RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 27 Man-Sensory Organs (Sense Organs)

August 17, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 27 Man-Sensory Organs (Sense Organs)

RBSE Class 12 Biology Chapter 27 Multiple Choice Questions

Question 1.
The shape of ear ossicle, malleus is:
(a) Hammer like
(b) Stirrup like
(c) Oval
(d) None of the above
Answer:
(a) Hammer like

Question 2.
The function of cones of the eye is:
(a) Secretion, balance
(b) Vision in darkness
(c) Monocular vision
(d) Vision in the high beam of light and colour differentiation
Answer:
(d) Vision in a high beam of light and colour differentiation

Question 3.
Man in the eye defect Myopia:
(a) Unable to easily see nearer objects
(b) Unable to easily see distant objects
(c) Unable to be the monocular vision
(d) None of the above
Answer:
(b) Unable to easily see distant objects

Question 4.
The balance est lavish by the part of the internal ear is
(a) Incus
(b) Sacculus
(c) Sacculus, Utriculus, semicircular canals
(d) Organs of Corti
Answer:
(c) Sacculus, Utriculus, semicircular canals

RBSE Class 12 Biology Chapter 27 Very Short Answer Type Questions

Question 1.
Write the names of bones present in the middle ear.
Answer:
Malleus, Incus & Stapes

Question 2.
Where the tectorial membrane is found?
Answer:
Scala Media of the cochlea of the ear.

Question 3.
Write functions of rods and cones found in the eye.
Answer:
Rods function even in dim-light and are responsible for general vision.
Cones are functional from moderate to bright light and are responsible for colour vision.

Question 4.
Which structure of the ear completes the mechanism of body balance?
Answer:
Cristae & Macullae

RBSE Class 12 Biology Chapter 27 Short Answer Type Questions

Question 1.
What is the role of the eustachian tube in the ear?
Answer:
It helps to maintain air pressure in the middle ear.

Question 2.
Name the places found in the retina in the sequence where vision is the best and where nothing is seen.
Answer:

Site of Best Vision – Yellow spot
Site of no vision – Blindspot

Question 3.
Write the names of muscles found in the human eyes.
Answer:

External rectus muscle
Internal rectus muscle
Superior rectus muscle
Inferior rectus muscle
Superior oblique muscle
Inferior oblique muscle

Question 4.
What is myopia?
Answer:
Myopia:

It is an eye disease which may occur to any person at any age.
The person is unable to see distant objects clearly.
It is caused either due to increased focal power of the lens (lens become more convex) or eyeball becomes larger.
It is also called a short-sightedness.
It can be corrected by use of biconcave lenses, which are also called as “minus lenses”.

Question 5.
What do you mean by colour blindness?
Answer:
Colourblindness:

It is a hereditary eye disease. It was first described by Horner. It is of 2 types –
1. Red-green colourblindness
2. Blue colourblindness
The red-green colourblindness is most common & it is also called a proton defect. Red-green colourblindness is caused due to the absence of red-green cones in the retina.
The person is unable to identify red, green, yellow & orange colours. These colours appear to be green in the absence of red cones and they appear to be red in the absence of green cones.
It is a sex-linked disease which is caused due to a recessive gene on the x-chromosome. The mother (female) acts as a carrier. The father of a colourblind daughter is always colourblind.
In the F₁ generation of carrier mother & normal father, 50% generation will be normal, 25% will be carrier & 25% will be colourblind.
In F₁ generation of carrier mother and colourblind father, 50% of daughters will be colourblind & 50% will be carrier likely, 50% of boys will be normal & 50% will be colourblind.
In the F₁ generation of normal mother & colourblind father, all males will be normal and all daughters will be a carrier.

RBSE Class 12 Biology Chapter 27 Essay Type Questions

Question 1.
Describe various eye diseases in detail.
Answer:
Common Eye Diseases:
The various eye diseases are as follows
1. Myopia:

It is an eye disease which may occur to any person at any age.
The person is unable to see distant objects clearly.
It is caused either due to increased focal power of the lens (lens become more convex) or eyeball becomes larger.
It is also called as short-sightedness. It can be corrected by use of biconcave lenses, which are also called as “minus lenses”.

2. Hypermetropia:

It is also called farsightedness.
The person is unable to see near objects clearly.
It is caused either reduced focal power of the lens (lens becomes less convex) or the eyeball becomes smaller.
It can be corrected by the uses of biconvex lenses which are also called as plus lenses.

3. Cataract:

It is an eye disease of old age. The eye lens gradually becomes opaque, as a result, the light rays fail to enter into the eye. Cataract disease is also known as salad motia.
Surgical removal of the lens and either implantation of the artificial lens or use of spectacles with the convex lens is a corrective measure.

4. Astigmatism:

It is due to the abnormal shape of the cornea, in which different curvature in different regions of cornea appears.
It can be corrected by using cylindrical lenses.
It can be corrected by the use of cylindrical lenses.

5. Conjunctivitis:

This eye disease is caused due to viral or bacterial infection.
There is inflammation & irritation in the conjunctiva. The conjunctiva also becomes red.

6. Colourblindness:
Colourblindness:

It is a hereditary eye disease. It was first described by Horner. It is of 2 types –
1. Red-green colourblindness
2. Blue colourblindness
The red-green colourblindness is most common & it is also called a proton defect. Red-green colourblindness is caused due to the absence of red-green cones in the retina.
The person is unable to identify red, green, yellow & orange colours. These colours appear to be green in the absence of red cones and they appear to be red in the absence of green cones.
It is a sex-linked disease which is caused due to a recessive gene on the x-chromosome. The mother (female) acts as a carrier. The father of a colourblind daughter is always colourblind.
In the F₁ generation of carrier mother & normal father, 50% generation will be normal, 25% will be carrier & 25% will be colourblind.
In F₁ generation of carrier mother and colourblind father, 50% of daughters will be colourblind & 50% will be carrier likely, 50% of boys will be normal & 50% will be colourblind.
In the F₁ generation of normal mother & colourblind father, all males will be normal and all daughters will be the carrier.

Question 2.
Describe in detail the mechanism of hearing.
Answer:
Mechanism of Hearing:

The pinna collects sound waves & transmit the waves into the external auditory meatus. These waves strike to the tympanum, as a result, the tympanum begins to vibrate.
The air imbalance caused in the middle ear due to vibrating tympanum is balanced by the air displaced through the eustachian tube.
The ear ossicles convey the sound vibrations to the fenestral ovals. They amplify the vibrations by 20 times.
The pressure imbalance caused by the vibrating fenestra ovalis membrane is balanced by the membrane of fenestra rotunda.
The vibrating membranes of the fenestra ovalis & fenestra rotunda cause the perilymph to vibrate which in turn makes the Reissner’s membrane to vibrate.
The vibrating Reissner’s membrane causes the endolymph in the scala media to move. As a result, the tectorial membrane moves & strikes to the organ of Corti.
Hence, the sound waves reach to the organ of Corti in the form of mechanical stimuli.
The organ of Corti converts these mechanical stimuli into electromagnetic waves and transmits these waves to the cerebellum through the VIII cranial nerve.

Question 3.
Describe the type of sense organs and describe them.
Answer:
Type Of Sense Organs:
There are many sense organs in the body to receive various stimuli.
Sherington classified the sense organs into three groups –

Exteroceptors:
Example: Tangoreceptors, Photoreceptors, Rheoreceptors, Algisoreceptors, Skin receptors etc.
Interoceptors:
Example: Sense organs found in the visceral organs
Proprioceptors:
Example: Sense organs found involuntary muscles, tendons, joints, ligaments etc.

Parker classified the sense organs into 3 groups on the basis of nature of stimuli –

Chemoreceptors
Example: Gustoreceptors, Olfactoreceptors
Mechanoreceptors
Example: Algisoreceptors, Tangoreceptors, Pressuroreceptors etc.
Radioreceptors:
Example: Thermoreceptors, Photoreceptors etc.

Both the exterior & interceptors remain interconnected by somatic & visceral sensory nerve fibres.
The receptors send the stimuli to the central nervous system.
Following sense, organs are found in human beings –

Photoreceptors	Eyes	Vision
Photoreceptors	Ears	Hearing
Tangoreceptors	Skin	Touch
Gustoreceptors	Tongue	Taste
Olfectoreceptors	Nose	Smell

Question 4.
Explain the structure of the inner ear.
Answer:
Internal Ear:

It is a semi-transparent structure which is also called the membranous labyrinth. It is situated in a bony labyrinth.
The internal ear is surrounded by perilymph.
It has 2 types of parts

1. Vestibule:

It has three semicircular canals which originate from utriculus at 90°. They are called anterior, posterior & external semi-circular canals.
The anterior & posterior semi-circular canals have a common origin which is called as crus commune.
The distal end of the semi-circular canal forms an ampulla. The ampullae of anterior & external semicircular canals remain closely situated.
The semi-circular canals & the ampullae are filled with a viscous fluid called endolymph.
Each ampulla has a sensory crista. The crista is lined by a sensory epithelium provided with stereocilia & kinocilia. The kinocilium is a movable flagellum.
The crista has an otoconium or otolith in the centre which is made up of CaCO₃. The otoconia move here & there in the endolymph. The cristae are sensory to the dynamic balance.
There is a sacculus near the utriculus which remains connected by a duct. Both the utriculus & sacculus has 1-1 macula which is sensory to static balance.
The sacculus gives out an endolymphatic duct. It remains closed in the mammals. In fishes, it opens outside the body.

2. Cochlea:

It is a spiral tube which originates from the sacculus. Both remain connected by a small duct called ductus reunions.
The cochlea has 2.5 coils in rabbit and 2.75 coils in man. The coils remain linked by a soft ligament.
Internally, the cochlea has 3 chambers viz. scala vestibule, scala media and scala tympani.
There is a basilar membrane between scala media & scala tympani. Similarly, there is a Reissner’s membrane between scala vestibule & scala media.
The scala vestibule & scala tympani remain united by a duct called helicotrema & they are full of perilymph fluid.
The scala media is full of endolymph fluid.

The basilar membrane modifies to form an organ of Corti in the scala media. The organ of Corti has three types of cells viz., Deiter’s cells., Hensen’s cells & pillar cells. These cells are provided with stereocilia & their nerve fibres form VIII cranial nerve.
There is a tectorial membrane just above the organ of Corti in the scala media which is formed by the modification of the Reissner’s membrane.
The organ of Corti is sensory to hearing.

Question 5.
Describe organs of taste and olfactory organs.
Answer:
Tongue & Nose:

The tongue gives the sense of taste and nose gives a sense of smell. Both these are chemical senses and they depend on the nature of a chemical substance.
Sense of taste is perceived by chemoreceptors present on the tongue. The human tongue has about 10,000 taste buds present on the top and sides of the tongue.
Odours are detected by olfactory hairs (sensory hair) that emerge from the receptors. When molecules of chemicals enter to nose with air. they stimulate sensory epithelium of nose, which send impulses to the olfactory region of the brain.

