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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

18301.

Question : Deuteromycetes reproduce only by

Answer»

Gametes
Motile spores
Conidia
Somatogamy

ANSWER :C
Discussion
18302.

Question : Deuteromycetes are imperfect fungi - Justify.

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Solution :The fungi belonging to DEUTEROMYCETES lack SEXUAL REPRODUCTION and are CALLED imperfect fungi.
Discussion
18303.

Question : Detoxification is one of the functions of ......

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SMALL intestine
liver
pancreas
stomach

Solution :N/A
Discussion
18304.

Question : Despite made of two different monomers amylose and amylopectin, starch is a homopolysaccharide-Comment.

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SOLUTION :STARCH is made up of amylose and amylopectin. Both are GLUCOSE POLYMERS, hence starch is considered as HOMOPOLYSACCHARIDES.
Discussion
18305.

Question : Despite the great heterogeneity of representative protistans,are there features that all members share ?

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Solution :Diverse forms of PREDOMINANTLY waterdwelling, EUKARYOTIC, UNICELLULAR micro-organisms.
Discussion
18306.

Question : Despite availability of plenty of water, leaves of certain plants wilt during the day and recover in the evening,

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Solution :Sometimes during the day, the rate of transpiration is comparatively more in respect to absorption of water by the PLANT. HENCE the plant wilts during the day. In the EVENING, the rate of transpiration is LESS in respect to absorption and hence the plant recovers in the evening.
Discussion
18307.

Question : Desmosome is the modification of

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GOLGI bodies
plasma membrane
ER-nucleus complex
endoplasmic membrane

Answer :B
Discussion
18308.

Question : Desert which receives the lowest rainfall ………….

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ANSWER :Ataeama DESERT (Chilc)
Discussion
18309.

Question : Deseribe the various types of sexual reproduction observed in algae.

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Solution :Sexual REPRODUCTION in algae are of THREE type
1. Isogamy: FUSION of morphologically and Physiologically SIMILAR gametes E.g. Ulothrix.
2. Anisogamy: Fusion of cither morphologically or physiologically DISSIMILAR gametes E.g. Pandorina
3. Oogamy: Fusion of both morphologically and physiologically dissimilar gametes. E.g. Sargassum.
The life eycle shows distinct alternation of generation.
Discussion
18310.

Question : Describleinternal structureof heart in detail .

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Solution :Internally heart is divided into four CHAMBERS. In that there are two atriums and two ventricles.
Two atriums are separated by inter artial septum.
Thick walled inter ventricular septum separates the right and left ventricles.
The atrium and ventricle of the same are separated by artrio ventricular septum. How EVER each of these septa are provided with an opening through which the two chambers of the same side are connected.
The opening between the right atrium and right ventricle is guarded by a valve FORMED of three muscular flaps or CUSPS, the TRICUSPID valve whereas a bicuspid or mitral valve guards the opening between left atrium and the left ventricle.
The openings between the right atrium and right ventricle is guarded by a valve formed of three muscular flaps or cusps, the tricuspid valve whereas a bicuspid or mitral valve guards the opening between left atrium and the left ventricle.
The opening of the right and the left ventricles into the pulmonary artery and the aorta respectively are provided with the semilunar valves.
the valves in the heart allows the flow of blood only in one direction. i.e. from atria to the ventricles and from the ventricles to the ventricles to the pulmonary artery or aorta. These valves prevent any backward flow.
Discussion
18311.

Question : Describlecardiac cycle .

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Solution :Heart beats about 72 times in a minute. It ACTS as a pump and its events show regular sequential events. This is called cardiac cycle.
The contraction phase of heart is called diastole and EXPANSION phase is called systole.
Followingphase are seen during cardiac cycle :
(i) Initially all four chkkambers are in diastole cycle :
(ii) Tricuspidand bicuspid valves are opened. Blood from the pulmonary VEINS and vena cava flows into the left and right ventricale respectively.
(iii) The semi lunar valves are closed at this stage. the SAN now generates an action potential which STIMULATES both the atria to undergo a simultaneous contraction the atrial systole. This increases the flow of blood into the ventricales by about 30 percent.
(iv) The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of HIS tranmits it through the entire ventricular musculature.
(v) This causes ventricular muscles to contract , the atria undergoes relaxatrion (diastole) coinciding withthe ventricular systole. Ventricular systole causes the closure of tricuspid and bicuspid valves that prevents the back flow of blood towords ventricles.
(vi) As ventricular pressure increases futher, the similunar valves guarding the pulmonary artery and the aorta are forced open, ALLOWING the vlood in the ventricles to flow through these vessels into the circulatory pathways.
(vii) The ventricals now relex and the ventricular pressure falls causing the causing the closure og semilunar valves which prevents the back flow of blood into the ventricles.
(viii) Now bicuspid and tricuspid valves are pushed open by the pressure inthe atria exerted by the blood which was being emptied them by veins.
(ix) The ventricles and atria are now again in a relaxed (joint diastole) state.
Discussion
18312.

Question : Describe water relations of a plant.

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Solution :Water is essential for all physiological activities of the plant.
Water plays a very important role in all living organisms.
It provides the MEDIUM in which most SUBSTANCES are dissolved.
The protoplasm of the cells is nothing but water in which different molecules are dissolved and suspended.
A watermelon has over 92 percent water.
Most herbaceous plants have only about 10 to 15 percent of its fresh weight as dry matter.
Woody plants have relatively very little water.
While soft parts of plants mostly contain water.
A seed may appear dry but it still has water - otherwise it WOULD not be alive and respiring.
Terrestrial plants take up huge amount of water daily but most of its lost to the air through evaporation from the leaves.
A MATURE corn plant absorbs ALMOST three litres of water in a day.
While a mustard plant absorbs water equal to its own weight in about 5 hours.
Because of this high demand for water, it is not surprising that water is often the limiting factor for plant growth and productivity in both agricultural and natural environments.
Discussion
18313.

Question : Describe vascular cambium responsible for secondary growth in detail.

