Selina ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions The Nervous System

by

Selina Concise Biology Class 10 ICSE Solutions The Nervous System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System

Exercise 1

Solution A.1.
(b) neurolemma

Solution A.2.
(d) Pons – consciousness

Solution A.3.
(b) Contains both sensory and motor fibres

Solution B.1.
(a) Cerebrospinal fluid
(b) Synapse
(c) Cerebrum
(d) Hypothalamus

Solution B.2.
(a) Stimulus: Receptor:: Impulse: Effectors
(b) Cerebrum: Diencephalon:: Cerebellum: Medulla oblongata
(c) Receptor: Sensory nerve:: Motor nerve: Effector

Solution B.3.
(a) Sensory
(b) Maintaining posture and equilibrium
(c) Spinal cord

Solution C.1.
(a) Corpus Callosum – It is located located in the forebrain. It connects two cerebral hemispheres and transfers information from one hemisphere to other.
(b) Central canal – It is located in centre of the spinal cord. It is in continuation with the cavities of the brain. It is filled with cerebrospinal fluid and acts as shock proof cushion. In addition, it also helps in exchange of materials with neurons.

Solution C.2.
(a) False
(b) False
(c) True
(d) True

Solution C.3.
(a)

Cerebrum Cerebellum
The cerebrum controls all voluntary actions. It enables us to think, reason, plan and memorize. The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(b)

Sympathetic Nervous System Parasympathetic Nervous System
Sympathetic nervous system prepares the body for violent action against the abnormal condition. Parasympathetic nervous system is concerned with re-establishing normal conditions after the violent act is over.

(c)

Sensory Nerve Motor Nerve
Sensory nerve brings impulses from the receptors i.e. sense organs to the brain or spinal cord.

Motor nerve carries impulse from the brain or spinal cord to effector organs such as muscles or glands.

(d)

Medulla Oblongata Cerebellum
Medulla oblongata controls the  activities of internal organs and many other involuntary actions The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(e)

Cerebrum Spinal Cord
The grey matter containing cytons lies in the cortex (outer region) while the white matter containing axons lies in the medullary region (inner region). The grey matter containing cytons lies in the medullary region i.e. inner side while the white matter containing axons lies in the cortex i.e. the outer region.

Solution C.4.
(a) Cerebellum maintains balance of the body and coordinates muscular activity.
(b) Myelin sheath acts like an insulation and prevents mixing of impulses in the adjacent axons.

Solution C.5.
(a) Synapse: It is a gap between the axon terminal of one neuron and the dendrites of the adjacent neuron. It transmits nerve impulse from one neuron to another neuron.
(b) Association Neuron: It interconnects sensory and motor neurons.
(c) Medullary sheath: It provides insulation and prevents mixing of impulses in the adjacent axons.
(d) Medulla Oblongata: It controls activities of internal organs such as peristalsis, breathing and many other involuntary actions.
(e) Cerebellum: It maintains balance of the body and coordinates muscular activity.
(f) Cerebrospinal Fluid: It acts like a cushion and protects the brain from shocks.

Solution C.6.
(a) Sensory, motor and mixed nerves
(b) Somatic and autonomic nervous system
(c) Natural and conditioned reflexes
(d) Sensory, motor and association neurons
(e) Gray and white matter

Solution C.7.
(a) Stimulus — receptor — sensory neuron — central nervous system — motor neuron — effector — response
(b) Resting — depolarization — repolarization
(c) Dendrites — Dendron — perikaryon — nucleus — axon — axon endings
(d) Cerebrum — diencephalon — mid-brain — cerebellum — pons — medulla oblongata

Solution D.1.
(a) Reflex action  is an autonomic, quick and involuntary action in the body brought about by a stimulus.
(b)

Example Type of Reflex
(i) Sneezing Simple
(ii) Blushing Simple
(iii) Contraction of eye pupil Simple
(iv) Lifting up a book Conditioned
(v) Knitting without looking Conditioned
(vi)  Sudden application of brakes of the cycle on sighting an obstacle in front Conditioned

Solution D.2.
The advantages of having a nervous system are as follows:

  1. Keeps us informed about the outside world through sense organs.
  2. Enables us to remember, think and reason out.
  3. Controls and harmonizes all voluntary muscular activities such as running, holding, writing
  4. Regulates involuntary activities such as breathing, beating of the heart without our thinking about them.

Solution D.3.
The brain and the spinal cord lie in the skull and the vertebral column respectively. They have an important role to play because all bodily activities are controlled by them. A stimulus from any part of the body is always carried to the brain or spinal cord for the correct response. A response to a stimulus is also generated in the central nervous system. Therefore, the brain and the spinal cord are called the central nervous system.

Solution D.4.
Reflex actions are involuntary actions which occur unknowingly. Voluntary actions on the other hand are performed consciously.
Picking up an apple and eating it is an example of voluntary action whereas withdrawal of hand on touching a hot object is an example of reflex action.

Reflex Action Voluntary Action
Reflex actions are involuntary actions which occur unknowingly. Voluntary actions on the other hand are performed consciously.
Commands originate in the spinal cord, autonomic nervous system and a few in the brain as well. Commands originate in the brain.

Solution D.5.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 1

Solution D.6.

Organ

 Sympathetic System Parasympathetic System
e.g. Lungs Dilates bronchi and bronchioles

Constricts bronchi and bronchioles

1. Heart

 Accelerates heartbeat Retards heartbeat
2. Pupil Dilates

Constricts

3. Salivary gland

 Inhibits the secretion of saliva causing the drying of the mouth

Stimulates the release of saliva

Solution E.1.
Salivation is an example of conditioned reflex that develops due to experience or learning. Saliva starts pouring when you chew or eat food. Therefore, this reflex will occur not just on the sight or smell of food. The brain actually needs to remember the taste of food. Boy B started salivating because he must have tasted that food prior unlike boy A.

Solution E.2.

Situation

 Organ/body part Change/action Part of autonomic nervous system involved
1. You have entered a dark room Eye Pupil dilates

Sympathetic
2. Your body is consuming lot of glucose while running a race Liver Glycogen is converted into glucose in liver Sympathetic
3. You are chewing a tasty food Salivary gland Salivation increases

Parasympathetic
4. You are running a race Adrenal gland Release of adrenaline and noradrenaline increases Sympathetic
5. You are retiring to bed for sleep Heart Heart rate slows down

Parasympathetic
6. You are shivering in intense cold Body hairs Hair raised

Sympathetic

Solution E.3.
Fill in the following information in the diagram.

  1. Central Nervous System
  2. Autonomic
  3. 12
  4. spinal
  5. 31
  6. dilates
  7. constricts
  8. liver

Exercise 2

Solution A.1.
(b) Cornea

Solution A.2.
(b) Cochlea

Solution A.3.
(c) Eustachian tube, tympanum and utriculus

Solution A.4.
(a) Retina

Solution B.1.
(a) Rhodopsin
(b) Eustachian tube
(c) Hammer
(d) Dura mater
(e) Eustachian tube
(f) Cornea
(g) Auditory nerves
(h) Rods and cones
(i) Hypermetropia

Solution B.2.
(a) Cones: Iodopsin:: rods: rhodopsin
(b) Sound: ear drum:: dynamic balance: semi-circular canals

Solution B.3.

Column I Column II
i. The blind spot (h) no sensory cells
ii. The yellow spot (g) centre of the retina
iii. Ciliary muscle (b) Shape of the lens
iv. Iris (e) free of rod cells, (a) colour of the eye
v. Meninges (c) Protective covering of the brain

Solution C.1.
(a) Myopia results when the eye ball is lengthened from front to back or the lens is too curved.
Hyperopia results from either too shortening of the eyeball from front to back or when the lens is too flat.
(b) Rods are sensitive to dim light but do not respond to colour.
cones are sensitive to bright light and are responsible for colour vision.
(c) cochlea is responsible for hearing; it can perceive the senses of hearing.
Semicircular canals are responsible for perceiving the senses to maintain the body balance.
(d) Rod cells contain rhodopsin whereas the cone cells contain iodopsin.
(e) Dynamic balance is when the body is in motion whereas static balance is positional balance with respect to gravity.

Solution C.2.
(a) False
Correct statement: Deafness is caused due to rupturing of the eardrum.
(b) False
Correct statement: Semicircular canals are concerned with dynamic balance.

Solution C.3.
(a) Fovea centralis is located at the back of the eye almost at the centre of the eyeball. It is the region of the brightest vision and also of the colour vision.
(b) Organ of corti is located in the inner ear. It contains sensory cells which process hearing.

Solution C.4.
(a) True
(b) False/ Ciliary muscles regulate the size of the lens.
(c) True
(d) False/The auditory nerve responsible for sound as well as for the body balance.
(e) True
(f) False/ flavour is a combination of taste and smell.
(g) False/ short-sightedness is myopia and hyperopia is long-sightedness.
(h) True

Solution C.5.
(a) Auditory canal, tympanum, ear ossicles, oval window, cochlea
(b) Conjunctiva, cornea, lens, retina, optic nerve

Solution C.6.
(a) Organ of Corti and hearing
(b) Olfactory nerve and smell
(c) Retina and vision
(d) Taste bud and taste

Solution C.7.
(a) Lacrimal gland is a tear gland located at the upper sideward portion of the eye orbit. Its secreation lubricates the surface of the eye, washes aways the dust particles and kills germs
(b) Yellow spot is the region of brightest vision and contains maximum sensory cells whereas a blind spot contains no sensory cells and this is the point of no vision.
(c) Presbyopia is an age-old eye defect. In this condition, the lens loses flexibility resulting in far-sightedness.
Cataract is also very common in old people, the cornea becomes opaque and the vision is cut down even to blindness.
(d) The process of focusing the eye at different distances is called the power of accommodation.
(e) The image formed on the retina is inverted and real.

Solution C.8.
An optical illusion is the life-like continuous movement on the screen. Television is an example of optical illusion, where the scanning beam of a picture frame of the TV camera moves so rapidly on the viewing screen of the TV set that our eyes cannot keep pace with it.

Solution C.9.
(a) Oval window is located in the middle ear. It helps in setting the fluid in the cochlear canals into vibration.
(b) Cochlea is located in the inner ear. It helps in transmitting impulses to the brain via the auditory nerve.
(c) Semicircular canals are located in the inner ear. These help in maintaining the dynamic equilibrium of the body.
(d) Utriculus is located in the inner ear. It joins the semi-circular canals to cochlea. It also helps in maintaining static balance of the body.