Question 6.
Describe the structure of the eye with diagrams.
Answer:
Structure of Eye Orbits:

Each eye is attached movably with the help of 6 eye muscles in the eye orbit.
These eye muscles are voluntary & striated.
There are 4 rectus & 2 oblique eye muscles which are as follows –

1.Anterior or external rectus (Lateral Rectus):

It is attached to the outer lateral side of the eyeball.
It receives Abducens cranial nerve.

2. Posterior or internal rectus (Medial Rectus):

It is attached to the inner (medial) side of the eyeball.
Oculomotor cranial nerve innervates it.

3. Superior rectus:

It is attached to the mid-dorsal side of the eyeball.
It receives an oculomotor cranial nerve.

4. Inferior rectus:

It is attached to the midventral side of the eyeball.
It receives an oculomotor cranial nerve.

5. Superior oblique muscle:

It is attached between the external rectus & superior rectus.
It receives a trochlear cranial nerve.

6. Inferior oblique muscle:

It is attached between the inferior rectus and external rectus.
It receives an oculomotor cranial nerve.

Anterior rectus & posterior rectus muscles rotate the eyeball inside & outside.
Superior rectus & inferior rectus rotate the eyeballs upside & downside.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 26 Man-Nervous System

August 17, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 26 Man-Nervous System

RBSE Class 12 Biology Chapter 26 Multiple Choice Questions

Question 1.
What is the unit of the nervous system?
(a) Nephron
(b) Neuron
(c) Brain
(d) Spinal cord
Answer:
(b) Neuron

Question 2.
Temperature controlling centre in the brain of man is:
(a) Pituitary gland
(b) Diencephalon
(c) Hypothalamus
(d) None of the above
Answer:
(c) Hypothalamus

Question 3.
A function of the parasympathetic nervous system is:
(a) Dilation of pupil
(b) Secretion of sugar in the liver
(c) Increase heartbeat
(d) Stimulate secretion of saliva
Answer:
(d) Stimulate secretion of saliva

Question 4.
The scientist who worked on conditioned reflex action:
(a) Mendel
(b) Pavlov
(c) Darwin
(d) Evan Wilmut
Answer:
(b) Pavlov

Question 5.
Number of cranial nerves present in Man are:
(a) 10 Pairs
(b) 10
(c) 12 Pairs
(d) 12
Answer:
(c) 12 Pairs

RBSE Class 12 Biology Chapter 26 Very Short Answer Type Questions

Question 1.
Name the found in brain.
Answer:
Brain Ventricles.

Question 2.
How many cranial nerves are found is a man?
Answer:
12 Pairs.

Question 3.
Which is the longest cell?
Answer:
Nerve cell (Neuron)

RBSE Class 12 Biology Chapter 26 Short Answer Type Questions

Question 1.
Which are two parts of the autonomous nervous system?
Answer:

Sympathetic Nervous System
Parasympathetic Nervous System

Question 2.
Why the name peripheral nervous system is given?
Answer:
Because the peripheral Nervous System originates from the central nervous system and it innervates all body organs.

Question 3.
Write one function of the following:

Cerebrum
Cerebellum
Medulla oblongata
Hypothalamus

Answer:

Cerebrum: It controls thinking, Argument, Planning, Memory etc.
Cerebellum: It controls body balance & Muscles coordination.
Medulla oblongata: It controls most of the involuntary activities of the body such as heartbeats, breathing, BP etc.
Hypothalamus: It has centres for hunger, satiety, thirst, temperature control etc.

Question 4.
Write the main parts of the brain.
Answer:

Forebrain: Cerebrum, Diencephalon
Midbrain: 4 optic lobes
Hindbrain: Cerebellum, Pons, Medulla

Question 5.
Write the name of the fluid-filled in the cavities of the brain.
Answer:
Cerebro – Spinal Fluid (CSF)

RBSE Class 12 Biology Chapter 26 Essay Type Questions

Question 1.
Explain simple reflex action (arch) with the help of well-labelled diagram depicting a transverse section of the spinal cord and nervous pathway of reflex action.
Answer:
Spinal Reflex Arch:

The nerve pathway involved in a reflex action is called a reflex arch. It includes the following 5 parts –
1. Sensory organ: Receptors on the body which receive an external or internal stimulus.
2. Sensory Nerve: Sensory nerve carries impulse from sensory organs to the spinal cord.
3. The nerve centre of spinal cord or Interneurons Centre of actions.
4. Motor nerve: Transmit, impulse away toward the spinal cord.
5. Effector organs: These organs complete their actions with a response to impulses received through a motor nerve.

Question 2.
Describe the structure of the brain of a man with a diagram.
Answer:
Structure of Brain:
The human brain is divisible into 3 parts –

Forebrain
Cerebellum
Brain stem

1. Forebrain:
In human beings, the forebrain includes two parts –

Cerebral hemispheres or cerebrum
Diencephalon

2. Cerebrum:

It forms 2/3 part of the brain and a longitudinal fissure divides it into two hemispheres.
Both the hemispheres remain connected internally by the corpus callosum.
Each cerebral hemisphere consists of five lobes –
1. Frontal lobe
2. Parietal lobe
3. Temporal lobe
4. Occipital lobe
5. Insula lobe
The insula lobe is not visible externally.
The frontal lobe is the largest. A central sulcus separates the frontal and parental lobes which are also called as the fissure of Ronaldo. Similarly, a lateral cerebral fissure separates the frontal and the temporal labels which are also called as the fissure of Sylvius.
The parental and the occipital lobes remain separated by a parieto-occipital fissure. The outer 2 to 4 mm thick part of the cerebrum is called cortex or grey matter. It consists of 10% of the total neurons of the brain. The cortex exhibits distinct gyri (ridges) and sulci (furrows).
These gyri and sulci increase the surface area of the cortex by 3-times. The inner mass of the cerebrum is called as white matter.

Functions of the cerebrum:

Frontal lobes: They control memory, reasoning, thinking, determination, emotions, judgment, planning, experience, concentration, will power and personality. They also control voluntary muscles and interpret many sensations.
Parietal lobes: They interpret sensory perception of touch, pain, cold, heat etc. They also control speech and handicraft. They also associated with the understanding of the thoughts.
Temporal lobes: They interpret and decode sound. In addition to language comprehension, they also interpret smell and auditory sensations.
Occipital lobes: The interpret and decode visual sensations particularly sensations concern with colours and shapes. They also control eye movements.
Insula lobes: They coordinate memory and other parts of the cerebrum.

2. Diencephalon:

It is divisible distinctly into epithalamus, thalamus and hypothalamus.
Thalamus forms 80% of the diencephalon and it functions as relay centre of the sensory perception of taste, touch, hearing, vision, pain, heat etc. It interprets these sensations and sends motor drives concerning to the voluntary movement.
The hypothalamus includes about 12 large nuclei and it is divisible into four parts –
1. Supraoptic part: It is situated just above the optic chiasma. Its axons enter into the pars nervosa of the pituitary gland.
2. Tuberal part: It is the broadest and the middle part. It is attached with the pituitary through a stalk called infundibulum.
3. It has a projection, the tuber cinerarium around the infundibulum.
4. Mammillary part: It is the posteriormost part which bears two small and round projections, the mamillary bodies. It controls reflex actions concern to smell.
5. Preoptic part: It is a small part which is situated anterior to the supraoptic part. It controls some autonomic reflex actions.

The function of the hypothalamus:

It controls coordination between psyche & body and neurons & endocrine systems.
It has centres of hunger, thirst, satiety, and temperature control.
It includes nuclei secreting oxytocin. and vasopressin hormones.
It controls a diurnal rhythm and sleep.

2. MidBrain:
It is the middle part of the brain and also called as mesencephalon. It consists of two parts –

1. Corpora Quadri-gamine:

It includes four optic lobes which are collectively called as Corpora Qudrigemina.
Out of 4 lobes, two lobes are superior optic lobes and two are inferior optic lobes.
The optic lobes remain joined to the dorsal surface of the midbrain with the help of Tatum.
The optic lobes receive stimuli related to vision & hearing.

2. Cerebral peduncle:

It is also called as Crura Cerebri.
It is a bundle of nerve fibres which connects the cerebral cortex to the spinal cord & other parts of the brain.

3. HindBrain:

It is posterior or terminal part of the brain.
It consists of three parts:
1. Cerebellum
2. Pons Varolli
3. Medulla Oblongata

1. Cerebellum: It is the second largest. It part consists of two lobes. Dorsal side of cerebellum is well developed. Its two main functions are –

Maintenance of posture or equilibrium.
Coordination of muscular activities.

2. Pons Varolli: It lies just above medulla oblongata and below cerebral peduncle. It joins both lobes of the cerebellum and also other parts of the brain.

It is made up of white matter.
It contains Pneumotaxis centre for regulation of breathing. It also controls chewing, saliva secretion, the release of tears and eye movements.

3. Medulla Oblongata: It is the posterior-most part of the brain, It is triangular and situated between the pons and spinal cord. Its cavity is called metachronal and medulla continues as the spinal cord. Functions of medulla oblongata are:

It has controlling centres of breathing, coughing, swallowing etc.
It has controlling centres for heartbeats, blood pressure, peristalsis in alimentary canal & other involuntary activities.

4. Brain Ventricles:

The mammalian brain is hollow and the cavities are called ventricles. These ventricles are lined by ependymal epithelium and are full of cerebrospinal fluid (C.S.F.).
The brain has 4 ventricles viz., I, II, III & IV. I & II ventricles are found in the cerebral hemispheres and are also called as paracoels.
Each paracoels forms a rhinocoel in the olfactory lobe which is also called as zero ventricles.
Both the paracoels open into a third ventricle through a foramen of Monro.
III ventricle is also called as dioctyl & it is found in the roof of the diencephalon.
The 3rd ventricle opens into IV ventricle through a narrow duct, the iter or aqueduct of Sylvius. The item is found in the cerebellum.
IV ventricle is also called as metachronal & it is found in the medulla.
There are two choroid plexus in the ventricles viz.,
1. Anterior choroid plus: It is found in the roof of the diaconal.
2. Posterior choroid plus: It is found in the roof of the metacoel. The choroid plus consists of pia mater & the network of blood capallaries. The anterior choroid plus secretes CSF and the posterior choroid plus absorb CSF & secrete it into the subarachnoid space.
There are three apertures in the posterior choroid plus to absorb the CSF viz.,
1. Foreman of Magendie – One
2. Foramen of Luschka – two
There are arachnoid villi in the subarachnoid space. These villi absorb CSF & return it into the venous blood.