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Solution :Vascular Cambium : Meristematic layer that is responsible for formation of two conducting tissues - xylem and phloem. It is called vascular cambium.
In mature/young stem it is present in patches as a single layer between the xylem and phloem. Later it forms a complete ring.
Formation of Cambial Ring : In dicot stems, the cells of cambium present between primary xylem and primary phloem is the intrafascicular cambium. The cell of medullary rays adjoining these intrafascicular cambium become meristematic and form the interfascicular cambium. Thus continuous ring of cambium is formed.
(B) Activity of Cambial Ring : The cambial ring becomes active and begins to cut off new cells, both towards the inner and the outer sides.
The cells cut off towards pith, mature into secondary xylem and the cells cut off towards periphery mature into secondary phloem.
The cambium is generally more active on the inner side then on the outer. As a result, the amount of secondary xylem produced is more than secondary phloem and soon forms a compact mass. At this point secondary xylem becomes the main part of the stem.
The primary and secondary phloem get gradually crushed due to the continued formation and accumulation of secondary xylem.
The primary xylem however remains more or less intact in or around the centre.
At some places, the cambium forms a narrow band of parenchyma which passes through the secondary xylem and the secondary phloem in the radial directions. These are the secondary medullary rays.

(c) Spring wood and AUTUMN wood:
The activity of cambium is under the control of many physiological and environmental factors. In temperate regions, the climatic conditions are not uniform throughout the year.
Spring wood : In the spring season, cambium is very active and produces a large number of xylary elements having vessels with WIDER cavities.
In spring season, the number of leaves increases. So for transport more number of vessels are needed.
The wood formed during this season is called spring wood or early wood.
The spring wood is LIGHTER in colour and has a lower density.
Autumn wood : In winter, the cambium is less active and forms xylary elements that have narrow vessels and this wood is called autumn wood or LATE wood.
Autumn wood is darker and has a higher density
Annual Rings : Spring wood and autumn wood that appear as alternate concentric rings, constitute the annual ring. Annual rings seen in a cut stem give an estimate of the age of the tree.
(d) Heartwood and sapwood : Heartwood : In old trees, the greater part of secondary xylem is dark brown due to deposition of organic compounds like tanins, resins, oils, gums, aromatic substance and essential oils in the central or innermost layers of the stem. These substance make it hard, durable and resistant to the attacks of micro organisms and insects. This region comprises dead elements with highly lignified walls and is called heartwood. The heartwood does not conduct water but it gives mechanical support to them.
Sapwood : The peripheral region of secondary xylem is lighter in colour and is known as the sapwood. It is involved in the conduction of water and minerals from root to leaf.
Discussion
18314.

Question : Describe various subphases of prophase - I of Meiosis - I.

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Solution :Prophase. I of Meiosis - I : It is longer and more complex. It has been sub divided into five sub phases based on chromosomal BEHAVIOUR.

(i) Leptotene : This is a FIRST phase of meiosis. During this phase the chromosomes become gradually visible. The compaction ofhromosomes continues throughout Leptotene.
Each chromosome is made up of two chromatids and CENTROMERE attached to them. But chromatids are tangled with each other so its dual form is not seen.
(ii) Zygotene : During this stage, chromosomes start pairing together and this process of association is called Synapsis. Such paired chromosomes are called homologous chromosomes.
This process proceeds further like a Zipper. Signal strands of chromosomes entangled or interwind around each other at some places. The complex formed by a pair of synapsed homologous chromosomes is called a Bivalent. Infact it is a tetrads.
(iii) Pachytene : During this stage bivalent chromosomes clearly APPEARS as tetrads. Single chromatid of the homologous chromosomes are entangled with each other.
This stage is characterised by the appearance of recombination nodules.
Crossing over occurs between non-sister chromatids of the homologous chromosomes.
Crossing over is the exchange of GENETIC material between two homologous chromosomes.
Crossing over is also an enzyme-mediated process and the enzyme involved is called recombinase.
Crossing over leads to recombination of genetic material on the two chromosomes.
Crossing over takes place at the exchange place of genes. This way new combinations are formed.
(iv) Diplotene : The beginning of diplotene is recognized by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the sites of crossing over.
These X-shaped structures are called Chiasmata.
The number of Chiasmata depend on length of chromosomes. They are more in long chromosomes. Exchange of genes takes place at site of formation of chiasmata.
(v) Diakinesis : Bispindle fibre separating homologous chromosomes is formed, single chromatids are separated at the site of formation of chiasmata.
By the end of diakinesis, the nucleolus disappears and the nuclear envelope also breaks down. Each chromosome is made up of two single chromatids and centromere attaching to it
Discussion
18315.

Question : Describe various stem modifications associated with food storage, climbing and protection.

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Solution :STEM Modification : (i) For food storage : Ginger (RHIZOME), Potato (Tuber), Onion (Bulb), COLOCASIA (Corm).
(ii) For climbing (support) : Sterm tendril (cucumber,grapevine,watermelon)
(III) For protection : Thorn (Bougainvillea, Citrus, Duranta) Description : Refer page 68, NCERT, Text Book of Biology for Class XI.
Discussion
18316.

Question : Describe various parts of a flowering plant with diagram.

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Solution :The morphology of angiosperms show broad DIVERSITIES
. They are CLASSIFIED by the presence of roots , stems , leaves and flowers . We also need to know about the POSSIBLE variations in different parts , found as ADAPTATIONS to various habitats , for protection , climbing , storage , etc.
.Flowering plants have roots , stems and leaves . They bear flowers and fruits also.
. The underground part of the flowering plant is the root system while the portion above the ground forms the shoot system.
Discussion
18317.

Question :Describe various phases of mitosis in detail with diagram.

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Solution :Two clear but continuous events take place in cell division.
(1) Karyokinesis
(2) Cytokinesis
(1) Karyokinesis :
If in nuclear division, the number of chromosomes are maintained then it is called mitosis and if the number of chromosomes becomes half then it is called meiosis.
Mitosis is divided into four stage of division.
Cell division is a progressive process and very clear cut lines cannot be grown between various stages. Mitosis is divided into main four stages :
(a) Prophase (b) Metaphase (c) Anaphase (d) Telophase
(a) Prophase :

Prophase which is the first stage of mitosis follows the S and `G_2`phases of interphase.
In the `S and G_2`phases the new DNA molecules formed are not distinct but interwined.
Prophase is marked by the initiation of condensation of chromosomal material. The chromosomal material becomes untangled during the process of chromatin condensation.
The centriole which had undergone duplication during S phase of interphase, now begins to move towards opposite poles of the cell.
The completion of prophase can thus be marked by the FOLLOWING characteristic events:
Chromosomal material condenses to form compact mitotic chromosomes. Chromosomes are seen to be composed of two chromatids attached together at the centromere.
Initiation of the assembly of mitotic spindle, the microtubules, the proteinaceous components of the cell cytoplasm help in the process.
Although plant cell lacks centriole, bispindle fiber is formed.
At the end of prophase NUCLEOLUS, golgi bodies, endoplasmic reticulum, nuclear envelope are not seen.
Chromosomes spread in whole cell.
(b) Metaphase :