Solution C.10.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 2

Solution C.11.

Structure

 Function
Yellow Spot

Region of the brightest vision

Auditory nerve

 Transfers impulse from inner ear to brain
Ciliary muscle

Helps to change the focal length of the eye lens

Spinal cord

 Conducts impulses
Oval window

Sets fluid in cochlear canal into vibration

Semicircular canals

Dynamic equilibrium

Solution D.1.
While reading a book, the lens is more convex or rounded due to contraction of ciliary muscles because the book is usually read from a short distance. When we raise our head and look at a distant object, the ciliary muscles relax to build the tension on the suspensory ligament so that they can stretch the lens. This change in the curvature of the lens makes us focus on distant object.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 3

Solution D.2.
The brain sees the vivid picture of the dream through the eyes. Our eyes have actually never seen the vivid picture. This is an example of optical illusion. The area of dream is controlled by the cerebrum of the central nervous system. So sometime we can remember the vivid picture seen in the dream.

Solution D.3.
If we look at a bright object and then close our eyes, the sensation of light persists for a short period. This is known as persistence image or the after image. It lasts for one-tenth of a second. Therefore by closing the eyes and gently pressing them with your palms, you see some specs of brilliant light.

Solution D.4.
Adaptation is the ability to adjust vision in bright and dark areas. When we enter a dark room from bright light, the rhodopsin pigment broken down in bright light is regenerated. It dilates the pupil and allows more light to enter the eyes. This is called dark adaptation. On the other hand, if we enter bright area from a dark room, the rhodopsin pigment is bleached. This constricts the pupil and reduces the light entering the eyes. This is called ‘light adaptation‘

Accommodation is the process of focusing the eye at different distances. This is mainly brought about by a change in the curvature of the lens. When the ciliary muscles contract, the lens becomes thicker and we are able to focus a nearby object. On the other hand when the ciliary muscles relax, the lens remains stretched i.e. the normal condition and we are able to focus on distant object.

Solution D.5.
Our eyes are designed to focus at a great variety of distances. To focus constantly at a short distance can make the lens focusing muscles fatigued. Therefore, we do not enjoy watching a movie from a very short distance from the screen in cinema hall.

Solution D.6.

Defect of vision

 Cause Corrective measure
Myopia Lengthening of eye ball from front to back or the lens is too curved.

Using suitable concave lens

Hyperopia

 Shortening of eye ball from front to back or the lens is too flat. Using suitable convex lens
Astigmatism Uneven curvature of the cornea

Using suitable cylindrical lenses

Presbyopia

 Loss of flexibility of lens Using suitable convex lens
Cataract Lens turning opaque

Surgery or use of convex lens or implantation of plastic lens.

Colour blindness

 Genetic defect No control measure
Squint Formation of cross-eye

Surgery and suitable exercise

Solution D.7.
The three ear ossicles are: Malleus (hammer), Incus (anvil) and Stapes (stirr up).
The last ear ossicle, stapes, vibrates and transmits the vibration to the oval window.
The role of other two ear ossicles is to magnify the vibration of stapes as a result of their lever like action.

Solution D.8.
The process of focusing the eye at different distances is called the power of accommodation. The ciliary muscles are responsible for the power of accommodation.

Solution E.1.
a. The ability of the eye to focus sharply on things which are near to the eye as well as far off is known as the power of accommodation.
b. Shape of the eye:
Near vision – flattened
Distant – rounded or more convex
c. Ciliary muscles and suspensory ligament
d. In the dark: Cells – rod cells, Pigment – rhodopsin
In the light: Cells – cone cells, Pigment – iodopsin

Solution E.2.
a. The middle ear or membranous labyrinth has two structures inside it, the cochlea and the semi-circular canals.
b. Malleus, incus and stapes
c. Static balance – Utriculus and sacculus (inner ear)
Hearing – Internal ear
Dynamic balance – Semi-circular canals (inner ear)
d. Collectively they are termed as ossicles.

Solution E.3.
(a) Cornea is comparable to the lens cover of the camera.
The iris and pupil act like the aperture of a camera.
(b) The cornea is the eye’s main focusing element. It takes widely diverging rays of light and bends them through the pupil; the rays are further converged by the lens.

Solution E.4.
(a) Myopia
(b) The two possible reasons for myopia are either the eye ball is lengthened from front to back or the lens is too curved.
(c) 1 – vitreous humour, 2 – blind spot, 3-lens, 4-pupil
(d) Concave lens
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 4

Solution E.5.
(i) Ear
(ii) m – malleus, i – incus and s – stapes respectively. These are collectively called as ear ossicles.
(iii) Cochlea. The vibrating movements in the hair of the sense cells of cochlea transmit the impulse for hearing to the brain via auditory nerve.
(iv) Tympanic membrane. It vibrates and then sets the ear ossicles into vibration in the process of hearing.

Solution E.6.
(i) Ear ossicles
(ii) A – Cochlea, B – Semicircular canals, C – Ear ossicles
(iii) Cochlea helps in transmitting impulses to the brain via the auditory nerve. Semicircular canals help in maintaining dynamic equilibrium of the body.
(iv) Organ of Corti

Solution E.7.
Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes) image - 5

Solution E.8.
(a) Myopia
(b) A-Normal eye, B-Myopia
(c) Looking glasses with the concave lens are required here.

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes)

by

Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes). You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes)

Exercise 1

Solution A.1.
(c) Removal of nitrogenous wastes

Solution A.2.
(a) Proximal convoluted tubule

Solution A.3.
(c) Sweating

Solution A.4.
Liver

Solution B.1.
(a) Liver
(b) Cortex
(c) Glomerulus
(d) Collecting duct
(e) Renal artery (Renal vein has urea but renal artery has higher concentration of urea as compared to renal vein).

Solution B.2.
(a) Afferent arteriole, glomerulus, efferent arteriole, capillary network, renal vein
(b) Renal artery, kidney, ureter, urinary bladder, urethra

Solution B.3.
(a) Ultrafiltration
(b) Excretion
(c) Osmoregulation
(d) Excretion

Solution C.1.
(a) Glomerulus is involved in the process of ultrafiltration.The liquid part of the blood which is plasma including urea, salts, glucose filters out from the glomerulus into the renal tubule.
(b) Henle’s loop is involved in reabsorption of water and sodium ions.
(c) Ureter carries urine to the urinary bladder by ureteral peristalsis.
(d) Renal artery supplied blood to the kidney.
(e) Urethra is involved in the process of micturition i.e. expelling urine out of the body.

Solution C.2.
Excretion helps in removing toxic wastes from our body and it also plays an important role osmoregulation i.e. the maintenance of the homeostasis of the body.
Carbon dioxide, water, nitrogenous compounds such as urea, uric acid and excess salts are some common excretory products.

Solution C.3.
A uriniferous tubule also known as the kidney tubule is the structural and functional unit of the kidney.
It takes in impure blood from the renal artery and removes wastes in the form of urine. It also provides a larger surface area for reabsorption of salts and water.

Solution C.4.
Maintaining a normal osmotic concentration in the body means regulating the percentage of water and salts. If this regulation mechanism fails we either end up losing vital salts and water or may accumulate unwanted salts and excess water in our body.

Solution C.5.
If one kidney is donated to a needy patient, the other kidney alone is sufficient for removing wastes or excretion. Thus, the donor can live a normal life.

Solution C.6.
During summer, a considerable part of water is lost through perspiration so the kidneys have to reabsorb more water from the urine. This makes the urine thicker in summer than in winters.

Solution C.7.

(a) Bowman’s capsule is a thin walled cup containing the glomerulus. This Bowman’s capsule along with the glomerulus is known as malpighian capsule.

(b) The renal cortex is the outer darker region of the kidney whereas the renal medulla is the inner lighter region of the kidney.

(c) Renal pelvis is the expanded front end of the ureter in the kidney whereas the renal papilla is the apex of the renal pyramid which projects into the pelvis.

(d) Urea is the chief excretory product which is excreted in the form of urine whereas urine is the filtrate left after reabsorption and tubular secretion which contains 95% water and 5% solid wastes.

(e) Excretion is the process of removal of chemical wastes especially nitrogenous wastes from the body.
Catabolism on the other hand is the set of metabolic pathways which break down molecules into smaller units and release energy.

Solution C.8.
Urea, creatinine, uric acid

Solution C.9.

Column I Column II 
(a)   Bowman’s Capsule Glomerulus
(b)   Contains more CO2 and less urea Renal Vein
(c)    Anti-diuretic hormone Regulates amount of water excreted
(d)   Contains more urea Renal artery

Solution C.10.
In a nephron, the blood flows through the glomerulus under great pressure. The reason for this great pressure is that the efferent (outgoing) arteriole is narrower than the afferent arteriole (incoming). This high pressure causes the liquid part of the blood to filter out from the glomerulusinto the renal capsule.

Solution D.1.

(a) Ultrafiltration – The process of the filtration of blood in the glomerulus under great pressure during which the liquid part of the blood i.e. plasma along with urea, glucose, amino acids and other substances enter the renal tubule.

(b) Micturition – The process of expelling urine out of the body through urethra by opening the sphincter muscles passing of urine involving relaxation of sphincter muscles between the urinary bladder and urethra.

(c) Renal pelvis – Renal pelvis is the expanded front end of the ureters into the kidney.

(d) Urea – A nitrogenous waste produced primarily in the liver due to the break down dead protein remains and extra amino acids.

(e) Osmoregulation – It is a process of maintaining the blood composition of the body i.e. the normal osmotic concentration of water and salts in the body.

Solution D.2.

Ultrafiltration – Ultrafiltration involves filtration of the blood which takes place in the glomerulus. The blood containing urea from the afferent arteriole enters the glomerulus under high pressure. The high pressure is created because the efferent arteriole is narrower than the afferent arteriole. The high pressure causes the liquid part of the blood to filter out from the glomerulus into the renal tubule. This filtrate is known as ‘glomerular filtrate’.
Glomerular filtrate consists of water, urea, salts, glucose and other plasma solutes. Blood corpuscles, proteins and other large molecules remain behind in the glomerulus. Therefore the blood which is carried away by the efferent arteriole is relatively thick.

