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18501.

Question : Describe briefly the tunica-corpus theory.

Answer»

Solution :Tunica-corpus theory . Schmidt proposed the Tunica-Corpus Theory (1924) to explain the structure of APICAL meristem. He explained that intercalary meristem differentiated into a central mass undifferentiated large cells CALLED corpus and outer layer of small cells called tunica. The cells of tunica undergo anticlinal divisions while the cells of corpus DIVIDE in different planes to produce PROCAMBIUM , cortical cells and leaf PRIMORDIA.
Discussion
18502.

Question : Describe briefly : (a) Arithmetic growth (b) Geometric growth (c) Sigmoid growth curve (d) Absolute and relative growth rates

Answer»

Solution :(a) ARITHMETIC growth : In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while the other differentiates and matures.
The simplest expression of arithmetic growth is exemplified by a root elongating at a constant rate.
(b) Geometric growth : Geometric growth is characterised by a slow growth in the INITIAL stages and a rapid growth during the later stages. The daughter cells derived from mitosis retain the ability to divide but slow down because of a limited nutrient supply.
(c) Sigmoid growth curve : The growth of living organisms in their natural environment is characterised by an .S. shaped curve called sigmoid growth curve. This curve is divided into three phases lag phase, log phase or exponential phase of rapid growth and stationary phase.
Exponential growth can be expressed as `W_(1)=W_(0)e^(rt)`
Where `W_(1) =` Final size
`W_(0) =` Initial size
r = Growth rate
t = TIME of growth
e = Base of natural logarithms

An idealised sigmoid growth curve typical of cells in culture and many higher plants and plant organs
(d) Absolute and relative growth rates : Absolute growth rate refers to the measurement and comparison of total growth per UNIT time.
Relative growth rate refers to a particular system per unit time, expressed on a common basis.
Discussion
18503.

Question : Describe briefly

Answer»

Solution :(a) Arithmetic growth, If the length of a plant organ is plotted against time shown a LINEAR curve, the growth is called arithmetic growth.
(b) Geometric growth. Such type of growth, that occurs in many higher plants and plant organss and measured in size or weight, is called geometric growth.
(c) Sigmoid growth curve. If the GEOMETRICAL growth is plotted against time, it SHOWS a sigmoid growth curve.
(d) Absolute and RELATIVE growth rates. Increase in total of two organs measured and COMPARED per unit time is called absolute growth rate. The growth of the given system per unit time expressed on a common basis is called relative growth rate.
Discussion
18504.

Question : Describe biological nitrogen fixation with reference to Rhizobium and Legume.

Answer»

Solution :The process of converting Atmospheric Nitrogen into Ammonia is termed as Nitrogen fixation.
Biological nitrogen fixation :
Symbiotic bacterium LIKE Rhizobium fixed atomspheric nitrogen. CYANOBACTERIA found in Lichens, ANTHOCEROS, Azolla and CORALLOID roots of Cycas also fix nitrogen. Non - Symbiotic (Free living bacteria ) like Clostridium also fix nitrogen.
a. Symbiotic nitrogen fixation :
Nitrogen fixation with nodulation
(i) Rhizobium bacterium is found in leguminous plants and fix atmospheric nitrogen.
(ii) This kind of symbiotic association is beneficial for both the bacterium and plant. Root nodules are formed due to bacterial infection.
(iii) Rhizobium enters into the host cell andproliferates, it remains separated from thehost cytoplasm by a membrane
b. Stages of Root nodule formation :
1. Legume plants secretes phenolics which attracts Rhizobium.
2. Rhizobium reaches the rhizosphere and enters into the root hair, infects the root hair and leads to curling of root haris.
3. Infection thread grows INWARDS adn separates the infected tissue from normal tissue.
4. A membrane bound bacterium is formed inside the nodule and is called bacteroid.
5. Cytokinin from bacteria and auxin from host plant promotes cell division and leads to nodule formation.
Nitrogenase enzyme complex, Minerals (Mo, Fe and S ), anarobic 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 asleghaemoglobin is synthesized in the nodules which acts as oxygen scavenger and removes the oxygen. Nitrogen fixing bacteria in root nodules appears pinkish due to the presence of this leghaemoglobin pigment.
Discussion
18505.