The complete disintegration of the nuclear envelope marks the start of the second phase of mitosis.
During this stage, morphology of chromosomes can be studied easily. chromosomes located in the middle are made up of two single chromosomes. They are held together by the centromere.
Small disc shaped structures at the surface of the centromeres are called kinetochores. These structures serve as the sites of attachment of spindle fibres.
Spindle fibres attached to the chromosomes are moved into position at the centre of the cell.
Hence, the metaphase is character used by alc the chromosomes coming to lie at the equator with ONE chromatid of each chromosome connected by its kinetochore to spindle fibres from the opposite pole.
The plane of alignment of the chromosomes at metaphase is referred to as the metaphase plate.
The key features of metaphase are :
(i) Spindle fibres attach to kinetochores of chromosomes.
(ii) Chromosomes are moved to spindle equator and get aligned along metaphase plate through spindle fibres to both poles.
(c) Anaphase :
At the onset of Anaphase, each chromosome arranged at the metaphase plate is split simultaneously and the two daughter chromatids are now referred to as chromosomes of the future daughter nuclei begin their migration towards the two opposite poles.

As each chromosome moves away from the equatorial plate, the centromere of each chromosome is towards the pole and hence at the leading edge, with the arms of chromosome trailing behind.
Thus, anaphase stage is characterised by the following key events :
(i) Centromeres split and chromatids separate.
(ii) Chromatids move to opposite poles.
(d) Telophase :

At this stage the chromosomes that have reached their respective poles decondense and lose their individuality. Their chromatin material tends to collect in a mass in the two poles.
This stage SHOWS the following key events :
(i) Chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements.
(ii) Nuclear envelope assembles around the chromosome clusters.
(iii) Nucleolus, golgi complex and ER reform.
Cytokinesis:
Mitosis accomplishes not only the segregation of duplicated chromosomes into daughter nuclei but the cell itself is divided into two daughter cells by a separate process called cytokinesis at the end of which cell division is complete.

In an animal cell, this is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.
Plant cells however, are enclosed by a RELATIVELY inextensible cell wall.
Therefore they undergo cytokinesis by a different mechanism.
In plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls.
The formation of the new cell wall begins with the formation of a simple precursor called the cell plate that represents the middle lamella between the walls of adjacent cells.
At the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.
In some organisms karyokinesis is not followed by cytokinesis as a result of which multinucleate condition arises leading to the formation of syncytium. e.g. liquidendosperm in coconut.
Discussion
18318.

Question : Describe various parts/components of axial skeletal system with diagram.

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Solution :Axial skeleton : Comprises 80 bones distributed along the main axis of the body.
The skull, vertebrae column, sternum and ribs constitute axial skeleto.

(a)Skull : It is composed of two sets of bones- cranial, facial, that totals 22 Bones.
Cranium : Cranial bones are 8 in number. They form the hard, protective outer covering, cranium for the brain.
Bones which forms the cranium are,
1 Frontal
2 Parietal
2 Temporal
1 Occipital
1 Sphenoid
1 Ethmoid
Total `bar (8)`
Cranial bones are joined with each other by wavy immovale joint caled sutures.
Th facial region is made up of 14 elements.
2 Nasal
2 Maxillae
2 Palatine
2 Zygomatic
2 Lacrimal
1 Vomer
1 Mandible
Total `bar (14)`
Conchae and vomer can not be seen from outer side.
A single U-shaped bone called hyoid is present at the base of buccal cavity.
Each middle ear contains three tiny bones- Malleus, Incus and STAPES collectively called Ear OSSICLES.
The skull region articulates with the superior region of the vertebrae column with the help of two occipital condyles. (skull is called dicondylic)
Vertebral column :
Vertebral column is formed by 26 serially arranged units called vertebrae and is dorsally placed.
It EXTENDS from the base of the skull and constitutes the main framework of the trunk.
Each vertebrae has a central hollow portion-neuralcanal through which the spinal cord passes.
Name of vertebrae is from the region of the body, where they are found. (1) Cervical vertebrae7
(2) thoracic vertebrae12
(3) lumbar vertebrae5
(4) Sacral vertebrae1 (fused)
(5) coccygeal vertebrae1 (fused)
`bar (26)`
Function : The vertebrae column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back.
It keeps the body posture straight while running or walking.
It provides attachmentsite for the muscles of the ribs and back.
(c) Sternum : It is a flat bone on the ventral midline of thorax, its upper part is called manubrium middle of clavicle bone, joins in the clavicular notch.
RIBS : There are 12 pairs of the rib is a thin flat bone connected dorsally to the vertebrae column and ventrally to the sternum.
It has two articulation surface on its dorsal end and is hence called bicephalic.
True ribs : First seven pairs of ribs are called true ribs. Dorsally, they are ATTACHED to the thoracic vertebrae and ventrally connected to thr sternum with the help of hyaline cartilage.
False ribs : The `8^(th)`, `9^(th)` and `10^(th)` pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. So they are called vertebrochondrial (false) ribs.
Floating ribs : `11^(th)` and `12^(th)` pair of ribs are not connected ventrally. hence called floating ribs.
Thoracic vertebrae, ribs and sternum together form the rib cage.

Function : They protects lungs, LARGE blood vessels and heart. It also provides attachments site for respiratory muscles.
Discussion
18319.

Question : Describe various forms of lipid with a few examples.

