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5.7: Summary Questions - Biology

5.7: Summary Questions - Biology


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1. Assign a crown class to each tree illustrated below.

2. Where on this tree should the live crown be measured?

3. Determine LCR for each tree illustrated.

  1. The data in Table 4-4 below are from an evenaged stand in the stem exclusion stage. They represent the average live crown ratios (LCR) of trees sampled in the stand by species. Data were collected by MHCC Forest Measurements I students in February 2010.
  • Do the LCR’s reflect a stand with high or low crown competition? Explain.
  • Do the LCR’s of each species seem reasonable given their shade tolerance?
  • These data were gathered in February. Does the time of year pose any problems for estimating LCR on deciduous species like red alder?
  • Which species would you expect to dominate this stand in the future? Why?

Answers to Summary Questions

  1. Trees on the ends of the illustrations are hard to determine, given that we do not know what is on the other side of them. I assume there are trees beyond what is shown, so am using crown size as a partial indicator of the amount of light they are receiving. Position in the crown is key. The following are what I would assign, but I think the starred one (*) is borderline, and could be assigned the next higher crown class.

  1. Since LCR is simply a ratio, any scale can be used to measure. I used the 20 scale on my triangular engineer’s ruler to obtain fairly good precision. So your values may differ, but LCR% should be similar.

Trees with slope measurements from left to right:

Tree 1: (70-14) = 70% Tree 2: (70-38) = 40% Tree 3: (70-22) = 60%

(70+10) (70+10) (70+10)

  1. The following answers refer to the data presented in Table 4-4.
  • The bulk of the trees in this stand are Douglas-fir, with an average live crown ratio of 40%. This indicates to me a canopy experiencing high crown competition. As crown closure occurs, trees drop low branches that are being shaded (particularly shade intolerant or intermediate species like red alder and Douglas-fir). Foresters often refer to a lower limit of 30% LCR as their cut-off for vigorously growing trees. If I were managing this stand, I would seriously consider a thinning to increase light availability to the trees I wanted to maintain on the site.
  • We would expect the live crown ratios to be the shortest on the shade intolerant species, and longest on the shade tolerant. Our measurements show this pattern; red alder
  • These trees were measured in the winter, with no leaves, and only buds to indicate where the bottom of the crown was. I would take the red alder LCR estimates with a grain of salt.
  • I would expect Douglas-fir to continue to dominate in the future, followed by western hemlock. Red alder is a short-lived species, and is already subordinate in the forest.

Biology Questions and Answers Form 2 - Biology Form Two Notes

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Biology Questions and Answers Form 2

KCSE Revision Questions and Answers

1. a) i) Define transport

ii) Explain the necessity of transport in plants and animals

b) i) Describe the structure and function of root hair

ii) State ways in which the root hairs are adapted to their functions

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

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

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

Monocotyledonous stem

i) Give the similarities and differences between a monocotyledonous and dicotyledonous stem

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

c) i) Name the transport structures of a flowering plant

ii) State the ways in which xylem vessels are adapted to their function

a) i) Why do flowering plants need water?

ii) Describe the movement of water from the soil to the leaves of a tall plant

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

i) Explain the forces that make water and mineral salts move through a plant

ii) Explain the uptake of mineral salts by plants

b) i) What is transpiration?

ii) Name the sites through which transpiration takes place in a plant

iii) State the importance of transpiration to plants

excess transpiration causes wilting

i) Explain the structural factors that affect the rate of transpiration in plants

ii) Explain the environmental factors that affect rate of transpiration in plants

iii) State the structural differences between xylem vessels and sieve tubes

iv) State the adaptations of plants which enable them to reduce water loss

v) State the factors that cause increase in the rate of transpiration from leaves

vi) Explain how drooping of leaves on a hot sunny day is advantageous to a plant

c) Explain how aquatic and terrestrial plants are adapted to deal with problems of transpiration

d) i) What is translocation

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

iv) Draw a labeled diagram to represent phloem tissue

ii) State the functions of the labbeled structures cytoplasmic strands

supply nutrients to sieve tube element

iii) name the compounds that are translocated in phloem

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

ii) Use the radio-active tracers

iii) Collecting exudate from stylets of aphids

e) Describe an experiment you would carry out to demonstrate that xylem transports water

2. a) i)List the components of animal transport systems

ii) Distinguish between closed and open circulatory systems

iii) What are the advantages of the closed circulatory system over open circulatory system?

iv) Distinguish between single circulatory system and double circulatory system Single circulatory

b) i) Describe the general layout of the transport system in mammals

ii) Describe the structure and function of the mammalian heart

iii) Explain how the mammalian heart is adapted to performing its functions

iv) Explain why blood leaving the lungs may not be fully oxygenated

e) Describe the structure and functions of the blood vessels

b) i) State the ways in which the composition of blood in the pulmonary arterioles differs from that in the pulmonary venules

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

c) i) State the functions of mammalian blood

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

iv) State the structural differences between a red blood cell and a white blood cell.

v) State the functional differences between a red blood cell and a White blood cell

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

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.

d) i) what is blood clotting?

ii) Name a protein, vitamin, an enzyme and a mineral element involved in blood clotting

iii) describe the blood clotting process

iv) State the role of blood clotting on wounds

v) Explain why blood flowing in blood vessels does not normally clot

iii. Explain the meaning of :

iii) What is the difference between rhesus positive and Rhesus negative blood samples?

vi) What is blood transfusion?

v) Under what conditions would blood transfusion be necessary in people?

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

ii) Distinguish between natural and acquired immunity

iii) What are allergic reactions?

vi) How does an allergic reaction occur?

ii) State the role of vaccination against certain diseases

3. a) i) What is gaseous exchange?

ii) Why is gaseous exchange important to organisms?

b) i) name the structure used for gaseous exchange by plants

ii) Briefly describe the structure of stomata

iii) State the factors which affect stomatal opening

iv) Name the theories suggesting the mechanism of opening and closing of stomata

v) Describe the mechanism of opening and closing of stomata

i) What is the advantage of having stomata open during daytime and having them closed at night?

c) i) State the ways in which leaves of plants are adapted to gaseous exchange

ii) Describe how gaseous exchange takes place in terrestrial plants

iii) State the ways in which floating leaves of aquatic plants are adapted to gaseous exchange

iv) How is aerenchyma tissue adapted to its function?

v) Explain stomatal distribution in plants of different habitats

d) i) List the types of respiratory surfaces of animals

ii) State the characteristics of respiratory surfaces in animals

iii) Describe gaseous exchange in protozoa

e) i) Make a labeled drawing of a fish gill

ii) How is a fish gill adapted to its function?

iii) Discuss gaseous exchange in bony fish

iv) What is counter-flow system?

vi) What is the advantage of counter-flow system?

f) i) Describe the mechanism of gaseous exchange in terrestrial insects

ii) State how traceholes are adapted to gaseous exchange

g) i) What is breathing?

ii) Name the structures in humans that are used in gaseous exchange

iii) Describe the mechanism of gaseous exchange in a mammal

iv) Explain how mammalian lungs are adapted to gaseous exchange

v) Name the features of alveoli that adapt them to their function

vii) How is the trachea of a mammal suited to its function?

viii) State the advantages of breathing through the nose rather than through the mouth

ix) Give the conditions under which the carbon iv oxide level rises above normal in mammalian blood

x) Explain the physiological changes that occur in the body to lower the carbon iv oxide level back to normal when it rises

h) i)Describe the factors which control the rate of breathing in humans

ii) Name the respirator diseases

4. a) i) Define respiration

ii) Explain the significance of respiration in living organisms

iii) Where does respiration take place?

b) i) Draw and label a mitochondrion

ii) State the most important function of mitochondria

iii) Give the functions of the labeled parts

c) Explain the roles of enzymes in respiration

d) i) What is aerobic respiration

ii) Give a word equation for aerobic respiration

iii) What are the end products of aerobic respiration?

e) i) What is anaerobic respiration

ii) What are obligate anaerobes?

iii) What are facultative anaerobes?

iv) State the Word equation representing anaerobic respiration in plants

v) Name the end products of anaerobic respiration in plants

g) i) Give a word equation of anaerobic respiration in animals

Glucose —> lactic acid + energy

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?

iv) Why does lactic acid level reduce after exercise?

v) State why accumulation of lactic acid during vigorous exercise lead to an increase in heartbeat

State the economic importance of anaerobic respiration

h) i) What is respiratory quotient(RQ)?