Question : Explain the three important levels of biodiversity.

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Solution :The earth hosts an immense variety of living organisms.
ACCORDING to a SURVEY the number of species that are known and described are between 1.7-1.8 million.
The number refers to the biodiversity on the earth.
The term Biodiversity or Biological DIVERSITY means the number and types of organisms present on the earth, form of life in the living world.
The living world includes all the living organisms such as micro organisms, plants, animals and humans.
Biodiversity is not LIMITED to the existing life forms.
If we explore new AREAS even old ones, new organisms are continously being added.
This huge available variety cannot be studied and identified without having a proper system of classification and Nomenclature.
Discussion
18506.

Question : Describe Binomial Nomenclature and what is its advantage ?

Answer»

Solution :Binomial nomenclature was developed by Carolus Linnaeus in 1751.
Binomial nomenclature is the system of providing distinct and appropriate names to organisms, each CONSISTING of two words, first generic name (ie, name of genus) and second specific epithet (i.e. name of species).
For example, scientific name of Mango is written as Mangifera indica.
In this name Mangifera represent the genus and indica is a particular species, or specific epithet.
Advantages of Binomial Nomenclature :
(1) Binomial names are universally acceptable and recognised.
(2) They remain same in all languages.
(3) The names are small and comprehensive.
(4) These is a mechanism to provide a scientific name to every NEWLY discovered ORGANISM.
(5) The names indicate RELATIONSHIP of a species with other species present in the same genus.
(6) A new organism can be easily provided with a new scientific name.
Discussion
18507.

Question : Describe anaerobic phase of aerobic respiration

Answer»

Solution : Definition : Glycolysis means the phase till formation of two molecules of pyruvic acid from one molecule of glucose takes place.
Origin : The term glycolysishas originated form the Greek words glucose form sugar and lysis for splitting.
The scheme of glycolysis was given by Gustav Embden, OttoMeyercdofand J. Parnas and is often referred to as the EMP pathway.
In anaerobic organisms, only glycolysisoccurs. Glycolysisoccurs in the cytoplasm of the cell and in thisprocess undergoes partial oxidation toform two molecules of pyruvic acid .
In plants, this glucose is derived from sucrose .
whichis theend productof photosynthesis or from storagecarbohydrates. Both these monosaccharidesreadilyenter the glycolytic pathway.
Glucose and fructose are phosphorylated to give rise to Glucose-6- phosphateby theactivity of the enzyme hexokinase.
Glucose + ATP `overset("Hexokinase")to ` Glucose-6-Phosphate
Thisphosphorylated form of glucose then isomerises toproduce fructose-6- phosphate.
Glucose-6- Phosphate `to` Fructose-6- Phosphate
Subsequentsteps of metabolism of glucoseand fructose are same.
Inglycolysis, a chain of TEN reactions, under the control of differentenzymes, takesplace toproducepyruvate from glucose.
Nowfructose-6- phosphate is converted into fructose-1, 6 -biphosphate in presence of ATP.
ATP is utilised at two steps. First in the conversion of glucose into glucose 6-phosphate an in second stepthe conversion of fructose 6-phosphate to fructose 1, e-biphosphate.
Now the fructose 1,6-biphosphate is split into dihydroxyacetone phosphate and3-phosphoglyceraldehyde (PGAL).
Fructose 1, 6-Biphosphate `to ` DHAP(3G) + PGAL (3G)
There isone STEP where NADH + `H^(+)`is formed from `NAD^(+)`,this is when 3-phosphoglyceraldehyde is converted to 1,3-biphosphoglycerate (BPGA).
3-phosphoglyceraldehyde `to` 1,3-biphosphoglycerate + NADH `+ H^(+)`
Two redox-equivalents are removed (in the form of hydrogen atoms) from PGAL, and transferred to a molecule of `NAD^(+)`
PGAL is oxidisedand with inorganic phosphate to get converted intoBPGA.
Theconversion of BPGA to 3-phosphoglyceric acid (PGA) ,is also an energy yielding process. This energy is trapped by the formation of ATP
Another ATP is SYNTHESISED during the conversion of PEP to pyruvic acid .
(Formationof ATP = 4ATP mode)
Pyruvicacid is then the key productof glycolysis. Whatis he metabolic fate of pyruvate ? depends on the cellular need.
There are three major ways in which differentcells handle pyruvic acid produced by glycolysis
(1) Lactic acid FERMENTATION.
(2) Alcoholic fermentation.
(3) Aerobic respiration.
fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes.
For the complete oxidation of glucose to `CO_(2)and H_(2)O`, however, organisms adopt Kreb, cycle which is also calledas aerobic respiration. This requires `O_(2)`supply.
Discussion
18508.

