disappearance and appearance of organisms i.e. predict characteristics of organisms

  1. d) i) Name the taxonomic units of classification in descending order

– Kingdom (largest unit)

– Phylum (animals)/division (plants)

– Class

– Order

  1. ii) What is a species?

iii) Name the major kingdoms used in classification

-monera

-protoctista/protista

-fungi

-plantae

-animalia

  1. e) i) Define the term binomial nomenclature

– a scientific system of naming organisms using the generic/genus and specific/species names

– e.g. for humans, Homo sapiens

  1. ii) State the principles followed during binomial nomenclature

iii) Give the advantages of using binomial nomenclature

  1. iv) Name the types of classification
  1. a) i) Define the term cell
  1. ii) What is cell biology?

b)i) What is a microscope?

– an instrument used to magnify objects and make them appear bigger.

  1. ii) Name the types of microscope

iii) State the purpose of using a light microscope

  1. iv) Draw a labeled sketch of a light microscope

 

 

 

 

 

 

 

 

 

 

 

  1. v) State the functions of the labeled parts

a.) Eyepiece  used to look through and to magnify the object

b.) Course adjustment knob raises or lowers body tube and focuses object roughly

c.)  Fine adjustment knob raises or lowers body tube by small distances to bring image into fine focus    d.)  objective lens brings image into focus and also magnifies object/image

  1. e) i) Explain the procedure followed when using a microscope

– put the microscope on the bench with the stage facing away from you (viewer)

– turn the lower power objective to click in line with the eyepiece

– Ensure that the diaphragm/iris is fully open

– Adjust the mirror until the stage is illuminated with enough light

– Place the slide containing the specimen on the stage for magnification

– Draw the image and indicate magnification of the drawing.

  1. ii) State the precautions that are necessary when handling a microscope
  1. d) i) What is magnification?

– The power of making an image larger

  1. ii) Give the formula used to calculate magnification in a light microscope

           

 

iii) Give the reasons for each of the following steps when preparing a cross-section of a stem or leaf for examination under the microscope

cutting very thin sections

Using a sharp razor blade during the cutting

Placing sections in water

Staining the sections with iodine before observing

  1. e) i) List the parts of a cell that can be seen under a light microscope

a   cell membrane

b   cytoplasm

c    cell wall

e   nucleus

d  vacuole

  1. ii) Draw the general structure of a plant and animal cell

iii) List the parts of a cell that can be seen under an electronic microscope and state the functions of each part.

Cell wall

Cell membrane

Cytoplasm

Nucleus

Nucleolus

Vacuole

Lysosomes

Golgi apparatus

Ribosomes

Mitochondrion

Chloroplasts

Endoplasmic reticulum

  1. iv) State the functions of cell sap
  1. e) Compare plant and animal cells
  1. f) Explain the meaning of each of the following
  2. Cell

– Basic unit of organization in an organism

– Specialized animal cells include sperm, ovum muscle

Specialized plant cells include epidermal, guard cell and palisade cell

  1. ii) Tissue

iii) Organ

  1. iv) Organ system
  1. g) i) Name the structures which are present in plant cells but absent in animal cells

– Chloroplast

– Cell wall

  1. ii) Name the structures which are present in animal cells but absent in plant cells
  1. h) Explain how to estimate cell size
  2. i) Materials

1000

 

  1. ii) Procedure

– Click to low power

 

 

 

 

 

 

 

 

 

 

 

  1. i) In a drawing of a giraffe, the height of the head from the ground was recorded as 10cm. the drawing also showed a magnification of 0.02. calculate the actual height of the giraffe

Drawing height  =  10cm   = 500cm

Magnification        0.02

  1. In a class experiment to estimate sizes of cells a student observed and obtained millimeter marks on the field of view of a microscope as shown in the diagram below.

 

 

Millimeter marks

– If the student counted 40 cells on the diameter of the field of view, what was the approximate size of the each cell in micrometers?

Diameter of field of view  =  3  x  1000  =  75 mµ

Number of cells               40

  1. Under which of the following light microscope magnifications would one see a larger part of the specimen? X40 or x400? Give a reason
  1. State the functions of the cell
  1. b) i) Describe the structure of cell membrane

– made up of three layers

– Lipid portion sandwiched between two protein layers

– Lipid portion enhances penetration of oil soluble substances

Pores present to facilitate inward and outward movement of water soluble substances

  1. c) i) What is diffusion?

iii) Explain the roles of diffusion in living organisms

 

  1. Suggest an experiment to demonstrate diffusion
  1. d) i) What is osmosis?

OR

  1. ii) State the factors affecting osmosis
  1. Explain the roles of osmosis in living organisms
  1. A group of students set up an experiment to investigate a certain physiological process. The set up is as shown in the diagram below.

 

 

 

 

After some time they observed that the level of sugar had risen.

What was the physiological process under investigation?

Why was there a rise in the level of sugar solution?

Suggest the results that the students would obtain if they repeated the experiment using cooked potato

What is the reason for your suggestion?

  1. Explain the following terms

Hypnotic

Isotonic

Hypertonic

Turgor pressure

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Plasmolysis

 

 

 

 

 

 

 

 

 

Wilting

Haemolysis

  1. e) A form one student placed red blood cells in different salt concentrations and obtained the following results:-

There was a gain (+) no change (0 zero) and a loss (-) in the volume of the cells as show below:-

 

 

 

 

 

 

 

 

 

 

 

 

Briefly explain the results of the experiment

  1. f) i) What is active transport?
  1. ii) State the factors affecting active transport

 

iii) Why is oxygen important in the process of active transport?

– Oxygen is required for respiration, which produces energy necessary for the process to occur.

2.0 ) the factors that affect the rates of the following process in living organisms.

– Diffusion gradient which refers to the difference in concentration of molecules between the region of high concentration and the region of low concentration. Increasing the concentration gradient causes an increase in rate of diffusion and vice versa.

-Surface area to volume ratio .is the ratio of total surface area exposed by an organism compared to its body volume. Small sized living organisms have a large surface area to volume ratio. The larger the surface area to volume ratio ,the high the rate of diffusion and vice versa. Small organisms like amoeba and paramecium can hence rely on diffusion for transport of substances into and within its body and removal of waste products

-thickness of membranes. Molecules take longer to diffuse across thick membranes than across thin membranes hence the thin the membrane the higher the rate of diffusion.

-Temperature. Increasing temperature increases the kinetic energy of diffusing molecules making them to spread faster. Increasing temperature increases the rate of diffusion and vice versa

-size of molecules/molecular weight.Small sized molecules/molecules of low molecular weight move/diffuse faster hence the rate of diffusion is high where the molecules involved are small or have low molecular weight and vice versa.

 B.)OSMOSIS

-Temperature. Increasing temperature increases the kinetic energy of water molecules making them to spread faster. Increasing temperature increases the rate of osmosis and vice versa

-concentration gradient/diffusion pressure deficit.  Refers to the difference in concentration on either side of a semi-permeable membrane. The higher the osmotic pressure difference the higher the rate of osmosis.

C.) ACTIVE TRANSPORT 

OXYGEN CONCENTRATION. It is required for respiration/to oxidize respiratory substrates to release energy required for active transport. an increase in oxygen concentration causes a simultaneous increase to the rate of active transport upto a certain level.

PH   Enzymes being protein in nature are PH  specific. Extreme change in PH   affect the  rate of respiration which is controlled by enzymes and may denature the enzymes reducing the rate of active transport.

GLUCOSE CONCENTRATION.  is the main respiratory substrate for energy production. An increase in glucose concentration in cells increase the rate of respiration and hence the rate of  active transport is increased upto a certain optimum level beyond which any additional increase in glucose concentration has no effect.

TEMPERATURE. The process of respiration by which energy for active transport is generated is controlled by enzymes. Enzymes work best at temperatures of between 350c-400c,usually called optimum temperature ranges. At very low temperatures enzymes are inactive lowering the rate of respiration hence low rates of active transport . increase in temperature above optimum ( above 400c)denatures enzymes slowing down respiration and ac tive transport until it finally stops.

ENZYMES INHIBITORS. They are substances which slow down (by competing with the enzyme for the active sites in the substrate) or stop ( by blocking the active sites of the enzyme) the activity/funtioning of enzymes  .this slows down or stops respiration and so is active transport.

CONCENTRATION OF CARRIER MOLECULES IN THE CELL MEMBRANE. They are substances that bind to the ions being transported actively and carrying them across the membrane . increase in concentration of carrier molecules increases the rate of active transport upto a certain level and vice versa.

4.) Explain briefly the role of osmosis in living tissues.

In plants:

Osmosis facilitates the absorption of water from the soil by plant roots, water is required for the process of photosynthesis.

Turgidity of cells contributes to support in herbaceous plants and helps plant to maintain shape.

Helps in closing and opening of stomata regulating the process of gaseous exchange and transpiration.

It facilitates feeding in insectivorous plants like venus fly trap.

In animals:

Enables reabsorption of water from the kidney tubules back to blood stream facilitating the process of osmoregulation.

It enables organisms in fresh water bodies like amoeba to absorb water.

it is applied in food preservation.

 

5.) explain what happens when plant and animal cells are put in hypotonic and hypertonic solutions.

  1. i)plant cells in hypotonic solution.

The concentration of the plant cell sap is hypertonic to the solution/water medium. the cell draws in water by osmosis through the cell wall, cell membrane into the cell cytoplasm. Water enters the cell vacuole by osmosis; it enlarges and exerts an outward pressure on the cell wall called turgor pressure. Increased turgor pressure pushes the cell cytoplasm against the cell wall until the cell wall cannot stretch any further. The cell becomes firm or rigid and is said to be turgid. As the cell wall is being stretched outwards, it develops a resistant inwards pressure that is equal and opposite to the turgor pressure and this is called wall pressure.

 

 

ii.) Plant cells in hypertonic solution.