RQ = volume of CO2 produced

volume of oxygen consumed

ii) Why are respiratory quotient important

iii) Name the respiratory substrates

iv) Why does anaerobic respiration of a given substrate yield a smaller amount of energy than aerobic respiration?

iv) Explain the disadvantages of anaerobic respiration

v) Mention the types of experiments carried out for respiration

5. a) i) Define the following terms

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

b) i) Describe how excretion takes place in green plants

ii) Why do plants lack complex excretory structures like those of animals?

ii) State the excretory products of plants and some of their uses to humans

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

ii) List excretory organs and products of mammals

d)i) Draw and label a mammalian skin

ii) Explain how the mammalian skin is adapted to its functions

the skin is made up of dermis and epidermis

e) What is the role of lungs in excretion?

f) State the functions of the liver

g) i) Draw a labeled diagram of mammalian nephrone

ii) Describe how the human kidney functions

iii) State the adaptations of proximal convoluted tubule to its function

iv) Name the common kidney diseases

6. a) i) Why is homeostatic control necessary?

ii) What is internal environment?

b) i) Why is constant body temperature maintained by mammals?

ii) Explain the advantage gained by possessing a constant body temperature

iii) How do mammals regulate body temperature?

iv) Why does body temperature of a healthy person rise up to 37 C on a hot humid day?

v) Name the structures in the human body that detect external temperature changes

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.

c) i) What is the meaning of osmoregulation?

ii) State the importance of osmoregulation

iii) State the ways by which desert mammals conserve water fewer glomeruli longer loop of Henle

iv) Explain why some desert animals excrete uric acid rather than Water

v) Explain why eating a meal with too much salt leads to production of a small volume of concentrated urine

vi) Explain how marine fish regulate their osmotic pressure

d) i) What is the biological significance of maintaining a relatively constant sugar level in a human body?

ii) Discuss the role of the following hormones in blood sugar control

e) Explain the part played by antidiuretic hormone in homeostasis

f) What is the role of blood clotting in homeostasis?

g) Describe the role of the following hormones in homeostasis

h) i) Distinguish between diabetes mellitus and diabetes insipidus

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?

iv) When is glycogen which is stored in the liver converted into glucose and released into the blood?

v) How would one find out from a sample of urine whether a person is suffering from diabetes mellitus?


Biology Questions and Answers Form 1 - Biology Form One Notes

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KCSE Revision Questions and Answers

Biology Notes Form 1 - Biology Form 1 Notes - Form 1 Biology Notes

Introduction to Biology

Biology is a branch of science that deals with the study of living things. There are diverse forms of life on earth ranging from the invisible microscopic living things to the gigantic life forms. It aims at explaining the living world in terms of scientific principles.

It is important to note, however, that living things interact with the non living things in the environment as Well. Biology, therefore also entails the study of non living things as well.

The role of human beings in shaping the environment is also investigated in biology.

In summary, biology deals with the study of origins, types, nature, growth, development, interactions and maintenance of all life forms on earth.

Biology is such a broad field of knowledge. It is divided into two broad branches

1. Zoology- This is a branch of biology that deals with the study of animal life.

2. Botany- This is a branch of biology that deals with the study of plant life.

Within the two branches, there exist even smaller branches because the branches (botany and Zoology) are very wide and complex.

The smaller branches of biology include:

a) Ecology- This is the study of the interrelationships between organisms and their environment. Ecology aims at establishing how organisms are related to each other and their environment.

Ecology is further subdivided into smaller branches. These can be forest ecology, marine ecology, rangeland ecology etc.

b) Genetics- This sub-branch of biology deals with the study of inheritance and variation. It deals with the study of how variations (differences) occur between parents and their offspring. It is also concerned with how various characteristics are passed on from parents to offspring.

c) Entomology- This is the study of insects.

d) Parasitology- This is the study of parasites.

e) Physiology- This deals with the study of the functions of various structures of an organism. It deals with the processes that take place in the body of organisms.

f) Anatomy- The study of the internal structure of organisms

g) Microbiology- This is the study of microorganisms

h) Bacteriology- The study of bacteria

i) Ornithology- This is the study of birds

j) Itchthology-This is the study of fishes

This list is in-exhaustive as there are very many other branches of biology.

Importance of biology

> Joint development of HIV/AIDS vaccine by Kenyan and British scientists.

> The coordinated fight against Severe Acute Respiratory Syndrome involving scientist all over the world.

> The fight to save the ozone layer from depletion through various international agreements such as the Kyoto protocol.

> Management of resources through international treaties such as the CITES (Convention against International Trade on Endangered Species).

Characteristics Of Living Things

Living things share a lot of characteristics in common. These characteristics are discussed below.

Nutrition is the process by which living things obtain and assimilate (utilize) nutrients. Living things require nutrients for various purposes growth, repair of worn out tissues and for provision of energy. Plants manufacture their own food using light energy, carbon (IV) oxide, water and mineral salts through the process of photosynthesis. Conversely, animals feed on already manufactured foods from plants and other animals.

Respiration is the process by which food substances are chemically broken down to release energy. During respiration, oxygen is used while energy, carbon (IV) oxide and water are released. Respiration occurs in all living cells.

The energy produced in living things is very useful as it enables the living things carry out some of their physiological processes.

The energy is also required for growth and development, movement and repair of worn out tissues.

Gaseous exchange refers to the process by which living things exchange oxygen and carbon (IV) oxide across the respiratory surfaces. Animals always take in air rich in oxygen and give out air rich in carbon (IV) oxide.

Carbon (IV) oxide is a waste product of chemical reactions in the body. Animals require oxygen for respiration. Gaseous exchange, therefore, enables animals obtain oxygen for respiration and get rid of carbon (IV) oxide, a waste product.

Plants, however, require carbon (IV) oxide for photosynthesis during the day. They give away oxygen as a by-product. The plants equally require oxygen for respiration and give away carbon (IV) oxide.

This is the process by which living things separate and eliminate the waste or harmful materials resulting from chemical reactions Within the cells. These harmful waste products of metabolism maybe toxic to the body if they are left to accumulate in the cells of the living things

e) Growth and Development

Growth refers to an irreversible increase in size and mass while development refers to the irreversible change in complexity of the structure of living things. Growth and development of living things is essential as it enables the living things to attain maximum size that can enable them to perform their functions and roles.

This is the process by which living things give rise to new individuals of the same kind. All living things reproduce. Reproduction is essential as it leads to perpetuation of species and it avoids extinction of certain animals and plants.

This is the ability of living things to perceive (detect) changes in their environment and respond to them appropriately. Living things respond to changes in temperature, humidity, light, presence or absence of certain chemicals.

Response of organisms to these changes is crucial as it enables them to escape from harmful stimuli. Ability to detect changes in the environment also enables organisms to obtain resources in their environment.

Movement refers to change is position (displacement) of a part or parts of an organism. Movement in plants includes folding of leaves, closing of flowers and growing of shoots towards light. The change of position of an entire organism from one position to another is locomotion.

a) Motor vehicles move, use energy and produce carbon dioxide and water. Similar characteristics occur in living organisms yet motor vehicles are not classified as living. List the other characteristics of living things that do NOT occur in motor vehicles.

b) Give the name to the study of:

Collection of Specimen

We have defined biology as the study of living things. For effective study, a biologist may have to collect some living things or some parts of living things for observation and analysis. The living things or parts of living things that are used for biological study are called specimens.

Biological studies always take place in laboratories. A laboratory is a building or a room that is designed and equipped for scientific studies.

Collections of living things especially animals may not be very easy. Some of the animals are not easy to catch while some are quite dangerous. Knowledge on proper specimen collection and handling of is very important. We will discuss some of the apparatus used in specimen collection.

a) Sweep net- This is used for catching flying insects.

b) Fish net- This is used for trapping small fish and other small Water animals.

c) Pooter- This is used for sucking small animals from rock surfaces or barks of trees.

d) Bait trap- This is used for attracting and trapping small animals including rats.

e) Pit fall trap- This is used for catching crawling animals.

f) Pair of forceps- This is an apparatus used for picking up small crawling animals e.g. stinging insects.

g) Specimen bottles- These are bottles used for keeping collected specimen. They are of different sizes depending on the size of the specimen being studied.

h) Magnifying lens- This is used to enlarge small objects. A hand lens is a common magnifying lens used in the laboratory. The magnifying power of the hand lenses is always indicated on the lens e. g. X10, X5, X8. The magnifying power of a lens shows how many times the image will be enlarged compared to the object.

How to use a magnifying lens

To use a magnifying lens, place the object to be enlarged on the bench. Hold the magnifying lens on one hand and while closing one eye, move the lens towards the object until the image comes into clear focus.

If a magnifying lens is used to make a drawing of a specimen, the magnification of the drawing will have no relation with the size of the drawing.

The magnification of the drawing can be calculated using the formula shown below.

Drawing magnification: Length of drawing divided by Length of the actual object

The sign of “times” must come before the magnification value e. g. X10, X5, X15 etc.

Precautions During Collection and Observation of Specimen

While collecting specimen for observation, a biologist should play close attention to the following:

Comparison Between Plants and Animals

Chapter Two: Classification 1

In particular, animals and plants are all living things yet they differ in many aspects. Amongst animals and plants also there exist a lot of differences.

There are millions of different plant and animal types exhibiting a range of differences. This created a need for a classification system of living things to make study of the living organisms easier.

