SS 3 BIOLOGY.

DATE: 12/08/2020 WEDNESDAY

CLASS ACTIVITY

1. Define test cross and mention its importance in genetics
2. Give reason(s) why blood group O can not receive blood from other blood group except from O.
3. A cross is made between a plant that has red flower and yellow seeds and a plant with white flower and green seeds. All the offspring of the Fi generation have red flower and yellow seeds. The Fi plants are allowed to self- fertilise. What are the phenotypic and genotypic ratios of F2 generation offspring?
4. If a woman is Rh-ve and her husband is heterozygous genotype Rh+ve. Can there be any problem in the course of child bearing?

Date: 11/08/2020

CLASS ACTIVITY

1. In an experiment, brown rabbits were interbred with white rabbits. It was observed that all the offspring in Fi generation had brown coat. The Fi offspring were allowed to self-breed, and the white coat trait reappeared in about one quarter of the offspring. Explain the results of the experiment.
2. When a certain variety of black chicken is crossed with white chicken, all the resulting offspring are checkered black and white. Explain this pattern of inheritance.
3. Four o`clock plant exhibit incomplete dominant for flower colours. When red flowered plants and white flowered plants are crossed, infer the flower colour of the resulting offspring.
4. Differentiate between incomplete dominant and codominant.

Tuesday Correction

1. Oviparous animals are animals that their female parents lay their eggs, which may be fertilized internally or externally into the environment.Its obtains nutrients only from the eggs. Examples of these animals are birds, fishes and insects.                                                                     Viviparous animals are animals that the fertilized eggs develop in the mother`s body. The embryo obtains its nutrients from the parent in addition to that present in the eggs.
2. Animals living in the polar regions have thick body covering to trap body heat near their body surface , thus providing insulation. They have a layer  of body fat  called Blubber, which insulates them against  the cold.
3. Strata.
4. i) Flying squirrel: it possesses muscular membrane along sides of its body that enable  it to glide from tree to tree     ii)Tree  frog : it has toes with stick  pads that help it to grasp onto the tree branches       iii) gecko: it has friction pads on their feet to help them climb up the smooth  tree barks.

Wednesday Correction

1. How plants and animals in savanna  are adapted to the environment conditions: Savanna experiences wet and dry season. During the dry season, water is scarce. Therefore many birds   and some nomadic animals  such as zebra ,buffalo migrate to regions  of plenty water .Their hoofed feet allow them to travel great distances. Some animals like elephant deal with annual water shortage by tearing  away bark  of baobab trees to get water found in the trunk. Some animals remain  dormant in their burrows during dry season.                                                              The challenges that grassland plants face during the dry season are shortage of water and occurrence of  fires. Some savanna plants  such as baobab are fire and drought resistance. Baobab has a very large trunk to store water during the dry season and  the succulent bark also  contains resins  that protect  the tree from bush  fires.

2. Functions of the blood

i)It transports materials like digested food, hormones, waste products and heat around the body. This is performed by the plasma.                                                                                                                                    ii)Transportation of the oxygen  by the Haemoglobin in the red blood cell.                                                                                                                  ii) Production of the Antibodies  by the White blood cell                                                                                          iii)Clotting or Coagulation of blood   by the Platelets .

3. Differences between the artery and vein

Artery	Vein
Carries blood away from the heart	Carries blood towards the heart
Blood is under great pressure, hence it flows fast	Blood is not  under great pressure, hence it flows slowly and smoothly
Has no valve	Has valve to prevent flow back
Carries red oxygenated blood except pulmonary artery	Carries bluish-red deoxygenated blood except the pulmonary vein

4. Function of the following:

i. Pulmonary artery :it carries deoxygenated blood from the heart to the lung

ii. Pulmonary vein: it carries oxygenated blood from the lung to the heart.

iii. Iilac artery: it supplies oxygenated blood to the pelvis, pelvic organs, reproductive organs, part of the thigh  and the lower leg.

iv. Renal artery: it carries oxygenated blood to the kidney.      

WEDNESDAY 05/08/2020

CLASS ACTIVITY

1. Explain how plants and animals have adapted to the environmental conditions in a savanna?
2. State five functions of the blood, hence, mention the types of blood that perform the function
3. State five differences between artery and vein
4. State the functions of the following blood vessels    i)pulmonary artery    ii) pulmonary vein             iii)lilac  artery         iv) renal artery

04/08/2020 TUESDAY

CLASS ACTIVITY

1. Explain the following terms: oviparous and viviparous animals and give two examples each.
2. Polar regions have characteristic animals such as polar bear, seals and penguins. Discuss the features that enable these animals to survive in  such temperatures. What is the purpose of these adaptations?
3. What is another name for different layers of a tropical rainfall?
4. Animals in rainforest have adaptations to help them move about. Describe the adaptation for each of the following organisms;  i) flying squirrel  ii) frog     iii) gecko