Question 3.
Describe the cranial nerves of mammals.
Answer:

Various vital activities of the body are controlled & coordinated by two systems –
1. Nervous System
2. Endocrine System
This chapter deals with the study of the Nervous System.
The nervous system is divided into 3 heads –
1. Central nervous system: It includes the brain & spinal cord.
2. Peripheral nervous system: It includes cranial & spinal nerves.
3. Autonomous nervous system: It includes the sympathetic & parasympathetic nervous system.
A nerve cell or neuron is the structural and functional unit of the Nervous System.
The space between the neurons is occupied by neuroglial cells in the nervous system.

Dendron	Axon
One or two or absent	Always one
Branched near the cyton	Branched distally & the branches are called as telocentric
Broad at the base & narrow at the ends	Almost uniform in diameter
Presence of Nissl’s granules	Absent
Carry nerve impulse towards the cyton	Carry impulse away from the cyton
Presence of cytoplasmic organelles in the basal part	Absence of cytoplasmic organelles

The cyton has a large & round nucleus, minute neurofibrils, cytoplasmic organelles & many Nissl’s granules. The Nissl’s granules are ribosomes & made up of RNA & proteins.
However, the centrosome is absent. Hence, the nerve cell never divides.
Nissl’s granules are the groups of lamellae which are attached with ribosomes. They synthesize proteins actively.
The nerve cells increase in size along with body growth.
The neurons are of 3 types on the basis of cell processes viz.,
- Unipolar:
  - It has only one axon.
  - They are found in invertebrates & vertebrate embryos.
- Bipolar:
  - It has one axon & one dendron.
  - They are found in eyes, cochlea & olfactory epithelium.
- Multipolar:
  - It has one axon & many dendrites
  - Most of the vertebrates have multipolar neurons.
On the basis of the structure of axon, the neurons are of two types
- Myelinated or Medullated: The axon is bounded by a myelin sheath
- Non-myelinated or Non-medullated: The axon is not bounded by a myelin sheath.

Question 4.
Describe reflex action.
Answer:
Reflex Action:

A reflex action is a rapid automatic response to a stimulus without conscious thought in the body. The response is instantaneous and involuntary.
There is no control of brain on reflex actions.
For example:
1. A person withdraws hand instantly on touch to a pointed pin or hot utensil.
2. Salivation in the mouth on right or smell or thought of good favourite food.
Reflex actions are to two types:
1. Simple or unconditional reflex action
2. Acquired or conditional reflex action

1. Unconditioned Reflex action:

They are natural or congenital.
Learning is not required.
Examples:
1. Instantly closing of eyelids when any object comes close.
2. Instantly coughing when any food particles/any liquid enters into the trachea.
3. Contraction of a pupil in strong light.

2. Conditioned or Acquired Reflex action:

This reflex action needs learning training or experience.
Russian Scientist of physiology Evan Pavlov explained it by his experiment on the dog.
Experiment on Dog – Pavlov starts ringing a bell just before the food is given to Dog. Later on, he observed that there is salivation in Dog’s mouth on ringing bell even without giving food.
Examples:
1. We instantly brake our vehicle when any object comes in front of it all of sudden.
2. Tying of shoelaces even though we are talking to somebody else.
3. Running away of dog, when we just bend down towards the ground as dog pretends that we are picking a stone to throw on him.
4. Standing up of student as soon as a teacher enters into class.

Mechanism of Reflex Action:

Fibres of the spinal cord and motor fibre of ventral root play a special role in a reflex action.
For example, pinprick on the skin stimulates sensations. Which stimulate dendrites of somatic sensory fibres. These fibres transmit stimulus to the neurons in the dorsal ganglia of the spinal nerve.
Terminal node of these cells sends stimulus to dendrites of nearest motor nerves. Here sensory induction becomes motor stimulus.
Axons of motor cells are fibres of ventral root and they carry stimulus up to legs, where contraction in muscles and produce movement of legs.
These actions are very fast and in total it is called reflex action.

Spinal Reflex Arch:

The nerve pathway involved in a reflex action is called a reflex arch.
It includes the following 5 parts –
1. Sensory organ: Receptors on the body which receive an external or internal stimulus.
2. Sensory Nerve: Sensory nerve carries impulse from sensory organs to the spinal cord.
3. The nerve centre of spinal cord or Interneurons Centre of actions.
4. Motor nerve: Transmit, impulse away toward the spinal cord.
5. Effector organs: These organs complete their actions with a response to impulses received through the motor nerve.

Question 5.
Describe spinal nerves in detail.
Answer:
Spinal Nerves:

These nerves arise from the spinal cord.
There are 31 pairs of spinal nerves.
All spinal nerves are of mixed type as they contain both sensory and motor neurons.
Every spinal nerve passes out through every two vertebrae of the vertebral column through an intervertebral foramen.
On the basis of their origin, the spinal nerves are named as follows below.

Description of Spinal nerves found in Man
Name of Nerves	Numbers	Location
Cervical spinal nerves	8 Pairs	1st vertebrae to below the cervical vertebrae
Thoracic spinal nerves	12 Pairs	Below each thoracic vertebrae on both side
Lumbar spinal nerves	5 Pairs	Below each lumbar vertebra on both sides
Sacral spinal nerves	5 Pairs	Near the second cervical level of the spinal cord
Coccygeal Spinal nerves	1 Pair	In the lower part of sacral plexus.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 25 Man-Excretory System

August 16, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 25 Man-Excretory System

RBSE Class 12 Biology Chapter 25 Multiple Choice Questions

Question 1.
The main excretory substance in human is:
(a) Uric acid
(b) Ammonia
(c) Urea
(d) Amino acid
Answer:
(c) Urea

Question 2.
The main excretory organ of human is:
(a) Lungs
(b) Kidney
(c) Skin
(d) Liver
Answer:
(b) Kidney

Question 3.
Henle’s loop contains:
(a) Urine
(b) Urea
(c) Blood
(d) Glomerular filtrate
Answer:
(d) Glomerular filtrate

Question 4.
A renal column of Bertini is related to:
(a) Kidney
(b) Urinary bladder
(c) Liver
(d) Testis
Answer:
(a) Kidney

Question 5.
Human kidneys are:
(a) Pronephric
(b) Metanephric
(c) Mesonephric
(d) All of the above
Answer:
(b) Metanephric

Question 6.
Ultrafiltration occurs in:
(a) Glomerulus
(b) Bowman’s capsule
(c) Urinary bladder
(d) Blood vessels
Answer:
(a) Glomerulus

Question 7.
The glomerular filtrate is:
(a) Water, ammonia and blood corpuscles
(b) Blood without blood corpuscles and plasma protein
(c) Blood without blood corpuscles
(d) Urine
Answer:
(b) Blood without blood corpuscles and plasma protein

Question 8.
The blood vessel which brings blood into the glomerulus:
(a) Efferent arteriole
(b) Renal artery
(c) Afferent arteriole
(d) Renal vein
Answer:
(c) Afferent arteriole

RBSE Class 12 Biology Chapter 25 Very Short Answer Type Questions

Question 1.
What is Excretion?
Answer:
Removal of metabolic waste products from the body is called excretion.

Question 2.
An animal which excretes ammonia is called?
Answer:
Ammonotelic.

Question 3.
Animals which excrete uric acid are called?
Answer:
Uricotelic.

Question 4.
What is Ultrafiltration?
Answer:
The process of filtration of blood from the glomerulus into the Bowman’s capsule because of increased glomerular blood pressure is called as Ultrafiltration.

Question 5.
Write the name of the excretory unit in the kidney.
Answer:
Nephron or uriniferous tubule.

Question 6.
Name the blood vessel coming out of the glomerulus.
Answer:
Efferent arteriole.

Question 7.
What is the term used for pain at the time of micturition?
Answer:
Dysuria.

Question 8.
Where the glomerulus is found? What is its role?
Answer:
The glomerulus is enclosed in Bowman’s capsule. It performs Ultrafiltration.

Question 9.
What is the malpighian body?
Answer:
The anterior part of the nephron is called a Malpighian body. It includes glomerulus and Bowman’s capsule.

Question 10.
Where the Henle’s loop is found?
Answer:
It is a U-shaped part of Nephron.

Question 11.
What are Renal columns of Bertin?
Answer:
The pushing of the cortex part of the kidney into the medulla is termed as a renal column of Bertin.

Question 12.
What is the main role of kidneys in the body?
Answer:
Removal of nitrogenous waste product (urea) from the body.

Question 13.
What is haemodialysis?
Answer:
Removal urea and other waste products from the blood using artificial kidney (Dialyser) are termed as haemodialysis.

Question 14.
Presence of Urea in the blood is called?
Answer:
Urania.

Question 15.
What is glycosuria?
Answer:
Presence of glucose in the urine is called glycosuria.

RBSE Class 12 Biology Chapter 25 Short Answer Type Questions

Question 1.
Describe the excretory organ in the human body other than kidneys.
Answer:
Other Excretory Organs in Man:

Skin: Sweat glands are found in the human skin, which secretes out sweat containing water and some nitrogenous excretory substances.
Lungs: Cellular respiration release CO₂as excretory substance, which is eliminated outside by respiratory process in the lungs.
Liver: Liver cells convert nitrogenous part of amino acids into ammonia and then ammonia into urea. The urea is released into the blood. The liver also forms bile pigments which are excreted along with the bile juice into the intestine.

Question 2.
What is Gout disease?
Answer:
Gout:

It is a hereditary disease in which blood uric acid is increased. It gets deposited in the synovial joints and in Kidney tissues. This disease may cause due to dehydration, fasting and diuresis.

Question 3.
Explain Bright’s disease.
Answer:
Bright’s disease or Nephritis:

This disease is caused due to infection of Streptococci bacteria in the glomeruli. As a result, they swell and their membrane becomes more permeable to RBC’s and proteins which filtered out and appears in the filtrate. If nephritis is not cured than fluid deposition occurs and the condition of swollen legs happens, which is called Edema or Dropsy.

Question 4.
What is the role of ultrafiltration and selective reabsorption in urine formation?
Answer:
Ultrafiltration:

It is a passive process which takes place from the glomerulus into the Bowman’s Capsule The glomerular epithelium has various micropores (diameter = 0.1μ) which increase the rate of filtration.
The ultrafiltration takes place due to increased blood pressure in the glomerulus which is increased due to the difference in the diameters of afferent & efferent arterioles.
There is ultrafiltration of all the components of the blood except the blood corpuscles, plasma proteins macromolecular foreign substance. The ultrafiltration forms glomerular filtrate in the Bowman’s capsule.
The glomerular filtrate is plasma without plasma protein.
Glomerular filtrate ⇒ Blood – Blood Cells + Plasma Proteins
Glomerular filtrate ⇒ Plasma – Proteins
The hydrostatic pressure in the glomerulus (GHP) is 65 to 75 mm of Hg. Out of it, the blood colloidal osmotic pressure (BCOP) due to plasma proteins is 30 mm of Hg.
On the contrary, the capsular hydrostatic pressure (CHP) is 20 mm of Hg. Hence, the net Effective glomerular filtration pressure (EFP) is 15 to 25 mm of Hg which is responsible for ultrafiltration.
In human beings, the glomerular filtration rate (GFR) is 125 ml per minute i.e. both the kidneys form 125 ml of filtrate per minute.
Hence, 180 litres of filtrate is formed per day. Out of it, only 1.5 litres of urine is produced per day which is 0.8% of the total filtrate.