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Solution :Lipids are the ESTERS of higher fatty acid with alcohol, such as glycerol, etc. These can be classified as
1 . Simple Lipids are esters of fatty acids with alcohol. These may be
(i) Fats These are esters of higher fatty acids with glycerol (triglycerides)·
(ii) Waxes These are esters of higher fatty acids with alcohol other than glycerol.
2. Compound or conjugated lipids, are those compounds which contain simple lipids and prosthetic (other additional) group. They include
(i) Glycerophospholipids, also known as phospholipids in which one of the fatty acid is replaced by phosphoric acid which is linked to NITROGENOUS groups like choline, ethanolamine, serine etc, e.g., Lecithin and cephalin, etc.
(ii) Sphingo lipids, are lipides having phosphoric acid with amine alcohol 4-sphinganine or sphingosine instead of glycerol in addition to fatty acid and choline.
(iii) Glycolipids, i.e., those which contain spinganine with a fatty acid and a MONOSACCHARIDE sugar, e.-g., cerebrosides and gangliosides.
3. Steroids are compounds with different chemical nature but similar physical properties. Their STRUCTURE is based upon a 4 ring cyclopentenoperhydro phenanthrene, e.g., cholesterol.
4. Prostoglandins are derivatives of arachidonic acid and contain 20 C-atoms. These are biologically active lipids.
Discussion
18320.

Question : Describe various conditions or factors necessary for growth.

Answer»

Solution :The GROWTH of plant is influenced by a variety of external and internal factors.
Growth of plant involves synthesis of protoplasm, cell division, cell enlargement and cell differentiation.
Some of the factors due to which growth of plants is influenced are mentioned below:
(a) Water : It is the FIRST and the foremost requirement of the plants for the enlargement of cell maintaining turgidity of growing cells, for extension of growth.
It also acts as medium for many enzymatic activities.
In water stress conditions growth of the plants seems to retarded.
(b) Oxygen : It helps in releasing metabolic energy essential for growth activities.
(c) Nutrients: These acts as (macro and micro essential elements nutrients) major raw materials for protoplasmic synthesis and also acts as a source of energy.
However under nutrient deficient conditions the growth of the plant is affected.
(d) Light: It helps in synthesis of food.
It also determines the root and shoot growth.
Along with light gravity it also serves as an ENVIRONMENTAL signal that affects CERTAIN phases/stages of growth.
(e) Temperature : For normal and appropriate growth of plant optimum temperature range is necessary i.e. `25-30^(@)C` (this happens because enzymatic reactions are ve fast at optimum temperature range).
Discussion
18321.

Question : Describe various components present in pancreatic juice.

Answer»

Solution :Semi digested food chyme comes to intestine by pyrrolic stomach. Chyme comes into contact of three different secretions in small intestine.
Pancreatic juice, bile juice and intestinal juice.
Various types of movements are generated by the muscularis layer of the small intestine. These movements help in a thorough mixing up of the food with various secretions in the intestine and there by facilitate digestion.
Pancreatic juice possesses inactive enzymes trypsinogen, chymotrypsinogen and procarboxypeptidases and AMYLASE, lipase and nuclease.
Trypsinogen is activated by an enterokinase, secreted by the intestinal mucosa into active trypsin which in turn activates the other enzymes in the pancreatic juice.
Proteases + PEPTONES `overset"Trypsin"to` polypeptide + Amino ACID
Trypsin convents chymotrypsinogen into chyme trypsin and procarboxy peptidases in to carboxyl peptidase.
proteoses + polypeptides + peptones `to`Chymotrypsin small polypeptide + Amino Acid.
Polypeptide `overset"Carboxypoptidose"to` Peptide, Amino acid
Amylase PRESENT in pancreatic, juice hydrolyses carbohydrates and disaccharide is formed. Polysaccharide (starch `overset"Amylase"to`Disaccharide Maltese).
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes.
Bile helps in EMULSIFICATION of fats i.e. breaking down of the fats into very small micelles.
Fats `overset"Lipases"to` Diglycerides `to` Monoglycerides
Nucleases of pancreatic juice reacts on nucleic acid and forms nucleotides and nucleosides.
Nucleic acids `overset"Nucleases"to` Nucleotides `to` Nucleosides
Digestion by intestinal juice in Jejunum and ileum: The enzymes of intestinal juice react on enzymes, final products react on them and convert into absorbable simple form.
Dipeptide `underset"(Erepsin)"overset"Dipeptidases"to` Amino Acids
Maltose `overset"Maltase"to` Glucose + Glucose
Lactose `overset"Lactose"to` Glucose + Galactose
Sucrose `overset"Sucrose"to` Glucose + Fructose
Nucleotides `overset"Nucleotidases"to` Nucleosides `overset"Nucleosidases"to` Sugars+ `N_2` Bases
Di and Monoglycerides `overset"Lipases"to` Fatty acid+Glycerol
Simple components produced from bio mega molecules are absorbed into jejunum and ileum of small intestine.
Undigested and unabsorbed components are passed into large intestine.
Discussion
18322.

Question : Describe various classes of Kingdom Fungi.

Answer»

Solution :The morphology of the MYCELIUM, mode of SPORE formation and fruiting BODIES form the basis for the DIVISION of KINGDOM into various classes. Phycomycetes:
Discussion
18323.

Question : Describe various categories of taxonomical studies.

Answer»

SOLUTION :(i) SPECIES :
Discussion
18324.

Question : Describe types of leaves.

Answer»

SOLUTION :There are teo types of leaves : (i) simpleleaf(ii) compound leaf.
(i) simple leaf : A leaf is SAID to be simple, when its lamina is entire or when incised, theincisions do not touch the midrib.
(ii) Compound Leaf : When the incisions of the lamina reach up to the midrib breaking called pund it into a number of leaflets, the leaf is called compound. The compound leaves may be of two types, pinnately compound leaf and palmately compound leaf.
More information:
Pinnately Compound Leaf:In a pinnately compound leaf a number of leaflets are present on a common axis, the RACHIS which REPRESENTS the midrib of the leaf as in neem.
Palmately Compound Leaf : In palmately compound leaves, the leaflets are attached at a common point. i.e., at the tip of petiole, as in SILK cotton.
Discussion
18325.

Question : Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?

Answer»

Solution :In tall trees, water rises with the help of the transpirational PULL generated by transpiration or loss of water from the stomatal pores of LEAVES. This is called the cohesion-tension model of water transport. During daytime, the water lost through transpiration (by the leaves to the surroundings) causes the guard cells and other epidermal cells to become flaccid. They in turn take water from the xylem. This creates a negative pressure or tension in the xylem vessels, from the surfaces of the leaves to the tips of the ROOTS, through the stem. As a result, the water present in the xylem is pulled as a single column from the stem. The cohesion and ADHESION forces of the water molecules and the cell walls of the xylem vessels prevent the water column from splitting.In plants, transpiration is driven by several environmental and physiological factors. The external factors affecting transpiration are wind, speed, light, humidity, and temperature. The plant factors affecting transpiration are canopy structure, number and distribution of stomata, water status of plants, and number of open stomata. Although transpiration causes water loss, the transpirational pull HELPS water rise in the stems of plants. This helps in the absorption and transport of minerals from the soil to the various plant parts. Transpiration has a cooling effect on plants. It helps maintain plant shape and structure by keeping the cells turgid. Transpiration also provides water for photosynthesis.
Discussion
18326.