TUESDAY    04/08/2020

CORRECTION TO LAST WEEK CLASS ACTIVITY  (Tuesday)

1. Let the allele for normal haemoglobin==A—                                                                                               Let allele for sickle-cell anaemia=S—                                                                                                                       Normal individual has –AA                                                                                                       Carrier of the disease has   AS                                                                                                                            Therefore cross between     2 carriers, ie      AS   X    AS   will give  offsprings with    —-                                  AA, AS,  AS, and SS.                                                                                                                                                     Therefore, the chances of the couple having a child with sickle-cell anaemia disease is 25%                b) I will advise the couple to go for genetic councelling. The risk of of having a child with sickle-cell anaemia will be calculated and this will enable them to make decision on whether  to have a child together. 
2. The mother genotype is  I^B  I^O  and the blood group is  B.  The child blood group is AB and his genotype is  I^A   I ^B  . Therefore, the father `s blood group can be  A or AB ( genotypically)

• Yes .

Father`s blood group—AB                                                                                                                                    Mother`s blood group—O                                                                                                                                         Child`s blood group—B                                                                                                                                                       Crossing father and mother    ie   AB  X OO                                                                                                                                               I                                                         I^A  I^{B    X}   I^O I^O                                                                                                                                                                                                                possible offsprings genotypes  are—I^A  I^O , I^A I^O  , I^B  I^{O ,} and  I^B  I^O and  their blood groups are  A and B  The genotype  of the mother is  I^O I^O.  and the suspected father is I^A  I^{B.  .}There is possibility that the child has received the allele  I^B  from the father and allele  I^O  from the mother..                                      So his genotype is   I^B  I^O    and his blood group is B.

Wednesday Correction

1. b) Functions of  i.Ciliary muscle—it alters  the thickness of the lens       ii)Suspensory ligament—it holds the lens in place    iii) Vitreous humour–  it aids refraction of light and also serves as nutrient   iv)Iris –it controls the size of the pupil and also controls the amount of light entering the eye.

2. b) Functions of the following     i)Sebeceous gland—it secretes an oily substansce called Sebum into the hair follicle, this lubricates the hair and keeps the skin soft and smooth.       Ii) Malpighian layer—it contains  a pigment which gives the skin its characteristic colour. The pigment also protects the the skin from ultra violent rays.        Iii)Conified layer—it prevents germs from entering the skin and preventing mechanical injury

29/07/2020 WEDNESDAY

1. a)Draw and label the structure of human eye( vertical section) b)State the functions of the following structures i)Ciliary muscle ii) Suspensory ligament iii)Vitreous humour iv) Iris v) Yellow spot
2. a) Draw and label the longitudinal section of the skin b) State the functions of the following structures : i)Sebaceous gland ii)Malpighian layer iii)Conified / Horny layer

1. Sickle cell anaemia is a receive disease. If a couple who are both carriers of the disease plan to marry and have a baby, what are the chances that their child will be born with sickle-cell anaemia? Use genetic diagram to illustrate your answer. What advice will you give the couple?
2. If the blood type of a woman is B, and her child`s blood

type is AB. What are the possible blood type of the father of the child?

3. Can a man with blood group AB possibly be the father of a boy whose blood group is B, given that the mother`s blood group is O?Give reason for your answer.

CORRECTION TO LAST WEEK CLASS ACTIVITY

mitosis	meiosis
Daughter cells contain same number of chromosome as the parent cell ie diploid	Daughter contain half number of the chromosomes as the parent cell ie haploid
Pairing of homologous chromosomes does not occur	Homologous chromosomes pair at prophase 1
No crossing over 	Crossing over occurs
Daughter cells are identical to parent cell. Two daughter cells are produced from one parent cell	Variationoccurs i9n the daughter cells Four daughter cells are produced
It occurs in soma body c ell during growth or repair of body parts	Occurs in the gonads during gamete formation.

. Tropism/. tropic movement:This is growth movement of plant part in response to an external unilateral stimulus and it is directl y dependent on the direction of the stimulus. Eg phototropism,, geotropism, hydrotropism

Nastism/nastic movement: this refers to the non-directional response of an organism to stimuli like touch, temperature, humidity. Eg Mimosa plant to touch.