External features of plants used in classification

External features of animals used in classification

Importance of Classification

Historical background of Classification

b) Flowering or non-flowering

Taxonomic Units of Classification

2. Phylum (animals)/division (plants)

All living organisms are classified into five major kingdoms:

a) Kingdom Monera- This is composed of microscopic unicellular organisms mainly bacteria e.g amoeba.

b) Kingdom Protoctista- This kingdom is comprised of members who are microscopic. Though, some are large enough to be seen with the naked eyes.

Members of this kingdom include algae and protozoa.

c) Kingdom Fungi- Members of this kingdom comprises the mushrooms, toadstools, moulds and yeast.

d) Kingdom Plantae- This kingdom comprises the moss plant, ferns, maize plants, hibiscus, meru oak tree etc.

e) Kingdom Animalia — Members of this kingdom include the tapeworms, hydra, fishes, human beings, lizards, earthworms etc.

In hierarchy of classification, a kingdom is further divided into several phyla (plural of phylum) or divisions (in plants). Within the phyla or divisions, organisms are further sorted out into groups known as classes based on their similarities and mode of life.

Each class is further subdivided into small groups called orders based on structural similarities. Orders subdivide into families which subdivide into genera (plural for genus).

Genera are then subdivided into smaller units of classification called the species.

Species is the smallest unit of classification whose members share many similarities and can freely interbreed to give rise to fertile or viable offspring.

Members of a particular species can, however, exhibit various differences e. g. differences in skin colour or body forms. Within the species, organisms can further be classified based on the differences in colour or forms.

In humans, this gives the races, in animals the term used is breed while in plants, variety is preferred. In bacteria, the term strain is used to describe the variant forms.

Members of different but very closely related species can breed but the resulting offspring will be sterile (infertile). In particular, a mule is a sterile offspring between a horse and a donkey.

Moving from kingdom to species, it is important to note that the number of organisms in each taxon decreases. The similarities, however, increase as one moves from kingdom to species.

Scientific Naming of Living Organisms

Rules of Binomial Nomenclature

Binomial nomenclature requires that:

a) The first part of the scientific name is that of the genus name which should begin with a capital letter. The second name is that of species. The species name should be written in small letters e. g.

c) Leopard- Panthera pardus

d) Domestic dog- Canisfamiliaris

e) Human being- Homo sapiens

b) When printed in books and other printed works, the scientific names should be printed in italics. However, in handwritten manuscripts and typed works, the genus and species names should be lined separately.

Printed work- Homo sapiens

c) The specific name is frequently written with the name of the scientist who first adequately described and named the organism e. g. Balanus balanoides Linneaus. d) Scientists must give a latinised name for a newly described animal or plant species where a Latin name is missing e,g.

Aloe kilzfiensis- A type of aloe found in kilifi

Meladogyne kikuyuensis- A nematode found in kikuyu.

Origin of scientific names

Scientific names assigned to organisms can be:

Chapter Three: The Cell

Magnification=Eyepiece magnification X Objective lens magnification

Figure 1. The light microscope

Handling and Care of the Microscope

Part of the microscope: Limb

function: supports the body tube and stage

function: provides firm and steady support to the microscope

function: holds the eyepiece and the revolving nose piece

Coarse adjustment knob:

function: raises or lowers the body tube through longer distances to bring the image into sharper focus

function: raises or lowers the body tube through smaller distances to bring the image into sharper focus. it is mostly used with the high power objective lens

function: an aperture that regulates the amount of light passing through the condenser to illuminate the specimen

function: contains a lens which contributes to the magnification of the specimen under review

function: bring image into focus and magnifies it.

function: reflects light through the condenser to the object on the stage

function: holds the objective lenses in place and enables the change from one objective lens to the other

function: concentrates light on the object on stage

function: flat platform where specimen on the slide is placed.it has two clips to hold the slide into position

The following rules should be observed when handling the microscope:

> Always use both hands when carrying the microscope. One hand should hold the base to provide support while the other hand holds the limb.

> Never place the microscope too close to the edge of the working bench or table.

> Do not touch the mirror or the lenses with your fingers.

> Dirty lenses should be cleaned using a special soft lens tissue paper or tissue paper moistened with ethanol. The other parts of the microscope may be cleaned using a microscope.

> Do not wet any part of the microscope.

> Make sure the low power objective lens clicks into position in line with the eye piece before and after use.

> After use, always clean and store the microscope in a safe place, free from moisture and dust.

How to use the Microscope

Cell Structures as seen under the Light Microscope

Figure 2 Plant and animal cells as seen under the light microscope

The cell as seen under the Electron Microscope

Structure and Functions of the Cell Organelles

> Mitochondria are self replicative that is they can divide to form new ones.

Figure 5. The Mitochondrion (Animal)

Figure 6, (generalize(lmage23 mitochondrion Structure)

d) Endoplasmic Reticulum

g)Golgi bodies/Golgi apparatus

1) They package and transport glycoproteins.

2) They are involved in secretion of synthesized proteins and carbohydrates.

3) They manufacture lysosomes.

Note: Golgi bodies are abundant in cells that are active in secretion. For instance pancreatic cells which secrete enzymes and the nerve cells which secrete neurotransmitter substances.

1. It gives plant cells their definite shape

2. It provides mechanical support and protection against mechanical injury.

3. The cell wall allows gases, water and other substances to pass through it.

Comparison between Plant Cells and Animal Cells

While there exist many similarities between plant and animal cells, there are a number of differences.

Estimation of Cell Size

The light microscope can be used to estimate the size of a cell. Most cells have diameters smaller than a millimeter. Due to this, cell sizes are always measures in smaller units.

These are micrometres and nanometers. These units of measurements are related as shown below.

1 millimeter (mm) = 1000 micrometres (pm).

1 micrometer (pm) = 1000 nanometres (nm).

Procedure in cell size estimation

cell diameter = diameter of the field of view in micrometers divided by number of cells.

Cell Specialization. Tissues. Organs and Organ Systems

Cell Specialization/Cell Differentiation

a) Tissue types in animals

1. Epithelial tissue- This is a thin continuous layer of cells for lining and protection of internal and external surfaces.

2. Skeletal muscle- This is a bundle or sheets of elongated cells with fibres that can contract. Its contraction and relaxation brings about movement.

4. Blood tissue- This is a fluid containing red blood cells, white blood cells and platelets.

The main functions of blood tissue are transportation of nutrients and gases as well as protection of the body against infections.

5. Connective tissue- This tissue consists of strong fibres that connects other tissues and organs thereby holding them in position.

b)Tissue types in plants

1. Epidermal tissue- This is a single thin layer of cells covering the outer surfaces. It protects inner tissues of plants from mechanical damage and infection.

2. Palisade tissue- This is a group of cells rich in chloroplasts containing chlorophyll. It has a site for the absorption of light energy and manufacture of food by photosynthesis.

3. Parenchyma tissue- This tissue consists of special thin walled irregularly shaped cells. They form packaging and storage cells.

4.Conducting tissue/Vascular bundle- This tissue consists of xylem and phloem. Xylem conducts water and dissolved mineral salts in a plant while phloem conducts food substances in solution.

a) Heart- composed of connective, muscle, epithelial and blood tissues.

b) Kidney- Composed of connective, epithelial and muscle tissues

c) Brain- Composed of epithelial, connective tissues

d) Lungs- Composed of epithelial, connective tissues.

a) Roots- composed of epidermal, conducting and parenchyma tissues.

b) Flowers- This is composed of epidermal, conducting tissues.

c) Stem- Composed of conducting, parenchyma, and epidermal tissues and palisade tissues in some cases

d) Leaves- Composed of palisade, conducting and epidermal tissues.

This is a group of organs Whose functions are coordinated and synchronized to perform the same function.

Organ systems are more pronounced in animals than in plants

Organ systems in animals include

a) Digestive system composed of organs such as oesophagus, stomach, intestines and their associated glands.

b) Circulatory system composed of the heart, blood vessels (arteries, veins, capillaries). c) Excretory this is composed of kidney, liver, and blood vessels.

d) Respiratory system composed of trachea, bronchus, and lungs.

e) Reproductive system composed of the reproductive organs and associated glands.

f) Nervous systems composed of the brain, spinal cord, eye, ear organs.

Chapter Four: Cell Physiology

a) Chloroplasts play a vital role in carbohydrate synthesis.

b) Mitochondrion produces energy required to carry out life processes.

c) Ribosomes manufacture of proteins.

Structure of the membrane

Properties of the cell membrane

a) The cell membrane is semi permeable- The pores that occur on the cell membrane allows the passage of the small size molecules but does not allow the passage of the large sized molecules.

Such a membrane is said to be selectively permeable or semi-permeable. In particular, when a cell is surrounded by a dilute sugar solution, the small sized water molecules will enter the cell but the larger sugar molecules will not pass through the cell membrane.

In contrast, the cell wall is permeable as it allows both sugar and water molecules to pass through it it has larger pores. This property of selectively permeability enables the cell membrane to select what enters and leaves the cell.

b)The cell membrane is sensitive to changes in temperature and pH- Cell membranes are made up of protein. Proteins are adversely affected by extreme changes in temperature and pH.