Question : Describe anatomy of body of Earthworm in short

Answer»

Solution :Skin (Anatomy of body WALL) :The body wall of the earthworm is covered externally by a thin non-cellular cuticle below which is the EPIDERMIS, TWO MUSCLE layers (circular and longitudinal) and an innermost coelomic epithelium.The epidermis is made up of a single layer of columnar epithelial cells which contain secretory gland cells.
Discussion
18509.

Question : Describe Allium cepa in botanical terms.

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Solution :Vegetative Characters :
1. Habit: Perennial herb with bulb.
2. Root: Fibrous adventitious root system.
3. Stem: Underground bulb.
4. Leaf: A cluster of radical leaves emerges from the underground bulb, cylindrical and fleshy having sheathy leaf bases with parallel venation.
Floral Characters :
1. Inflorescence: Scapigerous i.e. the inflorescence axis (peduncle) arising from the ground bearing a cluster of flowers at its apex. Pedicels are of equal lenth, arising from the apex of the peduncle which brings all flowers at the same level.
2. Flower : Small, white, bracteate, ebracteolate, pedicellate, complete, trimerous, actinomorphic and hypogynous. Flowers are protandrous.
3. Perianth : Tepals 6, white, arranged in two whorls of three each, syntepalous showing valvate aestivatikon.
4. Androecium: STAMENS 6, arranged in two whorls of three each, epiphyllous, apostamenous /FREE and opposite to tepals. Anthers dithecous, basifixed, introse, and dehiscing longitudinally.
5. Gynoecium: Tricarpellary and syncarpous. OVARY superior, TRILOCULAR with two ovules in each locule on axile PLACENTATION. Style simple, slender with simple stigma.
6. Fruit: A loculicidal capsule.
7. Seed: Endospermous.
8. Floral Formula:
Discussion
18510.

Question : Describe Amphibia.

Answer»

Solution :As the name indicates (Gr., Amphi : dual, bios : life), amphibians can live in aquatic as well as terrestrial habitats.
Most of them have two pairs of limbs.
Body is divisible into head and trunk.
Tail may be present in some.
The amphibian skin is moist (without scales).
The eyes have eyelids.
A tympanum represents the ear.
Alimentary canal, urinary and reproductive tracts open into a common chamber called cloaca which OPENS to the exterior.
Respiration is by GILLS, LUNGS and through skin.
The heart is three-chambered (two auricles and one ventricle).
These are cold blooded animals.
Sexes are separate.
Fertilisation is external.
They are OVIPAROUS and development is direct or indirect.
Examples : Bufo (Toad), Rana (Frog), Hyla (Tree frog), Salamandra (Salamander). Ichthyophis (Limbless amphibia).
Discussion
18511.

Question : Describe about nucleus in detail.