Tropism	Nastism
Response leads to growth	It does not
Response is slow	Response is fast
Response is towards the direction of the stimulus	Response is not towards the direction of the stimulus
Movement is not easily reversed	Movement is reversible

WEDNESDAY 22/07/2020

CLASS ACTIVITY

Study the diagram above, and answer the questions that follow; a) the point of crossing over is called what? b) mention two importance of the crossing over

2. Why is a reduction division important in organisms that reproduce sexually?

3. State five differences between mitosis and meiosis

4. Name the products of aerobic respiration

5. Describe an experiment to show that yeast can respire anaerobically

TUESDAY 21/07/2020

CLASS ACTIVITY

1. Draw and label mammalian heart
2. Explain briefly pulmonary circulation in mammals
3. Staye five processes by which the mammalian body reduces its body temperature
4. Define tropic and nastic movement,
5. State four differences tropic and nastic movement, give two examples each

16/07/2020

CORRECTION TO TUESDAY CLASS ACTIVITY

1. Water molecules will leave the bacterial cell by Osmosis and sugar molecules will enter the bacterial cell by Diffusion.
2. The water potential of the distilled water is higher than the water potential of the red blood cells. Hence, water enters the red blood cells by Osmosis. The cells swell and burst, releasing the red pigment(haemoglobin). The red pigment molecules diffuse from the bottom of the beaker to the region of lower concentration throughout the beaker.
3. Active transport is the process in which energy is used to move particles of a substance against a concentration gradient from the bregion of lower conc. to a region of higher conc. Examples absorption of dissolved mineral salts by the root hairs from the soil and the absorption of glucose and amino acids by the cells in the small intestine.

Diffusion	Osmosis
It involves the movement of particles of solute, gaseous or liquid	It involves movement of water molecules only
Absence of semi-permeable membrane	Presence of selective permeable membrane

5. Cytolysis is the bursting of the cytoplasm due to excess water entering the cytoplasm of the cell while Cyclosis is the streaming /movement of the cytoplasm inside a living cell in order to ensure distribution of materials.

6 Hydrolysis is the process of adding water to chemical reaction while condensation is removal of water molecules from chemical reaction.

CORRECTION TO WEDNESDAY ACTIVITY

1. Cortex, medulla, renal pelvis, renal artery, renal vein, ureter.
2. It transports urine to the urinary bladder

Renal artery	Renal vein
It carries oxygenated blood	It carries deoxygenated blood
It contains more urea	It has less urea .

1. Mention five parts of a section of kidney
2. State the function of the ureter
3. State two differences between renal artery and renal vein
4. What would be the effect on urine volume: a) On a hot day? b) On a cold day? c) If the blood pressure in the body falls?
5. Name the substance normally present in urine of a diabetic person
6. The darker the colour of urine, the more concentrated it is. Explain how you could monitor how dehydrated your body is by examing the colour of your urine

TUESDAY, 14/07/2020

CORRECTION TO TUESDAY CLASS ACTIVITY

1. a) Pancreas. b) Adrenal gland
2. Diabetis mellitus
3. i) Blood glucose level rises. ii.) Glucose appears in the urine.
4. By injecting insulin into the veins
5. a) Adrenaline. b) i.There is an increased heart beat rate . ii) Glucose is transported from the liver to the muscles at faster rate. A iii. Metabolic rate increases and more energy is released for muscles contraction
6. The higher water potential in the blood cells compared to that of the blood will cause water molecules to move out from the blood cells. This causes the blood cells to shrink.

Wednesday

1. In the small intestine, starch which is not digested in the mouth,VI’s digested by the pancreatic amylase to maltose. Maltose is broken down to glucose by maltase, lactose to glucose and galactose by lactase and sucrose to glucose and fructose by sucrase.
2. Liver secretes bile which is involved in the digestion of fat. It converts excess glucose to glycogen for temporary storage, and converts glycogen to glucose when the blood glucose level drops.
3. Bile emulsifies fats. It lowers the surface tension of the fats, ie,it reduces the attractive forces between the fat molecules. This causes the fats to break into tiny fat globules suspended in water, forming an emulsion. This increases the surface of the fats, speeding up the digestion.
4. This is the process by which excess amino acid is converted to urea and excreted with urine.
5. So that the digested food will not accumulate in the villi. If the digested products absorbed are allowed to accumulate inside the villi, diffusion gradient will decrease . This will slow down the absorption. The continuous flow of blood through the capillaries transport absorbed food substances away from the villi.
6. The length of the small intestine slows down movement of food through the small intestine, giving sufficient time for digestion and absorption of digested products.

Today’s Class Activity (14/07/2020).

1. When a bacterium finds itself in a strong , hypertonic solution of sugar, which substance will leave the cell and which will move into the cell? What are the process involved?
2. A beaker of distilled water was placed on a bench. Using pipette,a drop of red blood cells was placed at the bottom of the beaker. Explain why the water in the beaker was uniformly red after two hours.
3. What is active transport?
4. How does diffusion differ from osmosis?
5. Differentiate between cytolysis and cyclosis
6. Compare hydrolysis and condensation

WEDNESDAY. 08/07/2020

CLASS ACTIVITY

1. Describe how carbohydrates are digested in the small intestine of a man
2. What is the role of the liver in the digestion of and utilization of food substances?
3. What is the role of the bile in the small intestine?
4. What is deamination of amino acid?
5. Small intestine is long . How does this help in digestion and absorption?
6. The villus is highly supplied with blood capillaries. How does the help to maintain a diffusion gradient for absorption of digested products?