Selective absorption – The Glomerular filtrate entering the renal tubule contains a lot of usable materials such as glucose and sodium. As this filtrate passes down the renal tubule, a lot of water along with these usable materials is reabsorbed. Such reabsorption is called ‘selective absorption’. The reabsorption occurs only to the extent that the normal concentration of the blood is undisturbed.

Solution D.3.
Dialysis involves the use of artificial kidney or a dialysis machine. The patient’s blood is from the radial artery is led through the machine where excess salts and urea is removed. The purified blood is then returned to a vein in the same arm.
Dialysis is carried out in case of failure of both the kidneys. In case there is a permanent damage, then the dialysis is to be repeated for about 12 hours twice a week.

Solution E.1.
(a) The image shown can be left or right kidney. As the right kidney is slightly lower than the left one, so we need to have the images of both the kidneys for comparison.
(b) It is a longitudinal section of the kidney.
(c) 1-renal artery, 2-renal vein, 3-ureter, 4-cortex, 5-pelvis
(d) (i) 4/cortex
(ii) medulla
(iii) 5/pelvis

Solution E.2.
(a) Excretory system and Circulatory system.
(b) 1-kidney, 2-renal artery, 3-ureter, 4-urinary bladder, 5-urethera
(c) Nephron
(d) Urea and ammonia
(e) Ultrafiltration and selective reabsorption

Solution E.3.
(a) 4/Glomerulus
(b) 2/Efferent arteriole
(c) 1/ Afferent arteriole from renal artery
(d) 7/Collecting tubule
(e) 5/ Proximal convoluted tubule with blood capillaries

Solution E.4.
(a) The process of removal of chemical wastes especially nitrogenous waste from the body is known as excretion.
(b) Nephrons
(c) As the cortex region contains numerous nephrons or kidney tubules, therefore, it shows a dotted appearance.
(d) Kidneys help in removing wastes or excretion and osmoregulation.
(e) The blood vessel ‘B’ is renal artery and the blood vessel ‘A’ is renal vein.

So the blood vessel ‘B’ contains oxygenated blood with high concentration of urea and glucose whereas the blood vessel ‘A’ contains deoxygenated blood with low concentration of urea and glucose as compared to renal artery.

Solution E.5.
a. The structure is a Bowman’s capsule, which is part of the nephron. The Bowman’s capsule is found in the cortex of the kidney.

b.

  1. Afferent arteriole
  2. Glomerulus
  3. Bowman’s capsule
  4. Efferent arteriole

c. Urine formation occurs in two steps – ultrafiltration and reabsorption.

d. The process occurring in 2 and 3 is known as ultrafiltration.
In the glomerulus, the blood flows under high pressure because of the narrow lumen of the capillary network of the glomerulus. This forces most of the components (both waste and useable materials) of the blood out of the capillaries. This process of the filtration of blood under high pressure in the Bowman’s capsule is known as ultrafiltration.

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions The Circulatory System

by

Selina Concise Biology Class 10 ICSE Solutions The Circulatory System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 7 The Circulatory System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 7 The Circulatory System

Exercise 1

Solution A.1.
(a) lymphocytes and monocytes

Solution A.2.
(b) phagocytosis

Solution B.1.
(a) Blood platelets and blood coagulation
(b) Neutrophils and phagocytosis
(c) Erythrocytes and transportation of gases
(d) Lymphocytes and Produce antibodies
(e) Bone marrow and destruction of old and weak RBC’s/production of RBCs and WBCs.

Solution B.2.
(a) Red Blood Cells
(b) Blood Platelets

Solution C.1.
Structural Differences between White Blood Cells and Red Blood Cells:

White Blood Cells Red Blood Cells
1. White blood cells are amoeboid. Red blood cells are minute biconcave disc-like structures.
2. They are nucleated cells. They anucleated cells.
3. Haemoglobin is absent in red blood cells. Haemoglobin is present in red blood cells.

Solution C.2.
During blood transfusion it is necessary to know the blood groups before transfusion because it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death.

Solution C.3.
(a) Antibodies are produced by lymphocytes in response to the entry of pathogens in the blood stream.
Antibiotics are the medicines extracted from some bacteria and fungi. Antibiotics destroy or inhibit the growth of pathogens.

(b) RBC: RBC is enucleated, biconcave, disc-like structure, flat in the centre while thick and rounded at the periphery.
WBC: WBC is nucleated and amoeboid in shape.

(c) Serum: The plasma from which the protein fibrinogen has been removed is called serum.
Vaccine: Vaccine is killed or living weakened germs which are introduced in the body to stimulate the production of antibodies against pathogens for a particular disease.

Solution C.4.
Heparin

Solution D.1.
The functions of blood plasma are:

  • Transports of digested food from the alimentary canal to tissues.
  • Transports excretory materials from tissues to excretory organs.
  • Distributes hormones from the glands to their target site.
  • Distributes heat in the body to maintain the body temperature.

Solution D.2.
Blood clotting or coagulation occurs in a series of the following steps:

  • The injured tissue cells and the platelets disintegrate at the site of wound to release thromboplastin.
  • The thromboplastin with the help of calcium ions converts inactive prothrombin into active thrombin.
  • Thrombin in the presence of calcium ions converts soluble fibrinogen into insoluble fibrin which forms a mesh or network at the site of wound.
  • The blood cells trapped in this network shrink and squeeze out the plasma to leave behind a solid mass known as the clot.
    Selina Concise Biology Class 10 ICSE Solutions The Circulatory System image -1

Solution D.3.

(a) Rh factor – It is an inherited antigen often found on the blood cells. Some individuals have these antigens and are thus Rh positive (Rh+) while others who do not have this antigen are Rh negative (Rh-).

(b) Universal donor – The person with blood group O is a universal donor as this type of blood can be given to persons with any blood group i.e. O, A, B, AB.

(c) Diapedesis – It is the squeezing of leucocytes through the wall of capillaries into the tissues.

Solution D.4.
Blood clotting is not dependent on the exposure of blood to air. In fact, clotting can be caused by the movement of blood over a rough surface such as on cholesterol deposit inside of a blood vessel of the skin.

Solution D.5.
The functions of the blood are:

  1. Transport of digested food from the alimentary canal to tissues. These substances are simple sugars like glucose, amino acids, vitamins, mineral salts, etc.
  2. Transport of oxygen in the form of an unstable compound ‘oxyhaemoglobin’ from the lungs to the tissues.
  3. Transport of carbon dioxide from the tissues to the lungs.
  4. Transport of excretory materials from the tissues to the liver, kidney or the skin for elimination.
  5. Distribution of hormones from glands to the target sites.
  6. Distribution of heat to keep the body temperature uniform.
    (Any five)

Solution E.1.
(a)

  1. Red Blood Cell (RBC),
  2. White Blood Cell (WBC),
  3. Blood Platelet
  4. Blood Plasma.

(b) The red blood cells are minute biconcave disc-like structures whereas the white blood cells are amoeboid.

(c) Function of part 1 (RBC): Transport of respiratory gases to the tissues and from the tissues, transport of nutrients from the alimentary canal to the tissues.
Function of part 2 (WBC): WBCs play major role in defense mechanism and immunity of the body.
Function of part 3 (Blood Platelet): Blood platelets are the initiator of blood clotting.

(d) The average life span of a red blood cell (RBC) is about 120 days.

(e) Thromboplastin

Exercise 2

Solution A.1.
(d) Heart itself

Solution A.2.
(c) artery

Solution A.3.
(a) tricuspid valve

Solution A.4.
(b) renal artery

Solution A.5.
(b) inadequate supply of oxygen to the heart muscle

Solution A.6.
(d) destroy pathogens

Solution A.7.
(a) closure of tricuspid and bicuspid valves
(b) closure of aortic and pulmonary valves
(c) rushing of blood through valves producing turbulence

Solution B.1.
The average values of blood pressure in a normal adult human are 100-140 mm for systolic pressure and 60-80 mm for diastolic pressure.

Solution B.2.
Yes, the heart beats approximately 1, 03,680 times in a day.

Solution B.3.
(a) Hepatic portal vein
(b) Blood Capillaries
(c) Pulmonary artery
(d) White blood cells
(e) Venules
(f) Portal vein
(g) Atrial systole
(h) Tricuspid valve
(i) Atrial systole
(j) Pericardial fluid

Solution B.4.

(a) Blood platelets are involved in blood clotting or coagulation. Blood platelets integrate at the site of injury to release thromboplastin which initiates the process of blood clotting.

(b) Neutrophils perform phagocytosis i.e. they engulf pathogens that enter the blood stream and destroy them.

(c) Erythrocytes transport oxygen from the lungs to the tissues in the form of an unstable compound oxyhaemoglobin and transport carbon dioxide from the tissues to the lungs.

(d) Lymphocytes produce antibodies against pathogens which enter the blood stream. In some cases they also perform phagocytosis.

(e) Bone marrow is involved in formation of RBCS and WBCs. It is also involved in the destruction of old and weak RBCs.

Solution B.5.
(a) The blood vessel that begins and ends in capillaries is the hepatic portal vein.
(b) A blood vessel which has small lumen and thick wall is artery.
(c) The valve which prevents the back flow of blood in the veins and lymph vessels is semilunar valve.

Solution B.6.
(a) Lubb: Atrio-ventricular valve:: Dup: Semilunar valves
(b) Coronary artery: Heart::Hepatic artery: Liver

Solution B.7.
A matured mammalian erythrocyte lacks a nucleus and mitochondria. The lack of a nucleus increases the surface area-volume ratio of RBCs, thus increasing the area for oxygen absorption. Also, the lack of a nucleus reduces the size of the cell, making it easy to flow through the blood vessels and more cells can be accommodated in a small area.
The lack of mitochondria implies that the cell does not use any oxygen absorbed for respiration, thus increasing the efficiency of the cell to transport oxygen as all the oxygen absorbed is transported without any loss.

Solution C.1.
Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. In pulmonary circulation, blood enters the lungs through pulmonary arteries. Pulmonary veins collect the blood from the lungs and carry it back to the left atrium.
In systemic circulation, blood from the left ventricle enters the aorta through which the blood is sent to the body parts. From the body parts blood is collected by veins and sent back to the heart. Therefore, the blood circulation in the human body is called double circulation.