Answer»

Solution :Nucleus as a cell organelle was first described by Robert BROWN in 1831. Later the material of the nucleus stained by the basic dyes was given the name chromatin by Flemming
The interphase nucleus has highly extended and elaborate nucleoprotein fibres called chromatin, nuclear matrix and one or more spherical bodies called nucleoli.
The nuclear envelope which consists of two parallel membranes with a space between (10 to 50 nm) called the perinuclear space.
This forms a barrier between the materials present inside the nucleus and that of the cytoplasm. The outer membrane usually remains continuous with the ENDOPLASMIC reticulum and also bears ribosomes on it.

At a number of places the nuclear envelope is interrupted by minute PORES, which are formed by the fusion of its two membranes. These nuclear pores are the passages through which movement of RNA and protein molecules takes PLACE in both directions between the nucleus and cytoplasm.
Nucleoli : The nucleoli are spherical structures present in the nucleoplasm. The CONTENT of nucleolus is continuous with the rest of the nucleoplasm. It is a site for active ribosomal RNA synthesis.
Large and more numerous nucleoli are present in cells actively carrying out protein synthesis.
Discussion
18512.

Question : Describe about Endoplasmic Reticulum.

Answer»

Solution :Tiny tubular structures scattered in the cytoplasm of eukaryotic cells is called Endoplasmic RETICULUM (ER).
ER divides the intracellular space into TWO distinct compartments, i.e., luminal (inside ER)
The ER often shows ribosomes attached to their outer surface.
The endoplasmic reticulum bearing ribosomes on their surface is called ROUGH endoplasmic reticulum (RER)
In the absence of ribosomes they appear smooth and are called smooth endoplasmic reticulum (SER).

RER is frequently observed in the cells actively involved in protein synthesis and secretion.
They are extensive and continuous with the outer membrane of the nucleus.
The smooth endoplasmic reticulum is the major site for synthesis of lipid. In animal cells lipid-like STEROIDAL HORMONES are synthesised in SER.
Discussion
18513.

Question : Discuss "The respiratorypathway is an amphibolicpathway "

Answer»

Solution :GLUCOSE is the favoured substratefor respiration.
All carbohydrates are usually first converted into glucose before theyare USED for respiration.
Othersubstrates can also be respired, but then they do not enter the respiratory pathwayat the first STEP .

As per shownin figure, Fatswould need to be broken down into GLYCEROL and fatty acids first.
If fattyacids were to be respired they would first be degraded to acetyl CoAand enter thepathway.
Glycerol would enterthe pathway after being converted to PGAL.
The proteinwould be degraded by proteases ad the individual amino acids depending on their would enter thepathway at some stagewithin the Krebs. cycle or even as pyruvate of acetyl CoA .
Since respiration breakdownof substrates,the respiratory process has traditionallybeen considered a catabolicprocess and the respiratory pathway as a catabolic pathway.
Hence fatty acids would be broken down to acetyl CoA beforeentering the respiratorypathway when it is used as substate.
These vary compounds thatwould be withdrawn from the respiratory pathwayfor thesynthesisof the said substrates.
When the organism needsto synthesize fatty acid acetyl CoA would be withdrawnfrom the respiratory pathway for it
Hence, the respiratorypathway comes into the pictureboth during breakdown and synthesisor fatty acids.
Similarly during breakdownand synthesisof protein too, respiratoryintermediates forthe link.
Breaking down PROCESSES withinthe living organism is catabolism and synthesis is anabolism .
Because the respiratory pathway is involved in bothanabolism andcatabolism.
It wouldhence be better to consider therespiratorypathway an an amphibolic pathwayratherthenas a catabolic one.
Discussion
18514.

Question : Describe a papilionaceous corolla.

Answer»

Solution :1. Petals 5, apopetalous, unequal and PAPILIONACEOUS, vexillary or descendingly imbricate aestivation. All petals have a claw at the base.
2. The outer most petal is large called standard petal or vexillum. Lateral 2 petals are Ianceolate and curved. They are called wing petals or alae.
3. Anterior two petals are PARTLY fused and are called keel petals or carina which encloses the stamens and PISTIL.
It is SAID to be butterfly shaped.
Discussion
18515.