The plant cell sap is hypotonic to the solution medium. Water molecules are drawn out of the plant cells by osmosis into the hypertonic solution through the semi-permiable membrane of the plant cells. As a result the plant cell will start to shrink/less rigid and become flabby. The cell membrane/plasma membrane is pulled away from the cell wall and the cell is said to be flaccid. this process by which a plant cell lose water, shrink and become flaccid is called plasmolysis. However the shape of the plant cell is maintained by the tough rigid cellulose cell wall which prevents crenation in plant cells.

b.i) Animal cells in hypotonic solution.

The concentration of water in the cytoplasm of the plant cells is hypertonic to the solution medium in the test. Water molecules are then drawn into the animal cell cytoplasm from the surrounding medium by osmosis through the semipermiable membrane.  the cell swells as water is drawn into them by osmosis .as water continues to enter into the cell,the weak animal cell membrane bursts a process called lysis. in red blood cells  this process is called haemolysis. However in unicellular organisms like amoeba and paramecium,bursting of their cells does not take place because they have specialized organelles called contractile vacuoles for removal of exces water out of their bodies/cells.

ii.) Animal cells in hypertonic solution.

The concentration of the animal cells cell cytoplasm is hypotonic to the solution medium in the test. the surrounding hypertonic solution will draw water out of the animal cells by osmosis through the semi-permeable membrane. Continued loss of water causes the cells to be smaller in size and their membranes become wrinkled.  This process will continue until the concentration of the cell sap and the surrounding medium is equal i.e isotonic.the process by which animal cells lose water and shrink is called crenation.

6.) Explain briefly the role of active transport in living organisms.

It is involved in active reabsorption of glucose and mineral salts in kidney tubules during formation of urine. it enables the absorption of digested food from the alimentary canal/small intestines into the blood stream. Excretion of waste products from body cells for eventual removal.  Involved in transmission of nerve impulses within the nerve cells through the sodium pump which maintains a balance between sodium and potassium ions. It facilitates accumulation of substances in the body cells to offset osmotic pressure of organisms in dry and marine environment allowing them to absorb water by osmosis and avoid desiccation. In plants it enables plant roots to absorb water from the soil against the concentration gradient. It’s involved in translocation of manufactured food in the phloem tissue within the plant body. It’s involved in the opening and closing of the stomata through the sodium-potassium pump mechanism.

  1. iv) Outline the roles of active transport in living organisms
  1. ii) State the importance of nutrition
  1. b) Differentiate the various modes of feeding
  2. Autotrophism
  1. Heterotrophism
  1. State the importance of photosynthesis

Structural adaptation of the leaf to its function

 

 

The leaf has a broad and flattened lamina to provide a large surface area for trapping optimum light forphotosynthesis and allow maximum gaseous exchange.

The leaf epidermis is thin ( one cell thick) to reduce the distance across which diffusion of carbon (iv) oxide gas to palisade cells and oxygen gas from palisade cells takes place.

The leaf has numerous stomata that allows easy diffusion of gases into and out of the palisade tissue.

The leaf cuticle and epidermis are transparent to allow easy penetration of light to the photosynthetic tissue.

The palisade cells are numerous,elongated and contain numerous chloroplasts to trapping optimum light for photosynthesis.

The palisade tissue is just beneath the upper epidermis exposing them to trap optimum light for photosynthesis.

The leaf has numerous leaf veins consisting of    a.) xylem vessels and tracheids for transporting water and dissolved mineral salts from the soil to the photosynthetic tissue

b.) phloem tissue for translocation of of manufactured food from the leaf to storage organs and other parts of the plant .

Numerous and large air spaces in the spongy mesopyl layer for optimum gaseous exchange with the photosynthetic tissue.

Phylotaxy which is regular arrangement of leaves on the stem minimizes overshadowing and overlapping exposing all leaves to light for photosynthesis.

The prominent midrib and leaf veins reduces chances of rolling of leaves maintaining a large surface area for trapping optimum light for photosynthesis.

 

  1. C) Describe the structure and function of chloroplast
  2. Structure

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Function
  1. d) i) Give a word equation for photosynthesis

Carbon (iv)Oxide  +  water      sunlight         sugar   +  oxygen

 

Chlorophyll

 

NB  6CO2  +6H2O            C6H1206   + 6O2

.) Describe briefly the process of photosynthesis in plants.

The process of photosynthesis takes place in green plants allowing them to make their own food. The process is controlled by enzymes and involves a series of reactions that take place in chloroplasts. The raw materials required are water and carbon (iv) oxide.the process takes place in two consecutive stages i.e

Light reaction stage

It’s also called the light dependent stage as it requires light energy . the reactions take place in the granna of the chloroplast. light energy from the sun is trapped by chlorophyll in the chloroplast and converted into chemical energy. This energy splits water molecules into hydrogen ions and oxygen atoms a process is called photolysis. The oxygen atoms are released as aby product or used up in the process of respiration. The hydrogen ions formed are used in the dark stage of photosynthesis.

Water                                          hydrogen ions  +  oxygen atoms

2H2O                                                   4H+                             +   o2 g

Some of the light enrgy is used to combine a molecule called adenosine diphosphate(ADP) with a phosphate group to form the rich energy molecules called adenosine Tri-Phosphate(ATP)

ADP   +  P                                          ATP

Dark reaction stage

It’s also called the light independent stage of photosynthesis since light is not required because it can take place both in presence and absence of light. the reactions are controlled by enzymes. the hydrogen atoms released in the light stage are combined with carbon(iv)oxide to form simple sugars mainly glucose. The process uses energy from ATP. This is reffered to as carbon (iv) oxide fixation. The reactions take place in the stroma of chloroplast. The excess glucose is converted into starch or lipids for storage.

The general process of photosynthesis can be summarized by the following word and chemical equitions.

 

Water  +   carbon (iv) oxide        light & chlorophyll              glucose  +  oxygen

6H2o       + 6co2                                                                                           C6H12O6     + 6O2

9.) Factors that cause high rate of photosynthesis.

High water availability in the soil. Water  a raw material  for photosynthesis is split in presence of light to provide the hydrogen ions required in carbon (iv) fixation. When water is readily available more hydrogen ions are produced hence high rate of photosynthesis.

High light intensity. Light splits water molecules to hydrogen ions and oxygen atoms. Increasing light intensity increases the rate of photosynthesis up to a certain level beyond which other factors become limiting and rate of photosynthesis becomes constant.

rate

of

photosynthesis

 

 

 

Increasing light intensity

 

Day length.  Long day length especially at high latitudes (temperate regions) provides more light for photosynthesis causing an increase in the rate of photosynthesis.

Light quality. The preferred wavelengths for photosynthesis range between 400nm-700nm.  the rate of photosynthesis is higher in red and blue light and lower in all other types of light.

Concentration of carbon (iv) oxide. It’s a raw material required to combine with hydrogen ions to form simple carbohydrate molecules. Increasing the concentration of carbon (iv) oxide increases the rate of photosynthesis up to an optimum level beyond which other factors limit the rate of photosynthesis.

 

 

rate

of

photosynthesis

 

 

 

 

concentration of co2

 

  1. f) Give the differences between the light and dark reactions during photosynthesis
Light reactions Dark reactions

 

·      occurs in grana ·      occurs in stroma
  1. ii) What are organic compounds?
  1. b) i) What are carbohydrates?

– Compounds of carbon, hydrogen and oxygen

– The elements are in the ratio of I carbon: 2 hydrogen: 1 oxygen

  1. ii) Name the groups of carbohydrates
  1. State the general functions of carbohydrates
  1. c) i) what are proteins?
  1. ii) Name the types of amino acids
  1. State the classes of proteins
  1. Give the functions of proteins

d ) i) What are lipids

– Fats and oils

– They contain carbon, hydrogen and oxygen

– However, they contain a higher proportion of carbon and hydrogen but less oxygen that in carbohydrates

  1. ii) Name the types of lipids

iii) What are the building blocks of lipids?

  1. State the functions of lipids
  1. e) i) What are enzymes?
  1. ii) State the properties of enzymes

iii) State the factors that affect enzyme action

  1. Name the types of enzyme inhibitors
  1. What are the functions of enzymes?
  1. Holozoic
  1. Saprophytism
  1. Symbiosis
  1. b) Differentiate between omnivorous, carnivorous and herbivorous modes of nutrition
  2. Herbivorous
  1. Omnivorous
  1. c) i) What is dentition?
  1. ii) Distinguish between the terms homodont and heterodont
  1. Name the types of teeth found in mammals
  1. d) Describe the adaptations and functions of various types of mammalian teeth

Incisors

  1. Canines
  1. Premolars
  1. e) i)Draw a labeled diagram to represent internal structure of a mammalian tooth.

 

 

 

 

 

 

 

  1. ii) State the functions of the labeled structures labeled

Dentine

Enamel

Crown

Root

Cement

Neck

Pulp cavity

  1. f) i) What is dental formula?

 

  1. ii) Give examples of dentition in named mammals

iii) How would one use dental formula to identify the following?

Herbivores

 

 

 

 

Carnivore

 

 

 

 

 

 

  1. State the functions of the following structures in mammals

Carnassials

Pad of gum

  1. g) Name the common dental diseases
  1. ii) Explain the types of digestion

Intercellular

Extra cellular

  1. b) i) Draw human digestive system

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. ii) Describe the process of digestion in the various parts of the human digestive system

Mouth

Oesophagus

Stomach

Duodenum

Ileum

– produces intestinal juices

– Intestinal juice contains maltase, sucrase, lactase, erepsin, lipase, and several other peptidases

– Maltose is broken down to glucose and galactose by lactase

– Sucrose is acted upon by sucrase to glucose

Polypeptides are broken down into amino acids by erepsin

– Mucus secretion is to protect the ileum wall from digestion/autolysis

Colon

– Commonly called the large intestine

– Wider than the ileum

– has several mucus-producing cells

Highly folded for water absorption

– Also prepares food for egestion

– egestion is the process by which the insoluble parts of food are discharged from the body in form of faeces.