Changes in temperature and pH will alter the structure of the cell membrane thereby hindering the normal functioning of the cell membrane. High temperature denatures (destroys) the proteins thereby impairing the functions of the cell membrane.

c)The cell membrane possesses electric charges- The cell membrane has both positive and negative charges. These charges affect the manner in which substances move in and out of the ells. The charges also enable the cell to detect changes in the environment.

Physiological Processes of the Cell membrane

Demonstration of the process of diffusion using_potassium manganate (VII)

Requirements: potassium manganate (VII) crystals, glass tubing, 100 cm3 beaker and water.

a) Hold the glass tubing vertically in a beaker so that one end of the tubing rests on the bottom of the beaker.

b) Cautiously and quickly drop a crystal of potassium manganate (VII) through the upper opening of the glass tubing.

c) Close the upper hand of the glass tubing with the thumb.

d) Half fill the beaker with water.

e) Carefully withdraw vertically the glass tubing so that the crystal is left undisturbed at the bottom of the beaker.

f) Record your observations for the first 15 minutes.

g) Explain your observations.

Expected observations

The Role of Diffusion in Living Organisms

Diffusion plays an important role in plants in that:

In animals diffusion plays the following important roles

Factors affecting the rate of Diffusion

b) Surface area to volume ratio

c) Thickness of membranes and tissues

Demonstration of Osmosis Using a Visking Tubing

5OOcm3 beaker, visking tubing, a piece of thread, glass rod, concentrated sugar solution, 500 cm3 distilled Water.

1. Into the beaker, put 350 cm3 of the distilled water.

2. Dip the visking tubing in water to moisten it. Open the visking tubing and tie one end with the thread provided.

3. Half fill the visking tubing with the sugar solution provided and then tie the open end of the tubing. Ensure no sugar solution spills out of the tubing.

4. Immerse the visking tubing into the distilled water in the beaker and suspend it using the glass rod provided.

5. Leave the set up for about 30 minutes.

6. Record your observations.

7.Explain the observations made.

Even though there is a higher concentration of sugar molecules in the visking tubing, they were not able to diffuse out of the visking tubing due to their large molecular sizes. The visking tubing is semi permeable.

Water Relations in Animals

a) Red blood cell in hypotonic solution e. g. distilled water

When a red blood cell is placed in a hypotonic solution, water will move into the cell through osmosis. The cell will swell and burst. Swelling of red blood cell when placed in a hypotonic solution is referred to as haemolysis. The cell is said to be haemolysed.

b) Red blood cell in hypertonic solution

Water will, therefore, be drawn out of the cell into the hypertonic solution. The cell will shrink and become small. The cell is said to be crenated.

The process by which animal cells shrink and become smaller when placed in hypertonic solutions is referred to as crenation.

c) Red blood cell in isotonic solution

When placed in an isotonic solution, the cell remains unchanged. This is because there will be no net inflow or outflow of water between the cell and the solution.

This will prevent bursting or shrinking of the cells that would otherwise impair their physiology.

Water Relations in Plants

a) Plant cell in hvpotonic solution e. g. distilled water

Role of Osmosis in Organisms

As glucose accumulates in the guard cells, the osmotic pressure of the guard cells increase making them to draw water from adjacent cells through osmosis. When the guard cells become turgid, they bulge outwards leading to opening of the stomata.

Opening of the stomata is crucial as it allows for gaseous exchange in plants. At night, there is no glucose synthesis.

The glucose available in the guard cells is respired on leading to reduction of glucose and consequently reduction in osmotic pressure. The guard cells lose turgidity and close the stomata.

Factors Affecting the Rate of Osmosis

Role of active transport in living organisms

Factors affecting the rate of Active Transport

a) Oxygen concentration

Oxygen is required in respiration process that yields energy for active transport. Under low oxygen concentration, the rate of respiration will be low hence there will be production of little energy leading to low rate of active transport. Increase in oxygen concentration translates into a higher energy production leading to high rate of active transport.

Change in pH affects the respiratory process which is enzyme controlled. Respiratory enzymes require optimum pH for their efficient activity. Extreme pH conditions will increase lower the rate of active transport since the enzymes controlling respiration Will be denatured.

c) Glucose concentration

Glucose is the chief respiratory substrate. At low glucose concentration, there will b less production of energy leading to decreased rate of active transport. Rate of active transport increases with increase in glucose concentration due to increase in the rate of energy production.

Temperature affects the enzyme controlled respiration process. At low temperatures, the enzymes are inactive hence the rate of respiration will be low resulting into low rate of active transport since there will be less production of energy. An increase in temperature increases the rate of respiration since the enzymes become more activated. At temperatures beyond 40 degrees celcius, the enzymes become denatured, respiration stops and so does active transport.

e) Presence of metabolic inhibitors e. g. cyanide.

These are substances which act as metabolic poisons. They stop the rate of respiration leading to production of no energy. Active transport is, thus, stopped.

Nutrition Plants And Animals

a) The nutrients are required for growth and development of the living organisms.

b) The nutrients are required for energy provision as they are broken down to release energy.

c) They nutrients are also required for repair of worn out tissues

d) Nutrients are required for synthesis of very vital macromolecules in the body such as hormones and enzymes.

There are two main nutrition modes:

a) Autotrophism mode of nutrition through which living organisms manufacture their own food from simple inorganic substances in the environment such as carbon (IV) oxide, water and mineral ions. Organisms that make their own food through this mode are autotrophs.

b) Heterotrophism mode of nutrition in which living organisms depend on already manufactured food materials from other living organisms. Heterotrophs are the organisms that feed on already manufactured food materials.

In this mode of nutrition, organisms manufacture their own food from readily available materials in the environment. These organisms use energy to combine carbon (IV) oxide, water and mineral salts in complex reactions to manufacture food substances. Depending on the source of energy used to manufacture the food, there are two types of autotrophism:

a)Chemosynthesis

This is the process whereby some organisms utilize energy derived from chemical reactions in their bodies to manufacture food from simple substances in the environment. This nutrition mode is common in non green plants and some bacteria which lack the sun trapping chlorophyll molecule.

b) Photosynthesis

Importance of Photosynthesis

1. Photosynthesis helps in regulation of carbon (IV) oxide and oxygen gases in the environment.

2. Photosynthesis enables autotrophs make their own food, thus, meet their nutritional requirements.

3. Photosynthesis converts sunlight energy into a form (chemical energy) that can be utilized by other organisms that are unable to manufacture their own food.

External leaf structure

This helps reduce the rate of water loss in such plants. However, the plants in areas of water abundance have broad leaves to enable them lose the excess Water.

Functions of the cuticle

a) Being waterproof, it minimizes water loss from the leaf cells to the environment through transpiration and evaporation.

b) It protects the inner leaf tissues from mechanical damage.

c) It prevents entry of pathogenic microorganisms into the leaf.

Functions of the epidermis:

a) It protects the leaf from mechanical damage.

b) It also protects the leaf from entry of disease-causing microorganisms.

c) It secretes the cuticle.

Adaptations of the guard cells

c) Palisade mesophyll

d) Spongy mesophyll layer

e) Vascular bundle/tissue

Adaptations of the leaf to photosynthesis

Raw materials for photosynthesis

Conditions for photosynthesis

Photosynthesis Process

a)Light reaction/Light stage

i) Photolysis of water

Water-- Hydrogen atoms + Oxygen gas

ii) Formation of adenosine triphosphate (ATP)

b) Dark reaction/Dark stage

Testing for starch in a leaf

Factors affecting the rate of photosynthesis

a) Carbon (IV) oxide concentration

At this point, other factors such as light intensity, water and temperature become limiting factors.

Rate of photosynthesis is optimum at (35-40) °C. Beyond 40°C the rate of photosynthesis decreases and eventually stops since the enzymes become denatured.

Experiment to investigate the gas produced during photosynthesis

a) Set up the apparatus as shown in the figure below

b) Place the set up in the sunlight to allow photosynthesis to take place.

c) Leave the set up in the sun until sufficient gas has collected in the test tube.

d) Test the gas collected with a glowing splint.

e) Record your observations.

1) Carbon (IV) oxide concentration: Carry out the experiment using different amounts of dissolved sodium hydrogen carbonate e. g 5 g, 10g, 15g, 20g and examine the rate at which the gas collects.

2) Light intensity: An artificial light source can be used. Illuminate the plant and vary the distance between the set up and the light source While recording the time it takes for the gas jar to fill or counting the number of bubbles peer unit time.

3) Temperature: carry out the experiment at varying temperatures and record the rate at which the gas collects.