Reabsorption:

In the renal tubules, 99.2% of the glomerular filtrate is reabsorbed into the blood.
The reabsorption takes place by active transport and most of the reabsorption takes place in the PCT. However, water is absorbed by passive transport.
The reabsorption of the substance is according to the renal threshold value. The maximum capacity of both the kidneys to absorb a particular substance is called a renal threshold of the substance. The renal threshold value of the substances is according to their requirement in the body.
The substances are of 3 types on the basis of their renal threshold value –
1. High threshold substances:
  - Such substances are absorbed almost all.
  - Example: Sugar, amino acids, vitamins etc.
2. Low threshold substances:
  - They are absorbed in low concentration.
  - Example: Urea, creatine, creatinine, phosphate etc.
3. A threshold substances:
  - They are not absorbed.
  - Example: Uric acid etc.
If any substance exceeds its renal threshold value, it will appear in the urine.
The renal threshold value of glucose is 350 mg per minute which is also called as TmG.
The reabsorption of the substances is under the control of hormones which are as follows:
1. Glucocorticoids:
  - They are secreted by adrenal cortex & they control reabsorption of sugars, amino acids & vitamins in the PCT.
2. Mineralocorticoids:
  - They are secreted by the adrenal cortex and they control reabsorption of mineral ions in the PCT
3. Calcitonin & Parathormone:
  - They control the absorption of Cat+ in the DCT.
4. Vasopressin or ADH:
  - It is a hormone of the Pituitary gland which controls the absorption of water in the DCT. Most of the water (about 80%) is absorbed in the PCT along with the substances.
  - It is called as obligatory water reabsorption. Remaining water (about 20°.) is absorbed in the DCT under the control of ADH according to the body requirement. It is called facultative water reabsorption.
Excess release of the ADH result in a decreased volume of the urine

Question 5.
What do you mean by kidney transplantation? Explain in brief.
Answer:
Kidney Transplantation:
In a human being when kidneys stop functioning and are not curable than a kidney of other healthy person is to be transplanted. The Kidney is obtained from a donor who may be decreased or live. Kidney donor should be the nearest relative whose blood and tissue structures are to be similar.

If unmatched kidney transplantation is done than the immune system of patient will not accept the transplanted kidney and it will not work properly so the patient may die. Although special medicines are given to a patient to inactivate the immune system and the possibility to accept the kidney of the donor is increased.

RBSE Class 12 Biology Chapter 25 Essay Type Questions

Question 1.
Explain the excretory system in man by giving a suitable diagram.
Answer:
Excretory System of Man:

Kidneys are the main excretory organs in man. In addition, the excretory system also includes ureters, urinary bladder and urethra.

Kidney:

In man, the kidneys are metanephric which originate from mesoderm.
There is a pair of kidneys which are attached with the help of peritoneum in the dorsal side of the middle part of the coelom. The peritoneum is found only on the ventral side. Such kidneys are called retroperitoneal.
The kidneys are dark red in colour. They are unequally situated. In human beings, the right kidney is somewhat posteriorly situated.
The weight of each kidney is 120 – 170 gm.
The kidney is bean-shaped i.e., concavo-convex. The concave inner surface is called as hilus which gives out a ureter.
From this hilus surface, the renal artery enters into the kidney, the renal vein comes out and the renal nerves enter into the kidney.
The dimensional of a human kidney is 4″ × 2.5″ × 1″ inch.
An adrenal gland is attached to the anterior end of each kidney like a cap.

Ureters:

The inner hilum (hilus) surface of the kidney gives out a ureter.
The anterior end of the ureter is funnel-shaped which is called as the pelvis.
Both the ureters open separately into the urinary bladder.
The wall of the ureter is thick and muscular. Its peristalsis movement helps to move the urine.

Urinary Bladder:

The urinary bladder is baglike which is made up of smooth & involuntary muscles. The lumen of the urinary bladder is lined by transitional epithelium which has great power of stretching. Hence, the urinary bladder can store more urine.
The urinary bladder leads into male urethra & its opening is regulated by a sphincter. In human beings, the sphincter is made up of voluntary muscles. The sphincter is relaxed only at the time of micturition.
The act of passing urine is called micturition which is a partly voluntary process in human beings.
In a male, the urethra leads outside through the penis and in female through the vulva.
In woman urethra is absent. Hence, the urinary bladder opens outside directly through the urinary orifice. In man, the urethra can be divided into three parts –
1. Prostatic or Urethral Part: Anterior most part having length 2.5 cm. It receives prostate gland, urinary bladder and both the vas deferens.
2. Membranous part: It is the middle part which receives the Cowper’s glands.
3. Penile part: It is situated inside the corpus sporangium of the penis. Its length is about 15 cm.

Question 2.
Describe the functional structure of Uriniferous tubule.
Answer:
Structure of Nephron:

It is a unit of excretion in which urine is formed independently.
Its an anteriormost part is called a Malpighian body which includes a cup-like Bowmann’s capsule & a glomerulus.
The glomerulus is a bunch of blood capillaries. It receives the blood through an afferent arteriole and the blood comes out through the efferent arteriole. The diameter of the afferent is comparatively more.
The Bowman’s capsule opens into a proximal convoluted tubule (PCT) through a short & straight neck. The anterior part of the PCT is more coiled whereas its posterior part is almost straight. The PCT opens into a Henle’s loop.
The Henle’s loop is a U-shaped structure which has a distinct descending limb & an ascending limb. The ascending limb opens into the DCT.
The descending convoluted tubule (DCT) is a coiled duct.
Many DCT unites to form a collecting duct. The collecting ducts of one pyramid unite to form a duct of Bellini. The ducts of Bellini lead into the pelvis part.
The Malpighian body and a part of PCT & DCT are situated in the cortex. Most of the part of PCT & DCT, Henle’s loop and collecting ducts are found in the medulla.
The efferent arteriole forms a peri-tubular capillary network around the PCT, DCT & Henle’s loop which is called as Vasa recta.
The capillaries of Vasa recta join to form venules which finally open into the renal vein.

Histology of Nephron:

The wall of the Bowman’s capsule is made up of simple squamous epithelium.
Its inner wall has special cells, called podocytes.
The podocytes have finger-like processes which bind the glomerular-blood capillaries.
The neck of the nephron & collecting duct is made up the ciliated epithelium.
PCT is made up of simple cuboidal epithelium. It has brush-border of microvilli.
The Henle’s loop is made up of simple squamous epithelium.
DCT is made up of simple cuboidal epithelium. It is without microvilli.

Question 3.
Explain other excretory organs in Man.
Answer:
Other Excretory Organs in Man:

Skin: Sweat glands are found in the human skin, which secretes out sweat containing water and some nitrogenous excretory substances.
Lungs: Cellular respiration release CO₂as excretory substance, which is eliminated outside by respiratory process in the lungs.
Liver: Liver cells convert nitrogenous part of amino acids into ammonia and then ammonia into urea. The urea is released into the blood. The liver also forms bile pigments which are excreted along with the bile juice into the intestine.

Question 4.
Describe various diseases associated with the excretory system.
Answer:
Disorders related to Excretion:

Irregularities in excretion may cause various diseases in human. Some of the diseases are as follows:

Uremia: When the amount of urea in the blood becomes more than 10 – 30 mg/100 ml, the condition is called Uremia. Excess amount of urea in blood is harmful and may be fatal.
Gout: It is a hereditary disease in which blood uric acid is increased. It gets deposited in the synovial joints and in Kidney tissues. This disease may cause due to dehydration, fasting and diuresis.
Kidney stones: Normally crystals of substance like calcium oxalate, phosphate, uric acid etc get deposited in the renal pelvis as renal stones. They cause pain & difficulty in micturition.
Bright’s disease or Nephritis: This disease is caused due to infection of Streptococci bacteria in the glomeruli. As a result, they swell and their membrane becomes more permeable to RBC’s and proteins which filtered out and appears in the filtrate. If nephritis is not cured than fluid deposition occurs and the condition of swollen legs happens, which is called Edema or Dropsy.
Glycosuria: Presence or excretion of glucose in the urines is called Glycosuria. It is caused due to deficiency of insulin hormone. This disease is called diabetes mellitus.
Dysuria: Pain at the time of micturition is called dysuria.
Polyurea: The increased amount of urine due to less absorption of water is called as polyurea.
Cystitis: The swelling of the urinary bladder by the infection of bacteria, chemical or mechanical damage is called cystitis.
Diabetes Insipidus: Due to hyposecretion of antidiuretic hormone (ADH), the water absorption does not occur in the distal convoluted tubules and collecting tubule results into an increase in the volume of Urine. There is frequent and excess micturition.
Oliguria: Formation of less quantity of urine as compared to normal.
Proteinuria: Presence of more protein in urine is called proteinuria.
Albuminuria: Presence of more amount of albumin protein in the Urine.
Ketonurea: Increase the number of ketone bodies in urine such as acetone acetic acid.
Haematonurea: Elimination of red blood corpuscles (RBC’s) along with urine is called haematuria.
Haemoglobuinurea: Presence of haemoglobin in urine is called haemoglobinuria.
Pyuria: Presence of pus cells in urine is called pyuria.
Jaundice: Presence of bile pigments in huge amount in urine is a symptom of jaundice. It is seen during hepatitis or due to blockage of the bile duct.
Alkaptonuria: Presence of alkaline or homogentisic acid is termed alkaptonuria. When alcapton comes in contact with air than urine becomes black in colour, it is also known as Black urine disease.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 38 Human Population

August 16, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 38 Human Population

RBSE Class 12 Biology Chapter 38 Multiple Choice Questions

Question 1.
Who has first of all given the “Theory of Population” in context to the human population.
(a) Malthus
(b) Lamarck
(c) Bodenhamer
(d) Darwin
Answer:
(a) Malthus

Question 2.
What percentage of the population is contributed by India globally?
(a) 12.4%
(b) 17.85%
(c) 16.2%
(d) 15.13 %
Answer:
(b) 17.85%

Question 3.
The main reason for the increasing world population is –
(a) Increased Natality
(b) Improvement in living standard
(c) Global warming
(d) Decreased Mortality
Answer:
(d) Decreased Mortality

Question 4.
If there are a number of old persons in population it becomes –
(a) Increase in future
(b) Decrease in future
(c) Stationary
(d) None of the above
Answer:
(b) Decrease in future

Question 5.
Which is not a contraceptive?
(a) Condom
(b) Saheli pill
(c) Vault
(d) Steroid pill
Answer:
(d) Steroid pill

RBSE Class 12 Biology Chapter 38 Very Short Answer Type Questions

Question 1.
Define Population.
Answer:
A group of animals of a species at a specific place and at a specific time is called a population.