Rectum

– Muscular and enlarged

– it produces mucus

– used for storage and removal of faeces

Anus

– found at the exterior end of the rectum

The rectum opens into the anus

– The anus has anal sphincter to control egestion

– Anus is used for egestion of faeces

  1. c) Explain how mammalian intestines are adapted to perform their function

– The mammalian intestines are relatively long and coiled. This allows food enough time and increases surface area for digestion and absorption of products of digestion

– The intestinal lumen (inner wall) has projections called villi to increase surface area for absorption

– The villi have projections called micro-villi which lead to further increase of surface area for absorption

– The walls have glands which secrete enzymes for digestion e.g. maltase, sucrase, lactase, peptidase and enterokinase.

– Goblet cells (mucus secreting cells or glands) produce mucus which protects the intestinal wall from being digested and reduces friction.

– Intestines have openings of ducts which allow bile, a pancreatic juice into the lumen

– The intestines have circular and longitudinal muscles whose contraction and relaxation (peristalsis) leads to mixing of food with enzymes (juices) helps push food along the gut.

– The intestines are well supplied with blood vessels that supply oxygen and remove digested food.

– Intestines have lacteal vessels for transport of lipids (fats and oils)

– Intestines have thin epithelium to facilitate fast/rapid absorption/diffusion

  1. d) What is the function of hydrochloric acid in digestion?

– kills bacteria

– activates trypsinogen to trypsin which digests proteins to peptones and peptones to soluble amino acids

– provides acidic medium for gastric enzymes

  1. e) i) What is assimilation?

– The process by which digested food is taken up by cells and used in the body for various purposes.

  1. ii) State the uses of digested food in the bodies of animals

– Protection

– Repair

– Growth

– Energy production

  1. f) Name the types of food substances in the food that do not undergo digestion in human digestive system

Vitamins

  1. Mineral salts
  1. Roughage
  1. Water
  1. Basal Metabolic Rate (BMR)
  1. Occupation
  1. Age
  1. Body size
  1. Sex
  1. Climate

 

Test Procedure Observation Conclusion
Starch – add iodine solution – colour changes to blue black/dark blue Present
Reducing sugar Benedicts solution heat/boil/warm in hot water bath – colour changes to Green to yellow to orange to brown to red Present
Non-reducing sugar Dilute HCL, NaHCO3, heat/boil, warm in hot water bath – colour changes to Green to yellow to orange to brown to red Present
Proteins 1% CuSO4, 5% NaOH — colour changes to purple/violet Present
Ascorbic acid (Vitamin C) DCPIP drop wise DCPIP decolorized Present
Fats/oils (lipids) – rub on filter paper

– ethanol

– translucent mark

– white emulsions

present

 

FORM TWO TOPICS

 

  1. a) i) Define transport
  1. ii) Explain the necessity of transport in plants and animals
  1. b) i) Describe the structure and function of root hair
  1. ii) State ways in which the root hairs are adapted to their functions

 

  1. c) i) Compare the internal structure of a dicotyledonous root and a monocotyledonous root

Dicot root

 

 

 

 

 

 

Monocot root

 

 

 

 

 

 

  1. ii) State the similarities and differences between a dicotyledonous and monocotyledonous root

Similarities

Differences

Monocotyledonous Dicotyledonous
·      phloem and xylem are arranged in ring form alternately

·      pith present

·      phloem lies between radial rays of central xylem(star shaped)

·      pith absent

iii) Compare the internal structure of a monocotyledonous and dicotyledonous stem

Monocotyledonous                                Dicotyledonous

 

 

 

 

 

  1. Give the similarities and differences between a monocotyledonous and dicotyledonous stem

Similarities

Differences

Monocotyledonous Dicotyledonous
·      vascular bundles are many and scattered

·      some have hollow pith or pith is absent

·      no cambium layer therefore cannot undergo secondary growth

·      very little cortex

·      vascular bundles are few and arranged in a concentric ring near the epidermis

·      pith large and well developed

·      presence of cambium therefore undergoes secondary growth

·      cortex has several layers of cells

 

  1. State the differences between the internal structure of a root and a stem.
Root Stem

 

·      has root hairs

·      no cuticle

·      xylem and phloem arranged alternately

·      in xylem, the small vessels are towards the outside

·      cortex is the widest tissue

·      no root hairs

·      cuticle present

·      xylem and phloem arranged on the same radii

·      in xylem, the smallest vessels are towards the inside

·      pith is the widest tissue

  1. c) i) Name the transport structures of a flowering plant
  1. ii) State the ways in which xylem vessels are adapted to their function
  1. i) Why do flowering plants need water?
  1. ii) Describe the movement of water from the soil to the leaves of a tall plant

Soil

Stem

Leaves

iii) Name the process by which mineral salts enter into a plant

  1. Explain the forces that make water and mineral salts move through a plant
  1. i) What is transpiration?
  1. ii) Name the sites through which transpiration takes place in a plant

iii) State the importance of transpiration to plants

  1. Explain the structural factors that affect the rate of transpiration in plants
  1. explain the environmental factors that affect rate of transpiration in plants
  1. State the factors that cause increase in the rate of transpiration from leaves
  1. Explain how drooping of leaves on a hot sunny day is advantageous to a plant
  1. Explain how aquatic and terrestrial plants are adapted to deal with problems of transpiration
    1. Mesophytes
  1. i) What is translocation
  1. ii) Name the tissue which is responsible for translocation of manufactured food in flowering plants

iii) Name the processes that bring about the translocation of manufactured food

-active transport

Diffusion

Mass flow

Cytoplasmic streaming

  1. iv) Draw a labeled diagram to represent phloem tissue

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. State the functions of the labeled structures

cytoplasmic strands

Companion cell

Sieve tubes element

Describe an experiment you would carry out in order to demon-strate that phloem transports manufactured food substances in a plant

  1. Ringing experiment
  1. ii) Use the radioactive tracers

iii) Collecting exudate from stylets of aphids

  1. Describe an experiment you would carry out to demonstrate that xylem transports water
  2. Either
  1. OR
  1. ii) Distinguish between closed and open circulatory systems

iii) What are the advantages of the closed circulatory system over open circulatory system?

  1. iv) Distinguish between single circulatory system and double circulatory system

Single circulatory

Double circulation

-blood enters the heart twice in a complete circulation

– Pulmonary circulation from the heart to lungs and back

– Systemic circulation from the heart to body systems and back

  1. b) i) describe the general layout of the transport system in mammals
  1. ii) Describe the structure and function of the mammalian heart

 

 

 

 

 

 

 

 

 

 

iii) Explain how the mammalian heart is adapted to performing its functions

  1. iv) Explain why blood leaving the lungs may not be fully oxygenated
  1. e) Describe the structure and functions of the blood vessels
  2. Arteries
  1. Capillaries
  1. i) State the ways in which the composition of blood in the pulmonary arterioles differs from that in the pulmonary venules
Pulmonary arterioles Pulmonary venules
·      deoxygenated

·      high carbon IV oxide

·      low oxygen

·      more nutrients

·      oxygenated

·      low carbon IV oxide

·      high oxygen

·      less nutrients

  1. ii) Give the reasons why pressure of blood is greater in the arterioles than I the veins of mammals

iii) Name the common heart diseases in humans

  1. i) State the functions of mammalian blood
  1. ii) Describe how mammalian blood components carry out their functions

Plasma

Red blood cells (Erythrocytes)

White blood cells (leucocytes)

Blood platelets (thrombocytes)

iii) State the ways in which the red blood cells are adapted to their functions

  1. iv) State the structural differences between a red blood cell and a white blood cell.

Red blood cells                              white blood cells

 

 

  1. v) State the functional differences between a red blood cell and a white blood cell

Red blood cell                                     White blood cell

 

– Transports oxygen and carbon IV oxide      – protects body against harmful pathogens

  1. How does the heart increase blood flow to some parts of the body during exercise

Explain how oxygen and carbon Iv oxide are transported in the blood

Oxygen

Carbon IV oxide

Most carbon IV oxide is transported from tissues to lungs within the red blood cells and not in the blood plasma. Give the advantages of this mode of transport.

– PH of blood is not altered/homeostasis is maintained

– Within the red blood cell  is an enzyme, carbonic anhyrase which helps in fast loading(combining) and offloading of carbon Iv oxide

  1. i) what is blood clotting?

 

  1. ii) Name a protein, vitamin, an enzyme and a mineral element involved in blood clotting

Protein – fibrinogen/prothrombin

Vitamin –  k/quinine

Enzyme – thrombokinase/thromboplatin/thrombin

Mineral element – calcium

iii) Describe the blood clotting process

calcium ion

Fibrinogen thrombin   fibrin clot

vitamin K

  1. iv) State the role of blood clotting on wounds
  1. v) Explain why blood flowing in blood vessels does not normally clot

– Presence of anticoagulant in blood

  1. list the major types of human blood groups

Universal donor

Universal recipient

iii) What is the difference between rhesus positive and Rhesus negative blood samples?

  1. vi) What is blood transfusion?

– Introduction of blood from one person to another

  1. v) Under what conditions would blood transfusion be necessary in people?
  1. vi) How can low blood volume be brought back to normal?

How may excessive bleeding result in death?

State the precautions that must be taken before blood transfusion

  1. j) i) What is immunity?

– Resistance to disease by organisms

  1. ii) Distinguish between natural and acquired immunity

iii) What are allergic reactions?