Experiments on factors necessary for photosynthesis

Carbon (IV) oxide

Properties of Monosaccharides

Properties of Disaccharides

Properties of polysaccharides

Examples of polysaccharides

a) Starch- Made by linking numerous glucose molecules. It is a form in which carbohydrates are stored in plants.

b) Glycogen- Is a storage carbohydrate in liver and muscles of animals. It is broken down to glucose in animals when blood glucose falls.

c) Cellulose- This is a structural polysaccharide in plants. It is a component of the cell wall

d) Chitin- A structural carbohydrate found in cell wall of fungi and arthropod exoskeletons

Functions of polysaccharides

a) Essential- These are those amino acids that cannot be synthesized by the body systems hence have to be supplied in the diet.

b) Non essential- These are amino acids that can be synthesized by the body mechanisms hence do not need to be supplied in the diet.

a) First class proteins- Contain all essential amino acids

b) Second class proteins- Proteins lack one or more essential amino acids

Properties of Proteins

Functions of proteins

a) They are structural compounds of the body. Cell membrane is protein in nature. Hair, nails and hooves are made up of protein keratin.

b) Proteins are broken down to release energy during starvation when all carbohydrate and lipid reserves are depleted.

c) Functional proteins play vital roles in metabolic regulation. Hormones are chemical messengers while enzymes regulate the speed of metabolic reactions.

d) Proteins such as antibodies provide protection to the body against infections

e) Some protein molecules are transport molecules. Haemoglobin molecule plays a crucial role in transportation of respiratory gases.

f) Proteins play a vital role in blood clotting e. g. fibrinogen.

g) Contractile proteins such as actin and myosin bring about movement.

What are enzymes?

a) Extracellular: Are produced within the cells but used outside the cells e. g. digestive enzymes.

b) Intracellular: Are enzymes produced and used within the cells e. g. respiratory enzymes.

Importance of Enzymes

Hydrolysis . ..hydrolase

Reduction . ..reductase

Oxidation . ..oxidase

Mechanism of action of Enzymes

Properties of Enzymes

Factors affecting enzyme activity

d)Substrate Concentration

e) Enzyme Concentration

NAD- Nicotine Adenine Dinucleotide.

FAD- Flavine Adenine Dinucleotide.

NADP- Nicotine Adenine Dinucleotide Phosphate.

Competitive inhibitors

Non competitive inhibitors

Examples of non competitive inhibitors

Heavy metals (such as lead, mercury, silver), Cyanide, organophosphates such as malathion.

Modes of Heterotrophism

> Rhizobium and leguminous plants: rhizobium fixes nitrogen for the legume while the bacteria obtains manufactured food from the legumes.

> Lichen: association of fungi (absorbing water and nutrients) and algae (manufacturing food for the association.

> Catalase digesting bacteria and ruminants.

b. Premolar and molar

Classes of Holozoic Heterotrophs

a) Herbivores: heterotrophs that exclusively feed on vegetation.

b) Carnivores: heterotrophs exclusively feed on flesh.

c) Omnivores: heterotrophs that feed on both flesh and vegetation.

a) Write down its dental formula.

b) State its mode of feeding.

b) Periodontal Diseases

a) Gingivitis- Characterized by reddening of gums, bleeding and pus in the gums.

b) Pyorrhea- The teeth become loose due to infection of the fibres holding the teeth in the sockets.

Digestion in the mouth

a) Sublingual salivary gland beneath the tongue

b) Sub mandibular gland: under the jaw

c) Parotid gland: Found in the cheeks in front of the ears.

Digestion in the stomach

a) Pepsinogen-This is activated to pepsin which breaks down proteins to peptides.

b) Rennin- Digests caseinogens protein in milk to casein (curd).

e) Hydrochloric acid- This:

d) Mucus- Forms a protective barrier to the stomach wall against corrosion by the HC1. Mucus is secreted by goblet cells in the epithelial membrane of the alimentary canal.

a) Gall bladder in the liver- Secretes bile.

b) Pancreas- Secrete hormones and digestive enzymes.

i. Secretin hormone from the pancreas: Secretin stimulates secretion of pancreatic juice into the duodenum

ii. Cholecystokinin from the duodenal wall: This stimulates secretion of bile from the gall bladder.

a) Pancreatic amylase- This facilitates breakdown of the remaining starch into maltose

b) Trypsin- Digests proteins into peptides.

c) Pancreatic juice-Digests lipids into fatty acids and glycerol

d) Sodium hydrogen carbonate- This:

Provides alkaline medium for activity of the duodenum enzymes.

i. Aid in emulsification (breakdown of fat molecules into tiny fat droplets to increase surface area for digestion).

ii. The salts also provide a suitable alkaline medium for action of the duodenal enzymes.

iii. In addition they neutralize the acidic chyme.

Digestion in the ileum

a) Maltase: speeds up breakdown of maltose to glucose

b) Sucrase: speeds breakdown of sucrose to glucose and fructose

c) Peptidase: speeds breakdown of peptides to amino acids

d) Lipase: speeds breakdown of lipids to fatty acids and glycerol.

e) Lactase: speeds breakdown of lactose to glucose and galactose.

f) Polypeptidase: speeds breakdown of plypeptides into amino acids

The mucus secreted by the goblet cells lubricates food along the alimentary canal and also protect the canal from being digested by enzymes.

At the end of digestion in the ileum, the resulting watery emulsion is called chyle it contains soluble end products of digestion ready to be absorbed.

a) It is long to provide a large surface area for absorption

b) It has a narrow lumen so as to bring the digested food into close contact with the walls of the ileum for easier absorption

c) It is highly coiled to slow down movement of food thus allowing more time for digestion and absorption of food.

d) The inner surfaces have numerous villi and microvilli to increase surface area for absorption of end products of digestion.

e) The epithelial lining is one cell thick to reduce the distance through which digested food diffuses.

f) Has a dense network of blood capillaries into which digested food materials diffuse to increase transport and thus maintain a steep concentration gradient.

g) Have lacteal vessels in the villi for absorption of fatty acids and glycerol.

> This is the process through which the undigested and indigestible food substances are eliminated from the body. Caecum and Appendix

b) Fatty acids and glycerol

a) Fat soluble vitamins- They dissolve in fats and are often stored in the liver. Include Vitamins A, D, E, K.

b) Water soluble vitamins- Dissolve in water. Include vitamins B1, B2, B5, B12 and C.

deficiency disease symptoms

Vitamine B2 (riboflavine and nicotinic acid)

deficiency disease symptoms

Vitamin B5 (pentathonic acid)

deficiency disease symptoms

deficiency disease symptoms

Vitamin C (absorbic acid)

deficiency disease symptoms

deficiency disease symptoms

deficiency disease symptoms

deficiency disease symptoms

a) Macro-nutrients: Nutrients required in large quantities. These include nitrogen, sulphur, phosphorous, calcium, sodium, iron and magnesium.

b) Micro-nutrients: Nutrients required in small quantities. Include copper, manganese, boron, iodine and cobalt.

Importance of roughage

a) It rubs against the walls of the alimentary canal stimulating secretion of digestive enzymes and mucus to lubricate the epithelial lining.

b) Roughage enhance peristalsis since as they rub against the walls of the alimentary canal, they stimulate contraction and relaxation of the muscles.

c) Roughage is an absorbent it extracts water from the alimentary canal making the fecal matter bulky and moist hence can be easily propelled by peristaltic movements. This prevents constipation.

Factors affecting energy requirements in humans

Discuss how the following factors affect energy requirements in humans:


Genetics

Is the study of inheritance of characteristics or the study of transmission of characteristics/traits from the parents to the off springs.

sexual union
parents Male X female
↓ ↓
Gametes sperms eggs
↘ ↙
fertilization (Zygote)
↓ Growth and development
Off springs (sexually mature individual

Transmission of characters from the parents to off springs is by gametes. The male organism if animal contributes sperms and in plants pollen grains

The Female animal contributes ova and in plants it will contribute ovules. Animals have to mate to bring the two gametes together.

In human beings, they have sexual intercourse, in plants the two gametes come together by pollination (self/cross pollination). When the female and male gametes fuse, fertilization is said to have occurred. Fertilization is the fusion of the male and female nuclei to form a zygote.

In animals the zygote is formed in the fallopian tube and it begins to undergo growth (growth is an irreversible increase in the size of an organism.) Growth occurs by cell division.

The type of cell division which leads to growth is mitosis. The zygote also undergoes development. Development is the change in shape and form. Growth and development eventually lead to a sexually mature organism.

Sexual maturity in flowering plants is evidenced by the on set of flowering. In female animals its evidenced by ovulation and in male animals by sperm development.

Gametes are formed by meiosis (meiosis is the type of cell division resulting in formation of gametes and takes place in the reproductive cells while mitosis is the cell division that leads to growth of an organism and occurs in the non reproductive cells/somatic cells/body cells

Examples of somatic cells

Liver cells, check cells, ovary cells, testis cells etc

Examples of reproductive cell

Sortollic cells in the testis which form sperms

Follicles in the ovary that give rise to ova

With in the cell nucleus is the observable genetic material known as the chromosomes. In body cells chromosomes are found in pairs (Half from the female parent and half from the male parent.) In reproductive cells/gametes, they occur in a single set called haploid and represented by n but in body cells they occur in a double set called diploid and represented by 2n.