Solution C.2.
The first sound LUBB is produced when the atrio-ventricular valves i.e. tricuspid and bicuspid valves close at the start of ventricular systole.
The second sound DUP is produced at the beginning of ventricular diastole, when the pulmonary and aortic semilunar valves close.

Solution C.3.
People have a common belief that the heart is located on the left side of the chest because the narrow end of the roughly triangular heart is pointed to the left side and during its working the contraction of the heart is more powerful on the left side which can be felt.

Solution C.4.
(a)

Erythrocytes Leucocytes
They function in the transport of oxygen throughout the body and in the removal of carbon dioxide from the body. They help in the defense of the body against disease-causing pathogens.

(b) An artery carries blood away from the heart whereas a vein brings blood towards the heart.
(c) An artery generally contains oxygenated blood whereas a vein generally carries deoxygenated blood.
(d) Tricuspid valve is located between the right atrium and right ventricle of the heart whereas a bicuspid valve is located between the left atrium and left ventricle of the heart.

Solution C.5.

Column A Column B
SA node Pacemaker
Defective hemoglobin in RBC Sickle cell anemia
Muscle fibres located in the heart Purkinje fibres
The liquid squeezed out of blood during clotting Serum
Never tires, keep on contracting and relaxing Cardiac muscles
Cardiac cycle 0.85 sec
Liquid part of the blood without corpuscles Plasma

Solution C.6.

Substance From To
Oxygen Lungs Whole body
Carbon dioxide Whole body Lungs
Urea Whole body Kidneys
Digested carbohydrates Intestines Whole body
Hormones Endocrine glands Target organs

Solution D.1.
(a) Endothelium– It is the innermost layer of the muscular wall of an artery or a vein which faces the lumen.
(b) Lymph nodes– The structures from which fresh lymph channels arise which pour the lymph into major anterior veins.
(c) Venule– The smallest common blood vessel formed by the union of capillaries.
(d) Diastole– The relaxation of muscles of ventricles or atria.

Solution D.2.

Artery Vein
An artery is a vessel which carries blood away from the heart towards any organ. A vein is a vessel which conveys the blood away from an organ towards the heart.
Artery has thick muscular walls. Vein has thin muscular walls.
It has a narrow lumen. It has a broad lumen.
There are no valves. Thin pocket-shaped valves are present in the veins.
Arteries progressively decrease in size and branch to form arterioles. Arterioles further breaks up to form capillaries. Capillaries unite to form branches called Venules. Venules further unite to form veins.

Solution D.3.
Tonsils: Tonsils are lymph glands located on the sides of the neck. They tend to localize the infection and prevent it from spreading it in the body as a whole.
Spleen: The spleen is a large lymphatic organ. The spleen acts as a blood reservoir in case of emergency such as haemorrhage, stress or poisoning. It produces lymphocytes and destroys worn out RBCs.

Solution D.4.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity while the right ventricle pumps the blood only up to the lungs. Therefore, the left ventricle has thicker walls than the right ventricle.
(b) The right ventricle pumps blood to the lungs for oxygenation whereas the right auricle receives the blood from vena cavae and passes it to the right ventricle. Therefore, walls of the right ventricle are thicker than those of the right auricle.

Solution D.5.
(a) The left ventricle pumps blood to the farthest points in the body such as the feet, the toes and the brain against the gravity. Thus, it requires greater force to push the blood. In order to with stand with the force applied the walls of the left ventricle are thicker than the walls of all the chambers.

(b) The blood from stomach and intestines enters the liver via hepatic portal vein because the liver monitors all the substances that have to be circulated in body. The excess nutrients such as glucose, fats are stores in the liver. Excess amino acids are broken down by the process deamination. Toxic substances are detoxified.

(c) During blood transfusion it is important that the blood groups of the donor and the recipient are compatible. In case of an incompatible blood transfusion, the recipient develops antibodies that attack the antigens present on the RBCs of the donor causing the blood cells to clump together which may result in death. The examination of Rh factor is also necessary for the blood transfusion. Therefore, the blood groups of both the donor and recipient must be known before transfusing blood.

(d) Veins carry the blood from the body part towards the heart while the arteries carry the blood from the heart. Veins carry the blood against the force of gravity. Therefore, only the veins and not the arteries are provided with valves.

(e) Atrial wall is less muscular than the ventricular wall because the major function of atria is to receive blood from the body and pump in into very next ventricles. While the ventricles pump the blood out of the heart. Right ventricle to the lungs and the left ventricle to all the body parts.

(f) Arteries are responsible to carry oxygenated blood from the heart to the tissues. The blood flows in the artery under high pressure and in spurts. If arteries are located superficially then there is a high possibility of their damage which could lead to a lot of blood loss. To prevent this damage and blood loss, the arteries are deep seated in the body.

Solution E.1.
(a) A is artery, B is vein.
(b)

  1. Endothelium of the artery,
  2. Middle layer of smooth muscles and elastic fibres of the artery,
  3. External layer of connective tissue of the artery,
  4. Endothelium of the vein,
  5. Middle layer of smooth muscles and elastic fibres of the vein,
  6. External layer of connective tissue of the vein.

(c) An artery has thick muscular walls and a narrow lumen. It carries blood away from the heart towards any organ.
A vein on the other hand has thin muscular walls and a wider lumen. It carries blood away from an organ towards the heart.

Solution E.2.
(a) The structure 3 represents the heart. It forms the centre of double circulation and is located between the liver and the head (as per the diagram). Also the blood circulation (indicated by 1) begins from heart to lungs.
(b)

Aorta 5
Hepatic portal vein 7
Pulmonary artery 1
Superior vena cava 9
Renal vein 8
Stomach 10
Dorsal aorta 11

Solution E.3.
(a)

1 Systemic Circulation
2 Vena Cava
3 Aorta
4 Right Ventricle
5 Left Atrium
6 Pulmonary Artery
7 Pulmonary Vein
8 Pulmonary Circulation

(b) Blood flows twice in the heart before it completes one full round. The full round thus includes pulmonary and systemic circulation. For this reason, the blood circulation in the human body is called double circulation.
(c) The relaxation of muscles of ventricles or auricles is known as diastole.

Solution E.4.
(a) Tissue Fluid
(b) Red blood cells
(c) Lymph
(d) The lymph supplies nutrition and oxygen to those parts where blood cannot reach. The lymph drains away excess tissue fluids and metabolites and returns proteins to the blood from tissue spaces.

Solution E.5.
(a) Hepatic portal vein (4)
(b) Hepatic portal vein (4)

Solution E.6.
(a) A- Vein, B-Artery, C-Capillary
(b)

  1. External layer made of connective tissue
  2. Lumen
  3. Middle layer of smooth muscles and elastic fibres
  4. Endothelium

(c) An artery has thick muscular walls and a narrow lumen. It does not have any valve.
A vein on the other hand has thin muscular walls and a wider lumen. It has valves to prevent back flow of blood.

(d) A (Vein)- deoxygenated blood, B (Artery)- oxygenated blood

(e) At the capillary level the actual exchange of gases takes place.

Solution E.7.
(a) Atrial Diastole and Ventricular Systole

(b) Ventricular muscles are contracting during this phase because the valves between the two ventricles and pulmonary artery and aorta are open while the atrio-ventricular valves are closed.

(c)

1 Pulmonary Artery
2 Aorta
3 Pulmonary Vein
4 Left Atrium
5 Bicuspid Valve
6 Right Ventricle

(d) Part 1 (Pulmonary artery) – Deoxygenated blood
Part 2 (Aorta) – Oxygenated Blood

(e) Two i.e., bicuspid and tricuspid valves are closed in this phase.

Solution E.8.
a.

  1. Arteriole
  2. Artery
  3. Venule
  4. Capillaries
  5. Vein

b. Such an arrangement can be observed in the lungs.

Solution E.9.
a. 1 – Red blood cell
b. Diapedesis
c.

RBC WBC
They lack a nucleus. They have a nucleus.
They are biconcave and disc-shaped. They are spherical and have different sizes.

d. The process which occurs in B and C is phagocytosis. In this process, the WBCs engulf the foreign particles and destroy them, thus preventing the occurrence of disease.

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

by

Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All

Exercise 1

Solution A.1.
b) glucose formed in photosynthesis soon gets converted into starch

Solution A.2.
b) twelve

Solution A.3.
b) humidity

Solution A.4.
c) trapping light energy

Solution A.5.
a) continue to live, but will not be able to store food

Solution A.6.
a) Carbon dioxide is reduced and water is oxidised

Solution A.7.
c) activate chlorophyll

Solution A.8.
d) ensure that the leaves are free from starch

Solution A.9.
a) CO2

Solution B.1.
(a) Producers / Autotrophs
(b) Chloroplasts
(c) ATP (Adenosine triphosphate)
(d) Glucose
(e) Green plants
(f) Carbon dioxide dissolved in water
(g) Stroma
(h) Phloem

Solution C.1.

(a)

Respiration Photosynthesis
The gas released during respiration is carbon dioxide. The gas released during photosynthesis is oxygen.

(b)

Light Reaction Dark Reaction
Hydrogen and oxygen are produced here, along with release of electrons, which converts ADP into ATP. Glucose is the main product formed during dark reaction.

(c)

Producers Consumers
Producers show autotrophic mode of nutrition i.e. they are able to produce their own food from basic raw materials.
For example: green plants

Consumers show heterotrophic mode of nutrition i.e. they depend directly or indirectly on the producers for their food.
For example: Animals

(d)

Grass Grasshopper
Green grass being a producer is capable of producing its own food by photosynthesis. Grasshopper is a primary consumer (herbivore) and directly feeds on producers like grass.

(e)

Chlorophyll Chloroplast
Chlorophyll is the green pigment present in cell organelles called chloroplasts. Chloroplasts are cell organelles, situated in the cytoplasm of plant cells. They are present mainly in the mesophyll cells and in the guard cells of stomata.

Solution C.2.
(a) False
Correct Statement: Dark reaction of photosynthesis is independent of light and occurs simultaneously with light reaction.

(b) True

(c) False
Correct Statement: Starch produced in a leaf is stored temporarily in the leaf until the process of photosynthesis. At night it is converted back into soluble sugar and translocated to different part of the body either for the utilization or for the storage.

(d) True

(e) False
Correct Statement: Green plants are producers.

(f) False
Correct Statement: Respiration results in loss of dry weight of the plants.

(g) False
Correct Statement: Photosynthesis stops at a temperature of above 40oC.