  1. vi) How does an allergic reaction occur?

vii) State the role of vaccination against certain diseases

  1. a) i) What is gaseous exchange?
  1. ii) Why is gaseous exchange important to organisms?
  1. b) i) name the structure used for gaseous exchange by plants

 

  1. ii) Briefly describe the structure of stomata

iii) State the factors which affect stomatal opening

  1. iv) Name the theories suggesting the mechanism of opening and closing of stomata
  1. v) Describe the mechanism of opening and closing of stomata

 

  1. c) i) State the ways in which leaves of plants are adapted to gaseous exchange
  1. ii) Describe how gaseous exchange takes place in terrestrial plants
  1. How is aerenchyma tissue adapted to its function?
  1. Explain stomatal distribution in plants of different habitats
  1. d) i) List the types of respiratory surfaces of animals
  1. ii) State the characteristics of respiratory surfaces in animals

iii) Describe gaseous exchange in protozoa

  1. e) i) Make a labeled drawing of a fish gill

 

 

 

 

 

 

  1. ii) How is a fish gill adapted to its function?

iii) Discuss gaseous exchange in bony fish

  1. iv) What is counter-flow system?
  1. What is the advantage of counter-flow system?
  1. f) i) Describe the mechanism of gaseous exchange in terrestrial insects
  1. ii) State how traceholes are adapted to gaseous exchange
  1. g) i) What is breathing?
  1. ii) Name the structures in humans that are used in gaseous exchange

iii) Describe the mechanism of gaseous exchange in a mammal

Breathing in

Breathing out

  1. iv) Explain how mammalian lungs are adapted to gaseous exchange
  1. v) Name the features of alveoli that adapt them to their function
  1. Give the conditions under which the carbon iv oxide level rises above normal in

mammalian blood

  1. Explain the physiological changes that occur in the body to lower the carbon iv oxide level back to normal when it rises
  1. h) i)Describe the factors which control the rate of breathing in humans
  1. ii) Name the respirator diseases
  1. a) i) Define respiration
  1. ii) Explain the significance of respiration in living organisms

iii) Where does respiration take place?

  1. b) i) Draw and label a mitochondrion

 

 

 

 

 

 

 

 

 

  1. ii) State the most important function of mitochondria

iii) Give the functions of the labeled parts

Outer membrane

Cristae

Matrix

  1. c) Explain the roles of enzymes in respiration
  1. d) i) What is aerobic respiration
  1. ii) Give a word equation for aerobic respiration

iii) What are the end products of aerobic respiration?

 

  1. e) i) What is anaerobic respiration
  1. ii) What are obligate anaerobes?

iii) What are facultative anaerobes?

  1. iv) State the word equation representing anaerobic respiration in plants

– Glucose   ethanol +  carbon iv oxide  + energy

  1. v) Name the end products of anaerobic respiration in plants
  1. g) i) Give a word equation of anaerobic respiration in animals

– Glucose    lactic acid + energy

  1. ii) Name the end products of respiration in animals when there is insufficient oxygen supply

iii) Why is there a high rate of lactic acid production during exercise?

  1. iv) Why does lactic acid level reduce after exercise?
  1. v) State why accumulation of lactic acid during vigorous exercise lead to an increase in heartbeat

State the economic importance of anaerobic respiration

What is oxygen debt?

 

  1. h) i) What is respiratory quotient(RQ)?

RQ  =   volume of CO2 produced

Volume of oxygen consumed

  1. ii) Why are respiratory quotient important

iii) Name the respiratory substrates

  1. iv) Why does anaerobic respiration of a given substrate yield a smaller amount of energy than aerobic respiration?

– Some energy locked up in intermediate products like ethanol in plants and lactic acid in animals

  1. iv) Explain the disadvantages of anaerobic respiration
  1. v) Mention the types of experiments carried out for respiration
  1. a) i) Define the following terms

Excretion

Secretion 

Egestion

Homeostasis

  1. ii) Explain why excretion is necessary in plants and animals

-products of excretion are usually harmful while some are toxic

– if allowed to accumulate in the cells they would destroy tissues and interfere with normal metabolism

– They are therefore removed through excretion

 

  1. b) i) Describe how excretion takes place in green plants
  1. ii) Why do plants lack complex excretory structures like those of animals?
  1. ii) State the excretory products of plants and some of their uses to humans
  1. c) i) Describe excretion in unicellular organisms

-examples are amoeba and paramecium

-They have to remove waste products such as carbon IV oxide and nitrogenous substances e.g urea and ammonia

– These diffuse from the body surface into the surrounding water

– Diffusion is due to large surface area

 

  1. ii) List excretory organs and products of mammals

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d)i) Draw and label a mammalian skin

  1. ii) Explain how the mammalian skin is adapted to its functions

Epidermis

Dermis

  1. e) What is the role of lungs in excretion?
  1. f) State the functions of the liver
    1. Excretion
  1. ii) Homeostasis

Deamination

Detoxification

Thermal regulation

  1. g) i) Draw a labeled diagram of mammalian nephrone

 

  1. ii) Describe how the human kidney functions

iii) State the adaptations of proximal convoluted tubule to its function

  1. iv) Name the common kidney diseases
  1. a) i) Why is homeostatic control necessary?
  1. ii) What is internal environment?
  1. b) i) Why is constant body temperature maintained by mammals?
  1. ii) Explain the advantage gained by possessing a constant body temperature

iii) How do mammals regulate body temperature?

  1. iv) Why does body temperature of a healthy person rise up to 37oC on a hot humid day?
  1. v) Name the structures in the human body that detect external temperature changes
  1. vi) State the advantages that organisms with small surface area to volume ratio experience over those with larger

Explain why individuals with smaller sizes require more energy per unit body weight than those with larger sizes.

 

  1. c) i) What is the meaning of osmoregulation?
  1. ii) State the importance of osmoregulation

– Maintenance of constant level of water and slats (osmotic pressure) for optimum/suitable conditions for metabolism suitable for cellular functions

iii) State the ways by which desert mammals conserve water

  1. iv) Explain why some desert animals excrete uric acid rather than water
  1. v) Explain why eating a meal with too much salt leads to production of a small volume of concentrated urine
  1. vi) Explain how marine fish regulate their osmotic pressure
  1. d) i) What is the biological significance of maintaining a relatively constant sugar level in a human body?

 

  1. ii) Discuss the role of the following hormones in blood sugar control

Insulin

Glucagon

  1. e) Explain the part played by antidiuretic hormone in homeostasis
  1. f) What is the role of blood clotting in homeostasis?
  1. g) Describe the role of the following hormones in homeostasis
    1. Aldosterone
  1. Adrenaline
  1. h) i) Distinuish between diabetes mellitus and diabetes insipidus
  1. ii) How can high blood sugar level in a person be controlled?

iii) Why does glucose not normally appear in urine even though it is filtered in the mammalian Bowman’s capsule?

  1. iv) When is glycogen which is stored in the liver converted into glucose and released into the blood?
  1. v) How would one find out from a sample of urine whether a person is suffering from diabetes mellitus?

 

 

 

 

 

FORM III TOPICS

 

  1. a) i) What is meant by the term binomial nomenclature?
  1. ii) State briefly the general principles of classification of living organisms
  1. b) State the main characteristics of the five kingdoms of organisms
    1. Monera
  1. c) Describe the economic importance of:
  2. Fungi

Bacteria

  1. d) State the main characteristics of the following division of kingdom plantae
  2. Bryophyte
  1. Pteridophyta
  1. e) Name sub-divisions of spermatophyte and state the characteristics of each class
  2. Gymnospermae (cornifers)
  1. Angiospermae (flowering plants)

Dicotyledonae

Monocotyledonae

  1. iv) State the importance of plants
  1. f) i) Give the general characteristics of phylum arthropoda
  1. State the characteristics of the following classes of arthropoda

Diplopoda

Chilopoda

Insecta

Crustacean

Arachnida

iii) State the economic importance of insects

 Beneficial effects

Harmful effects

  1. g) i) State the general characteristics of chordate

Give the characteristics of the following classes of chordate

Pisces

Amphibian

Reptilia

Aves

Mammalian

  1. i) What is a dichotomous key?
  1. iv) Describe the procedure of using a dichotomous key. Make a list of major features of the characteristics to be identified

ivYou are provided with a specimen kale leaf. Use the dichotomous key below to identify the taxonomic group to which the specimen belongs. Show the steps (number and letter) in the key that you followed to arrive at the identify of the specimen

1  a) leaf broad                              go to 2

  1. b) leaf narrow Araicaria

 

2  a)  leaf parallel veined                Cynodon

  1. b) leaf net-veined                             go to 3

3 a) leaf with one lobe (simple)              go to 4

  1. b) leaf with many lobes (compound)  Grevellea

4 a)  leaf fleshy                              Kalanchoa

  1. b) leaf not fleshy go to 5

5  a)  leaf petiole modified to form sheath      go to 6

  1. b) leaf petiole not modified to form sheath Brassica

6 a)  leaf purple                             Tradescantia

  1. b) leaf green Commelina

steps – 1a, 2b, 3a, 4b, 5b

Identify – Brassica

  1. v) You have been provided with four animals labeled K (mature adult housefly), L (mature adult grasshopper, M(maize flour beetle) and N(worker termite) use the dichotomous key below to identify the specimens. Write down in the correct order, the steps (number and letter) in the key that you followed to arrive at your answer.

Dichotomous key

1 a)  animal with wings                          go to 2

  1. b) animal without wings go to 7

2 a) with two pairs of wings                          go to 3

  1. b) with one pair of wings Diptera

3 a) with membranous wings                         go to 4

  1. b) hind pair of membranous wings                     go to 6

4 a) with long abdomen                         Odontata

  1. b) medium sized abdomen go to 5

5 a) wings with coloured scales                      Lepidoptera

  1. b) wings without scales                       hymenoptera

6  a) forewings hard and shell-like                 coleopteran

  1. b) forewings hard but not shell-like Orthoptera

7  a) body horizontally flattened                    Isoptera

  1. b) body laterally flattened Symphonoptera

Identify the orders o the various specimens as per the table below

Specimen Order Steps followed
K-housefly

L- grasshopper

M- beetle

M-termite

Diptera

Orthoptera

Coleoptera

Isoptera

 

1a,2b

1a, 2a, 3b, 6b

1a, 2a, 3b, 6a

1b, 7a

 

2 a) Define the following ecological terms

  1. Ecology
  1. b) i) What are abiotic factors?
  1. ii) Explain how abiotic factors affect living organisms

Wind

Temperature

Light

Humidity

PH

Salinity

Topography

Rainfall (water) or precipitation

Pressure

Mineral salts (trace elements)

  1. c) i) What are biotic factors?
  1. ii) Give examples of biotic factors affecting ecosystems
  1. d) Discuss how the various biotic factors affect living organisms
  2. Competition
  1. Predation
  1. Diseases and parasites
  1. Symbiotic
  1. Human activities

Saprophytism

e)i) What is nitrogen cycle?