TUESDAY 07/07/2020

Class activity

(1) In human body, mention the gland that secret insulin and adrenaline.

(2) Name the disease caused by a failure oftue gland to produce insuline.

(3) State two possible signs of the disease named in (2.) above.

(4.) How can the disease be treated.

(5) When you see an angry dog barking and charging towards you, you will run away fast as soon as you can.

(a) what hormone is secreted in such an emergency?

(b) State three effects of the hormones that help you to run away.

(6) If the blood glucose level rises sharply above the normal, the red and the white blood cells will shrink and become dehydrated. Explain why.

01/07/2020

CORRECTION TO TUESDAY QUESTIONS

1. Cones and Rods.
2. Rods
3. The focal point of the eye ,yellow spot/fovea, is packed with the cones while the rods are distributed around the retina. Consequently, the image that is focused on the fovea will activate the rods which is around the fovea, producing a slightly blurred image. The brain compensates to some extent for this blurring.
4. Colour vision depends upon cones which are only activated at high light intensity /during day time. Cones are therefore not useful to animals that are active at night.
5. Rods contain Visual purple, the pigment concerned with dim light. Bright light bleaches the visual purple of the rods. Thus, when a person enters a dark room from a brighter one, he may not be able to distinguish the objects around him for some time. This because it takes a while for the visual purple to be formed again in the rods. In dim light , radial muscles in his eyes( iris) contract while the circular muscles relax. His pupils dilate, increasing the amount of light entering this eyes and allowing him to see.
6. Accommodation/ Focusing : is defined as the ability to focus both near and far objects on the retina. It is the adjustment of the lens of the eye so that clear images of objects at different distances are formed on the retina.
7. Eye Defects are:

a. Long-sightedness/ Hypermetropia: A person with long sight is able to see distant objects clearly but can not see near objects. This is because the eye ball is too small so images of objects are formed after the retina.

b. Short-sightedness/ Metropia: A person with short sight can see near objects clearly but cannot see distant objects clearly. This is because the is longer than normal, causing parallel rays from far objects to focus at a point in front of the retina. The person see the a blurred image.

c. Astigmatism: The person sees blurred/ distorted images. This due to the curvature of the cornea or lens which becomes oblong in shape instead of being spherical.

d. Presbyopia/Old-sightedness; Presbyopia affects almost everyone from 40 years old. The symptoms include difficulty in reading fine print, requiring brighter for reading and fatigue. The shape of eyeball is not affected but the cilliary muscles weakens and the lens loses its elasticity. With increasing in age, the lens of the eye loses its elasticity and power of accommodation. Light rays from near objects cannot be focused on the retina and images are blurred.

e. Night blindness: Night blindness is the inability to see clearly in dim light or at night which may be due to a deficiency in Vitamin A which is essential for the formation of Visual purple.

f. Cataract: The formation of cataract is a condition in which the lens loses its transparaency and becomes opaque. This reduces the amount of light entering the eye, often resulting in symptoms such as blurred vision, poor night vision.

g. Conjunctivitis: Conjunctivitis is caused by the inflammation of the conjunctiva. It is characterised by redness and swelling in the eye along with the feeling of discomfort and sometime discharges of pus and blurred vision.

CORRECTION TO WEDNESDAY

1. A
2. C
3. A
4. C
5. B
6. A
7. A
8. A
9. B
10. C

ESSAY

a. Active transport

b. Seaweeds would not be able to take in essential mineral salts from seawater without the process of active transport .When seaweeds do not obtain these essential nutrients/ mineral salts, they will die of nutrient deficiency.

WEDNESDAY, 01/07/2020.

OBJECTIVE QUESTIONS

1. The process by which complex substances are broken down into simpler one is referred to as  a)catabolism   b) metabolism c) anabolism
2. The organ which is sensitive to light in euglena is   a) gullet b)chloroplast  c)eyespot
3. The organelles present in cell that are actively respiring and photosynthesizing are   a) mitochondrion and chloroplast  b) lysosome and mitochondrion  c) nucleus and golgi apparatus
4. Which of the following distinguishes a butterfly from moth?  a) Both are active during the day b) the wing of butterfly rest horizontally but those of moth rest vertically  c) the abdomen of moth is fatter than that of the butterfly
5. Which of the following types  of feathers is used for flight?  a) covert b) down  c) quill
6. The structural adaptation of desert plants for water conservation is a) spiny leaves  b) sponge mesophyll  c)prominent stomata in leaves
7. The plants that grow in desert areas are referred to as  a) xerophytes b) hydrophytes  c) mesophytes
8. Proboscis is a structure that is mostly found in  a) insects  b)tapeworms  c) molluscs
9.  A plants tissue that carries water and mineral salts is  a) phloem  b) xylem  c)cortex
10. Which of the following helps in blood clotting? a) plasma b) white blood cell c) platelete

ESSAY QUESTION

The concentration of salts in the cells of some seaweeds may be much higher than that of seawater. Yet such plants are able to absorb salts from the seawater.

a. How do you think seaweeds are able to do this?

b. In what ways is the process important to life of seaweeds?