Question : Descibe external structure of hert.

Answer»

Solution :Hert is the mesodermally derived organ. It is situated in the thoracic cavity, in between the two LUNGS, SLIGHTLY tilted to the left.
It has the size of a clenched fist. It is protected by a double walled membranous bag, PERICARDIUM
It is enclosed with pericardial fluid.
HUMAN heart has four chamber, two relativly small upper chambers are called atria and two longer LOWER chambers called ventricles.
Discussion
18516.

Question : Descibe Chemiosmotic Hypothesis.

Answer»

Solution :To describe the process of synthesis of ATP in chloroplast, the chemiosmotic hypothesis has been put forward to explain the mechanism.
Like in respiration, in photosynthesis too, ATP synthesis is linked to the development of a proton gradient ACROSS a membrane.
These are membrances of the thylakoid.
Main difference : In respiration, protons `(H^(+))`accumulate in the intermembrane space of the mitochondria when electron move through the ETS.
Formation of protein gradient :
(a) Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions `(H^(+))` that are produced by the splitting of water accumulate within the lumen of the thylakoids.
(B) As electrons move through the photosystems, protons are transported across the membrane.
This happens because the primary acceptor of electron which is LOCATED towards the outer side of the membrane transfer its electron not to an electron carrier but to an H carrier.
Hence, this molecule removes a proton from the stroma while transporting an electron.
When this molecule passes on its electron to the electron carrier on the inner side of the membrane, the proton is released into the inner side or the lumen side of the membrane.
Reduction of NADP :
(c ) The NADP reductase enzyme is located on the stroma side of the membrane. Along with electrons that come from the acceptor of electron of PS I, protons are necessary for the reduction of `NADP^(+) "to" NADPH + H^(+)`. These protons are also removed from the stroma.
`underset("NADP Reductase")(NADP^(+)2H^(+)rarrNADPH+H^(+))`
Hence, within the chloroplast, protons in the stroma decreases in number, while in the lumen there is accumulation of protons. This creates a proton gradient across the thyakoid membrane as well as a measurable decreases in pH in the lumen.
Importance : This protein gradient is important because it is the breakdown of this gradient thatleads to release of energy. The gradient is broken down due to the movement of protons across the membrane to the through the transmembrance channel.

Stucture of ATPase enzyme :
`F_(0)` : The ATPase enzyme consists of two PARTS: ONE called the `F_(0)` is embedded in the membrane and forms a transmembrane channel that carreis out faciltated diffusion of protons across the membrane.
`F_(1)` : The other portion is called `F_(1)` and protrudes on the outer surface of the thylakoid membrane on the side that faces the stroma.
The break dwon of the gradient provides enough energy to cause a coformational change in the `F_(1)` particle of the ATPase, which makes the enzyme synthesize serveral molecules of energy - packed ATP.
Necessary elements of Chemoismosis :
(a) Chemiosmosis requires a membrane, a proton pump,a proton gradient and ATPase.
Energy is used to pump protons acorss a membrane,to create a gradient or a high concertration of protons within the tylakoid lumen.
ATOase has a channel that allows diffusion of protons back across the membrane, this releases enough energy to activate ATPase enzyme that catalyes the formation of ATP.
Requirement of Chemiosmosis :
Along with the NADPH produced by the movement f electrons, the ATP will be used immediately in the biosynthetic reactio taking place in the stroma, responsible for fixing `CO_(2)`, and synthesis of sugar `(C_(3) "pathway")`.
Discussion
18517.

Question : Descending limb of Henle's loop is permeable to water due the presence of ...

Answer»

SOLUTION :AQUAPORINS
Discussion
18518.