Chromosomes

These are threadlike structures found in the nucleus of the cell and they contain the genetic material responsible for inheritance. They form the physical basis for inheritance since their structure can be observed under a high power microscope.

Each chromosome is made up of two longitudinal strands called the chromatids.

Each chromatid has a double helical DNA molecule. The two chromatids are held together by a structure called the centromere. During cell division, the spindle fibres are attached to the centromeres.

The chromosomes are present in pairs. The pairs are called the homologous pairs (they must be similar in structure and also have the same chemical composition). A species will always have the same number of chromosomes. This is called the chromosome number and it will always be an even number. This number is called the diploid number. During gamete formation, the homologous chromosomes separate and the gametes will have only half the number of chromosomes. This number is called the haploid number. Thus the somatic or the vegetative cells of all organisms are diploid and the gametes are haploid.

Number of chromosomes per cell nucleus varies from species to species. In man there are 46 chromosomes per nucleus (23 pairs) of somatic cells and 23 chromosomes in the gametes.

Is the chemical compound responsible for inheritance of characters.

DNA is one of the nucleic acids and the other is RNA.

DNA: Deoxyribo nucleic acid (less oxygen)

RNA: Ribo nucleic acid (more oxygen)

Both are similar in composition but different in structure.

Chemical composition of DNA

  • Contains a 5 carbon sugar (ribose sugar) in its structure and there fore has five corners.
  • It has a nucleic acid. An example is the phosphoric acid
  • Nitrogen/organic base is also present. There are four types namely

i) Adenine ii) Thymine iii) Cytosine iv) Guanine.

  • In both DNA and RNA adenine, Cytosine and Guanine are found.
  • Thymine is found in DNA while Uracil is found in RNA.

The following combine

  1. Adenine (A) + Thymine (T) for DNA or Adenine (A) + Uracil (U) in case of RNA.
  2. Cytosine (C) + Guanine (G)

Always the pairing of bases as indicated above is due to the matching of their structure (i.e complimentary base pairing rule)

Formation of the DNA molecule

Formed from nucleotide units to form long chains

DNA replication

It’s the ability of DNA to produce a copy of its self.

DNA replication occurs in three stages and its catalyzed by a series of enzymes.

i) Twisted strands un wind giving two strands. The strands are still joined.

ii) Weak bonds between the nitrogen bases will be broken down to give two separate strands.

iii) One strand will induce the formation of another strand which is complimentary to it and the same thing will happen to the other strand.

After replicating, two new DNA molecules similar to the original DNA molecule are formed.

Why is DNA suitable for inheritance?

i) Because of its ability to replicate

ii) DNA is capable of carrying large a mount of genetic information

iii) DNA is a very stable chemical and therefore can not easily be changed.

Cell division

Cell division is a process which leads to cell multiplication.

It occurs in both plants and animals. Original cells which undergo division are known as parent cells and the new on ones resulting from division are known as daughter cells.

There are two types of cell division i.e Mitosis (mitotic cell division) which occurs in somatic cells and Meiosis (meiotic cell division) which occurs in reproductive cells.

Stages of mitosis

1) Interphase (resting stage of the parent cell) During this stage the following happens to prepare a cell for nuclear division)

In this stage the cell builds up energy reserve in form of ATP

It also builds up food/nutrient reserve

Replication of DNA also takes place in the chromosomes. i.e the amount of DNA is doubled

There is synthesis/replication of new cell organelles/structures eg mitochondria, endoplasmic reticulum, centrioles, chloroplasts etc

Chromosomes become visible as long thin entangled threads.

The nucleolus begins to shrink and centrioles move to the opposite ends of the cell

Chromosomes shorten and they can be seen to comprise of 2 chromatid joined at the centromere

Nuclear membrane breaks up

Mendel’s contribution in genetics:

He was an Austrian and by practice he was a monk. He carried out experiments about inheritance in plants over 120 years ago.

The 1 st experiment he carried out was referred to as monohybrid inheritance. The experiment considered one type of contrasting characters at a time. Hybrid is as a result of crossing between two different characteristics. In the first experiment mendel used pure breeding seeds

E.g tall plants crossed with tall plants Tall off springs only (no short plants)

He planted garden pea (Pisum sativum). The garden pea showed a variety of characteristics e.g colour of flowers, colour of pods, height of stems, nature and texture of pods, and shaped of pods.

The pattern of transmission of different characteristics was interesting eg when a plant showing one set of characteristic is cross pollinated with that showing opposite characteristic, the first generation off spring will be showing one parent’s characteristic.

When the first generation plants are self pollinated, a mixture of both parental characteristics is shown.

Parents: Tall cross pollinated with short

1 st generation off springs: All tall

1 st generation off springs self pollinated

2 nd Generation off springs: ¾ tall and ¼short.

For colour of pods

Parent plants Green pods X Yellow pods

Gametes pollen grains ovules

Fertilization

1 st Generation off springs green coloured pods

The seeds from the first generation are planted again and after flowering, self pollination was carried out.

1 st Generation off springs Green pods X (self pollinated) green pods

Gametes pollen grains ovules

Fertilization

2 nd Generation off springs ¾ green pod ¼ yellow pod plants

A mixture of green poded and yellow poded plants was got.

Mendel referred to what is responsible for the characteristic as genes carried by chromosomes.

A gene is a unit of inheritance:

There fore, A gene responsible for green pods is dominant (green pod is a dominant character) and is represented by letter G and yellow pod is a recessive character and the gene is represented by letter g.

Random fertilization

1 st gen Gg Gg Gg Gg All off springs green poded

1 st generation off springs self pollinated.

Parents Green pods X Green pods

(GG, Gg, Gg) = ¾ 2 nd generations plants with greens pods and (gg) = ¼ 2 nd generation plants with yellow pods.

Tallness is dominant character and shortness is recessive character.

Let the gene for tallness be T

Let the gene for shortness be t

There fore the genotype for the tall plant is TT and for the short plant is tt

Parents Tall plant X Short plant

1 st generation Tt Tt Tt Tt First generation plants, all tall

First generation off springs self pollinated

1 st generation Tall plant X Tall plant

(TT, Tt, Tt) = ¾ 2nd generation off springs tall and tt = ¼ 2 nd generation off springs short.

Working out fertilization using the punnet/chi square

Genetic terms

  1. 1. Genotype: is the genetic make up of an organism. From the illustrations above (Tallness), we see 3 genes TT, Tt and tt which gives 1:2:1 as the genotypic ratio.

TT and tt are known as homozygous genes

They are called so because they were formed from fusion of the same gene.

TT is homozygous dominant (tall) and tt is homozygous recessive (short)

TT and tt are pure breeds.

Tt is known as heterozygous (tall) created from different genes fusing together. Its not a pure breed.

  1. 2. Phenotype: is the external expression of a gene present in an organism. When expressing its self its known as an allele which is a short of allelomorph.

When not expressing its self, its simply termed as agene.

A dominant gene is one that over shadows a weaker gene known as a recessive gene.

A recessive gene is one that is over shadowed by a dominant gene.

This happens when both the recessive and dominant genes for a particular trait/ characteristic are present in an organism i.e heterozygous (Tt )

  1. 3. Filial generation: The off springs that result from fusion of gametes in various generations eg

F1 Generation 1 st generation.

F2 Generation 2 nd generation.

F3 Generation 3 rd generation.

Its used to determine the genotype of either homozygous dominant (TT, GG, HH) or Heterozygous (Tt, Gg, Hh). Since genotypes TT and Tt both produce tall plants, its not possible to know from the phenotype whether the tall plants are homozygous dominant or heterozygous.

The test or back cross is done by crossing the tall plants from the F1 generation with a true recessive plant (tt).

The proportion of tall and short off springs in F2 will determine the genotype in the tall F1 plants.

In case of homozygous dominant (TT), when crossed with homozygous recessive, the off springs are 100%tall

Parents (F1) Tall plant X Short plant

Genotype T T t t (Homozygous recessive)

2 nd generation Tt Tt Tt Tt 2 nd generation plants, all tall

An in case of Heterozygous (Tt)

Parents (F1) Tall plant X Short plant

Genotype T t t t (homozygous recessive)

2 nd generation Tt tt tt Tt F2 generation plants, 50% tall and 50% short

Incomplete dominance/partial/co-dominance

This is a condition where genes controlling contrasting characteristics have equal influence when in heterozygous genotype. Such gene are said to be co-dominant genes.

E.g in hibiscus plants genes responsible for the red and white flower colours are co-dominant.

If a plant with red flowers is cross pollinated with that of white flowers, what are the possible genotype and phenotype of the F1 off springs?