TUESDAY,    30/06/2020.

1. Mention the two photoreceptors on the retina.
2. Which type of photoreceptor is used in the dim light condition?
3. Why do objects appear slightly blurred or out of focus when you look at them in the dim light?
4. Why do animals which hunt at night not have  colour  vision
5. A  person suddenly entered a cinema from a bright light place. At first,  he could not even see pathway. He waited for some minutes before he could see. Explain what happened in his eyes that enable him to see in dim light.
6. What is  Accommodation?
7. Mention and describe six eye defects

Wednesday( correction)

1.Study from your textbook

2. If the testes are located  inside the body cavity, it will affect the production of sperms because sperm production requires a low temperature, low than the body temperature. This low temperature can be achieved outside the body cavity.

3. A- Vagina

    B- Oviduct/ fallopian tube

    C- Uterus/ womb

    D- Ovary

  E- Urethra.

23/06/2020

CORRECTION  FOR TUESDAY’S  QUESTIONS (NOTE; DO NOT FORGET TO CHECK SS 2 TUESDAY AND WEDNESDAY’S NOTES)

1. Osmoregulation in man.

Osmoregulation is defined as  the maintenance of water to salt contents in the blood.Osmoregulation in man is  carried out by the kidneys.  When concentration of water is high in the blood, the brain sends signal to the pituitary  gland to secrete small amount of ADH , so that small amount of water is re-absorbed  back from the filtrate into the blood. Therefore, diluted urine is formed( large amount of urine is excreted). On the other hand, if the concentration of salts is high, the brain stimulates the pituitary gland to secrete  large amount of ADH, so that large amount of water is re-absorbed back in the tubule into the body. Hence, concentrated urine is formed( small amount of urine is excreted). This is how the kidney regulates the water to salts content.

• 2..Importance of Contractile  vacuole in Amoeba.

Osmoregulation is carried out by contractile vacuole in Amoeba. The contractile vacuole removes the excess water from the cytoplasm in order to prevent cytolysis.The cytoplasm of amoeba has higher solute concentration than the surrounding water, therefore, water molecules move into the cell via osmosis. If this excess water is not removed, it will accumulate in the cell( cytoplasm)  and eventually lead to the bursting of the cytoplasm, this is called Cytolysis. Excess water gradually enters into the contractile vacuole until it is fully filled with water.The contractile vacuole contracts and discharges the water through a pore in the membrane into the water body.

• 3. The two substances are glucose and amino acids.  They  are selectively reabsorbed by active transport in the nephron.
• 4. The blood in the renal artery is oxygenated and enters the kidney. The blood in the renal vein is deoxygenated  and exits the kidney.
• 5. Ultrafiltration occurs in the Bowman’s capsule.

WEDNESDAY, 24/06/2020

CLASS ACTIVITY.

1. Study the structure of mammalian male reproductive system and identify the following structures: scrotal sac, testes, epididymis, sperm duct,seminal vesicle, urinary bladder, ureter, prostate gland, and penis. Also, learn their functions.
2. The testes lie inside the scrotal sac outside the body. What do you thick might happen if the testes are situated inside the body cavity?
3. Make a drawing of the female reproductive system of a mammal. Label on your drawing the following structures, using the names of the structures and appropriate letters:

A-the place where sperms are deposited during intercourse

B- the place where fertilization occurs

C- the structure where the foetus develops

D- the structure which produces eggs and hormones

E- the tube which carries urine out of the body.

TUESDAY, 23/06/2020

Hello students! How have been your class activities? Hope you are going through them and checking your corrections?

CORRECTION FOR TUESDAY ACTIVITY.

1. Internal fertilization is the fusion of male and female gametes inside the female organism, eg man,goat, hen etc . While external fertilization is the fusion of male and female gametes outside the female organism, the eggs are laid outside by the female organism and the male organism releases sperms on them for fertilization to occur externally, eg fishes andamphibians.