  1. ii) Draw a simplified diagram representing the nitrogen cycle

 

 

 

 

 

 

 

iii) Describe the nitrogen cycle

  1. f) i) Describe how energy flows from the sun through the various trophic levels in an ecosystem

Give the reasons for loss of energy from one trophic level to another in a food chain

Why are green plants referred to as primary producers in an ecosystem?

– They utilize the energy from the sun to manufacture food for themselves and for subsequent trophic level (consumers) and other organisms

  1. vi. Explain the following terms giving suitable examples

Food chain

 

Food web

 

 

Pyramid of numbers

 

 

 

 

Pyramid of biomass

 

 

 

 

Account for the decrease of biomass in the successive trophic levels

  1. h) i) Describe the three characteristics of a population growth

Dispersion

Density

  1. ii) Explain how the following methods are used to estimate population of organisms

quadrat method

Line transect

Belt transect

Capture-recapture method

  1. b) i) What is pollution?
  1. ii) Explain the various human activities that have caused pollution

Causes and effects of air pollution

 

Control of air pollution

State the causes, effects and methods of controlling and  prop roots for support water pollution

Causes and effects

Control of water pollution

-pollution caused by domestic effluents may be controlled by treating domestic waste, using biotechnology, banning the use of phosphate-based detergents, using plastic pipes instead of those made from lead, recycling gabbage, using biodegradable detergents.

Pollution caused by industrial waste may be controlled by treating/cooling industrial waste, carrying out environmental impact assessment before establishing industries

Oil spillage may be controlled by cleaning spilled oil biotechnology and penalizing the industry individual or companies which cause oil spills/water pollution

Pollution caused by agrochemicals may be controlled by using mechanical control of weeds, biological control of weeds and pests, biodegradable organic fertilizer herbicides, insecticides pesticides, organic farming educate farmers on the use of correct amount of agrochemicals

  1. iv) State the causes /effects and control methods of soil pollution

Causes and effects

Control of soil pollution

  1. v) Define biological control give suitable examples
  1. vi) What is eutrophication?

vii) What are the effects of eutrophication?

– The plants block light from reaching plants underneath hence no photosynthesis

The plants die and decompose leading to lack of oxygen hence animals also die

  1. c) Describe the symptoms, mode of transmission and control of cholera, typhoid malaria and amoebic dysentery in humans
Disease Causative agent Transmission Symptoms Control
Cholera Vibrio cholerae(bacterium) Contaminated food or water

Spread by flies form faeces

Intestinal pain

Diarrhea

Vomiting

Dehydration

Proper hygiene e.g. boiling drinking water

Vaccination

Typhoid Salmonella typhi (bacterium) ·      contaminated food or water

·      spread by flies from faeces

·      fever

·      rashes

·      diarrhea + blood from bowels

·      proper sanitation

·      vaccination

Malaria Plasmodium (protozoa) – bite by infected female anopheles mosquito – fever

Joint pains

Vomiting

Headache

Anaemia

·      killing mosquito

·      killing the mosquito larvae

·      draining stagnant water

·      clearing bushes

·      treatment

·      sleep under mosquito net

Amoebic dysentery (amoebiasis) Entamoeba hystolytica (bacterium) – contaminated food or water due to improper disposal of faeces – intestinal pain

Diarrhea

Vomiting

Dehydration

·      sanitation

·      personal hygiene

·      cook food well

·      treatment using drugs

  1. d) Discuss Ascaris lumbricoides under the following sub-headings
  2. Mode of transmission
  1. Effects of parasite on the host
  1. Control and prevention
  1. e) Discuss schistosoma under the following sub-headings
  2. Mode of transmission
  1. Effects on host

 

  1. iv) Control and prevention
  1. ii) Why is reproduction important?

iii) Name the types of reproduction

  1. b) i) What is cell division?
  1. ii) What are chromosomes?
  1. c) i) What is mitosis?
  1. ii) Describe the five stages of mitosis

Interphase

 

 

Prophase

Metaphase

 

 

 

 

 

 

Anaphase

 

 

 

 

 

 

Telophase

 

 

 

 

 

 

 

  1. ii) State the significance of mitosis

 

  1. d) i) What is meiosis?
  1. ii) State the significance of meiosis

iii) Give a summary of the stages of meiosis

First meiotic division

Interphase

 

 

 

Prophase I

 

 

 

Metaphase I

 

 

 

Anaphase I

 

 

Telophase I

 

 

 

Second meiotic division

Prophase II

 

 

 

Metaphase II

 

 

 

Anaphase II

-chromatids of each chromosome separate to the poles

 

 

 

 

Telophase II

 

 

 

  1. iv) Give the similarities between mitosis and meiosis
  1. v) What are the differences between mitosis and meiosis?
Mitosis Meiosis
·      maintenance of chromosome number (diploid)

·      takes place in somatic cells/growth

·      no crossing over/no variations

·      results into 2 daughter cells

·      no pairing/no synapsis/no bivalent formed

·      a one division process of four stages

·      reduction/halving of chromosomes (haploid

·      occurs in reproductive cells/gonads/produces gametes

·      crossing over takes place/variation occurs

·      results into 4 daughter cells

·      there is paring/synapsis/bivalent

·      a two division process of four stages each

 

  1. d) i) What is asexual reproduction
  1. ii) What is the significance of sexual reproduction in living organisms?

iii) State the advantages of sexual reproduction

  1. iv) Give the disadvantages of sexual reproduction
  1. e) i) What is asexual reproduction?
  1. ii) State the advantages of asexual reproduction

iii) Give the disadvantages of asexual reproduction

  1. iv) Explain how reproduction occurs by the following methods of asexual reproduction

Sporulation

Budding

Binary fission

  1. f) i) What is a flower?
  1. ii) Draw a longitudinal section of a labeled diagram of a flower

 

 

 

 

iii) Give the functions of the parts of a flower

Receptacle

Calyx

Corolla

Androecium

Gynaecium

  1. iv) What is inflorescence?
  1. v) Explain the meaning of the following terms which describe flowers

Hermaphrodite

Unisexual

Carpelate

Staminate

Dioecious plants

 

Monoecius plants

Complete flower

Incomplete flower

  1. vi) Explain the meaning of the following types of ovary

Superior

Inferior (epigynous)

  1. g) i) What is pollination?
  1. ii) Explain the types of pollination

iii) State the advantages of pollination

  1. iv) List the agents of pollination
  1. v) How are flowers adapted to wind and insect pollination?

Insect pollinated flowers (entomophilus)

 

Wind pollinated flower (anemophilus)

  1. vi) State the ways in which plants prevent self-pollination

vii) Give the characteristics that ensure cross pollination takes place in flowering plants

viii State the advantages of cross pollination

  1. h) i) What is fertilization?

– Fusion of male and female gametes to form a zygote

  1. ii) Describe how fertilization takes place in a flower

 

 

 

 

 

 

iii) What is double fertilization?

  1. iv) Name the changes that Occur in a flower after fertilization
  1. i) Distinguish between a fruit and a seed
  1. ii) How is a seed formed?

iii) Draw a labeled diagram of a seed

  1. iv) Describe the main parts of a seed

Testa

Hilum

Micropyle

Radicle

Cotyledons

  1. v) Draw a labeled diagram of a fruit

 

 

 

 

 

 

 

  1. vi) How is a fruit formed?

vii) Explain the importance of fruits in the survival of plants

  1. iv) State the differences between a seed and fruit
Seed Fruit
·      fertilized ovule

·      attached to placenta through funicle

·      one scar called hilum

·      has seed coat/testa

·      seed wall undifferentiated

·      fertilized ovary

·      attached to branch through a stalk

·      two scars (style scar and stalk scar)

·      has fruit wall/pericarp

·      fruit wall is differentiated

  1. i) What is placentation?
  1. ii) Explain the following types of placentation

Marginal

 

 

 

Basal

 

 

 

Parietal

 

 

 

 

 

 Axile

 

 

 

Free central placentation

 

 

 

 

 

 

  1. i) How are fruits grouped?

Simple fruits

Aggregate fruits

Multiple (compound) fruits

  1. ii) What are succulent fruits?

iii) Give types of juicy fruits

-Berry

 

 

 

 

 

 

Drupe

 

 

 

 

Pome

  1. iv) What are dry fruits?
  1. v) What are dehiscent fruits?
  1. vi) Give types of dehiscent fruits

Legumes

Follicle

Capsule

vii) What are indehiscent fruits?

  1. Give main types of indehiscent fruits

Nut

Achene

  1. i) What is seed and fruit dispersal?
  1. ii) Why is dispersal of seeds and fruits necessary?

iii) Explain how seeds and fruits are adapted to various methods of dispersal

Adaptations for wind dispersal

Water dispersal seeds

Animal dispersal seeds

Self dispersal/explosive

  1. ii) State the advantages and disadvantages of external fertilization

Advantages

Disadvantages

iii) State the advantages and disadvantages of internal fertilization

Disadvantages

Advantages

iii) Give a reason why it is necessary for frogs to lay many eggs

  1. iv) Compare external and internal fertilization
External Internal
·      occurs in water outside the bodies of animals

·      many eggs are laid

·      usually less contact between male and female

·      both fertilized eggs exposed to danger

·      occurs inside the body of the female animal

·      fewer eggs released from ovary

·      very close contact in form of copulation between male and female

·      Fertilized eggs are enclosed hence highly protected inside females’ body.