Question 2.
What is the population growth curve?
Answer:
Population growth in animals exhibits a fixed pattern which can be exhibited by a population growth graph. The population growth graph is drawn between population size and time.

Question 3.
Write anyone important factor for decreased mortality in India?
Answer:
Advanced medical facilities and time bond vaccination.

Question 4.
What will happen if the population if natality and mortality rate will be equal?
Answer:
The size of the population of that species will be stationary.

Question 5.
What is IUD in contraception devices?
Answer:
Intra-Uterine Device (IUD):

Intra-uterine devices are small devices that are inserted into the uterus by doctors.
Examples:
1. Non-Medicated IUD – Lippes Loop.
2. Copper Releasing IUD – Cut, Cu7, Multiload 375 etc.
3. Hormone Releasing IUD – LNG 20, Protestant etc.
These devices promote phagocytosis of sperms to prevent fertilization. They are used to delay pregnancy. Cut, Cu7 are devices containing copper. The release of copper ions inhibits the motility of sperms & fertilisation. These are considered as ideal contraceptives for females. IUD hormones prevent implantation in the uterus two children IUD is an ideal method. It is widely used in India.

Question 6.
What are contraceptive pills?
Answer:
Contraceptive Pills:

These Pills are used as oral contraceptives. These pills contain small doses of estrogen or progesterone or a combination of both. These oral contraceptives should be taken for a period of 21 days.
After a gap of 7 days, that is, at the time of menstruation, these pills should be avoided. After this gap, these pills should be taken again in the same pattern until and unless female desires conception or fertilization.
Progesterone or combination with estrogen may also be given to females via injections. They work similarly to orally taken pills.
Saheli is a new oral contraceptive without any steroids hormone. It contains a chemical called centchroman. It is a highly effective contraceptive pill.
Emergency pills Contain a chemical called Levenorgastril. This pill is to be taken after forced sex or unplanned sex within 72 hours. It prevents ovulation. It is also called a pill – 72.

Question 7.
What is Vasectomy?
Answer:
Vasectomy:

Vasectomy is one of the methods used in males. In this method, a small incision is made and vas deferens get tied.
It is commonly known as “Male Nasbandi”.

Question 8.
What is Tubectomy?
Answer:
Tubectomy:

Similarly, in females, Tubectomy is done. In tubectomy, a small part of Fallopian tube is removed via an incision or it gets tied.
This techniques are effective and have reversibility. These techniques are commonly known as “female Nasbandhi”.

RBSE Class 12 Biology Chapter 38 Short Answer Type Questions

Question 1.
Write the difference between mortality and natality.
Answer:
Natality or Fertility:

It determines the growth of any population. In a population, the ability to produce a new baby from active reproducing organisms is called fertility. In growing, population natality is one of the important factors.
The birth rate is the number of births per 1000 individuals per year. It can be calculated by the following formula –
N = \(\frac { B }{ t } \)
where N = Natality/Birth rate
B = Birth of children per 1000 people
t = Time period in a year
Bodenheimer divided the human population into 3 groups –
1. Pre-reproduction population: Population below 18 years of age.
2. Reproductive population: Population between 18 to 45 years of age.
3. Post-reproductive population: Population above 45 years of age.
A number of the reproductive population indicates an increase in population growth. Similarly, number post-reproductive population indicates a decrease in the population growth.

Mortality:

Mortality or Death rate is the number of deaths per thousand persons in a year.
M = \(\frac { D }{ t } \)
where M = mortality, D = Number of death in one thousand persons and t = time in years.
In many countries, a decline in death rates is recorded which resulted in an increase in population. This could be because of the following reasons:
1. Developed medical facility including Immunization vaccination.
2. Level of cleanliness has increased.
3. Awareness towards personal health & cleanliness has increased.
4. Availability of food and water is in a sufficient amount.
Mortality and Natality are the important factors which determine the population. In other words, using mortality and natality rate, population pattern of any country can be determined.
1. Crude Birth Rate: The crude birth rate is the rate of births among a population of 1000.
2. Crude death Rate: It is the rate of death among a population of 1000.
3. Annual Rate of Natural Increase/Growth:
  1. Difference between mortality and natality is called Annual growth rate.
  2. Annual growth rate = Natality/Birth rate – Mortality rate.

If mortality and natality are equal, population shape is stationary means the growth rate of the population will be zero. However, it occurs very rarely. It can be seen in some developed society.

Question 2.
Define population density. Write the methods to determine it.
Answer:
Population Density:

Population density is a measurement of the population of organisms in per unit area or unit volume. Population density is a numerical expression.
For Example, 100 trees per acre or 200 fishes per km or 50 billion diatoms m³.
It shows the total number of individuals of specific species lived in a habitat. It can be calculated using the formula –

where d = density, N = No. of organisms, a = Area, t = Time

The population is changing from time to time i.e. it is not permanent. Change in population causes a change in density.

Types of Population Density:
Population Density is of two types:

Crude Density:
Crude density includes all the organisms within the land range. For example, if 1000 rabbit lived in 1 mile than population density will be 1000 rabbit per mile. It is calculated in per mile, per acre, or per meter. It is also known as Habitat density.
Ecological Density:
Ecological density includes only that portion of land which is actually colonized by the species. For example, if 500 rabbits are lived in the one-mile area but they needed only half mile to live than ecological density will be 1000 rabbit per mile.

Determination of Population Density:
Population density can be calculated by several methods, some of them are as follows –
1. Direct count method:

This method is used to count big size organisms which live in groups. This method is suitable for birds and big sized mammals (elephant, panther, and rabbit are counted by this method).
In this method, every individual is counted. This method showed an exact population density of species.
Human Population census is the best example of it. In India, the human population is counted in every 10 years.

2. By Sampling Method:

This method is used to determine population density in ecological habitat. In this method, samples have been collected from different parts and the number is counted. Average of all the samples is calculated to know the population density.
Example 1: To calculate the population of the density of paramecium, the Culture medium of Paramecium is shaken well and Paramecium is counted in cm on a watch glass. Sum of the result is counted and the average of that is calculated in per 1 cc.
Example 2: To calculate the population density of terrestrial animals, Quadrats may be used to determine population size and density. Each quadrat marks off an area of the same size – typically, a square area, within the habitat. In the end, the data can be used to estimate the population size and population density within the entire habitat. Sum of all the results is calculated and the average is to be taken out. This method indicates the almost correct population density.

3. By marking and recapture Method:

For organisms that move around, such as mammals, birds, or fish, rabbit, a technique mark-recapture method is often used to determine population size.
This method involves capturing a sample of animals and marking them using tags, bands, paint, or other body markings. Then, the marked animals are released back into the environment and allowed to mix with the rest of the population.
Later, a new sample is collected. This new sample will include some individuals that are marked and some individuals that are unmarked. Using the ratio of marked to unmarked individuals, scientists can estimate how many individuals are in the total population.

Suppose, we capture 100 organisms and release them back into the forest after marking. After some time, the second sample is taken and find that 10 are already marked. We can formulate the following relationship:

4. By Indirect Method:
In a laboratory, the population density is calculated using an indirect method. Such as:
On the basis of consumption of O₂and release of CO₂.

On the basis of the consumption quantity of food.
On the basis of the remaining food material present in the stomach.
On the basis of dead animal remaining of carnivorous animals.

5. By Sample Plot Method:

In ecological habitat, this is the most simple and commonly used method.
In this method, the area in which density is to be determined is divided into an equal plot.
Numbers of organisms are counted in some plot and average of number is taken out.
The average number is multiplied by a total number of plots which indicate the population density of that area.
For example, We have to calculate population density in 100 km². Now, 100km² is divided into 100 plots of 1Km². Suppose the average of 5 plots is 125 than the average of 1 plot is 25. Population density of 100Km² field is 25 × 100 ⇒ 2500.

Question 3.
Write the different methods to determine population density.
Answer:
Determination of Population Density:
Population density can be calculated by several methods, some of them are as follows –
1. Direct count method:

This method is used to count big size organisms which live in groups. This method is suitable for birds and big sized mammals (elephant, panther, and rabbit are counted by this method).
In this method, every individual is counted. This method showed an exact population density of species.
Human Population census is the best example of it. In India, the human population is counted in every 10 years.

2. By Sampling Method:

This method is used to determine population density in ecological habitat. In this method, samples have been collected from different parts and the number is counted. Average of all the samples is calculated to know the population density.
Example 1: To calculate the population of the density of paramecium, the Culture medium of Paramecium is shaken well and Paramecium is counted in cm on the watch glass. Sum of the result is counted and the average of that is calculated in per 1 cc.
Example 2: To calculate the population density of terrestrial animals, Quadrats may be used to determine population size and density. Each quadrat marks off an area of the same size – typically, a square area, within the habitat. In the end, the data can be used to estimate the population size and population density within the entire habitat. Sum of all the results is calculated and the average is to be taken out. This method indicates the almost correct population density.

3. By marking and recapture Method:

For organisms that move around, such as mammals, birds, or fish, rabbit, a technique mark-recapture method is often used to determine population size.
This method involves capturing a sample of animals and marking them using tags, bands, paint, or other body markings. Then, the marked animals are released back into the environment and allowed to mix with the rest of the population.
Later, a new sample is collected. This new sample will include some individuals that are marked and some individuals that are unmarked. Using the ratio of marked to unmarked individuals, scientists can estimate how many individuals are in the total population.

Suppose, we capture 100 organisms and release them back into the forest after marking. After some time, the second sample is taken and find that 10 are already marked. We can formulate the following relationship:

4. By Indirect Method:
In the laboratory, the population density is calculated using an indirect method. Such as:
On the basis of consumption of O₂ and release of CO₂.

On the basis of the consumption quantity of food.
On the basis of the remaining food material present in the stomach.
On the basis of dead animal remaining of carnivorous animals.

5. By Sample Plot Method:

In ecological habitat, this is the most simple and commonly used method.
In this method, the area in which density is to be determined is divided into an equal plot.
Numbers of organisms are counted in some plot and average of number is taken out.
The average number is multiplied by the total number of plots which indicate the population density of that area.
For example, We have to calculate population density in 100 km². Now, 100km² is divided into 100 plots of 1Km². Suppose an average of 5 plots is 125 than the average of 1 plot is 25. Population density of 100Km² field is 25 × 100 ⇒ 2500.

Question 4.
Write a note on

‘S’ – shaped growth form
Factors affecting population
Natality and Mortality

Answer:
1. ‘S’ – Shaped Growth or Sigmoid growth curve:
It is also called a sigmoid curve.
It has 3 phases:
1. Early or Lag phase: There is little or no growth.
2. Middle or Log or Exponential phase: There is fast growth in the population.
3. Stationary or Zero or Plateau phase:

The population growth is zero.
Mortality and natality are equal.
The population becomes equal to the environmental carrying capacity.