Let the gene responsible for red flower colour be R and the gene for white flower colour be W. There fore the Genotype for plant with red flowers is RR and for the plant with white flowers is WW

Parents Red flower X White flower

1 st generation RW RW RW RW First generation: all pink flowered plants

F1 off springs are self pollinated, state the possible genotype, phenotype, genotypic ratio and phenotypic ratio

F1 off springs Pink flowered plant X Pink flowered plants

2 nd generation RR WW RW RW

Phenotype = Red flower plant (RR), Pink Flower Plants (RW and RW) and white Flower plant (WW)

Blood groups in human beings

A and B are co-dominant genes but dominant over O

Genes Possible genotype Phenotype
A AA or AO Blood group A
B BB or BO Blood group B
O O Blood O
A and B AB only Blood group AB

What is the possible genotype of off springs from a marriage between a man of blood group A and a woman of blood group B?

Possible genotype of the father: AA and AO and that of the mother: BB and BO

Blood group B Blood group A
Parents Mother X Father Off springs
BB AA → all AB
BO AA → AB, AB, AO,AO
BB AO → AB, AB, BO, BO
BO AO → AB, BO, AO, O

There fore the possible blood groups of the F1 children are: A, B, AB and O

  1. In a mixed day school, Angela got pregnant and she is of blood group B, Kapere a fellow student was accused to be responsible for her condition, which he denied. Angela gave birth to bouncing baby boy of blood group O. As an investigation was done Kapere was un cooperative and his blood group would not be discovered, but both his parents were of blood group A. Work out to find whether kapere would be the likely father of the baby.
  2. A woman of blood group A claims that a man of blood group AB is the father of her child. A blood test reveals that the child’s blood group is O. is it possible that the woman’s claim is correct? Could the father have been of blood group B? Explain your reasoning.

Multiple allele.

Multiple allele is a situation where by more than two alleles are controlling a certain characteristic. For example alleles A, B and O control the ABO blood group system in man.

Other conditions in Man transmitted in mendellian fashion. (Monohybrid inheritance in man)

Is a condition which results when the pigment for normal skin colour fails to form and this due to a recessive gene a.

Characteristics of albinism

White skin, Pink eyes and Golden hair.

To obtain an albino the child must receive recessive genes from both parents. This implies that an albino is homozygous recessive.

Gene for normal skin pigment: A and gene for Albino: a

Homozygous dominant: (AA) normal skin colour

Heterozygous: (Aa) Normal skin colour

Homozygous recessive: (aa) Failure of formation of normal skin pigment (albino)

To get a child who is an albino:

I) Both parents must be carriers

Parents Mother (carrier) X father (carrier)

Aa: Normal skin colour but carrier

II) One parent is a carrier and the other is an albino

Parents Mother (albino) X father (carrier)

Aa: normal skin colour but carrier

From the above two marriages, the mode of transmission of genes is similar to the mendellian fashion.

In this condition the person doesn’t possess bi concave shaped red blood cells but the shape is like that of a new moon. A person with sickle cells doesn’t have a large surface area so that sufficient oxygen can be transported through haemoglobin found in a normal person. People with sickle cell anaemia have short breath, tend to sleep when tired and have retarded growth.

Gene S is responsible for abnormal Haemoglobin.

Dominant gene H is responsible for normal haemoglobin.

NB: this is not an example of incomplete dominance.

Homozygous dominant HH: Normal haemoglobin

Heterozygous HS: sickle cell carrier (shows mild signs but never gets attacks)

Homozygous recessive SS: sickler

SEX DETERMINATION

Sex is determined by a special type of chromosome found in the sperms and the ova and they are termed as sex chromosomes. In man, since one chromosome is X and the second is Y, they may be referred to as heterozomes and those that are similar are autosomes. Ova can only carry the X chromosome Sperms may either carry the X chromosome or Y chromosome. The sex of the child depends on which sperm fertilizes the egg. If its an X sperm, the off spring is XX (girl) and if it’s the Y sperm, the off spring is XY (boy). Each off spring has characteristics limited to it. These are termed as sex limited characteristics

Man Female
Have a penis Have a clitoris
Have beards No beards
Have narrow tips and nipples Have wide hips and breasts

Sex linkage: sex linked genes

Is a condition where the genes controlling a trait/characteristic have to be transmitted on a sex chromosome. Such traits are referred to as sex linked characters controlled by sex linked genes. Most of the sex linked genes are recessive and commonly found on the X chromosome and in rear cases on the Y chromosome

When the X chromosome has a recessive gene in males, normally the Y chromosome is empty. Since the chance for the Y chromosome to be empty is common, there fore males can inherit a recessive gene from a carrier mother and inherits an empty Y from the father and becomes a sufferer.

There fore sex linked characters are common in males than females. (in most cases females end up as carriers if they have a sex linked gene)

Examples of sex linked genes:

X H X H -Normal female homozygous dominant:

X H X h -Carrier female (heterozygous)

X h X h – Female Haemophilia sufferer

X h Y-Male Haemophilia sufferer

  1. Colour blindness: is the inability to distinguish between primary colours ie red, green and blue.

X C X C -Normal female homozygous dominant:

X C X c -Carrier female (heterozygous)

X c X c – Colour blind female

  1. Premature balding
  2. Browning of teeth
  3. Porcupine man: the growth of thick hair at the entrance of the auditory canal. Its suspected to be associated with the Y chromosome because its found only in male.
  1. Explain what is meant by sex linked characters
  2. List examples of sex linked characters found in man
  3. In a marriage, a woman who is heterozygous for the sex linked character of colour blindness marries a colour blind man. What would be the result of this marriage?

Show the working if the sons from this marriage got married to carrier females, what would be the result?


Questions in Review Questions

(ii) Which of the following is not an insectivorous plant:

(iii) This leaf shows a parallel venation:

(iv) The point on the stem from where the leaf arises is

(v) Which of the following is essential for photosynthesis:

The part of the plant which grows under the ground.

The part of the plant which grows above the soil.

Q3) Differentiate between the following:

(i) Tap Root and FIbrous Root

(ii) Simple Leaf and Compound Leaf

(iii) Parallel Venation and Reticulate Venation

Q4) What are the four functions of the roots?

Q5) Mention the functions of the following:

Q6) Define venation. What are the different types of venation found in the leaves?

Q7) Describe the modification of leaf in any one insectivorous plant.

Q8) Write the two main functions of leaves.

Q9) What is the modification seen in the Bryophyllum. Explain.

Q11) Name the wide flat portion of the leaf.

Q12) What purpose is served by the spines horned on the leaves of cactus?

Q13) Explain why leaf survival is so important to the plant?

Q14) Give an example of the following and draw generalized diagrams for the same:

(i) Simple leaf and Compound leaf

(ii) Parallel Venation and Reticulate Venation

Q15) In list some of the advantages of transpiration to green plants.

Q16) Why do some plants have to trap insects?

Q17) Explain some of the modifications of leaves found in plants.

Q18) What is a tendril? Explain its use to the plant.

Q19) Complete the cross word using the clues given below. Check your performances with the correct solutions given at the end of the chapter.


The Help Summary and Analysis of Chapters 5 - 7

(Written from Skeeter's point of view.) A college graduate who has recently moved back to her parents' home, Skeeter is facing a number of difficulties. She is hurt and angry about her best friend Hilly's threat to kick her out of the League, which indicates just how far apart the two women have grown. Skeeter's mother is concerned that her daughter has not gotten married yet at the late age of twenty-three, so she constantly nags Skeeter to do whatever she can to get a man. Skeeter, however, is much more interested in finding a job where she can pursue her passion for writing. The only job to which she has applied (an editor position at Harper & Row) has not gotten back to her. Flipping through the employment listings in the local paper, Skeeter notices that many jobs offer women lower pay than men for the same work.

Skeeter misses her childhood maid, Constantine, who was mysteriously dismissed from the household during Skeeter's senior year of college. Skeeter reminisces about her close relationship with Constantine, remembering a time when a boy called her ugly and Constantine comforted her by saying that true ugliness comes from being a mean and spiteful person. Skeeter's mother constantly harangued her to brush her hair or wear more flattering clothes, but Constantine was interested in what Skeeter's thoughts and ideas. Still, Skeeter understands that there's much about Constantine's life that she doesn't know for example, Constantine once casually mentioned that her father is white, which surprises Skeeter and raises a number of questions. Skeeter and Constantine remained close even when Skeeter goes off to college, writing letters to each other frequently. In Constantine's last letter to Skeeter, she mentions that she has a surprise for her, but Skeeter's mother says that Constantine left to live with her people up in Chicago. Skeeter is certain that there is more to the story than this.

(Written from Skeeter's perspective.) One September morning, Skeeter is delighted to see that she has received a reply from Harper & Row. She hasn't gotten the job, but one of the editors, Elaine Stein, has decided to give her some career advice. Stein tells Skeeter that she should get an entry-level job at a newspaper, and should write about the things that disturb her. She offers to read some of Skeeter's ideas as well, and Skeeter sends her a list of topics such as drunk driving, illiteracy, and so on. Skeeter heads to the Jackson Journal, asking about any reporter positions that they have open. Though the editor, Mr. Golden, is rather rude (he asks if she managed to have any fun in college), he offers her a position writing Miss Myrna's weekly cleaning advice column.