Question : Describe : Tricarboxylicacid cycleORdescribe Kreb's cycle

Answer»

Solution :The TCA cycle start with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yieldcitric acid. The reaction is catalysed by the enzyme citratesynthase and a molecule of CoAis released.
`OA A (4C) + "Acetyl "CoA + H_(2) O (2C) underset("Citrate synthatase")to` Citric acid 6C + CoA
Citrate is then isomerised to isocitrate.
It is followingby two successive steps ofdecarboxylation, leading to theformation of `alpha`-ketoglutaric acidand then succinyl-CoA
Citric acid `to`Iso Citrate `to alpha` - ketoglutaric acid
Remaining steps of citric acid cycle :
Succinyl-Coa is oxidised to OAAallowing the cycleto continue.
During the conversion of succinyl-CoAto succinic acid a molecule of GTP is SYNTHESIZED .
This is a substrate levelphosphorylation.
INA coupled reaction GTP is converted to GDPwith thesimultaneoussynthesis of ATPform ADP.
`GTP to GDP + Pi to ADP + Pi to ATP`
Also there are three points in the cyclewhere `NAD^(+)`is reduced to NADH + `H^(+)`and onepoint where `FAD^(+)`is reduced to `FADH_(2)` .The continued oxidation of acetyl CoAVia theTCA cyclerequiresthe continued replenishmentof oxaloaceticacid
It ISTHE first member of the cycle . INADDITION it alsorequires regeneration of `NAD^(+) and FAD^(+)`from NADH and `FADH_(2)`respectively
The summary equation for this phase of respiration may be written as follow :
Pyruvic acid `+ 4NAD^(+) + FAD^(+) + 2H_(2) O + ADP + Pi underset("Matrix")overset("Mitochondrial")to `
`3CO_(2) + 4NADH + 4H^(+) + FADH_(2) + ATP`
Glucose has been broken down to RELEASE `CO_(2)`and eightmolecules of `NADH + H^(+)` , two of `FADH_(2)` have been synthesised besidesjusttwo molecules of ATP in TCAcycle .
Discussion
18327.

Question : Describe : Transpiration Pull.

Answer»

Solution :In HIGHER PLANTS, the flow of water upward through the xylem in plants can achieve fairly high rates up to 15 metres per hour.
In this either water is pushed or pulled through the plant. Most researchers agree that water is mainly pulled through the plant. and that the driving FORCE for this PROCESS is transpiration from the leaves.
This is referred to as the cohesion-tension-transpiration pull model of water transport.
Water is transient in plants. Less than 1 percent of the water reaching the leaves is used in photosynthesis and plant growth. Most of it is lost through the stomata in the leaves. This water LOSS is known as transpiration.
By enclosing a healthy plant in polythene bag and observing the droplets of water formed inside the bag.
You could also know water loss from a leaf using cobalt chloride paper, which turns colour on absorbing water.
Discussion
18328.

Question : Describe tissue structure of wall of digestive tract.

Answer»

Solution :The wall of alimentary canal from oesophagus to rectum possesses FOUR layers :
(a) Scrosa : Serosa is the OUTERMOST layer and is made up of a thin mesothelium.
(b) Muscularis : It is formed by smooth muscles usually arranged into an inner circular and outer longitudinal layer.
(c) Sub-mucosa : The sub-mucosal layer is formed of loose connective tissues containing NERVES, blood and lymph vessels. In doudenum, gland are also present in sub-mucosa.
(d) Mucosa : Inner most layer lining the lumen of the alimentary canal is mucosa. This layer forms irregular folds and small finger-like projections called villi in the stomach.

In the intestine, small finger like folding called villi are located.
The CELLS lining the villi produce numerous microscopic projections called microvilli.
It appears like brush border. These MODIFICATIONS increase the surface area enormously.
Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal are located.
Mucosal epithelium has goblet cells which secrete mucus.
That help in lubrication.
Mucosa also forms glands in the stomach and crypts in between the bases of villi in the intestine. (Crypts of Lieberkuhn)
All the four layers show modification in different parts of this alimentary canal.
Discussion
18329.

Question : Give a Schematic representation of Three domain classification.

Answer»

Solution :Three domain classification was proposed by Carl Woese (1977) and his co-workers. They classified organisms based on the difference in 16S rRNA genes. This adds the taxon domain higher than the kingdom. In this SYSTEM, prokaryotes are DIVIDED into two domains-bacteria and Arachaea. All eukaryotes are placed under the domain Eukarya. ARCHAE appears to have common features with Eukarya. Archaea DIFFER from bacteria in cell wall composition and differ from bacteira and eukaryotes in membrane composition and rRNA types.
Discussion
18330.

Question : Describe the vascular bundle of a monocot root.

Answer»

Solution :Vascular bundle of monocol root. It CONSISTS of several ( 8 or more) alternate radial xylem and phloem bundles. The vascular bundles are arranged in the form of ring with a PITH in the centre.
Xylem is exarch with protoxylem towardsoutside and metaxylem towards centre.
Function. Xylem provides MECHANICAL strength and is MEANT for CONDUCTION of water and mineral.
Phloem alternates with xylem. These two are separated by conjunctive parenchyma.
Function. Phloem is meant for translocation of organic food .
Discussion
18331.

Question : Describe the various types of placentations found in flowering plants.

Answer»

SOLUTION :For ANSWER SEE section-A, Q. No. 17
Discussion
18332.

Question : Describe the various stages of nirogen cycle.

Answer»

Solution :The nitrogen cycle CONSISTS of following states:
1. Fixation of atmospheric nitrogen : DI - nitrogen molecule from the atmosphere progressively gets reduced by addition of a pair of hydrogen atoms. Triple bond between two nitrogen atoms `(N -= N)` are cleaved to produce ammonia. Nitrogen fixation process requires Nitrogenase enzyme complex, Minerals (MO, Fe and S), anaerobic condition, ATP, electron and glucose 6 phosphate as `H^+` donor. Nitrogenase enzyme is active only in anaerobic condition. To create this anaerobic condition a pigment known as leghaemoglobin is synthesized in the nodules which acts as oxygen scavenger and removes the oxygen. Nitrogen fixing bacteria in root modules appears pinkish due to the presence of this leghaemoglobin pigment.