3. Four functions of placenta:

a. It allows dissolved food( glucose, amino acid, mineral salts) to diffuse from mother’s blood into the foetus’s blood.

b. It allows dissolved oxygen to diffuse from mother’s blood into the foetus’s blood

c. It allows metabolic waste products to diffuse from the foetus’s blood into mother’s bloodstream

d. It allows antibodies to diffuse from mother’s blood into foetus’s blood.

e.It produces progesterone which maintains the uterine lining during pregnancy

f. It attaches the foetus to the uterine wall of the mother. etc

4. Four functions of the umbilical cord:

a. It attached the foetus to the placenta

b. The vein in the umbilical cord transports digested food and oxygenated blood from placenta tothe foetus.

c. The artery transports waste products and deoygyenated blood from foetus to the placenta.

d. It suspends foetus in the amniotic cavity

5. Two functions of amniotic fluid

a. It supports and cushions the embryo before birth.

b. It serves as shock absorber.

c . During birth, it lubricantes and reduces friction in the vagina (birth canal).

d. It protects the embryo against mechanical injury.

6. Two hormones:

i.Follicle-stimulating hormone

ii. Oestrogen

iii. Luteinizing hormone

iv. Progesterone

7. Amniotic fluid protects the foetus against mechanical injury.

8. Placenta and umbilical cord.

Correction for Wednesday Activity.

1. Parthenocarpic in flowering plants is a phenomenon in which fruit develops without fertilization. They are seedless fruits Such fruits are called Parthenocarpic fruits. E.g banana and pineapple.
2. Double fertilization in flowering plants means the fusion of the two male gametes present in the pollen tube after pollination. The first male nucleus fuses with the egg cell or ovum to form the zygote, which develops into the embryo of the seed. This is the first fertilization. The second male nucleus fuses with the secondary nucleus or definitive nucleus to go form the primary endosperm nucleus, which develops into the endospermic. This is the second fertilization.
3. Endospermic seeds are monocotyledon seeds that contain endosperms which store food. The endosperm is not completely absorbed by the embryo after fertilization. E.g. maize,millet etc. While endospermic seeds are dicotyledon seeds which do not contain endosperms. The endosperm has completely absorbed by the embryo after fertilization. They only store good in the cotyledons. E.g beans, melon.
4. Conditions for Germination:

Water ,suitable temperature, air(oxygen), enzymes, and food( cotyledon/energy).

5. Role of enzymes in germination: With absorption of water, the cotyledon produces enzymes. The enzymes digest the stored food(in the cotyledon) so that the growing embryo can use it. The digested food is transported to the growing regions of the embryo I.e the plumule and radicle.

6. Seed dormancy: This is the ‘resting’ period of seed when they do not germinate as soon as they are shed, no matter how favourable the environmental conditions are.

7. Characteristics of dormant seeds

i. They are usually dry.

ii. They can withstand adverse environmental conditions

iii. Their vital activities are reduced

iv. They respire anaerobically.

8. The role played by the cotyledon in green plants: the cotyledons take up and store food from the endosperm for use by the embryo.

TODAY’S ACTIVITY. (Note ;You can view ss 2 today and tomorrow’s notes and study them for revision).

1. Write briefly on osmoregulation in man.
2. What is the importance of contractile vacuole in Amoeba?
3. Name two substances that pass out of the glomerulus into the Bowman’s capsule but are not normally absent in the urine of man. What happens to these substances in the nephron?
4. State differences between the blood in the renal artery and the blood in the renal vein
5. In which part of the kidney tubule does ultrafiltration occur?

17/06/2020. WEDNESDAY

TODAY’S ACTIVITY

1. What do you understand by the term Parthenocarpic in flowering plants?Give two examples of such plant.
2. Explain the term double fertilization in flowering plants
3. Differentiate between endoplasmic seeds and non endospermic seeds
4. Mention five conditions necessary for germination of seeds
5. What is the role of enzymes in germination of seeds
6. What is seed dormancy?
7. List two characteristic of dormant seeds
8. Explain the part played by the cotyledons in the life of green plants

TUESDAY 16/06/2020

CORRECTION

B .Specimen A- Terrestrial/Grassland

Specimen B- Terrestrial

Specimen C- Terrestrial/ termitarium

Specimen D- Terrestrial

Specimen E- Terrestrial

C. A food chain

Grass-Grasashopper – Lizard

D. Specimen A – adaptive features

i. Possession of wings for flight

ii. Possession of mandible and maxillae for cutting a and chewing of food materials

iii. Possession of long hind limbs for landing.

Specimen B-

Possession of fibrous roots for absorption of water and mineral salts.

Possession of chlorophyll in the leaves for photosynthesis

Specimen C-

Possession of strong mandible to attack the enemies

Possession of projected mouthparts used for injection of pioson into any intruder.