Skeeter is elated with her new job, but her mother comments about how ironic it is that Skeeter - who has never cleaned anything in her life - is writing a cleaning column. Her mother is dismayed that this job will give Skeeter no opportunities to meet a potential husband. Skeeter realizes that her mother is right about her lack of knowledge of cleaning, and decides to ask her friend Elizabeth Leefolt's maid, Aibileen, some questions about cleaning in order to write the column. Elizabeth grudgingly allows this, and so Aibileen answers Skeeter's questions about scrubbing bathtubs and getting stains out of clothing. Skeeter casually mentions Constantine and how she moved up to Chicago, but Aibileen says that Skeeter’s mother fired Constantine. Skeeter confronts her mother about this, and her mother refuses to explain further, saying only that Skeeter will understand when she hires help of her own.

Skeeter continues asking Aibileen about cleaning advice for her column, and the two begin to grow close. Aibileen casually mentions Constantine's very light-skinned daughter, who appeared at Skeeter's mother's house shortly before Constantine was fired. Aibileen stops herself and says she is scared to say anything more.

Skeeter receives a reply from Elaine Stein, who is not impressed by her dull and passionless list of ideas. Skeeter is devastated, but she also realizes that she should have chosen ideas that interest her rather than ideas that sound impressive. The seed of an idea begins to grow in her mind, a dangerous but compelling one that won't go away.

(Written from Aibileen's perspective.) Little Mae Mobley is growing up fast, and Aibileen reminisces about her other "used-to-be" babies while Mae plays in the yard. Mae Mobley's mother steps in to snap at her for not eating in her high chair, and Mae Mobley calls herself a bad girl. Aibileen denies this, and tells Mae that she is smart and kind she prompts her to repeat this and remember it.

Aibileen realizes it's time to potty-train Mae Mobley. This is made more difficult by the fact that Miss Leefolt refuses to let Mae Mobley watch her using the toilet, which would set an example for the young girl to follow. Mae has a great deal of trouble figuring out how to use the toilet, so Aibileen realizes that she will have to show the girl what to do. Aibileen faces a conundrum: show Mae Mobley on the white toilet, or on the "colored toilet" in the garage set aside for the black maids? She decides that the best course of action is to take the girl to the colored toilet Aibileen sits down and uses the toilet. After she is done, Mae immediately jumps up and uses the toilet herself for the first time, to Aibileen's great pride.

When Miss Leefolt comes home, Aibileen proudly tells her that Mae Mobley has learned to use the toilet. Eager to show her mother, Mae runs to the bathroom. the "colored" bathroom in the garage. Miss Leefolt flies into a rage, spanking her daughter harshly, and telling her never to use that bathroom because it is dirty and diseased.

Aibileen continues to give Skeeter information about cleaning for her newspaper column, and she finds herself talking to Skeeter about things she has never spoken to any white person about, such as the politics of her church or her son Treelore's excellent grades. Still, she can't bring herself to talk to Skeeter on the third anniversary of Treelore's tragic death.

But life goes on. Aibileen goes grocery shopping to supply the Leefolt family's Thanksgiving, and while there she runs into another maid who tells her about another tragic event: a young black man named Robert (who is the grandson of Louvenia, one of the matriarchs of the black community) used an unlabeled white restroom, and two white men beat him with a tire iron until he was blind in one eye. Aibileen is horrified.

Aibileen is shocked to find Skeeter waiting for her when she gets home. Skeeter says she wants to interview Aibileen about what it's like to be a maid, and what it's like to work for a white family. Aibileen says that this is extremely dangerous, and points out the violence meted to black people who challenge the status quo, most recently Louvenia's grandson. Skeeter asks her to think about it, but Aibileen says no and walks away.

Analysis:

Thus far we have mainly heard the about the difficulties faced by black women (which are substantial), but chapters 5 and 6 indicate that white women also face some difficulties. Though she is only twenty-three, Skeeter is under enormous pressure to marry all of her friends have done so, her mother is constantly badgering her about getting a boyfriend. Skeeter knows that marriage will mean the end of her professional ambitions, and moreover, it will mean subordinating her intellect to that of her husband. But though she tries to avoid marriage, her family, friends, and society do not let her forget that she will be considered a failure until she marries.

As a woman, Skeeter's position in the white world is inferior. She is discouraged from getting a job despite her education and talents, and she discovers that she will be paid less than a man for performing the same work. When she appears for a job interview at the local paper, she receives inappropriate comments from the managing editor about her appearance and social activities. She persists despite this disrespect, and manages to get a minor position writing a column about women's work. Still, it's a start.

Despite her gender, Skeeter still has a great deal of power over the maids. She essentially plagiarizes Aibileen's experience to write her column and doesn't think twice about it. However, Skeeter also begins to question some of the things she has been taught since childhood, such as the idea that black people are all dirty and diseased.

Chapters 5 and 7 offer the reader a more detailed perspective on the relationship between black maids and the white children they raise. Skeeter treasures her memories of Constantine, and had a more loving relationship with her than she did with her own mother. However, Skeeter is stunned to learn that Constantine had a daughter despite their close relationship, Constantine never felt safe telling Skeeter about this major aspect of her personal life. Though maids might seem to be family or even closer than family, this is still a tenuous economic arrangement that does not favor the maids. The black women are dependent for their livelihood on this work, whereas white families know they can always hire another maid.

The black maids struggle with having so little official influence over the children they care for this is made clear through the relationship between Aibileen and Mae Mobley. After Miss Leefolt spanks her daughter for using the colored bathroom, Aibileen feels deep guilt for indirectly causing harm to the girl, but she also hates that Mae Mobley is being taught to believe that black people are dirty and diseased. Aibileen is responsible for daily activities like feeding, cleaning, and playing with Mae Mobley, and she is the one the girl spends most of her time with. However, she has absolutely no control over the major decisions in Mae's life, and she has a limited capacity to instill important values.

The vicious beating perpetrated against Louvenia's grandson marks a major turning point in the narrative. We begin to see an indication that racism can be even more serious than separate bathroom facilities or snide remarks from employers - racism can literally kill. Dread of similar treatment will haunt all of the black characters throughout the rest of the novel. They know very well that the price for challenging racism can be violence or death.


How to solve Aptitude Ratio and Proportion problems?

You can easily solve all kind of Aptitude questions based on Ratio and Proportion by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Ratio and Proportion problems.

Exercise :: Ratio and Proportion - General Questions

A and B together have Rs. 1210. If of A's amount is equal to of B's amount, how much amount does B have?

B's share = Rs. 1210 x 2 = Rs. 484.
5

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

Let the third number be x.

Then, first number = 120% of x = 120x = 6x
100 5

Second number = 150% of x = 150x = 3x
100 2

Ratio of first two numbers = 6x : 3x = 12x : 15x = 4 : 5.
5 2

A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

140 x 5x , 150 x 7x and 175 x 8x
100 100 100

7x, 21x and 14x.
2

The required ratio = 7x : 21x : 14x
2

14x : 21x : 28x

In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is:


5.7: Summary Questions - Biology

While questions should be posed in the protocol before initiating the full review, these questions should not become a straitjacket that prevents exploration of unexpected issues (Khan 2001) . Reviews are analyses of existing data that are constrained by previously chosen study populations, settings, intervention formulations, outcome measures and study designs. It is generally not possible to formulate an answerable question for a review without knowing some of the studies relevant to the question, and it may become clear that the questions a review addresses need to be modified in light of evidence accumulated in the process of conducting the review.

Although a certain fluidity and refinement of questions is to be expected in reviews as a fuller understanding of the evidence is gained, it is important to guard against bias in modifying questions. Data-driven questions can generate false conclusions based on spurious results. Any changes to the protocol that result from revising the question for the review should be documented in the section ‘Differences between the protocol and the review’. Sensitivity analyses may be used to assess the impact of changes on the review findings (see Chapter 9, Section 9.7).When refining questions it is useful to ask the following questions:

What is the motivation for the refinement?

Could the refinement have been influenced by results from any of the included studies?

Are search strategies appropriate for the refined question (especially any that have already been undertaken)?


Biology Model Test Questions: (MCQs) for NEET

The multiple choice format is widely used in educational testing. This section of our website, focuses on Biology MCQs with subject wise and cumulative approach. The prime objective is to make you familiar with the competitiveness of Entrance Examination by means of publishing these question papers in a organized format. As you scan these MCQ question papers it will instigate in you the understanding and confidence of all subjects of Biology.

Solve hundreds of Biology question papers in Multiple Choice Question (MCQs) format. If you are pursuing your intermediate science studies or targeting for entrance examinations, you can solve these question papers online. We have taken all the necessary care to provide you with genuine question papers and the answers from reliable sources. Hope our labors server student’s community.

Remember the technique of solving MCQ questions can only be mastered by regular practice. Read each question carefully. Have a second look and rephrase if you don’t understand it. It is recommended don’t read the answers. Solve the problem and see if your answer is listed amongst the other choice. Another way of solving MCQ could be the “Process of Elimination”. You know each MCQ has a correct option. So check each option which one best works.

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