(h) True
(i) True
(j) True

Solution C.3.
(a) grana
(b) iodine solution
(c) chloroplast
(d) Calvin cycle
(e) Sucrose

Solution C.4.

(a) False
Photosynthesis increases with the light intensity up to a certain limit only and then it gets stabilized.

(b) False
The atmospheric temperature is an important external factor affecting photosynthesis. The rate of photosynthesis increases up to the temperature 35oC after which the rate falls and the photosynthesis stops after 40oC.

(c) False
Ice cold water will hamper the process of photosynthesis in the immersed leaf, even if there is sufficient sunshine because the temperature is an important factor for the rate of photosynthesis.

(d) False
For destarching, the potted plant can kept in a dark room for 24-48 hours.

(e) False
There is no start point or end point in the carbon cycle, the carbon is constantly circulated between the atmosphere and the living organisms.

(f) False
If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will continue to occur because the dark reaction is independent of light and it occurs simultaneously with the light dependent reaction.

(g) True

Solution C.5.
Photons, grana, water molecules, hydrogen and hydroxyl ions, oxygen

Solution C.6.

Photosynthesis Respiration
Carbon dioxide is used up and oxygen is released. Oxygen is used up and carbon dioxide is released.
Photosynthesis occurs in plants and some bacteria. Respiration occurs in all living organisms.
Photosynthesis results in gain of dry weight of the plants. Respiration results in loss of dry weight of the plants.
Glucose is produced which is utilized by the plants. Glucose is broken down to obtain energy.
The raw materials for the photosynthesis are water, carbon dioxide and sunlight. The raw material for respiration is glucose.

(Any 4)

Solution C.7.
Oxygen is released during photosynthesis. Some of this oxygen may be used in respiration in the leaf cells, but the major portion of it is not required and it diffuses out into the atmosphere through the stomata. However, in a sense, even this oxygen is not a waste because all organisms require it for their existence including the plants.

Solution C.8.
The presence of starch is regarded as evidence of photosynthesis. Hence before starting an experiment on photosynthesis, the plant should be placed in the dark for 24-48 hours to destarch the leaves. During this period, all the starch from the leaves will be sent to the storage organs and the leaves will not show the presence of starch. So the various experiments on photosynthesis can be carried out effectively.

Solution C.9.
Destarching means removal of starch. Destarching is carried out so that all the starch from the leaves will be sent to the storage organs. Hence all the leaves will not show the presence of starch and photosynthesis can be studied. Destarching ensures that any starch present after the experiment has been formed under experimental conditions.

Solution C.10.
If a green plant is kept in bright light, it tends to use up all the CO2 produced during respiration, for photosynthesis. Thus, the release of CO2cannot be demonstrated. Hence, it is difficult to demonstrate respiration as these two processes occur simultaneously.

Solution C.11.
The chloroplasts are concentrated in the upper layers of the leaf which helps cells to trap the sunlight quickly. Also the epidermis is covered by a waxy, waterproof layer of cuticle. This layer is thicker on the upper surface than the lower one. Hence most leaves have the upper surface more green and shiny than the lower one.

Solution C.12.

  • Place hydrilla plant (a water plant) in a beaker containing pond water and cover it by a short-stemmed funnel. (Make sure the level of water in the beaker is above the level of the stem of the funnel)
  • Invert a test tube full of water over the stem of the funnel.
  • Place the set up in the sun light for a few hours.

Observation:
Bubbles appear in the stem which rise and are collected in the test tube. When sufficient gas gets collected, a glowing splinter will be introduced in the test tube, which will burst into flames.

Inference:
The splinter glows due the presence of oxygen in the test tube which proves that the gas collected in the test is released by hydrilla during photosynthesis.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 1

Solution C.13.

(i) Light Reaction: 
The light reaction occurs in two main steps:

  1. Activation of chlorophyll – On exposure to light energy, chlorophyll becomes activated by absorbing photons.
  2. Splitting of water – The absorbed energy is used in splitting the water molecule into hydrogen and oxygen, releasing energy. This reaction is known as photolysis of water.

Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 2
The fate of H+, e– and (O) component are as follows:
The hydrogen ions (H+) obtained from above are picked up by a compound NADP (Nicotinamide adenine dinucleotide phosphate) to form NADPH.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 3
The oxygen (O) component is given out as molecular oxygen (O2).
2O → O2
The electrons (e–) are used in converting ADP into energy rich ATP by adding one inorganic phosphate group Pi.
ADP + Pi → ATP
This process is called photophosphorylation.

(ii) Dark reaction: The reactions in this phase does not require light energy and occur simultaneously with the light reaction. The time gap between the light and dark reaction is less than one thousandth of a second. In the dark reaction, ATP and NADPH molecules (produced during light reaction) are used to produce glucose (C6H12O6) from carbon dioxide. Fixation and reduction of carbon dioxide occurs in the stroma of the chloroplast through a series of reactions. The glucose produced is either immediately used up by the cells or stored in the form of starch.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 4

Solution C.14.
Complete the following food chains by writing the names of appropriate organisms in the blanks:
(i) Grass → Rabbit. → Snake → Hawk
(ii) Grass/Corn → Mouse → Snake → Peacock

Solution C.15.
Non-green plants such as fungi and bacteria obtain their nourishment from decaying organic matter in their environment. This matter comes from dead animals and plants. Fungi and bacteria break down the organic matter to obtain the nourishment and they release carbon dioxide back in the atmosphere.

Solution C.16.
Chlorophyll is the foundation site for the photosynthesis in green plants. The initiation of photosynthesis takes place when the chlorophyll molecule traps the light energy. The light energy is then converted into chemical energy in the form of glucose using carbon dioxide (CO2) from the atmosphere, and water (H2O) from the soil. All other organisms, directly or indirectly depend on this food for their survival. The starting point of any food chain is always a plant. If green plants were to suddenly disappear, then so would virtually all life on Earth. Thus, we can say that all life owes its existence to chlorophyll.

Solution C.17.
To test the leaf for starch, the leaf is boiled in water to kill the cells. It is next boiled in methylated spirit to remove chlorophyll. The leaf is placed in warm water to soften it. It is then placed in a dish and iodine solution in added. The region, which contains starch, turns blue-black and the region, which does not contain starch, turns brown.

Solution D.1.

a. The student wanted to show that sunlight is necessary for photosynthesis. / The role of sunlight in photosynthesis is being investigated.

b. Yes. The other uncovered leave of the potted plant act as a control.

c. Destarching ensures that any starch present after the experiment has been formed under experimental conditions. Therefore, the plant was kept in the dark before the experiment.

d.

  • The student dipped the leaf in boiling water for a minute to kill the cells.
  • Then he boiled the leaf in alcohol/methylated spirit over a water bath to remove chlorophyll. The leaf becomes hard and brittle.
  • He then places the leaf in hot water to soften it.
  • Next the student spreads the leaf in a dish and pours iodine solution on it. The presence of starch is indicated by a blue-black colour.
  • The uncovered portion (exposed to sunlight) turned blue-black colour and the covered portion showed brown colour. The difference in the colours of covered and uncovered part of leaves indicates the importance of sunlight in photosynthesis.

Solution D.2.
(a) Guard cells: They regulate the opening and closing of stomata and thus regulate the entry of carbon dioxide through the stomata.

(b) Cuticle: Cuticle is transparent and water proof due to which light can penetrate this later easily.

(c) Mesophyll cells: Mesophyll cells are the main sites for photosynthesis. Chloroplasts are mainly contained in the mesophyll cells. When sunlight falls on the leaf, the light energy is trapped by the chlorophyll of the upper layers of mesophyll, especially the palisade cells.

(d) Xylem Tissue in the Leaf Veins: Water is essential for photosynthesis to occur. Water is taken up by the roots from the soil, sent up through the stem and finally brought to the leaves (site of photosynthesis) through the xylem tissue. The water is then distributed in the mesophyll tissue.

(e) Phloem Tissue in the Leaf Veins: The prepared food is transported from leaves to all parts of the plant by the phloem tissue. The glucose is converted into insoluble starch and later into soluble sugar i.e. sucrose, which is transported in solution through the phloem in the veins of the leaf and down through the phloem of the stem.

(f) Stoma: The main function of stoma is to let in carbon dioxide from the atmosphere for photosynthesis. Also most of the oxygen produced during photosynthesis diffuses out into the atmosphere through the stomata.

Solution D.3.
a.

  1. Sunlight
  2. Oxygen
  3. Glucose
  4. Xylem

b. A – Transpiration
B – Translocation

Solution D.4.
a. Food chain
b. Hawk, eagle
c. Photosynthesis
d. Carbon

Solution D.5.
Test to determine the presence of starch in a leaf:

  • Dip a leaf in boiling water for a minute to kill the cells.
  • Boil the leaf in methylated spirit in a water bath to remove the chlorophyll, till the leaf turns pale blue and becomes hard and brittle.
  • Now place the leaf in hot water to soften it.
  • Place the leaf in a Petri dish and pour iodine solution over it.
  • The appearance of a blue-black colour on the leaf is indicative of the presence of starch.
  • The absence of starch is indicated by a brown colouration.

Solution D.6.
a. To demonstrate the importance of carbon dioxide in photosynthesis
b. No, the experiment will not work satisfactorily, as the beaker contains lime water and not potassium hydroxide to absorb CO­2.
c. Place potassium hydroxide in the beaker instead of lime water
d. Before starting the experiment, it is necessary to destarch the leaves of the plant by keeping the plant in complete darkness for 48 hours. This is because if the plant is not destarched, then the experiment will give false results because starch stored previously may be detected in the leaf placed in the beaker even if no starch is produced during the experiment.

Solution D.7.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 5

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions Transpiration

by

Selina Concise Biology Class 10 ICSE Solutions Transpiration

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 5 Transpiration. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 5 Transpiration

Exercise 2

Solution A.1.
(a) Open stomata, dry atmosphere and moist soil

Solution A.2.
(a) increase

Solution A.3.
(b) temperature is high

Solution A.4.
(c) sunken stomata

Solution A.5.
(d) hydathodes

Solution A.6.
(d) transpiration

Solution A.7.
(d) hot, dry and windy

Solution A.8.
(b) Lenticels

Solution A.9.
(b) evaporation of water from the aerial surfaces of a plant

Solution B.1.
(a) Lenticels
(b) Guttation
(c) Potometer
(d) Nerium
(e) Ganong’s photometer
(f) Stomata and cuticle
(g) Hydathodes
(h) Guttation

Solution B.2.
(a) vapour, aerial
(b) stomata, transpiration
(c) suction, water (heat)

Solution C.1.
(a) guttation
(b) protection and reduced transpiration
(c) transpiration
(d) reduced transpiration

Solution C.2.
(i) False
(ii) True
(iii) True
(iv) False
(v) Most transpiration occurs at mid-day.
(vi) Potometer is an instrument used for measuring the rate of transpiration in green plants.