Overall equation :
`N_2 + 8e^(-) + 8H^(+) + 16 ATP to 2NH_3^(+) + H_2 + 16 ADP + 16 P i`
2. Nitrification : Ammonia `(NH_3^+)` is converted into Nitrite `(NO_2^-)` by Nitrosomonas bacterium. Nitrite is then converted into Nitrate `(NO_3^-)` by Nitrobacter bacterium. Plants are more adapted to absorb nitrate `(NO_3^-)` than ammonium ions from the soil.
`2NH_3^+ + 3O_2 overset("Nitrosomonas)(RARR) 2NO_(2)^(-) + 2H^(+) + 2H_2O`
`2 NO_(2)^(-) + O_(2) overset("Nitrobacter")(rarr)2NO_(3)^(-)`
3. Nitrate Assimilation: The process by which nitrate is reduced to ammonia is called nitrate assimilation and occurs during nitrogen cycle.
`NO_(3)^(-) underset(Mo)overset("Nitrate reductase")(rarr) NO_(2)^(-)`
4. Ammonification : Decomposition of organic nitrogen (PROTEIN and amino acids) from dead plants and animals into ammonia is called ammonification. Organism involved in this process are Bacillus ramosus and Bacillus vulgaris.
5. Denitrification : Nitrates in the soil are converted back into atmospheric nitrogen by a process called denitrification. Bacteria involved in this process are Pseudomonas, Thiobacillus and Bacillus subtilis.
`underset((NO_3^-))("Nitrate") overset("Pseudomonas")(rarr) underset((N_2))("Molecular Nitrogen")`.
Discussion
18333.

Question : Describe the umbel type of inflorescence.

Answer»

Solution : Umbel is a simple RACEMOSE inflorescence in which pedicellate flowers are BORNE radially from a single point DUE to reduction in the size of PEDUNCLE e.g. CENTELLA.
Discussion
18334.

Question : What are the types of transpiration?

Answer»

Solution :Types of Transpiration : Transpiration is of following three types:
1. Stomatal transpiration: Stomata are microscopic structures present in high number on the lower epidermis of leaves. This is the most dominant form of transpirationand being responsible for most of thewater loss ( 90-95%) in plants.
2. Lenticular transpiration: In stems of woody plants and trees, the epidermis is replaced by PERIDERM because of secondary growth. In order to provide gaseous EXCHANGE between the living cells and outeratmosphere, some pores which looks like lens-shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1% of the TOTAL.
3. Cuticular transpiration : The cuticle is a waxy or resinouslayer of cutin, a fatty substance covering the epidermis of leaves and other plant parts. Loss of water through cuticle is relatively SMALL and it is only about 5 to 10% of the total transpiration. The thickness of cuticle increase in xerophytes and transpiration is very much reduced or totally absent.
Discussion
18335.

Question : Describe the transpiration pull theory .

Answer»

Solution :Cohesion tension or Cohesion and transpiration pull theory :
It is a physical force to explain ascent of sap. Cohesion tension theory was was originally proposed by Dixon and JOLLY and again put forward by Dixon. This theory is based on the following features :
1. Strong cohesive force or TENSILE strength of water:
Water MOLECULES have the strong mutual force of attraction called cohesive force due to which they cannot be easily sparated from one another.
Further, the attraction between a water molecule and the wall of the xylem element is called adhesion.
These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem.
The magnitude of the chesive force is much high (350 atm ) and is more than enough to ascent sap in the tallest trees.
2. Continuity of the water column in the plant:
An important factor which can break the water column is the introduction of air bubbles in the xylem.
Gas bubbles expanding and displacing water within the xylem element is called cavitation or EMBOLISM. However the OVERALL continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.
3. Transpiration pull or Tension in the unbroken water column :
The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top the entire rope will move upward.
In plants such a pull is generated by the process of transpiration which is known as transpiration pull.
Water vapour evaporates from mesophy 11 cells to the intercellular spaces near stomata as a result of active transpiration.
The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential.
So, water moves as a pull from cell to cell along the water potential gradient. This tension generated at the top (leaf) of the unbroken water column is transmitted downwards from petiole stem and finally reaches the roots . The cohesion theory is the most accepted among the plant physiologists today.
Discussion
18336.

Question : Describe the tap root modification for storage purpose with diagram.

Answer»

Solution :Tap ROOT modification - Storage ROOTS
1. CONICAL Root - These are cone like, broad at the base and gradually tapering TOWARDS the apex. e.g. Daucus carota.
2. Fusiform root - These roots are swollen in the middle and tapering towards both ends. eg. Raphamus sativus
3. Napiform root - It is very broad and suddenly tapers like a tail at the apex. e.g. Beta vulgaris
Discussion
18337.

Question : Describe the structures of olfactory receptors.

Answer»

Solution :The smell receptors are excited by air borne chemicals that DISSOLVE in fluids. The yellow coloured patches of olfactory epithelium form the olfactory organs that are LOCATED on the roof of the nasal cavity. The olfactory epithelium is covered by a thin coat of mucus layer below and olfactory glands bounded connective tissues, above. It contains three types of cells: supporting cells, BASAL cells and MILLIONS of PIN shaped olfactory receptor cells (which are unusual bipolar cells). The olfactory glands and the supporting cells secrete the mucus. The unmyelinated axons of the olfactory receptor cells are gathered to form the filaments of olfactory nerve [cranial nerve-I] which synapse with cells of olfactory bulb. The impulse, through the olfactory nerves, is transmitted to the frontal lobe of the brain for identification of smell and the limbic system for the emotional responses to odour.
Discussion
18338.

Question : Describe the structure of tRNA with a diagram.