Specimen D

Possession of gular fold for attracting the matesf

Possession of scale for body regulation of temperate

Possession of fore and hind limbs for movement

Specimen E-

Possession of poison to attack its enemies

Ability to lay ambush to catch its preys

E. Classes of the specimens:

Specimen A -Insects

B- Monocoty

C- Insects

D- Reptilia

E. Reptilia

F. Structural differences between specimen A and C

A	B
Possession of eyes	Absence of eyes hence it is blind
Presence of wings	Absence of wings
Possession of small	Possession of big head
Possession of short hind limbs	Possession of long hind limbs

G. Respiratory organs

Specimen A- Tracheal system

B- Stomata and lenticels

C- Tracheal system

D. Lungs

E. Lungs

Correction for Wednesday:

1. Evidence from Comparative anatomy
2. Evidence from Embryology
3. Evidence from Domesticated Organisms
4. Evidence from Geographic Distribution
5. Evidence from Vestigial Organs
6. fossil Records

TODAY’S ACTIVITY

1. Differentiate internal and external fertilization
2. Compare the egg of fish, toad, lizard and domestic fowl.
3. State four functions of placenta
4. Mention four functions of the umbilical cord
5. State two functions of the amniotic fluid
6. Mention two hormones that are involved in the menstrual cycle
7. How is the foetus protected against pressure from sudden movement of its mother’s body

8. Mention two structures that link foetus to its mother.

WEDNESDAY. 10/06/2020

Good day students, hope you are attempting all your class activities? It’s for your own success.

Today’s work: Questions

1. Mention five evidences of Evolution
2. State the theories of use and disuse and give examples.
3. Differentiate between Lamarckism and Darwinism theory of natural selection
4. Describe structural adatations, hence, describe how the structural adaptation of a named animals helps it to escape from predators

09/06/2020 Tuesday.

Hello my always ready to learn students , hope you are spending ur time wisely?

Correction on last week topic

1. Reason why an individual of blood group A can not donor blood to a blood group B is that :Blood group B – the recipient contains antigen B and produces anti-A antibody. The presence of blood group A introduces antigen B into blood group B ,this elicit an immune response in which the production of anti-A antibodies in the recipient’s blood is increased, leading to the agglutination of the red blood cells. That is anti-A antibody sees antigen B as a foreign body or pathogen, so it flights against the antigen A leading to agglutination.
2. Yes. A blood group O person can donate to all kind of blood group because he does not contain any antigen that can elicit an immune response in the recipient’s blood, so agglutination can not occur.
3. Questions on rhesus factor
4. 1. A Rhesus +ve person can not donate to a rhesus-ve because a rhesus +ve contains antigen on its red blood cells but has no antibody. So the presence of rh+ve in rh-ve recipient will produce antibodies that will result in the agglutination of the red blood cells. Rhesus positive van only donate to rhesus positive .
5. 2. A rhesus negative can easily donate to both rhesus positive and negative because rhesus negative has no antigen that can elicit an immune response in the recipient.

Today’s work : Alternative to practical

Specimens given are:

A-Grasshopper

B-Grass

C-Solder termite

D- Lizard

E. Snake.

1a.Identity the above specimens without reason

b. State the habitats of specimens A-E

c . Construct a food chain using three of the specimens above

d. State two adaptive features of these specimens.

e. State the class of specimens A-E.

f. State four structural differences between specimen A and

g . Mention the respiratory organs of specimens A,C,and D.

h. Mention four economic importance of specimen C..

June 2, 2020

Hello students, good afternoon.

Correction to the last class activity.

Discontinuous variation	Continuous variation
It is not affected by environmental conditions	It is affected by environmental conditions.
It is brought about by the only one or few genes	It is brought about by the combined effects of many genes
There are no intermediate forms between these trains	There are intermediate forms between there traits
Individual falls into distinct categories. E.g ABO blood group in man, ability to roll tongue, taste PTC etc.	Individual shows a continuous variation from one extreme to the other. E.g height, colour,weight, intelligence etc

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Blood group	Antigen	Antibody
A	A	Anti-B
B	B	Anti-A
AB	A and B	Absent
O	No antigen	Anti-A and Anti-B

Re-call:GENETICS

Genetics is defined as the study of heredity and variations.

What is Heredity?

Heredity: It is defined as the transmission of characters or traits from parents to their offspring. It ensures the continuity of characters in living organisms and also accounts for the differences or variation among individuals in a population. Heredity refers to inheritance of characters or traits. There are two types of inheritance:

1. Monohybrid Inheritance: It involves the inheritance of one pair of contrasting characters e.g height (tall and dwarf), colour(white or black)e.t.c. It is just the inheritance of only one pair of contrasting characters.
2. Dihybrid Inheritance: It involves the inheritance of two pairs of contrasting characters simultaneously. For instance, the height (tallness and dwarf) and the colour (white and black) I.e the two pairs of contrasting characters are inherited simultaneously.

VARIATIONS

Variations are differences in traits or characters between individuals of the same species.

Types of Variations.

1. Morphological Variations
2. Physiological Variations

Morphological Variation : It involves physical charateritics such as height, weight, eye,skin and hair colours and intelligence fingerprints. Morphological variation gives rise to continuous variation lt is also affected by environmental conditions.

Physiological Variation:It involves bodily functions or characteristics like behaviour, tongue rolling, ability to taste PTC and blood groups.It gives rise to discontinuous variation. It is not affected by environmental conditions.