Solution C.3.
(a) Transpiration increases with the velocity of wind. If the wind blows faster, the water vapours released during transpiration are removed faster and the area surrounding the transpiring leaf does not get saturated with water vapour.

(b) When the rate of transpiration far exceeds the rate of absorption of water by roots, the cells lose their turgidity. Hence, excessive transpiration results in wilting of the leaves.

(c) Plants absorb water continuously through their roots, which is then conducted upwards to all the aerial parts of the plant, including the leaves. Only a small quantity of this water i.e. about 0.02% is used for the photosynthesis and other activities. The rest of the water is transpired as water vapour. Hence water transpired is the water absorbed.

(d) There are more stomatal openings on the lower surface of a dorsiventral leaf. More the number of stomata, higher is the rate of transpiration. Hence more transpiration occurs from the lower surface.

(e) Cork and Bark of trees are tissues of old woody stems. Bark is thick with outermost layer made of dead cells and the cork is hydrophobic in nature. These properties make them water-proof and hence they prevent transpiration.

(f) In both perspiration and transpiration, water is lost by evapouration from the body of the organism as water vapour. This evaporation reduces the temperature of the body surface and brings about cooling in the body of the organism.

(g) On a bright sunny day, the rate of transpiration is much higher than any other days. The leaves of certain plants roll up on a bright sunny day to reduce the exposed surface and thus reduce the rate of transpiration.

Solution C.4.
(a) False
Reason: Potometer is used to measure the rate of transpiration in a plant. Demonstration of transpiration occurring from the lower surface of a leaf is done by analyzing the changes in colour of pieces of dry cobalt chloride paper attached (and held in place) to the two surfaces of a leaf.

(b) True
Reason: Transpiration carried out by the large number of trees in a forest. This increases the moisture in the atmosphere and brings rain.

(c) False
Reason: Hydathodes are special pores present on the ends of leaf veins through which guttation occurs and water droplets are given out. Their openings cannot be regulated. Stomata on the other hand are minute openings in the epidermal layer of leaves through which exchange of gases as well as transpiration occurs. Water is given out as water vapour. Stomatal opening is regulated by guard cells.

(d) False
Reason: Transpiration is reduced during high atmospheric humidity. High humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

(e) True
Reason: Desert plants need to reduce transpiration as much as possible so as to survive in the hot and dry environment. Hence some of them have sunken stomata as an adaptation to curtail transpiration.

(f) True
Reason: During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. During mid-day, the outside temperature is higher, due to which there is more evaporation of water from the leaves. Therefore more transpiration occurs during mid-day.

Solution C.5.

Guttation Bleeding
It is the removal of excess of water from the plants because of excess water buildup in the plant. It is the removal of water from the plant because of injury.
Water escapes from specialisedstructures called hydathodes. Water escapes in the form of sap from the injured part of the plant.

Solution D.1.
Wilting refers to the loss of cellular turgidity in plants which results in the drooping of leaves or plant as a whole because of lack of water.
During noon the rate of transpiration exceeds the rate of absorption of water by roots. Due to the excessive transpiration, the cells of leaves lose their turgidity and wilt.

Solution D.2.
The lower surface of leaf is sheltered from direct sunlight. If more stomata are on the upper surface of a leaf, then excessive transpiration would occur, resulting in quick wilting of the plant. Hence most plants have more numerous stomata on the lower surface of a leaf to control the rate of transpiration.

Solution D.3.
Take the small potted rose plant and cover it with a transparent polythene bag. Tie its mouth around the base of the stem. Leave the plant in sunlight for an hour or two.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 1
Drops of water will soon appear on the inner side of the bag due to the saturation of water vapour given out by the leaves. A similar empty polythene bag with its mouth tied and kept in sunlight will show no drops of water. This is the control to show that plants transpire water in the form of water. If tested with dry cobalt chloride paper, the drops will be confirmed as water only.

Solution D.4.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 2
Potometer is a device that measures the rate of water intake by a plant. This water intake is almost equal to the water lost through transpiration. Potometers do not measure the water lost due to transpiration but measure the water uptake by the shoot.

Solution D.5.

  • Transpiration occurring through lenticels i.e. minute openings on the surface of old stems is called lenticular transpiration.
  • Stomatal transpiration is controlled by the plant by altering the size of the stoma, where as this does not happen in case of lenticular transpiration. This is because the lenticels never close, but remain open all the time.
  • The amount of stomatal transpiration is much more than the amount of lenticular transpiration.

Solution D.6.
The factors that accelerate the rate of transpiration are:

  • High intensity of sunlight
  • High temperature
  • Higher wind velocity
  • Decrease in atmospheric pressure
    (Any three)

Solution D.7.
Forests have large number of plants especially trees. Each plant loses water in the form of water vapour everyday into the atmosphere through transpiration. A large apple tree loses as much as 30 litres of water per day. So huge amount of water is escaped into the atmosphere by forests. This increases the moisture in the atmosphere and brings more frequent rains.

Solution D.8.
The advantages of transpiration to the plants are:

  • Transpiration brings about a cooling effect to the plant body since evaporation of water reduces the temperature of leaf surface.
  • Transpiration helps in the ascent of sap by producing a suction force acting from the top of the plant.
  • Transpiration helps in distributing water and mineral salts throughout the plant body.
  • Transpiration helps in eliminating excess water.

Solution D.9.

  1. If the water content of the leaves decreases due any reason, the guard cells turn flaccid, thereby closing the stomatal opening and transpiration stops.
  2. Some plants have sunken stomata whereas others have reduced number of stomata to reduce transpiration.
  3. In some plants, leaves may be dropped or may be absent or changed into spines as an adaptation to reduce transpiration.
  4. The leaves may be covered by thick cuticle such as in Banyan tree, so as to reduce transpiration.

Solution D.10.
No, they are not dew drops.

This is water given out by the plant body through guttation. Since the banana plant is growing in humid environment, transpiration is hampered. But the roots continue to absorb water from the soil. This builds up a huge hydrostatic pressure within the plant and forces out the excess water from the hydathodes, which are pores present at the tips of veins in the leaf. This is observed especially during the mornings.

Solution D.11.
(a) Intensity of light – During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. At night they are closed. Hence more transpiration occurs during the day. During cloudy days, the stomata are partially closed and the transpiration is reduced.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 3
(b) Humidity of the atmosphere – When the air is humid; it can receive very less water vapour. Thus, high humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

Solution E.1.

(i) The leaf D would become most limp. This is because water would be lost through transpiration from upper as well as the lower surface of leaf D since it is uncoated.

(ii) The least limping would be shown by leaf C since its upper and lower surfaces have been coated with vaseline. So no water is lost from the leaf through transpiration since the stomatal openings get blocked by vaseline.

Solution E.2.
(a)
Guard Cell
Inner wall of the Guard Cell
Stoma/Stomatal Aperture
(b) Open state
(c) The structure of stoma remains same in monocots as well as in dicots. Hence, the stoma from the diagram can be of a monocot leaf or of a dicot leaf.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 4

Solution E.3.
(a) Transpiration
(b) Oil is put on the surface of water to prevent loss of water by evaporation.
(c) Yes, the transpiration rate will increase. Transpiration would occur faster. The observable changes will occur in less time.
(d) The spring balance progressively measures the change in weight of the set-up. This because as the plant transpires, it creates the suction force in plant which allows roots to absorb more water from the test tube. Hence, the water in the test will get reduced. Thus, the weight of the entire set will decrease.

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

by

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved

Exercise 1

Solution A.1.
c) Imbibition

Solution A.2.
c) Hypertonic salt solution

Solution A.3.
b) Turgidity

Solution A.4.
d) Grow downward into the soil

Solution A.5.
b) Diffusion

Solution A.6.
c) a selectively permeable membrane in between

Solution A.7.
a) Pure water

Solution A.8.
d) water

Solution A.9.
b) Root pressure

Solution A.10.
d) it allows a solvent to pass through freely but prevents the passage of the solute

Solution B.1.
(a) Turgidity
(b) Guttation
(c) Osmosis
(d) Xylem
(e) Endosmosis
(f) Diffusion
(g) Root pressure

Solution B.2.
(a) Turgor pressure
(b) Flaccidity
(c) Bleeding

Solution B.3.
(a) the fluids inside
(b) transported inside against their concentration gradient
(c) turgor movements

Solution B.4.
(a) shrink
(b) water
(c) opposite

Solution B.5.

Column I Column II
a Xylem (iv) upward flow of water
b Phloem (iii) downward flow of sap
c Cell membrane (i) semi-permeable
d Root pressure (v) guttation
e Cell wall (ii) permeable

Solution C.1.
(a)

Plasmolysis Deplasmolysis
1. It refers to the shrinkage of the cytoplasm and withdrawal of the plasma membrane from the cell wall caused due to the withdrawal of water when placed in a hypertonic solution.

2. In Plasmolysis, the cell becomes flaccid.

1.Deplasmolysis is the recovery of a plasmolysed cell when it is placed in water, wherein the cell’s protoplasm again swells up due to the re-entry of water.

2. In deplasmolysis, the cell becomes turgid.

(b)

Turgor pressure Wall pressure
Turgor pressure is the pressure of the cell contents on the cell wall. Wall pressure is the pressure exerted by the cell wall on the cell content.

(c)

Guttation Bleeding
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure. Bleeding is the loss of cell sap through a cut stem.

(d)

Turgidity Flaccidity
1. It is the state of a cell in which the cell cannot accommodate any more water and it is fully distended. 1. It is the condition in which the cell content is shrunken and the cell is not tight.