In S (sigmoid) form, the population increases slowly at first, then more rapidly but it soon slows down gradually as the environment resistance increases.

2. Factors affecting Population Growth:

The population of any country is effected by 4 factors
1. Natality
2. Mortality
3. Migration
4. Age Distribution
More natality in any country and less mortality increases population and the vice versa. Similarly, if human Immigration is more than migration, the population will increase and vice versa.

1. Natality or Fertility:

It determines the growth of any population. In a population, the ability to produce a new baby from active reproducing organisms is called fertility. In growing, population natality is one of the important factors.
The Birth rate is the number of births per 1000 individuals per year. It can be calculated by the following formula –
N = \(\frac { B }{ t } \)
where N = Natality/Birth rate
B = Birth of children per 1000 people
t = Time period in a year
Bodenheimer divided the human population into 3 groups –
1. Pre-reproduction population: Population below 18 years of age.
2. Reproductive population: Population between 18 to 45 years of age.
3. Post-reproductive population: Population above 45 years of age.
A number of the reproductive population indicates an increase in population growth. Similarly, the number of post-reproductive population indicates a decrease in population growth.

2. Mortality:

Mortality or Death rate is the number of deaths per thousand persons in a year.
M = \(\frac { D }{ t } \)
where M = mortality, D = Number of death in one thousand persons and t = time in years.
In many countries, a decline in death rates is recorded which resulted in an increase in population. This could be because of the following reasons:
1. Developed medical facility including Immunization vaccination.
2. Level of cleanliness has increased.
3. Awareness towards personal health & cleanliness has increased.
4. Availability of food and water is in a sufficient amount.
Mortality and Natality are the important factors which determine the population. In other words, using mortality and natality rate, population pattern of any country can be determined.
1. Crude Birth Rate: The crude birth rate is the rate of births among a population of 1000.
2. Crude death Rate: It is the rate of death among a population of 1000.
3. Annual Rate of Natural Increase/Growth:
  1. Difference between mortality and natality is called Annual growth rate.
  2. Annual growth rate = Natality/Birth rate – Mortality rate.
4. If mortality and natality are equal, population shape is stationary means the growth rate of the population will be zero. However, it occurs very rarely. It can be seen in some developed society.

3. Migration:

The movement by humans from one area to another is known as Migration. The humans who undergo migration are called Migrants.
The people who migrate into a territory are called immigrants, and they’re conveyed as Immigration.
Similarily the people who leave a territory are called Emigrants. and movement is Emigration. However, migrations can occur in different places in the same country. Only migrations between 2 countries affect the population.
Due to Emigration population density gets decreased and due to immigration, it increases. Due to migration increase in population is called Net Immigration. It can be calculated by –
- Net Immigration = Immigration – Emigration
- Net Immigration can be positive, zero or negative.

4. Age Distribution:

The age distribution is an important factor. It affects both Mortality and Natality. Distribution of various age groups in the population is called Age Distribution.
Natality and Mortality vary according to age. In a similar way reproduction also depend on age. Population growth depends on the ratio of all these factors. This ratio determines the future of any population.
On the basis of the economy, the human population is divided into 2 groups
1. Dependent age group (DAG): It includes children of 18 – 20 years of age which are economically dependent on their parents.
2. Productive age group (PAG): The population which is capable of earning.
  - The ratio of DAG & PAG is called a dependency ratio (DR)
  - DR ⇒ Number of DAG/Number of PAG

Age Pyramids:

An age pyramid is a graphical illustration that shows the distribution of various age groups in a population. Population pyramids often contain continuous stacked histogram bars, making it a horizontal bar diagram.
In ecology, 3 types of a pyramid can be imagined:
1. A pyramid with Broad Base: In this type of pyramid, the base part is wide. In these types of a pyramid, the natality rate is high and the population grow rapidly. A number of an organism in each generation increase from the previous generations. This types of pyramids are found in Yeast, Paramecium, housefly.
2. A Bell-shaped Polygon: In this type of population growth rate is slow and become stationary. Pre reproductive and reproductive age groups are an almost equal number. Post reproductive age groups will remain small. This type of population is bell-shaped.
3. An Urn Shaped Pyramid: In this pyramid child population is very less. The birth rate becomes very low. Since post – productive is dominating, the diminishing age pyramid is urn-shaped.

3. Natality or Fertility:

It determines the growth of any population. In a population, the ability to produce a new baby from active reproducing organisms is called fertility. In growing, population natality is one of the important factors.
The birth rate is the number of births per 1000 individuals per year. It can be calculated by the following formula –
N = \(\frac { B }{ t } \)
where N = Natality/Birth rate
B = Birth of children per 1000 people
t = Time period in a year
Bodenheimer divided the human population into 3 groups –
1. Pre-reproduction population: Population below 18 years of age.
2. Reproductive population: Population between 18 to 45 years of age.
3. Post-reproductive population: Population above 45 years of age.
A number of the reproductive population indicates an increase in population growth. Similarly, the number of post-reproductive population indicates a decrease in population growth.

Mortality:

Mortality or Death rate is the number of deaths per thousand persons in a year.
M = \(\frac { D }{ t } \)
where M = mortality, D = Number of death in one thousand persons and t = time in years.
In many countries, a decline in death rates is recorded which resulted in an increase in population. This could be because of the following reasons:
1. Developed medical facility including Immunization vaccination.
2. Level of cleanliness has increased.
3. Awareness towards personal health & cleanliness has increased.
4. Availability of food and water is in a sufficient amount.
Mortality and Natality are the important factors which determine the population. In other words, using mortality and natality rate, population pattern of any country can be determined.

Crude Birth Rate: The crude birth rate is the rate of births among a population of 1000.
Crude death Rate: It is the rate of death among a population of 1000.
Annual Rate of Natural Increase/Growth:
Difference between mortality and natality is called Annual growth rate.
Annual growth rate = Natality/Birth rate – Mortality rate.
If mortality and natality are equal, population shape is stationary means the growth rate of the population will be zero. However, it occurs very rarely. It can be seen in some developed society.

Question 5.
Write a note on age composition in any population.
Answer:
Age Distribution:

The age distribution is an important factor. It affects both Mortality and Natality. Distribution of various age groups in the population is called Age Distribution.
Natality and Mortality vary according to age. In a similar way reproduction also depend on age. Population growth depends on the ratio of all these factors. This ratio determines the future of any population.
On the basis of the economy, the human population is divided into 2 groups:
1. Dependent age group (DAG): It includes children of 18 – 20 years of age which are economically dependent on their parents.
2. Productive age group (PAG): The population which is capable of earning.
  - The ratio of DAG & PAG is called a dependency ratio (DR)
  - DR ⇒ Number of DAG/Number of PAG

Question 6.
What is age Pyramid?
Answer:
Age Pyramids:

An age pyramid is a graphical illustration that shows the distribution of various age groups in a population. Population pyramids often contain continuous stacked histogram bars, making it a horizontal bar diagram.
In ecology, 3 types of a pyramid can be imagined:
1. 1. A pyramid with Broad Base: In this type of pyramid, the base part is wide. In these types of a pyramid, the natality rate is high and the population grow rapidly. A number of the organism in each generation increase from the previous generation. This types of pyramids are found in Yeast, Paramecium, housefly.
  2. A Bell-shaped Polygon: In this type of population growth rate is slow and become stationary. Pre reproductive and reproductive age groups are an almost equal number. Post reproductive age groups will remain small. This type of population is bell-shaped.
  3. An Urn Shaped Pyramid: In this pyramid child population is very less. The birth rate becomes very low. Since post – productive is dominating, the diminishing age pyramid is urn-shaped.

Question 7.
Describe the various method of contraception used by females.
Answer:
Family Planning Welfare:

Population control can effectively be done by using family planning technique. It is nowadays called “Family Welfare”.
It involves the use of contraceptives and by making the public families aware of the methods of conceptions that they can decide when to give birth to young and what number they want.
The various family planning methods are as follows:

1. Barrier Methods:
1.1. Condom:

In this method, the process of fertilization between sperm and ovum is restricted using a physical barrier such as a condom. Condoms are barriers made up of rubber or latex sheath that are used to cover the penis in male and vagina in the female. This method is used to prevent semen from entering the female vagina.
The use of condoms had increased since the last few years. For males, the brand “Nirodhi”, a condom is very popular.
Use of condom also prevents sexually transmitted diseases such as AIDS besides preventing pregnancy.
Both male and female condom has to be discarded after use. It can be used by an individual by its own and in this way, it can remain personal.

1.2. The diaphragm, Cervical Cap & Vault:

The diaphragm, Uterus convex cap, vault, rings etc., are other contraceptives made of rubber and which are put into female reproductive tract to cover Uterus cervix and they stop the entry of sperm into the female tract.

1.3. Spermicidal:

Spermicidal serum, jelly and foam are also used as contraceptive measures.

2. Intra-Uterine Device (IUD):

Intra-uterine devices are small devices that are inserted into the uterus by doctors.
Examples:
1. Non-Medicated IUD – Lippes Loop.
2. Copper Releasing IUD – Cut, Cu7, Multiload 375 etc.
3. Hormone Releasing IUD – LNG 20, Protestant etc.
These devices promote phagocytosis of sperms to prevent fertilization. They are used to delay pregnancy. Cut, Cu7 are devices containing copper. The release of copper ions inhibits the motility of sperms & fertilisation. These are considered as ideal contraceptives for females. IUD hormones prevent implantation in the uterus two children IUD is an ideal method. It is widely used in India.

3. Contraceptive Pills:

These Pills are used as oral contraceptives. These pills contain small doses of estrogen or progesterone or a combination of both. These oral contraceptives should be taken for a period of 21 days.
After a gap of 7 days, that is, at the time of menstruation, these pills should be avoided. After this gap, these pills should be taken again in the same pattern until and unless female desires conception or fertilization.
Progesterone or combination with estrogen may also be given to females via injections. They work similarly to orally taken pills.
Saheli is a new oral contraceptive without any steroids hormone. It contains a chemical called centchroman. It is higly effective contraceptive pill.
Emergency pills Contain a chemical called Levenorgastril. This pill is to be taken after forced sex or unplanned sex within 72 hours. It prevents ovulation. It is also called a pill – 72

4. Surgical methods:

Surgical methods are also being used to control birth. These methods are also known as Surgical methods. These methods are used to terminate any further pregnancy. These methods block gamete transfer and thus prevent conception.
They are two types:

1. Vasectomy:

Vasectomy is one of the methods used in males. In this method, a small incision is made and vas deferens get tied.
It is commonly known as “Male Nasbandi”.

2. Tubectomy:

- Similarly, in females, Tubectomy is done. In tubectomy, a small part of Fallopian tube is removed via an incision or it gets tied.
- This techniques are effective and have reversibility. These techniques are commonly known as “female Nasbandhi”.