  1. b) i) Draw and label the human male reproductive system

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. ii) Describe how the mammalian male reproductive system is adapted to perform its functions

Penis

Scrotum

Epididymis

Vas deferens

Gametes

Accessory glands

 

iii) How is the sperm adapted to perform its function?

  1. c) i) Draw and label the human female reproductive system

 

 

 

 

 

 

 

 

  1. ii) Describe how the various structures of the human female reproductive system are adapted to their function

Ovaries

Oviducts (fallopian tube)

Uterus

Cervix

Vagina

Clitoris

iii) Explain how the ovum is adapted to its function

  1. iv) Explain the differences between sperm and ovum
Sperm Ovum
·      long with a tail and head

·      small

·      locomotory

·      stores little food

·      has  acrosome (tip with lytic enzymes)

·      nucleus prominent but cytoplasm negligible

·      spherical

·      large

·      stationary

·      a lot of food stored in yolky cytoplasm

·      lacks acrosome

·      a lot of cytoplasm

  1. d) i) Explain the process of fertilization

 

 

 

 

 

 

  1. i) Explain the process of implantation

 

iii)

 

 

 

 

 

 

 

 

 

 State the functions of umbilical cord

  1. iv) State the role of placenta
  1. e) i) What is gestation period?
  1. ii) Explain the functions of the membranes associated with placenta

Chorion

Amnion

Yolk sac

Allantois

iii) Explain the events that take place to facilitate parturition

  1. iv) State the reasons why later in pregnancy the ovary will b e removed without disturbing the pregnancy
  1. f) i) What are secondary sexual characteristics

– Characteristics (physiological and anatomical) that start developing at puberty due to the influence of male and female hormones

  1. ii) State the main secondary changes in Boys

Girls

iii) Describe the role of hormones in secondary sexual characteristics inBoys

Follicle stimulation hormone (FSH)

Testosterone

Girls

FHS

LH

Oestrogen

Progesterone

Prolactin

Oxytocin

  1. g) i) What is menstruation?

– vaginal discharge due to disintegration of endometrium

  1. ii) Describe the role of hormones in the human menstrual cycle

iii) What is menopause?

  1. h) Explain the symptoms, methods of transmitting and prevention (control) of the following sexually transmitted diseases
  2. i) Gonorrhea

–  caused by a bacterium called Neisseria gonorrhea

– transmitted through sexual intercourse,

– infects urethra and vaginal tract (epithelia)

– Symptoms include pain, discharge of mucus and bad smell (females)

– Effects include sterility, heart diseases, blindness of foetus and arthritis

– Treatment by antibiotics

– Control and prevention by proper sexual conduct

  1. ii) Herpes

iii) Syphilis

  1. iv) Trichomoniasis
  1. v) Hepatitis

– Viral disease

Affects the liver

Transmitted through sexual intercourse

No known treatment

  1. vi) Candidiasis

vii) HIV/AIDS

  1. Growth
  1. Development
  1. b) i) Differentiate growth in plants and animals
Plants Animals
·      confined to shoots/root tips (apical)

·      have definite growth regions(meristems)

·      often indefinite/continuous

·      growth results in branching

·      affected by light, auxins, hormones and temperature

·      occurs all over the body(intercalary)

·      different parts grow at different rates(allometric)

·      maximum growth on maturity(definite)

·      no branching

·      affected by nutrients

  1. ii) List the processes involved in growth
  1. List the parameters used to measure growth
  1. iv) Name the patterns of growth in organisms
  1. c) i) Name the different types of growth curves
  1. ii) Draw a sigmoid growth curve and explain its different phases/stages

 

 

 

 

A-lag phase

– Slow growth rate at first

Organism adapting to the environment

B-exponential phase

C- Stationery phase (plateau)

– Birth rate equals death rate (equilibrium)

Lack of nutrients, accumulation of toxic waste products

D-phase of decline

iii) Draw an intermittent curve and explain the various stages

 

 

 

 

 

A-growth

B-no growth

C- moulting/ecdysis

  1. d) i) What is seed dormancy?
  1. ii) State the biological importance of seed dormancy

iii) State the factors which cause seed dormancy

Internal factors

External factors

  1. iv) Give the conditions necessary to break seed dormancy
  1. e) i) What is seed germination?
  1. ii) What is viability

iii) Discuss the various conditions necessary for the germination of seeds

Water

Air

Suitable (optimum) temperature

Enzymes

Viability

  1. iv) Name and describe the types of germination

Epigeal

 

 

 

 

 

 

 

 

 

cotyledons are brought above the ground level during germination due to elongation of hypocotyls

Hypogeal

  1. v) Name the part of the bean seed that elongates to bring about epigeal germination

hypocotyl

  1. vi) Account for the loss in dry weight of cotyledons in a germinating bean seed

vii) Describe the physiological changes that occur in a seed during germination

viii) Explain the biological significance of cotyledons being brought above the ground in epigeal germination

  1. f) i) Distinguish between primary and secondary growth

Primary growth

Secondary growth

-occurs at the cambium meristems

Increases width (girth) of the stem

  1. ii) What are meristems

iii) State the characteristics of meristematic cells

  1. iv) State the location and function of the following meristematic tissues

Apical meristem

Intercalary meristem

Lateral meristems

  1. v) Describe primary growth
  1. vi) Describe secondary growth in plants

vii) State the significance of secondary growth

  1. g) i) Describe one method which can be used to measure the average growth rate of a single leaf of a plant

EITHER

Average rate of growth =    total increase in length

period taken to achieve full length

OR

Average rate of growth  =    total increase in area

period of time taken to achieve final area

  1. i) Describe how the growth of a root can be determined

Materials

Procedure

time observe the intervals with the marks

iii) A boy hammered a nail in the bark of a tree at a height of 1.5metres above the ground.

Four years later, the nail was found at the same height although the tree had grown 3 meters taller. Explain the above observation

The nail was hammered at a point where vertical growth had stopped/further growth was confined to increase in width/diameter.

Vertical growth is confined to tips/apex/vertical apical meristem

  1. h) i) Describe the role of hormones in growth and development of plants

indole acetic acid/IAA/ auxins

Cell division/increase in cell division

Tropic responses

Cell elongation/increases in ell elongation

Development of abscision layer

Growth of ovaries into fruits/parthenocarpy/initiates flowers

Inhibits growth of lateral buds/produces apical dominance

Stimulates adventitious/lateral roots

Gibberellins (Gibberellic acid/GA3

Promote cell elongation/rapid cell division/increase in length of the internodes

Promote fruit formation without fertilization/parthenocarpy

Reduces root growth

Breaks seed dormancy/promotes germination

Cytokinnins (Kinnins/Kinnetin/Zeatin)

Ethylene (ethynelC2H4)

Abscisic acid (ABA) abscisin hormone/dormin)

Traumatin

Florigen

  1. ii) State the applications of plant hormones in agriculture

iii) Explain apical dominance

  1. iv) Describe the role of hormones in the growth and development of animal

somatotrophin (growth  hormones)

Thyroxine

Androgens

Oestrogen

Ecdysone

  1. t) i) What is metamorphosis?

 

  1. ii) Explain complete metamorphosis

radical changes in the body during the life cycle of an organism

called holometabolous development

example is egg    larva  pupa   adult  (imago)

occurs in animals such as butterfly and bee

 

 

 

 

iii) What is the significance of each of the four stages in complete metamorphosis?

Larval stage

Pupa

Adult

  1. iv) Describe incomplete metamorphosis

 

  1. v) Name the hormones that control metamorphosis in insects
  1. vi) State the advantages of metamorphosis in the life of insects

 

FORM IV TOPICS

 

  1. a) i) Define the term genetics

iii) List some characteristics which are inherited

iii) State the importance of genetics

  1. b) i) Explain the meaning of the following terms

Heredity

Trait

Gene

Allele

Chromosomes

DNA

  1. ii) List the types of chromosomes
  1. c) i) What is variation?
  1. ii) State the causes of variation in organisms

iii) Name the types of variation

  1. iv) Explain the following terms

Acquired characteristics

Inherited characteristics

Genotype

Phenotype

Dominant gene (character)

Recessive gene

Homozygous

Heterozygous

F1 and F2

  1. d) i) Explain Mendels first law of inheritance
  1. ii) Give an example of this law

iii) What is monohybrid inheritance?

  1. What is complete dominance?
  1. e) i) What is meant by co dominance?
  1. ii) Give an example of co dominance

In a certain plant species, some individual plants may have only white, red or pink flowers. In an experiment a plant with white flowers was crossed with a parent with red flowers. Show results of F1 generation. Use letter R for red gene and W for white gene.

If the plants form F1 were selfed, work out the phenotype ratio for the F2 generation

Phenotypic ratio   1red:2pink:1white

Genotypic ratio 1:2:1

  1. f) i) What is a test cross?

– A cross between an individual showing a character for a dominant gene(that is homozygous or heterozygous) with a homozygous recessive individual

OR

  1. ii) State the importance of a test cross in genetics

iii) What are multiple alleles?

  1. iv) Explain the inheritance of ABO blood groups

Give a worked example using parents with heterozygous blood groups AO and BO

  1. Explain the inheritance of Rhesus factor (Rh) in human beings

Let the gene for dominant Rh factor be R while gene for recessive be r

  1. g) i) What does the term linkage mean?

– These are genes which occur together on a chromosome and are passed to offspring without being separated

  1. ii) Define the term sex-linked genes

iii) What is meant by the term sex linkage?