Answer»

Solution :(i) The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.
(ii) The two dimensional clover leaf model of tRNA was proposed by Robert Holley. The secondary structure of tRNA depicted in the FOLLOWING picture looks like a clover leaf.
(iii) In actual structure, the tRNA is a compact molecule which looks like an inverted L.
(iv) The clover leaf model of tRNA shows the PRESENCE of three arms namely DHU arm, middle arm and T`PSI`C arm.
(v) These arms have loops such as amino acyl binding loop, ANTICODON loop and ribosomal binding loop at their ends. In addition it also shows a small lump called variable loop or extra arm.
(vi) The amino acid is attached to one end (amino acid acceptor end) and the other end consists of three anticodon nucleotides.
(vii) The anticodon pairs with a codon in mRNA ensuring that the correct amino acid is incorporated into the growing polypeptide chain.
(viii) Four different REGIONS of double-stranded RNA are formed during the folding process. Modified bases are especially common in tRNA.
(ix) Wobbling between anticodon and codon allows some tRNA molecules to read more than one codon.
Discussion
18339.

Question : Describe the structure of the following with the help of labelled diagrams.(i) Nucleus(ii) Centrosome

Answer»

SOLUTION :
Discussion
18340.

Question : Describe the structure of stomata.

Answer»

Solution :The epidermis of leavesand green stems possess many small pores called stomata. The length and breadth of stomata is about `10 - 40 mu` and `3-10 mu`respectively. Mature leaves contain between 50 and 500 stomata per `mm^(2)`. Stomata are made up of two GUARD cells, special SEMILUNAR of kidney-shaped living epidermal cells in the epidermis. Guard cells are attached to surrounding epidermal cells KNOWN as subsidiary cells or accessory cells. The guard cells are joined together at each end but they arefree to separate to form a pore between them. The inner wall of the guard cell is thicker than the outer wall. The stome opens to the interior into a cavity called sub-stomatal cavity which remains CONNECTED with the INTERCELLULAR spaces.
Discussion
18341.

Question : Describe the structure of neuron with diagram.

Answer»

Solution :Neurons are functional unit of neural system.
In microscopic structure of neuron three main parts are observed :
(1) Cell body (2) Dendrites (3) Axon.
(1) Cell body : Body of neuron is called cyton. Cyton may be OVAL, round or stellate shaped. Cytoplasm of neuron is called Neuroplasm. Big round nucleus is present. In the centre of the cyton. Neuroplasm, consist of cellular organelles, small basophilic GRANULES (Nissl.s granules) Nissl.s granules are mostly near the nucleus.
(2) Dendrites : Divides, redivides and project out of the cyton. It also contains Nissl.s granules. Dendrites transmit impulses towards the cell body.
(3) Axon : Is a long fibre, the distal end of it is branched. Each branch terminates into bulb like structure called Synaptic knob.
Synaptic knob : It has synaptic vesicles containing chemicals called Neurotransmitter.

The axon transmit nerve impulses away from the cyton to a synapse or to a NeuroMuscular Junction.
Types of neurons : Based on the number of axon and dendrites, the neurons are divided into three types :
(1) Multipolar : One axon and two or many dendrites. i.e. : cerebral cortex.
(2) Bipolar : One axon and one dendrite e.g. : RETINA of eye.
(3) Unipolar : Cell body with one axon only e.g. : present in embryonic stage.
Two types of axons are present accordingly neuron is
(i) Myelinated neuron (ii) Nonmyelinated neuron.
(i) Medullated neuron : The myelinated nerve fibres are enveloped with Schwann cells which FORM a myelin sheath around the axon.
The gaps between two adjacent myelin sheaths are called nodes of Ranvier
Myelinated nerve fibres are found in cranial nerves and spinal nerves.
(ii) Non-myelinated neuron : They are enveloped by a Schwann.s cells but myelin layer is not formed around the axon.
It is found in SOMATIC neural system and autonomous neural system.
Discussion
18342.

Question : Describe the structure of pancreas, location and discuss its hormonal role.

Answer»

Solution :Pancreas is a composite gland which acts as both exocrine and endocrine gland.
The endocrine pancreas consists of .Islets of Langerhans..
There are about 1 to 2 million islets of langerhans in a normal human pancreas representing only 1 to 2 % of the pancreatic tissue.
The two main types os cells in the islet of Langerhans are called `alpha` - cells and `beat` - cells. The `alpha`- cells secrete a hormone celled glucagon, while the `beta` - cells secrete insulin.
Glucagon is a peptide hormone and plays an important role in maintaining the normal blood GLUCOSE LEVELS.
Glucagon acts mainly on the liver cells and stimulates glycogenoysis resulting in an increased blood SUGAR hyperglycemia.
In ADDITION, it also stimulates the process of gluconeogenesis which also contributes to hyperglycemia.
Glucagon reduces the cellular glucose uptake and utilisation. Thus, glucagon is a hyperglycemic hormone.
Insulin is a peptide hormone which plays a major role in the regulation of glucose homeostasis.
Insulin acts mainly on HEPATOCYTES and adipocytes and enhances cellular glucose up take and utilisation. As a result, there is a repid movement of glucose from blood to hepatocytes and adipocytes resulting in decreased blood glucose levels (hypoglycemia).
It also stimulates conversion of glucose to glycogen (glycogensis) in the target cells.
The also stimulates coversion of glucose to glycogen (glycogenesis) in the target cells.
The glucose homeostasis in blood is thus maintained jointly by the two - insulin and glucagons.
Prolonged hyperglycemia leads to a complx disorder celled diabetes mellitus which is associated with loss of glucose through urine and formation of harmful compounds known as ketone bodies.
Diabetic patients are successfully treated with insulin therapy.
Discussion
18343.

Question : Describe the structure of nephron with a neatly labelled diagram.

Answer»

Solution :Nephrons are highly coiled microscoping tublules. They are the structural the functional kidney. Each nephron has two distinct regions viz.
1. Bowman's capsule.
2. Tubule.
1. Bowman's Capsule: It is spherical and double walled cup shaped sac like structure, found in the cortex region of the kidney. It possesses a LUMEN which is lined by a thin walled squamous epitelium. The capsular lumen is continous with tubular lumen. The capsule infolds anteriorly to receive affeent and efferent arterioles. Afferent arteriole gives off fine branches and rorms a tuft of capillaries inside teh Bowman's capsule which is called Glomerulus, Bowman's capsule with glomerulus is called Malphigian body or malphigian corpuscle.
2. Tubule: Tubule of a Nephron shows'three distinct regions. They are:
a. Proximal Convoluted Tubule (PCT).
b. HENLE's Loop.
c. DISTAL Convoluted Tubule (DCT).
a. Proximal Convoluted Tubule : The proximal convoluted tubule is a coiled tubelying next to copsular. It is lined inside by CUBOIDAL epithelium having microvilli. These microvilli form a brush border within a the lumen of PCT.
b. Henel's Loop: The henel's Loop is 'U' shaped fine tube with descending and ascending lumbs. The lumen of the Henle's loop is lined by FLATTENED epithelium, Henle's loop is found the in the medulla region of the kidney.
c. Distal Convoluted Tubule: The structure of distal convoluted tubule resembles with that of PCT, it is a also lined by cuboidal epithelium. However the cuboidal epithelium of the DCT shows indistinct brush border. the DCT opens into a collecting tubule which receives a number of nephrons.
The efferent arteriole divides into a number of capillaries. These capillaries form a network all oer the three regions of tubule and this network is called peritubular network.
Discussion
18344.