Class activity:

1. Distinguish between continuous and discontinuous Variation giving two examples in each case.
2. Briefly explain how the knowledge of variation can be applied in :

a. Crime detection

b. Blood transfusion.

3. If the blood type of a woman is B, and her child’s blood type is AB, what are the possible blood types of the father of the child?

4. Can a man with blood group AB possibly be the father of a boy whose blood group is B, given that the mother’s blood is group O? Give reasons for your answer.

5. A cross is made between a plant that has red flowers and yellow seeds and a plant with white flowers and green seeds. All the offsprings of F1 generation have red flowers and yellow seeds. The F1 plants are allowed to self – fertilise. What are the phenotypic and genotypic ratios of the F2 generation offsprings?

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GROWTH 12/05/2020

Hello my dear students, how has been your holiday? Its a great moment with you and you will enjoy it. I will like to revise a topic; Growth.

Growth is one of the characteristics of living organisms. For growth to occur the rate of anabolic must exceed the rate of catabolic reaction ie rate of building up must exceed the rate catabbolic processes.

Can you still recall examples of catabolism and anabolism?

Growth is defined as the increase in size or complexity of an organism due to the addition of new protoplasm.

Basis of Growth.

They are ;Cell Division, Cell Enlargement and Cell Differentiation.

Cell division ;it is the division of a single cell into two identical daughter cells. it results in an increase in cell number.

Types of Cell division;

Mitosis and Meoisis.

Mitosis ;is a cell division which involves duplication of chromosome whereby each daughter cell has exactly the same chromosome content as the parent. For multicellular organisms to grow , new cells must be synthesized by mitosis. For example ,skin cells must divide regularly to replace dead cells that have been shed from the surface of the skin. Dead cells are constantly being replaced by the process of mitosis. Mitosis results into two daughter cells.

Hello my students, its a great moment with you and i hope you will enjoy it.

I will like to revise on the topic- Microoraganisms. You all know the pandermic disease around the world now is covid- 19. A question for you- What is the difference between COVID-19 and CORONAVIRUS? .

Coronavirus is a microrganism (a virus) which causes a disease called Covid-19. While Covid-19 is a disease caused by a microorganism called Coronavirus , i.e Coronavirus is a pathogen\microoganism while Covid-19 is a disease .

Microorganisms are organisms that can not be seen with our naked eyes but with the aids of Microscope. They are everywhere -in the air, water ,in and on our body . in our food etc. Most diseases are caused by microorganisms. Although not all microorgainsms are harmful, some are beneficial to us, e.g Rhizobium spp. found in the root nodule of leguminous plants which helps in nitrogen fixation.

There are five groups of microorganisms-Virus , Bacteria, Fungi, Protzoa and Algae.

STRUCTURE OF VIRUSES

Assignment -Study the structure of Viruses and Bacteria, then deduce the reasons why it has been very difficult to get COVID-19 Vaccine.

Likewise study modes of life of these two microorganisms and state three differences between them.

GROWTH 12/05/2020

Hello my dear students, how has been your holiday? Its a great moment with you and you will enjoy it. I will like to revise a topic; Growth.

Growth is one of the characteristics of living organisms. For growth to occur the rate of anabolic must exceed the rate of catabolic reaction ie rate of building up must exceed the rate catabbolic processes.

Can you still recall examples of catabolism and anabolism?

Growth is defined as the increase in size or complexity of an organism due to the addition of new protoplasm.

Basis of Growth.

They are ;Cell Division, Cell Enlargement and Cell Differentiation.

Cell division ;it is the division of a single cell into two identical daughter cells. it results in an increase in cell number.

Types of Cell division;

Mitosis and Meoisis.

Mitosis ;is a cell division which involves duplication of chromosome whereby each daughter cell has exactly the same chromosome content as the parent. For multicellular organisms to grow , new cells must be synthesized by mitosis. For example ,skin cells must divide regularly to replace dead cells that have been shed from the surface of the skin. Dead cells are constantly being replaced by the process of mitosis. Mitosis results into two daughter cells.

19/05/2020

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Hello students, its a previledge having you today, stay connected and enjoy your reading.

Revision on; How a Chromosome is formed.

Firstly, can you differentiate between Chromosome, Chromatin and Chromatid?

What is a chromosome?

A chromosome is a thin rod or thread like material found in the nucleus of a cell. It contains or houses the genes. Chromosome appears as a thin thread called Chromatin(DNA molecule).

At interphase, chromatin can not be distinguished individually. At this phase, chromatin thread(DNA molecule) replicate itself producing two identical Chromatin threads . These two identical chromatin threads are called Sister Chromatids, they coil and shorten to become Chromosomes. The sister chromatids are joined at the Centromere. A chromosome is a two sister chromatids.

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