  1. iv) Name the sex-linked traits in humans
  1. v) Give an example of a sex linked trait in humans on:

Y Chromosome

X Chromosome

  1. vi) In humans red-green colour blindness is caused by a recessive gene C, which is sex-linked. A normal man married to a carrier woman transmits the trait to his children. Show the possible genotypes of the children.

Let C represent the gene for normal colour vision (dominant)

Let c represent the gene for colour blindness

Parental phenotype Norman man x carrier woman

  1. Haemophilia is due to a recessive gene. The gene is sex-linked and located on the x chromosome. The figure below shows sworn offspring from phenotypically normal parents

What are the parental genotypes?

Work out the genotypes of the offspring

  1. h) i) What is mutation?
  1. ii) Describe how mutations arise

iii) State the factors that may cause mutation

Radiations                        Effects

X-rays                        gene/chromosome alteration

Ultra violet rays                  structural distortion of DNA

Chemicals                                 Effect

– colchicines               prevents spindle formation

Cyclamate                          chromosome aberrations

Mustard gas                chromosomes aberrations

Nitrous acid                adenine in DNA is deaminated so behaves like guanine

Acridone orange                 addition and removal of bases of DNA

Formaldehyde

  1. iv) State the characteristics of mutations
  1. v) Explain chromosomal mutation

– Change in nature, structure or number of chromosomes

  1. vi) Explain how the following types of chromosomal mutations occur

Duplication

Inversion

Deletion

Translocation

Non-disjuntion

Failure of homologous chromosomes/sister chromatids to separate/segregate during meiosis

Polyploidy

vii) What are gene mutations?

Deletion

Inversion

Insertion

Substitution

  1. State the practical applications of genetics
    1. Breeding programmes (research)

– legal questions of paternity knowledge of blood groups or blood transfusion

  1. iv) Genetic counseling
  1. v) Others

– Pre-sex determination

Understanding human evolution and origin of other species.

  1. a) i) Explain the meaning of evolution
  1. ii) Differentiate organic evolution from chemical evolution as theories of origin of life

iii) What is special creation?

  1. b) Discuss the various kinds of evidence for evolution
  2. Fossils

 

  1. Comparative anatomy
  1. Comparative serology/physiology
  1. Geographical distribution

Examples

  1. Cell biology (cytology)
  1. c) i) State the evolutionary characteristics that adopt human beings to the environment

– Brain

– Eyes

– Upright posture/bipedal locomotion

– prehensible arm/hand

– Speech

  1. ii) State the ways in which Homo sapiens differs from Homo habilis
  1. d) i) Explain Larmarck’s theory of evolution
  1. ii) Explain why Lamarck’s theory of evolution is not accepted by biologists today

iii) Explain Darwin’s theory of evolution

  1. e) i) What is natural selection?

– Organisms with certain characteristics are favoured by the environment

Such organisms tend to survive and produce viable offspring

Others not favored are eliminated from subsequent generations

  1. ii) With examples, explain how natural selection takes place

iii) State the advantages of natural selection to organisms

  1. iv) State the ways in which sexual reproduction is important in the evolution of plants and animals
  1. v) Explain the significance of mutation in evolution

– Mutation bring about variation which can be inherited

– Some of these variations are advantageous to the organism

– Others are disadvantageous

– The advantageous variations favour the organism to compete better in the struggle for survival

– This results into a more adapted organism to its environment or new species/varieties

– Those with disadvantageous characters will be discriminated against therefore eliminated from the population/death/perish

  1. vi) Plain why it is only mutations in genes of gametes that influence evolution

vii) How would you prove that evolution is still taking place?

  1. f) How has industrial melaninism i.e. peppered moth contributed towards the mechanism of evolution

– This is an example of natural selection

– The peppered moth exists in two distinct forms, the speckled white form (normal form) and a melanic form (the black/dark)

– They usually rest on leaves and barks of trees that offer camouflage for protection

– Originally the “speckled white” form predominated the unpolluted area of England

– This colouration offered protection against predatory birds

– Due to industrial pollution tree barks have blackened with soot

– The white form underwent mutation

– A black variety/mutant emerged suddenly by mutation

– It had selective advantage over the white forms that were predated upon in the industrial areas

– The speckled white form is abundant in areas without soot/smoke

  1. a) i) Define irritability, stimulus and response irritability

-also called sensitivity

– Responsiveness to change in environment

Stimulus

A change in the environment of organism which causes change in organism’s activity

Response

  1. ii) State importance of irritability to living organisms
  1. b) i) What are tactic responses?
  1. ii) What causes tactic responses?

iii) State the importance of tactic response to:

Members of kingdom protista

Microscopic plants

– escape injurious stimuli/seek favorable habitats

  1. Name the type of response exhibited by:

Euglena when they swim towards the source of light

  1. State the advantages of tactic responses to organisms
  1. c) i) Define the term tropism
  1. ii) Explain the various types of tropism in plants

Phototropism

Geotropism

Thimotropism/Haptotropism

Hydrotropism

chemotropism

iii) State the ways in which tropisms are important to plants

  1. iv) Explain the differences between tropic and tactic responses
Tropisms Taxes
–       growth curvature in response

–       slow

–       influenced by hormones

–       locomotory response

–       fast

–       external influence

  1. d) The diagram below represents growing seedlings which were subjected to unilateral light at the beginning of an experiment

 

 

 

 

 

 

R

  1. State the results of P, Q and R after 5 days
  1. Account for your results in (i) above

P- Growth substance/growth hormone/IAA/auxin are produced by the stem tip

Q- Source of auxin has been removed

R- The auxins cannot be affected by light because the tip has been covered

  1. State the expected results after 3 day is if the box were removed
  1. e) In an experiment to investigate a certain aspect of plant response, a seedling was placed horizontally as shown in diagram I below. After seven days the appearance of the seedling was as shown in diagram 2

Account for the curvature of the shoot and root after the seven days

  1. Shoot
  1. Root

 

  1. f) What is etiolation?
  1. a) i) What is coordination in animals
  2. – The linking together of all physiological activities that occur in the body so that they take place at the right time and in the correct place
  3. ii) Name the main systems for coordination in animals

– Nervous system/sensory system

– Endocrine (hormonal system)

iii) List the components of the mammalian sensory system

– Central nervous system (CNS), brain & spinal cord

– Peripheral nervous system (PNS) cranial and spinal nerves

– Sense organs

– Autonomic nervous system (ANS) nerve fibers and ganglia

iv).Explain the terms receptors, conductors and effectors

– Receptors are structures that detect stimuli i.e. sense organs

– Conductors transmit impulses from receptors to effectors e.g. neurons

– Effectors are the responding parts e.g. muscles, glands

  1. v) What are the functions of the central nervous system?

– provides a fast means of communication between receptors and effectors

– coordinates the activities of the body

  1. vi) State the differences between somatic and autonomic systems of peripheral nervous system

– Somatic is concerned with controlling the conscious or voluntary actions of the body i.e. skin, bones, joints and skeletal muscles

– the autonomic (automatic) nervous system controls involuntary actions of internal organs, digestive system, blood vessels, cardiac muscles and glandular products.

  1. b) i) What is a neurone?

 

 

 

 

 

 

 

  1. ii) Name the parts of a typical neurone and state the functions of each part
  1. Describe the structure and function of a motor neurone

 

 

 

 

 

 

 

 

 

  1. Describe the structure and function of sensory neurone

 

 

 

 

 

 

 

 

 

 

  1. Describe the structure and function of a relay neurone

 

 

 

 

 

 

 

 

 

 

  1. c) State the functions of the major parts of the human brain
  2. i) Cerebrum
  1. ii) Mid brain
  1. Hind brain
  1. i) What is reflex action?
  1. ii) Describe a reflex action that will lead to the withdrawal of a hand from a hot object

 

 

 

 

 

 

 

 

iii) Explain how an impulse is transmitted across the synapse (gap)

 

  1. Briefly describe the transmission of a nervous impulse across a neuro-muscular junction
  1. i) What is a conditioned reflex?
  1. ii) Explain a conditioned reflex

iii) Compare a simple reflex action with a conditioned reflex

Simple reflex Conditioned reflex
·      independent of experience

·      one stimulus to evoke response

·      some sensory and motor neurons used

·      reflex is simple

·      dependent on experience

·      both substitute and original reflex evoke response

·      sensory component replaced but motor remains unchanged

·      reflex is modified

  1. i) What are endocrine glands?
  1. ii) State the functions of hormones in animals

 

iii) Name the main endocrine glands, their secretions and functions in the human body

Gland Hormone Functions
Thyroid Thyroxine Increases rate of metabolism
Parathyroid Parathyroid hormone Regulates calcium and phosphate levels
Pituitary Hormone growth Regulate growth of body
  Gonadotrophic hormone Stimulates the development of male and female sex organs
Lactogenic hormone (prolactin) Stimulates secretion of milk after child birth
Thyrotropic hormone(TSH) – proper functioning of thyroid gland/thyroxine production
Adrenocorthicotropic hormone (ACTH) – stimulates release of adrenal cortex hormone
Oxytocin ·      regulates blood pressure

·      stimulates smooth muscles

·      stimulates contraction of uterus during childbirth

·      aids in flow of milk from mammary glands

Follicle stimulating hormone(FSH) ·      causes maturation of egg in females

·      stimulates sperm production in males

Vasopressin (ADH) Antiduretic hormone – regulates water balance by kidneys
Adrenal Adrenaline (epinephrine) ·      for emergency

·      prepares body to cope up with stress

  Aldosterone – maintains balance of salt and water in blood
  Cortisone ·      breaks down stored proteins to amino acids

·      aids in breakdown of adipose tissue

·      regulates sugar level in blood

·      prevents inflammation

 

 

 

  Sex hormones ·      supplements sex hormones produced by gonads

·      promotes development of sexual characteristics

Pancrease Insulin ·      regulates level of sugar in blood

·      enables liver to store sugar

Glucagons ·      regulates level of sugar in blood
Ovaries Oestrogen ·      causes sexual secondary characteristics in females