Question : Describe the structure of human stomach.

Answer»
Discussion
18345.

Question : Describe the structure of Gram positive and Gram negative bacterial cell wall using diagram.

Answer»

Solution :Most of the gram POSITIVE cell wall contain considerable AMOUNT of teichoic ACID and teichuronic acid. In addition, they may contain polysaccharide molecules. The gram negative cell wall contains three components that lie outside the peptidoglycan layer : 1. Lipoprotein 2. Outer membrane and 3. Lipopolysaccharide. Thus the DIFFERENT results in the gram stain are due to differences in the structure and composition of the cell wall.
Discussion
18346.

Question : Describe the structure of chloroplast.

Answer»

Solution :Chloroplastsare themain siteof photosynthesis and bothenergyyieldingprocess (Light reaction ) and fixationof carbon dioxide (Dark reaction ) that takes place in chloroplast . ITIS a doublewall membrance bounded organelle, discoid or lens shaped, 4 -10`mu m` is diamter and`1 -33 mu m` is thickness. The membranceis a unit membranceand space beween themis 100 to`200 Å` Acolloidal and proteinaceous MATRIX called stroma is presentinside.
A sac likedmembranous systemcalledthylakoidor lamellase is present in stroma and theyare arrangedone abovethe otherforminga stack of coin like structure called grnum( plural grana). Eachchloroplast contans40 t0 80grana and eachgraanaconsistsof 5 to 30 thyaliods .
Thylakoidsfoundin granum are calledgrana lamellaeand in stroma are calledstromalamellae.Thylakoid disc sizeis 0.25 to 0.8 micro in diameter. A thinnre lamella calledFretmembranceconnects grana. PigmentsystemI islocated on outer thykoidmembrancefacingstromaandPigment systemI is locatedon innermembrancefacinglumenof thylakoid . Grana lamellaehave both PS I and PS II whereas stroma lamellae haveonly PSI . Chloroplast contains 30-35 % Proteins , 20-30%phospholipids , 5 - 10% chlorophyll , 4 - 5 % , Carotenoids ,70 Sribosomes , circular DNA and strachgrains. Inner surfaceof lamellar membracne consists of small sphericalstructurecalled as Quantasomes . Presence of70 S rbosome and DNA gives them status ofsemi-autonomy and proves endosymbiotic hypothesis whichsayschloroplastevolved from bacteria . Thylakoid contains pigment systems whichproduces ATP and NADPH ` + H^(+)`usingsolar ENERGY . Stroma containsenzymes whichreduces carbondioxideinto carbohydrates .
Discussion
18347.

Question : Describe the structure of chloroplast along with its diagram.

Answer»

Solution :
It is green coloured plastid. It CONTAIN green pigment CHLOROPHYLL, double straded DNA and 70S ribosomes. It has stacks LIKE GRANA and stroma region. It carry out photosynthesis. It also gives colour to the fruit and flower.
Discussion
18348.

Question : Explain molecular structure of ATP?

Answer»

Solution :Respiration is responsible for generation of ATP. The discovery of ATP was made by Karl Lohman (1929). ATP is a nucleotide consisting of a base-adenine, a pentose sugar - ribose and THREE phosphate groups. Out of these phosphate groups the LAST two are attached by high energy rich bonds (figure). On hydrolysis, it releases energy (7.3 K cal or 30.6 KJ/ATP) and it is found in all living cells and hence it is called universal energy currency of the cell. ATP is an instant source of energy within the cell. The energy contained in ATP is used in synthesis carbohydrates, proteins and lipids. The energy transformation CONCEPT was established by LIPMAN. (1941).
Discussion
18349.

Question : Describe the structure and function of pancreas.

Answer»
Discussion
18350.

Question : Describe the steps involved in DNA finger printing.

Answer»

Solution :Steps in DNA Finger printing `:`
(i) Extraction of DNA `:` The process of DNA fingerprinting starts with obtaining a sample of DNA from BLOOD, semen, vaginal fluids, hair roots,teeth, bones, etc.,
(ii) Polymerase chain reaction (PCR) `:` In many situtations, there is only a small amount of DNA available for DNA fingerprinting. If needed many copies of the DNA can be produced by PCR (DNA amplification).
(iii) Fragmenting DNA`:` DNA is treated with restriction enzymes which cut the DNA into smaller fragments at specific sites.
(iv) Separation of DNA by electrophoresis`:` During electrophoresis in an agarose ge, the DNA fragments are separated into bands of different sizes. The bands of separated DNA are sieved out of the GEL using a nylon membrane (treated with chemicals that allow for it to break the hydrogen bonds of DNA so there are single strands).
(v) Denaturing DNA`:` The DNA on gels is denatured y using alkaline chemicals or by heating.
(vi) Blotting`:` The DNA band pattern in the gel is transferred to a thin nylon membrane placed over the .size fractioned DNA strand. by Southern blotting.
(vii) Using probes to identify specific DNA`:` A radioactive probe (DNA labeled with a radioactive substance) is added to the DNA bands. The probe attaches by BASE pairing to those restriction fragments that are complementary to its sequence. The probes can also be prepared by using either .fluorescent substance. or .radioactive isotopes..
(viii) Hybridization with probe `:` After the probe hybridizes and the excess probe washed off, a photographic FILM is placed on the membrane containing .DNA hybrids..
(ix) Exposure on film to make a GENETIC/DNA Fingerprint `:` The radioactive label exposes the film to form an image (image of bands) corresponding to specific DNA bands. The thick and thin dark bands form a pattern of bars which constitutes a genetic fingerprint.
Discussion