Question 8.
Give an account of castration.
Answer:
Vasectomy:

- Vasectomy is one of the methods used in males. In this method, a small incision is made and vas deferens get tied.
- It is commonly known as “Male Nasbandi”.

RBSE Class 12 Biology Chapter 38 Essay Type Questions

Question 1.
An increasing population in India is a big problem. Write your opinion.
Answer:
Population in India:

India is the second-largest country in the world concern to the human population. The India population was 134 crore as on Nov. 2017. Maximum population is found in China which was 141 crore in Nov. 2017.
In India, the population is increasing at a rate of 1.2%. If this growth in the population continues, India will become the country of the largest human population by 2030. India harbours 17.85% of the world population.
Population in Western countries is decreasing. On the country, it is increasing in India and other Asian countries.
In India, out of a total population, 69.7 crores are males and 65.2 crores are females. The sex ratio in India is 945 females per 1000 males. 50% of Indian population is in age slot of 0 – 25 years. 72.2% population inhabiting rural area and about 27.8% population living in an urban area.

Question 2.
Write a note on

Human age pyramid
The male and female ratio in Indian population

Answer:
1. Age Pyramids:

An age pyramid is a graphical illustration that shows the distribution of various age groups in a population. Population pyramids often contain continuous stacked histogram bars, making it a horizontal bar diagram.
In ecology, 3 types of a pyramid can be imagined:
1. A pyramid with Broad Base: In this type of pyramid, the base part is wide. In these types of a pyramid, the natality rate is high and the population grow rapidly. A number of an organism in each generation increase from previous generations. This types of pyramids are found in Yeast, Paramecium, housefly.
2. A Bell-shaped Polygon: In this type of population growth rate is slow and become stationary. Pre reproductive and reproductive age groups are an almost equal number. Post reproductive age groups will remain small. This type of population is bell-shaped.
3. An Urn Shaped Pyramid: In this pyramid child population is very less. The birth rate becomes very low. Since post – productive is dominating, the diminishing age pyramid is urn-shaped.

2. The male and female ratio in Indian population:

In India, out of a total population, 69.7 crores are males and 65.2 crores are females. The sex ratio in India is 945 females per 1000 males. 50% of Indian population is in age slot of 0 – 25 years. 72.2% population inhabiting rural area and about 27.8% population living in an urban area.

Question 3.
Describe the different population growth curve.
Answer:
Population Growth Form:

The population of either human or organisms always increases and the increase is more when the population reproduces actively. The population growths always grow in a particular manner which is called Growth Curve.
The growth curves are of two types:
1. ‘S’ Shaped Growth Curve
2. ‘J’ shaped Growth Curve

1. ‘S’ – Shaped Growth or Sigmoid growth curve:
It is also called a sigmoid curve.
It has 3 phases:
1. Early or Lag phase: There is little or no growth.
2. Middle or Log or Exponential phase: There is fast growth in the population.
3. Stationary or Zero or Plateau phase:

The population growth is zero.
Mortality and natality are equal.
The population becomes equal to the environmental carrying capacity.

In S (sigmoid) form, the population increases slowly at first, then more rapidly but it soon slows down gradually as the environment resistance increases.

2. ‘J’ – shaped curve:

It is J – shaped.
It is exhibited by the populations of insects & human beings.
It is also called abiotic potential curve.
According to it, the population grows speedily and this growth is more in large populations in comparison to small populations.
J – shaped curve has 3 phases –
1. Lag or Establishment phase
2. Exponential phase
3. Crash phase
The human population is in its exponential phase. The crash phase is yet to come.

Question 4.
Write the various factors involved to control the population.
Answer:
Factors to Control Population:

Increasing population in India has raised several socio-economical problems. We have to control this problem otherwise our economic growth will slow down and we will face the problem of food and place. Population control can be achieved in the following ways:

1. Increase Age of Marriage:

To control population child marriage should be strictly controlled and age of female and male should be increased to 21 and 24 years respectively for marriage.

2. Education:

The population problem can also be controlled by educating the people that poverty, scarcity of resources and low standard living are because of population explosion.
People should be made aware of the planned family & its benefits. Sex education should be included in an educational system.

3. Natural Method:

This technique works on the principle of stopping fusion of ovum and sperm. One of the methods is Abstinenea is a particular period. Couples avoid sexual intercourse between 10 to 17 days in which there is an ovulation process. In this period the chance of fertilization and pregnancy is more.
Coitus interruptus, withdrawal or pull – out method, is also a good method of birth control in which a man during sexual intercourse, withdraws his penis from a woman’s vagina prior to orgasm (ejaculation) and then directs his ejaculate (semen) away from the vagina in an effort to avoid insemination.
Lactational amenorrhea is the temporary postnatal infertility that occurs when a woman is amenorrheic (not menstruating) and fully breastfeeding. The time period is 46 month & the chance of pregnancy is zero. This technique is effective only until six months of child age.

4. Family Planning Welfare:

Population control can effectively be done by using family planning technique. It is nowadays called “Family Welfare”.
It involves the use of contraceptives and by making the public families aware of the methods of conceptions that they can decide when to give birth to young and what number they want.
The various family planning methods are as follows:

1. Barrier Methods:
1.1. Condom:

In this method, the process of fertilization between sperm and ovum is restricted using a physical barrier such as a condom. Condoms are barriers made up of rubber or latex sheath that are used to cover the penis in male and vagina in the female.
This method is used to prevent semen from entering the female vagina. The use of condoms had increased since the last few years. For males, the brand “Nirodhi”, a condom is very popular.
Use of condom also prevents sexually transmitted diseases such as AIDS besides preventing pregnancy.
Both male and female condom has to be discarded after use. It can be used by an individual by its own and in this way, it can remain personal.

1.2. The diaphragm, Cervical Cap & Vault:

The diaphragm, Uterus convex cap, vault, rings etc., are other contraceptives made of rubber and which are put into female reproductive tract to cover Uterus cervix and they stop the entry of sperm into the female tract.

1.3. Spermicidal:

Spermicidal serum, jelly and foam are also used as contraceptive measures.

2. Intra-Uterine Device (IUD):

Intra-uterine devices are small devices that are inserted into the uterus by doctors.
Examples:
1. Non-Medicated IUD – Lippes Loop.
2. Copper Releasing IUD – Cut, Cu7, Multiload 375 etc.
3. Hormone Releasing IUD – LNG 20, Protestant etc.
These devices promote phagocytosis of sperms to prevent fertilization. They are used to delay pregnancy. Cut, Cu7 are devices containing copper. The release of copper ions inhibits the motility of sperms & fertilisation. These are considered as ideal contraceptives for females. IUD hormones prevent implantation in the uterus two children IUD is an ideal method. It is widely used in India.

3. Contraceptive Pills:

These Pills are used as oral contraceptives. These pills contain small doses of estrogen or progesterone or a combination of both. These oral contraceptives should be taken for a period of 21 days.
After a gap of 7 days, that is, at the time of menstruation, these pills should be avoided. After this gap, these pills should be taken again in the same pattern until and unless female desires conception or fertilization.
Progesterone or combination with estrogen may also be given to females via injections. They work similarly to orally taken pills.
Saheli is a new oral contraceptive without any steroids hormone. It contains a chemical called centchroman. It is higly effective contraceptive pill.
Emergency pills Contain a chemical called Levenorgastril. This pill is to be taken after forced sex or unplanned sex within 72 hours. It prevents ovulation. It is also called a pill – 72

4. Surgical methods:

Surgical methods are also being used to control birth. These methods are also known as Surgical methods. These methods are used to terminate any further pregnancy. These methods block gamete transfer and thus prevent conception.
They are two types:

1. Vasectomy:

Vasectomy is one of the methods used in males. In this method, a small incision is made and vas deferens get tied.
It is commonly known as “Male Nasbandi”.

2. Tubectomy:

Similarly, in females, Tubectomy is done. In tubectomy, a small part of Fallopian tube is removed via an incision or it gets tied.
This techniques are effective and have reversibility. These techniques are commonly known as “female Nasbandhi”.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 42 Bio-Medical Technologies

August 14, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 42 Bio-Medical Technologies

RBSE Class 12 Biology Chapter 42 Multiple Choice Questions

Question 1.
Which of the following is used to measure blood haemoglobin?
(a) Haemocytometer
(b) Haemoglobinometer
(c) Westergren method
(d) Wintrobe method
Answer:
(b) Haemoglobinometer

Question 2.
In which of the following disease, number of leucocytes increases?
(a) Tuberculosis
(b) Typhoid
(c) Measles
(d) Blood cancer
Answer:
(d) Blood cancer

Question 3.
Which is used to diagnose the heart-related diseases?
(a) EEG
(b) ECG
(c) RIA
(d) CAT scan
Answer:
(b) ECG

Question 4.
Which of the following radiations are used in a CT Scan?
(a) α – rays
(b) β – rays
(c) γ – rays
(d) X – rays
Answer:
(d) X – rays

Question 5.
Full name of MRI is:
(a) Multiple Resonance Imaging
(b) Magnetic Radio Imaging
(c) Magnetic Resonance Indicator
(d) Multiple Radio Imaging
Answer:
(c) Magnetic Resonance Indicator

RBSE Class 12 Biology Chapter 42 Very Short Answer Questions

Question 1.
What is used for Total leucocyte count?
Answer:
Neubauer’s haemocytometer.

Question 2.
What is Leucocytosis?
Answer:
When the number of WBC becomes more than normal.

Question 3.
In which diseases, the value of ESR increases.
Answer:
Rheumatoid Arthritis, Multiple Myeloma, TB etc.

Question 4.
Name the instrument that records heartbeats.
Answer:
Stethoscope.

Question 5.
EEG is related to the diagnosis of which organ functioning?’
Answer:
Brain.

Question 6.
Which is used in place of X – rays in MRI.
Answer:
Magnetic Resonance:
Imaging uses the electric charge and low magnetic field of the nuclei of the hydrogen atom generated in an environment of a very strong magnetic field and radio waves. Hydrogen atoms are present in proteins and water molecules.

Question 7.
Which crystals are used in sonography?
Answer:
Lead Zirconate Crystals.

Question 8.
Which functions as labelling molecule in RIA?
Answer:
Radioisotopes substance.