Solution C.2.
(a)

  1. False
  2. False
  3. False
  4. True
  5. False
  6. False

(b)

  1. A plant cell placed in hypotonic solution gets turgid.
  2. Addition of salt to pickles prevents growth of bacteria because they turn flaccid.
  3. Cells that have lost their water content are said to be plasmolysed.
  4. The shrinkage of protoplasm, when a cell is kept in hypertonic solution.

Solution C.3.
The cell is said to be turgid when the plant cell wall becomes rigid and stretched by an increase in the volume of vacuoles due to the absorption of water when placed in hypotonic solution. On the other hand, the cell is said to be flaccid when the cell contents get shrunken when the cell is placed in hypertonic solution and the cell is no more tight. Flaccidity is the reverse of turgidity.

Example: Weeds can be killed in a playground by sprinkling excessive salts around their base.
Or
A plant cell when immersed in hypertonic solution like salt solution for about 30 minutes will become flaccid or limp.

Solution C.4.

(a) Common salt when sprinkled on the grass causes the Plasmolysis of grass cell ultimately leading them to death. Hence, if we sprinkle some common salt on grass growing on a lawn, it is killed at the spot.

(b) If a plant is uprooted, the leaves continue losing water by transpiration, but there is no more water absorbed the roots. This does not allow the compensation for the loss of water by transpiration; hence the leaves of the uprooted plant wilt soon.

(c) Transplantation causes stress to the seedlings. If the seedlings are transplanted in the morning, they would have to immediately bear the additional stress of excessive transpiration occurring during the hot afternoon. Transplantation in the evening helps the seedlings to adjust for a longer time during the night (cooler temperatures) because the quantity of water absorbed exceeds the loss of water through transpiration. Therefore, it is better to transplant seedling in a flower bed in the evening and not in the morning.

(d) In a hypertonic solution, the solution outside the cell has higher solute concentration than the fluids inside the cell. Therefore, water flows out from the plant cell due to exosmosis. The cytoplasm shrinks and the plasma membrane withdraws away from the cell wall and this the cell becomes flaccid. Hence a plant cell when kept in a hypertonic salt solution for about 30 minutes turns flaccid.

(e) Potato cubes contain excess of salts and sugars as compared to the water in which the cubes are placed. Hence, due to endosmosis, water from the surrounding enters the potato cubes making them firm and increasing their size.

Solution C.5.

(a) True.
Plasmolysis occurs due to outflow of water from the cell when placed in hypertonic solution due to which the cytoplasm shrinks away from the cell wall. On the other hand, deplasmolysis is the result of the re-entry of water into the plasmolysed cell when placed in hypotonic solution due to which the protoplasm again swells up pressing tight against the cell wall.

(b) False.
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure whereas bleeding is the loss of cell sap through a cut stem.

(c) False.
There is only one seed coat in a seed.

(d) False.
The leaves of the twig remain turgid since its xylem is intact and xylem is responsible for water conduction in plants.

(e) False.
Guttation occurs due to excessive root pressure. It is maximum when root pressure is maximum which occurs in the early mornings or at night. This is because during these times, transpiration is very low and water absorption is very high.

(f) False.
Dry seeds when submerged in water swell up due to imbibitions. On contact with water dry seeds imbibe water and swell up.

Solution D.1.
Examples of turgor movements in plants:

  1. In Mimosa pudica, a sensitive plant, the stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles called pulvinus. This causes the folding and drooping of leaves of the plant.
  2. The leaves of insectivorous plants close up to entrap a living prey. When the insect come in contact with the leaf, it loses it turgor hence closing the leaves of the plant.
  3. The bending movements of certain flowers towards the sun.
    (Any two)

Solution D.2.
The closing and opening of the stomata depends on the turgidity of the guard cells. Each guard cell has a thicker wall on the side facing the stoma and a thin wall on the opposite side. Guard cells contain chloroplasts. As a result of the synthesis of glucose during photosynthesis and some other chemical changes, the osmotic pressure of the contents of the guard cells increases and they absorb more water from the neighbouring cells, thus becoming turgid. Due to turgor, the guard cells become more arched outwards and the aperture between them widens, thereby opening the stoma.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 1
At night or when there is shortage of water in the leaf, the guard cells turn flaccid and their inner rigid walls become straight, thus closing the stomatal aperture.

Solution D.3.
If the concentration of mineral nutrient elements is higher inside the root-hairs than in the surrounding soil, then roots take them in from the soil by ‘active transport’. In active transport, the mineral ions are forcibly carried from the surrounding soil i.e. the region of their lower concentration into the roots i.e. the region of their higher concentration through the cell membrane by expenditure of energy. This energy is supplied by the cell in the form of ATP.

Solution D.4.
When soaked in water, the seeds swell up due to imbibition and endosmosis. During these two processes water enters the cell. Due to endosmosis, at some point, the seed coat is unable to bear the turgor pressure and hence, the seed coat bursts.

Solution D.5.
Leaves of the sensitive plant wilt and droop down on a slight touch due to turgor movement. Petiole of sensitive plant is held up by turgid pulvinus tissue. The stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles i.e. pulvinus. The cells of the lower side of pulvinus lose water and the petiole collapses. This causes the wilting and drooping of the leaves.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 2

Solution D.6.
As water is lost from the leaf surface by transpiration, more water molecules are pulled up due to the tendency of water molecules to remain joined i.e. cohesion. This produces a continuous column of water throughout the stem which is known as ‘transpiration pull’. A negative pressure or tension is produced in the xylem that pulls the water from the roots and soil. Transpirational pull is an important force which causes the ascent of sap.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 3

Solution E.1.
(a) The cell is flaccid i.e. it is plasmolysed.
(b) Plasma Membrane
(c) Plasmolysis would not occur and flaccidity would not be seen i.e. the protoplasm would not have shrunken away from the cell wall.
(d) Cell Wall is absent in animal cell.

Solution E.2.
(a) Flaccid Cell
(b) The liquid is hypertonic solution. It has higher solute concentration outside the cell than the fluids inside the cell.
(c)
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 4

Solution E.3.

(a) Osmosis

(b) Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

(c) After an hour or so, the level of sugar solution in the thistle funnel will rise and the level of water in the beaker will drop slightly.

(d) For control experiment, the beaker will contain the water. At the same time, instead of the sugar solution; the thistle funnel with the cellophane paper tied on its mouth and inverted in the beaker will also contain water.

(e)

  1. concentrated sugar solution – Cell sap (of higher concentration than that of the surrounding water) within the root hair.
  2. parchment paper – cell membrane of root hair.
  3. water in the beaker – water in soil.

(f) cellophane paper, egg membrane, animal bladder (any one)

(g)

  1. The roots of plants absorb water and minerals from surrounding soil due to osmosis.
  2. Osmosis allows plants to absorb water from the soil which helps plants to keep cells alive in roots, stems and leaves.
  3. Osmosis is also important in the opening and closing of stomata which is an important feature for the processes like transpiration and photosynthesis. (Any two)

Solution E.4.

a.
A – Cell wall
B – Cell membrane
C – Cytoplasm
D – Nucleus

b. A root hair gets turgid because of the absorption of water from the surrounding. Absorption of water by root hair is achieved by the process of osmosis. The concentration of water in the surrounding is more than that of the interior of the cell; this causes the water from the surrounding to move in because of endosmosis.

c.

Cell wall Cell membrane
The cell wall of a root hair is freely permeable and allows both salt and water to pass through. The cell membrane of a root hair is semi-permeable and does not allow large dissolved salt molecules to pass through.

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 5

Solution E.5.

(a) Water is hypotonic to the potato cells, due to which endosmosis occurs and water enters the potato cells. The protoplasm swells up pressing tight against the cell wall. The cells are fully distended i.e. turgid. This causes the firmness and increase in the size of the potato cubes when placed in water.

(b) Sugar solution is hypertonic to the potato cells, due to which exosmosis occurs and water flows out of the potato cells. The potato cell loses its distended appearance, the cytoplasm shrinks and the plasma membrane withdraws from the cell wall. The cells become limp or flaccid. This causes the softness and decrease in size of the potato cubes when placed in sugar solution.

(c) The process being investigated is osmosis. Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

Solution E.6.
(a) It is the diagrammatic cross-section of a part of a root.

(b)

  1. Root hair
  2. Epidermis
  3. Cortex
  4. Endodermis
  5. Phloem
  6. Xylem

(c) Cortex (label 3) is the ground tissue and is active in the uptake of water and minerals. It also helps in storage of photosynthetic products.
Phloem (label 5) helps in transporting the prepared food from leaves to different parts of the plant.

Solution E.7.
(a) The process of water absorption by plant roots through osmosis is being studied here.

(b) A root-hair contains cell sap which contains higher concentration of salts as compared to outside soil water. This difference sets off osmosis and outside water diffuses into the root-hair. From the cell bearing root-hair, water passes into adjoining cells one after another to finally the xylem vessels.

(c) The surface of water was covered with oil to prevent any loss of water by evaporation.

Solution E.8.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 6

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals

by

Selina Concise Biology Class 10 ICSE Solutions Genetics- Some Basic Fundamentals

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 3 Genetics Some Basic Fundamentals. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 3 Genetics – Some Basic Fundamentals

Exercise 1

Solution A.1.
d) Ascaris

Solution A.2.
a) 3 : 1

Solution B.1.
(a) – (iii) Study of laws of inheritance of characters
(b) – (v) Chromosomes other than the pair of sex chromosomes
(c) – (iv) A gene that can express when only in a similar pair
(d) – (ii) The alternative forms of a gene
(e) – (i) Chromosomes similar in size and shape

Solution B.2.
Lion, tiger, domestic cat (Any two)

Solution B.3.
Colour-blindness, Thalassaemia, Sickle cell anaemia and Haemophilia (Any two)

Solution B.4.
Homozygous dominant – RR
Homozygous recessive – rr

Solution C.1.

(a)

Phenotype Genotype
The observable Characteristic which is genetically controlled is called phenotype. The set of genes present in the cells of an organism is called its genotype.

(b)

Character Trait
Any heritable feature is called a character. The alternative form of a character is called trait.

(c)

Monohybrid cross Dihybrid cross
It is a cross between two pure breeding parent organisms with different varieties taking into consideration the alternative trait of only one character. It is a cross between two pure breeding parent organisms with different varieties taking into consideration the alternative trait of two characters.

Solution C.2.
The characteristics of a species such as physical appearance, body functions and behavior are not only the outcome of chromosome number, but these depend on the genotype of every organism. That means the set of genes present in the organisms may very and therefore lion, tiger and domestic cat have the same number of 38 chromosomes, their characteristics (like different appearances) are the result of the genes located on the chromosomes.