·      prepares uterus for pregnancy

Progesterone ·      growth of mucus lining of uterus

·      maintains uterus during pregnancy

Testes Androgens(testosterone) ·      causes secondary sexual characteristics in males
Stomach cells Gastrin ·      stimulates release of gastric juice
Intestinal cells Secretin ·      stimulate release of pancreatic juice
  1. iv) Give the differences between nervous and endocrine (hormonal) communication
Nervous Hormonal (endocrine)
Response confined to effector organs (localized targets) Response more widespread (various targets)
Speed of response is rapid Response less rapid
Nervous impulse thro\\ugh nerves/nerve cell/neurons Hormones transferred through blood
Duration of response is short Persist for long
Speed of transmission is rapid Speed of transmission is slower
Transmission is electrical Transmission is chemical
   
  1. v) State the effects of over secretion and under secretion of adrenaline and thyroxine in humans
Hormone Over secretion Under secretion
Adrenaline ·      thin toneless muscles

·      high blood pressure

·      weak bones

·      obesity

·      early onset of sexual development

·      low blood pressure

·      inability to withstand stress

·      fatigue

·      muscular weakness

·      muscle wasting

·      increased dark pigmentation of skin

Thyroxine ·      increased metabolism

·      increased heartbeat

·      physical restlessness

·      mental restlessness

·      protruding eyeballs

·      enlarged thyroid gland

·      cretinism(retarded growth and low mental development

·      lowered metabolism

·      low ventilation rate of lungs

·      low body temperature

·      lowered mental activity

·      coarse hair

·      puffy eyes

·      enlarged thyroid gland

 

  1. g) i) Define the following terms

Drug

Drug abuse

  1. ii) State the types of drugs, examples and side effects

Sedatives

Pain-killers

Hallucinogens

Stimulants

iii) State the general effects of drug abuse on human health

  1. h) i) List the special sense organs in mammals and the major function of each

– Eye for sight

– Ear for hearing and balance

– Nose for smell

– Skin for touch, temperature detection, pain detection

 

 

 

iii) What is accommodation of the eye?

  1. iv) Explain how an eye viewing a near object adjusts to viewing a far object
  1. v) What changes occur in the eye if it changes from observing an object at a distance to one at a closer range?

– ciliary muscles contract

– Tension in suspensory ligaments reduces/relax/ slackens

– Lens bulges/thickens/increases curvature

– Radial muscles contract

– Circular muscles relax

– Size of pupil becomes large to allow in more light.

  1. Explain how the eye forms an image
  1. Name the defects of the eye and state how they canbe corrected

Short sight (Myopia)

 

 

 

Long sight (Hypermetropia)

 

Presbyopia

Squinting

Astigmatism

  1. State the advantages of having two eyes in human beings

I i) What are the functions of the human ear?

  1. How are the structures of the human ear suited to perform the function of hearing?

 

 

 

 

 

 

 

 

 

iii) Explain how the structure of the human ear performs the function of balancing

  1. iv) State what would happen if the auditory nerve was completely damaged
  1. ii) What is locomotion?

iii) State the importance of support systems in living organisms

  1. iv) State the importance of locomotion in animals
  1. b) i) Name the tissues in higher plants that provide mechanical support
  1. ii) State the importance of support in plants

iii) Name the types of plant stems

  1. iv) Name the tissues in plants that are strengthened with lignin
  1. v) What makes young herbaceous plants remain upright?
  1. vi) State the ways by which plants compensate for lack of ability to move from one place to another
  1. c) i) Explain the ways in which erect posture is maintained in a weak herbaceous stem

– This is the function of turgidity and presence of collencyma

Cells take in water and become turgid

  1. ii) Explain how support in plants is achieved
  1. d) i) Give the reasons why support is necessary in animals
  1. ii) Why is movement necessary in animals?
  1. e) i) Name the organ used for support by animals

– Skeleton

  1. ii) Name the different types of skeletons in animals, giving an example of an animal for each type of skeleton named

iii) State the difference between exoskeleton and endoskeleton

  1. iv) State the advantages of having an exoskeleton
  1. v) Explain the importance of having an endoskeleton
  1. f) i) Explain how a fish is adapted to living in water
  1. ii) Explain how a finned fish is adapted to locomotion in water
  1. g) i) Name the main parts of the vertebral column giving the types of bones found in each part

Axial skeleton

Appendicular skeleton

  1. ii) What are the vertebrae?

iii) State the functions of the vertebral column

  1. iv) State the general characteristics of vertebrae

 

  1. v) Name the bones of the vertebral column

– Cervical vertebra

– Thoracic vertebra

– Lumbar vertebra

– Sacral vertebra

– Caudal vertebra

  1. vi) Describe how the various vertebrae are adapted to their functions
Bone Structure Function
Skull ·      cranium and jaw bones

·      made of several bones joined together

·      large box called cranium and smaller paired boxes for eyes, ears, nose, jaws

·      has large hole called foramen magnum for the passage of spinal cord

·      attachment of jaws

·      protect brain and other delicate parts

Cervical region

Atlas (first cervical)

 

 

 

 

 

 

 

 

 

 

 

·      ring shaped

·      no Centrum

·      broad, flat transverse processes

·      vertebraterial canal for passage of vertebral artery

·      facet for articulation of condyles of skull

·      protect spinal cord

·      attachment of muscles

·      allow nodding of head

Axis (second cervical)

 

 

 

 

 

 

 

 

 

 

 

·      adontoid peg projects from Centrum

·      large flattened neural spine

·      vertebrasterial canal

·      small transverse process

·      allows head to rotate

·      protects spinal cord

·      provides surface for muscle attachment

Cervical (others)

 

 

 

 

 

 

 

 

 

 

·      short neural spine

·      branched transverse process for neck muscles

·      vertebraterial canals

·      wide neural canal

·      support weight of head

·      protect spinal cord

·      neck muscle attachment

Thoracid

 

 

 

 

 

 

 

 

 

 

 

·      long backward pointing neural spine

·      transverse process that points sideways

·      facets for articulation of ribs

·      notch for spinal nerves to pass through

·      forms rib cage

·      articulation with one end of a rib

·      protects, spinal cord

·      muscle attachment

Lumbar

 

 

 

 

 

 

 

 

 

 

·      short neural spine

·      long transverse process pointing towards abdomen

·      large Centrum

·      extra processes e.g. prezygapophysis, hypapophysis, anapophysis, metapophysis

·      protect organs of abdomen

·      support upper part of body

·      protect spinal cord

·      muscle attachment

Sacral

 

 

 

 

 

 

 

 

 

·      fused bones to form sacrum

·      well developed transverse process of first vertebra

·      vertebraterial canals

·      short neural spine

·      protects alimentary canal

·      attachment of hip girdles

·      protect spinal cord

·      muscle attachment

Rib

 

 

 

 

 

 

 

·      long

·      flattened

·      attached to sternum from front

·      protect internal organs

·      muscle attachment

 

vii) Describe the bones that form the appendicular skeleton

Bone Structure Function
Pectoral girdle scapular (shoulder bone)

 

 

 

 

 

 

 

 

·      Broad i.e. Flattened blade

·      glenoid cavity to articulate with humerus

·      metacromion/acromion for muscle attachment

·      hard to provide support

·      socket with cartilage/smooth surface to reduce friction

Support

Muscle attachment

Articulates with humerus

Humerus

 

 

 

 

 

 

 

 

·      long shaft for muscle attachment

·      round head to articulate with glenoid cavity

·      trochlea for articulation with ulna

·      olecranon fosa to prevent arm bending the other way

·      movement

·      muscle attachment

Ulna and radius

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

·      ulna longer and on side of little finger

·      has sigmoid notch and olecranon process to form hinge joint with humerus

·      radius is smaller and lies along thumb side and does not join ulna

·      allows articulation with wrist bones

·      movement

·      muscle attachments

Pelvic girdle(hip bone)

 

 

 

 

 

·      composed of three fused bones (ilium, ischium, pubis)

·      upper end fused to sacrum

·      lower end has acetabalum for articulation with femur

·      has abturator foramen for passage of nerves and blood vessels

·      movement

·      muscle attachment

·      support

·      absorbs pressure exerted by ground when animal moves

Femur

 

 

 

 

 

 

 

 

 

 

 

 

 

·      rounded head to fit in acetabulum of pelvis

·      projections called trochanter for attachment of thigh muscles

·      condyles at lower end for articulation with tibi

·      patella that covers knee and prevents leg from bending backwards

·      movement

·      muscle attachment

Tibia and fibula

 

 

 

 

 

 

 

·      tibia is longer than fibula

·      tibia is outer bone and fibula is inner bone

·      tibia lies on side of large toe

·      fibula is fused to tibia (on outer side)

·      movement

·      muscle attachment

 

 

  1. a) What is a joint?
  1. ii) State the functions of joints

iii) Name the main types of joints

  1. iv) Give the features of movable joints
  1. b) Describe the synovial joints
  2. Ball and socket

 

 

 

 

  1. Hinge joint

 

 

 

 

 

 

 

  1. Pivot joint
  1. c) i) What is synovial fluid?
  1. ii) State the functions of synovial fluid
  1. Explain the following terms
  2. Ligament
  1. Cartilage
  1. i) What is a muscle?
  1. ii) State the functions of muscles
  1. Describe the structure and function of various types of muscles
  2. skeletal muscles
  1. Involuntary muscles
  1. Explain how muscles cause movement of the human arm
  1. i) State the structural differences between skeletal muscles e.g. biceps and smooth muscles e.g. gut muscle
Skeletal (biceps) Smooth (gut) muscle
·      multinucleated

·      striated/stripped

·      long muscle fibers

·      block/cylindrical

·      uninucleated

·      unstriated

·      short muscle fibers

·      spindle shaped

  1. ii) Name the cartilage found between the bones of the vertebral column
  1. What are the functions of the cartilage named in (d) ii) above
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