RBSE Class 12 Biology Chapter 42 Short Answer Questions

Question 1.
Explain in short about the Westergren method of ESR.
Answer:
Erythrocyte Sedimentation Rate (ESR):

If a blood sample mixed with an anticoagulant like Trisodium citrate and kept undisturbed for some time in an ESR measuring tube, blood cells being denser than plasma, moves towards the bottom of tubes and settler down. The rate of setting down of the blood cells is called Erythrocyte Sedimentation Rate. The ESR measurement can be done by two methods.
1. Westergren method
2. Wintrobe method

In most of the pathological laboratories, only the Westergren method is used for ESR testing. Hence, here this method is described.
Westergreen tube is filled up to zero points with the blood containing anticoagulants. This tube is kept legitimately in a verticle position in ESR stand.
After an hour the upper level of erythrocytes is read in the ESR tube. This is the value of Erythrocyte sedimentation rate.
The ESR value of a healthy person is as follows:
- Male = 0 – 16 mm per hour.
- Female = 0 – 20 mm per hour.
If the rate of ESR is more than the normal value, it indicates towards irregularities in the body. Value of ESR increases in many chronic diseases like tuberculosis and inflammatory diseases like Rheumatoid arthritis, multiple myeloma or allergy, lymphatic inflammation etc.
Moreover, the value of ESR increases during pregnancy, Anaemia and with an increase in Age. These days Automated mini ESR machine is used to enhance the accuracy in the value of ESR. The machine works at a controlled temperature of 18°C. In the traditional methods, temperature regulation is not possible and a chance of 25 – 30% error exists.

Question 2.
Write the medical significance of Differential Leucocyte Counts.
Answer:
Medical Significance:

Any irregularity observed in the differential counting of the leucocytes is indicative of specific disease; further specialized testing can give a proper diagnosis of the disease and thus helps in proper treatment.

Some of these conditions are described below.

Nature of irregularity in differential leucocyte counting	Symptoms of possible diseases
Increase in the number of Neutrophils	Infection leading to normal pus, Sign of inflammation & dermatitis.
Increase in Eosinophils	Hypersensitivity or Allergy and signs of Parasitic infection
Increase in Basophils	Chickenpox disease
Increase in Lymphocytes	Whooping cough
Increase in Monocytes	Signs of Tuberculosis
The decrease in T₄ Lymphocytes	Signs of AIDS

(In General differential leucocyte count do not include T₄ lymphocytes).

Question 3.
Write the use of ECG.
Answer:
Use of ECG:

It is used to detect cardiac disorders like arrhythmias, conduction disturbances; myocardial ischemia etc.
It reveals other findings related to life-threatening metabolic disturbances like hyperkalemia or increased, susceptibility to sudden cardiac death (Example: QT prolongation syndrome).
It is used to study the rate of heartbeat, and cardiac disorders such as myocardial infarction (MI), coronary artery diseases etc.

Question 4.
Write the significance of Sonography technique.
Answer:
It is used to assess foetal growth and also the abnormalities in an adult body. It can provide a picture of blood flow through the beating heart.

Question 5.
Explain the role of ESR in disease diagnosis.
Answer:
Value of ESR increases in many chronic diseases such as TB, Arthritis, Allergy etc. ESR value also increases during pregnancy, Anaemia and old age.

Question 6.
Why MRI technique is better and safe than a CT scan.
Answer:
MRI provides high contrast images from all body axis. This technique can clearly differentiate between the white and grey matter of the brain.

RBSE Class 12 Biology Chapter 42 Essay Type Questions

Question 1.
Explain in detail about the process of haemoglobin measurement in blood.
Answer:
Estimation of Haemoglobin in blood:

Estimation of haemoglobin in the blood is called Haemoglobinometry. Haemoglobin is a respiratory pigment found in red blood corpuscles. Chemically, it is a chemoprotection which is used for the transport of O₂ and CO₂.
Working efficiency of a person has affected adversely if due to some reason his/her haemoglobin content drops below the normal value. A value of haemoglobin less than the normal value is an indicator of Anaemia.
The normal value of Haemoglobin is as follows:

Age	Haemoglobin g/100 ml blood
Healthy Adult male	15.5 ± 2.5
Healthy Adult female	14 ± 2.5
Children (3 months)	11.5 ± 2.5
Children (3 – 6 years)	12 ± 1
Children (10 – 12 years)	3 ± 1.5

Haemoglobin content is measured with Haemoglobinometer. Traditionally Sahli’s Haemoglobinometer is used for this purpose.
Photohaemoglobinometer and Autoanalyzer are also used.
Sahli’s Haemoglobinometer:
This device contains a graduated tube and two standard matching tubes placed in a stand. Between the two standard tubes is placed a measuring graduated tube. In the measuring tube, N/10 HCl is filled up to zero (2 g%) point.
Now using the haemoglobin pipette 20 Jul (0.02 ml) blood is transferred to the measuring tube. Blood is thoroughly mixed with N/10 HCl already present in the tube.
The N/10 HCl converts haemoglobin into a brown coloured substance called haematin. The measuring tube is now kept in between in the two standard tubes.
Distilled water is added drop by drop mixing with continuous mixing to the measuring tube until the colour of the haematin solution matches with the colour of the standard tubes/comparison tubes.
When the colour matching between the tube is ensured, haemoglobin content is measured by observing the reading on measuring graduated tube.

Question 2.
Explain the diagram of ECG and write its uses.
Answer:
Salient Features of ECG:

The ECG is recorded on special graph paper which is divided into 1 mm²gride like boxes.
The normal ECG includes P wave, QRS complex and ST – T – U complex.
The P wave is a small upward deflection and it represents atrial depolarization.
The QRS Complex represents rapid ventricular depolarisation (systole). It includes a small downward deflection, a rapid upright stroke and a small downward deflection.
The ST – T – U complex includes ST regiment, T wave & U wave. It represents the ventricular repolarisation (diastole).
The J point is the junction between the end of the QRS complex and the beginning of ST-segment.
Trial repolarization is too low in amplitude to be detected.
There are four major ECG Intervals –
1. R – R interval: It is used to compute the heartbeats per minute.
2. P – R Interval: It is used to measure the time between atrial & ventricular depolarisation.
3. QRS interval: It reflects the duration of ventricular depolarisation & atrial repolarisation.
4. QT interval: It includes both ventricular depolarisation and repolarisation times.
The QRS Complex is subdivided into specific deflections or waves.
The initial negative deflection is termed as Q – wave.
The first positive deflection is termed as R – wave A negative deflection after the R – wave is termed as S – wave.
An entirely negative QRS complex is termed as QS wave.
Normal T wave is a dome-shaped upward deflection.
Normal U wave is a small round upward deflection that follows the T – wave.

Question 3.
Write an explanatory note on MRI.
Answer:
Magnetic Resonance Imaging (MRI):

MRI was discovered by F. Bloch and E.M. Purcell independently and they shared the Nobel prize in 1952.
In this technique, there dimensional images of body organs are obtained without using X – rays & other radiations.
MRI detects water because it focuses on the behaviour of hydrogen atoms in water molecules.
MRI distinguishes between water-poor & water abundant tissues. The tissues with little water such as bones & teeth do not appear in MRI.
This technique is based on the natural behaviour of protons of hydrogen atoms and the most abundant source of protons are the hydrogen atoms in the water molecule.
MRI is used mainly to study organs like brain & spinal cord, to examine joint injuries & slip discs and to visualize minute cancerous tumours.
MRI is based on the phenomenon known as nuclear magnetic resonance (NMR).
During MRI testing, the patient is positioned in a two-meter wide chamber of an MRI scanner.
This chamber is surrounded by a large cylindrical electromagnetic field that produces a very strong magnetic field and waves.
Because of the very high magnetic field, nuclei of the hydrogen atom get activated and release radio signals.
These signals are processed by a computer. Through the processing of radio signals, thin high contrast images are obtained.
The MRI images are much better than the images obtained by CT scan.
They provide high contrast, using MRI technique, images from all body axis can be obtained.
Although MRI is a costly technique it is very helpful for the diagnosis and study of the brain and spinal cord.
This technique can clearly differentiate between the white and grey matter of the brain.

Question 4.
What is RIA? Explain its procedure and uses.
Answer:
Radio Immuno Assay (RIA):

Radioimmunoassay is an analytic technique which is used from the last many years.
Here, the molecule which is to be analysed, which acts as an antigen is marked with a radiolabelled substance.
The labelled and normal antigen molecules are allowed to react with a specific antibody.
A comparative analysis of the reaction is done. In this technique, the radioisotope is used as a marker molecule.
This technique was invented by Rosalyn and Yalow. Now – a – days it has become an important diagnostic technique.
This is especially important for the analysis of those biochemical factors that are present in very minute concentration (microgram, nanogram or picogram) and cannot be analysed by the traditional gravimetric and volumetric processes.
The radioisotopes used in radioimmunoassay are high specificity molecules, which provide very high sensitivity to this technique.
In this technique, the standard solution of different concentration of normal molecules of the substance to be analysed is used with the solutions of the radiolabelled substance of similar concentrations.
This mixture is allowed to react with antibodies. At the stage of equilibrium, antigen-antibody complexes are absorbed by the suitable reagent.
The precipitated and supernatant parts are separated and by measuring the radioactivity, the concentration of the substance is estimated.
A major characteristic of Radio Immuno Analysis is that the patient does not suffer from any side effect.
He is never treated with the radioisotopic molecule because the whole process is executed outside the body.

Uses of RIA:

Using this technique, the concentration of important biological components like vitamins (B₁₂, Folic acid), hormones (Thyroxine, Triodothyroxine, T₃, Cortisol, Testosterone, Estrogen, Tropic hormone etc.) drugs (Digoxin, Digitoxin etc.) and
antigenic substances can be determined.
For the diagnosis of abnormalities of endocrine glands, radioimmunoassays are very useful. For example presence of an excessive amount of some hormones in the blood may be a result of hyperactivity of endocrine gland or due to the impact of the tropic hormone, such questions can only be answered by this technique.
This technique is useful for the diagnosis of insulinoma, tumour etc which will help in the proper treatment of these diseases.

RBSE Solutions for Class 12 Biology

Rajasthan Board RBSE Class 12 Biology Chapter 27 Man-Sensory Organs (Sense Organs)

RBSE Class 12 Biology Chapter 27 Multiple Choice Questions

RBSE Class 12 Biology Chapter 27 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 27 Short Answer Type Questions

RBSE Class 12 Biology Chapter 27 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 26 Man-Nervous System

RBSE Class 12 Biology Chapter 26 Multiple Choice Questions

RBSE Class 12 Biology Chapter 26 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 26 Short Answer Type Questions

RBSE Class 12 Biology Chapter 26 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 25 Man-Excretory System

RBSE Class 12 Biology Chapter 25 Multiple Choice Questions

RBSE Class 12 Biology Chapter 25 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 25 Short Answer Type Questions

RBSE Class 12 Biology Chapter 25 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 38 Human Population

RBSE Class 12 Biology Chapter 38 Multiple Choice Questions

RBSE Class 12 Biology Chapter 38 Very Short Answer Type Questions

RBSE Class 12 Biology Chapter 38 Short Answer Type Questions

RBSE Class 12 Biology Chapter 38 Essay Type Questions

Rajasthan Board RBSE Class 12 Biology Chapter 42 Bio-Medical Technologies

RBSE Class 12 Biology Chapter 42 Multiple Choice Questions

RBSE Class 12 Biology Chapter 42 Very Short Answer Questions

RBSE Class 12 Biology Chapter 42 Short Answer Questions

RBSE Class 12 Biology Chapter 42 Essay Type Questions

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