Solution C.3.

Character Dominant trait Recessive trait
Flower Colour Purple White
Seed Colour Yellow Green
Seed Shape Round Wrinkled
Pod Shape Inflated Constricted
Flower Position Axial Terminal

(Any 3)

Solution C.4.

  • Colour-blindness is caused due to recessive genes which occur on the X chromosome.
  • Males have only one X chromosome. If there is recessive gene present on X chromosome, then the male will suffer from colour-blindness.
  • Females have two X chromosomes. It is highly impossible that both the X chromosomes carry abnormal gene. Hence, if one gene is abnormal and since it is recessive, its expression will be masked by the normal gene present on the other X chromosome. Females are unlikely to suffer from colour-blindness.

Solution C.5.
Phenotypic Ratio – 3 (Black Fur) :1 (Brown Fur)
Genotypic Ratio – 1(Homozygous Black Fur):2 (Heterozygous Black Fur): 1 (Homozygous Brown Fur)

Solution D.1.

(a)  Heterozygous:  The condition in which a pair of homologous chromosomes carries dissimilar alleles for a particular character.

For example –

  1. A daughter (XXo) from a normal homozygous mother for colour vision (XX) and a colour blind father has one normal and one defective allele (XoY).
  2. Certain tongue rollers are heterozygous with Rr genotype.

(b)  Homozygous:  The condition in which a pair of homologous chromosomes carries similar alleles for a particular character.

For example –

  1. A colorblind daughter (XoXo) will have both the X chromosomes with defective alleles.
  2. A non-roller will have rr (homozygous) genotype.

(c)  Pedigree Chart:  A pedigree chart is a diagram that shows the occurrence and appearance or phenotypes of a particular gene or organism and its ancestors from one generation to the next. In the pedigree chart, males are shown by squares and females by circles.

Solution D.2.
Mendel’s laws of inheritance are:

  1. Law of Dominance: Out of a pair of contrasting characters present together, only one is able to express itself while the other remains suppressed. The one that expresses is the dominant character and the one that is unexpressed is the recessive one.
  2. Law of Segregation : The two members of a pair of factors separate during the formation of gametes. The gametes combine together by random fusion at the time of zygote formation. This law is also known as ‘law of purity of gametes’.
  3. Law of Independent Assortment: When there are two pairs of contrasting characters, the distribution of the members of one pair into the gametes is independent of the distribution of the other pair.

Solution D.3.

    • The sex of the child depends on the father. The egg contains only one X chromosome, but half of the sperms contain X-chromosome whereas the other half contains Y-chromosome. It is simply a matter of chance as to which category of sperm fuses with the ovum and this determines whether the child will be male or female.
    • If the egg fuses with X-bearing sperm, the resulting combination is XX and the resulting child is female.
    • If the egg fuses with Y-bearing sperm, the resulting combination is XY and the resulting child is male.
      Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals image - 1

Solution E.1.

Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals image - 2

Solution E.2.
(a) Black
(b) No

Solution E.3.

Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals image - 3

Solution E.4.
(a) Father
(b) Two sons and three daughters
(c) The child 1 (daughter) is colour blind
(d) X chromosome
(e) Haemophilia

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes

by

Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure Of Chromosomes. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure of Chromosomes

Exercise 1

Solution A.1.
(b) DNA and Histones

Solution A.2.
(c) Coloured bodies

Solution B.1.
(a) – Nucleotides.
(b) – Nucleosome.
(c) – Hydrogen Bond.
(d) – Phosphate, Sugar and Nitrogenous base.

Solution C.1.
Chromatin fibre is unfolded, uncondensed, extended DNA. It is only visible when cell under goes division whereas chromosomes are condensed DNA and they are visible when the cell is divided.

Solution C.2.
Rungs of DNA ladder is made of nitrogenous bases which includes Adenine (A), Guanine (G), Cytosine (C) and Thymine (T).

Solution C.3.
(a) The four nitrogenous bases in the DNA ladder are Guanine, Thymine, Adenine and Cytosine.
(b) Genes are specific sequences of nucleotides on a chromosome.
(c) A nucleotide is composed of a phosphate, sugar (pentose) and a nitrogenous base.
(d) Nucleosomes are groups of histone molecules surrounded by DNA strands.
(e) If there are 46 chromosomes in a cell there will be 46 chromatin fibres inside the nucleus during interphase.

Solution D.1.
Nucleosome is basic structural unit of DNA. Each strand of DNA winds around a core of eight histone molecules. This core can be imagined like a football, around which a long rope is wound with one or two loops. Each such complex structure is called a nucleosome. A single human chromosome may have about a million nucleosomes.

Solution D.2.
Gene is a structural and functional unit of heredity and variations. Genes are specific sequences of nucleotides on a chromosome that encode particular proteins which express in the form of some particular feature of the body. In other words, gene is the DNA segment of the chromosome and it controls the expression of characteristics.

Solution E.1.
(a) 2
(b) 2 on each strand
(c) 1- Phosphate, 2- Sugar, 3- Bases, 4- Hydrogen Bond, 5 – Base
(d)Nucleotide

Solution E.2.
B, C and A.

Exercise 2

Solution A.1.
(c) both ovary and testis

Solution A.2.
(c) Anaphase, telophase

Solution A.3.
(c) DNA

Solution B.1.
Cell A: 2
Cell B: 4

Solution B.2.
(a) – Metaphase.
(b) – Telophase.
(c) – Prophase.
(d) – Anaphase.

Solution B.3.
(a) Somatic (body).
(b) Four.
(c) Reproductive.
(d) 23 and 23.
(e) Haploid.
(f) Centriole.

Solution C.1.

(a) A chromosome is an organized structure of DNA and protein found in cells. It is a single piece of coiled DNA containing many genes, regulatory elements and other nucleotide sequences whereas a chromatid is one of the two copies of DNA making up a duplicated chromosome, which are joined at their centromeres, for the process of cell division (mitosis or meiosis).

(b) The centrosome is an area in the cell where microtubules are produced. Within an animal cell centrosome, there is a pair of small organelles called the centrioles. During animal cell division, the centrosome divides and the centrioles replicate (make new copies) whereas each chromosome in its condensed form consists of two chromatids joined at some point along the length. This point of attachment is called centromere.

(c) An aster is a cellular structure shaped like a star, formed around each centrosome during mitosis in an animal cell whereas spindle fibers are aggregates of microtubules that move chromosomes during cell division.

(d) A haploid cell is a cell that contains one complete set of chromosomes. Gametes are haploid cells that are produced by meiosis whereas a diploid cell is a cell that contains two sets of chromosomes. One set of chromosomes is donated from each parent.

Solution C.2.
In this statement, reduction means that the number of chromosomes are reduced to half i.e. out of the 23 pairs of chromosomes in humans, only single set of chromosomes are passed on to the sex cells.

Solution C.3.
Gametes must be produced by meiosis for sexual reproduction because the numbers of chromosomes are reduced to half during meiosis and then the normal diploid numbers of chromosomes are regained during the process of fertilization.

Solution C.4.

(a) F; Surface skin cells are continuously lost and replaced by the underlying cells.

(b) T; All types of human cells, have 46 chromosomes. The only type of cell which does not have 46 chromosomes are the sex cells, which have only half of the number, so they have 23 chromosomes. The egg cell is a sex cell (found in female). So it must have 23 chromosomes.

(c) F; Nuclear membrane disappears in Prophase itself, however it reappears during Telophase.

(d) T; Mitotic cell division can be a mode of asexual reproduction in unicellular organisms like amoeba or yeast cell which divides into two daughter cells.

(e) T; While the maternal and paternal chromosomes are separating, the chromatid material gets exchanged between the two members of a homologous pair resulting in genetic recombination.

Solution D.1.
a.

  1. Centromere
  2. Spindle fibres
  3. Chromatids

b. The stage described in the diagram is the late anaphase of mitosis in an animal cell. The stage can be identified by the presence of separated chromatids which are found at the two poles of the cell. The appearance of the furrow in the cell membrane classifies the stage as the late anaphase.
c. The division is mitotic division and this kind of cell division occurs in all the cells of the body except for the reproductive cells.
d. The stage before anaphase is metaphase.
Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 1

Solution D.2.

Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 2

Solution D.3.
The exchange of chromatids between homologous chromosomes is called crossing-over. This is the process by which the two chromosomes of a homologous pair exchange equal segments with each other.
Crossing over occurs in the first division of meiosis. At that stage each chromosome has replicated into two strands called sister chromatids. The two homologous chromosomes of a pair synapse, or come together. While the chromosomes are synapsed, breaks occur at corresponding points in two of the non-sister chromatids, i.e., in one chromatid of each chromosome.
Since the chromosomes are homologous, breaks at corresponding points mean that the segments that are broken off contain corresponding genes, i.e., alleles. The broken sections are then exchanged between the chromosomes to form complete new units, and each new recombined chromosome of the pair can go to a different daughter sex cell. It results in recombination of genes found on the same chromosome, called linked genes that would otherwise always be transmitted together.

Solution D.4.

(a) Late prophase. Because the nuclear membrane and nucleolus have disappeared.
(b) Centrioles.
(c)

  1. Centromere
  2. Chromatids.
  3. Spindle fibre.

(d) Metaphase. The centromeres of chromosomes are drawn to the equator by equal pull of two chromosomal spindle fibres that connects each centromere to the opposite poles, forming a metaphasic plate.

(e)

Mitosis Meiosis
(i) Two daughter cells are produced. (i) Four daughter cells are produced.
(ii) It is equational division i.e. the number of chromosome in the daughter cells or parent cells remains the same. (ii) It is reductional division i.e. the number of chromosomes is reduced to half in the daughter cells.

Solution D.5.
(a) Metaphase.
(b) 4.
(c) A – Animal
B – Animal
C – Plant
(d) (iv)

Solution D.6.
(a) This is an animal cell because:

  1. The outline is circular (in plants it would be angular {rectangular or polygonal}) and cell wall is absent.
  2. Centrosomes on centrioles are present. (These are found only in animal cells)

(b) Mitosis.
(c) B, C, D, A.
(d) Interphase.
(e)
Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 3

 

More Resources for Selina Concise Class 10 ICSE Solutions