Question : Descending limb of Henle's loop, hypertonic urine, reabsorption of water, ascending limb of Henle's loop

Answer»


ANSWER :ASCENDING LIMB of HENLE's LOOP
Discussion
18519.

Question : Dermatogen gives rise to

Answer»

CORTEX
Epidermis
Stele
Root cap

Answer :B
Discussion
18520.

Question : Dermatogen, periblem and plerome are associated with

Answer»

INTERCALARY MERISTEM
Promeristem
Lateral meristem
ROOT APICAL meristem

Answer :D
Discussion
18521.

Question : Dermal tissues, ground tissues and vascular tissues are formed from

Answer»

INTERCALARY MERISTEMS
Apical meristems
Fascicular vascular CAMBIUM
Inter fascicular cambium

Answer :B
Discussion
18522.

Question : Depolarization means reversal of polarity. Mention the ionic charge on both sides of axolemma.

Answer»

Solution :Depolarization - Reversal of polarity
When a nerve fibre is stimulated, sodium voltage-gate opens and makes the axolemma permeable to `Na^(+)` ions, meanwhile the potassium voltage gate closes. As a result, the rate of LOW of `Na^(+)` ions into the AXOPLASM exceeds the rate of flow of `K^(+)` ions to the outside fluid [ECF]. Therefore, the axolemma BECOMES POSITIVELY charged inside and negatively charged outside. This reversal of ELECTRICAL charge is called Depolarization.
Discussion
18523.

Question : Depolarisation of axolemma during nerve conduction takes place because

Answer»

EQUAL amount of `Na^+ and K^+` move out across axolemma
only `Na^+` move INSIDE
more `Na^+` moves OUTSIDE than `K^+` moving outside
None of these

Solution :NA
Discussion
18524.

Question : Depictthe tissuelineageduringsecondarygrowthin dicotstem and root .

Answer»

SOLUTION :
Discussion
18525.

Question : Dependent gametophytes on sporophyte are seen for the first time in

Answer»

GYMNOSPERMS
Angiosperms
Pteridophytes
BRYOPHYTES

ANSWER :A
Discussion
18526.

Question : Deoyribonuclease, ribonnuclease and Carboxypeptidase are secreted by

Answer»

LIVER
STOMACH
Pancreas
Stomach

Answer :C
Discussion
18527.

Question : Dental formula of permanent teeth in man is ………..

Answer»

2122/2122
2123/2123
2102/2102
2122/2022

Answer :B
Discussion
18528.

Question : Dental formula shows ………… .

Answer»

STRUCTURE of TEETH
Monophyodont or diphyodont CONDITION
Number and type of teeth in both jaws
Number and type of teeth in one HALF of both jaws

Solution :N/A
Discussion
18529.

Question : Dental formula in human beings is

Answer»

`(3,2,2,3)/(3,2,2,3)`
`(2,1,2,3)/(2,1,2,3)`
`(1,2,3,2)/(1,2,3,2)`
`(2,2,3,3)/(2,2,3,3)`

SOLUTION :`(2,1,2,3)/(2,1,2,3)` 2-Incisors, 1-Canine, 2-Premolar, 3-Molar.
Discussion
18530.

Question : Densely stained reticular structure found near the nucleus in a eukaryotic cell, is

Answer»

ENDOPLASMIC reticulum
Cytoskeleton
GOLGI apparatus
Vacuoles

ANSWER :C
Discussion
18531.

Question : Dengue fever can be thrombocytopenia due to direct infection of

Answer»

THROMBOCYTES 
Bone marrow MEGAKARYOCYTES 
LYMPHOCYTES 
WBC 

ANSWER :B
Discussion
18532.

Question : Dendrochronologydealswith determiningthe ………….

Answer»

GROWTHOF TREES
AGE of trees
Branchingpattern of trees
Growth and climaticinfluencein trees

Answer :B
Discussion
18533.

Question : Dendrochonology is used for determining the age of

Answer»

FOSSILS
ROCKS
TREES
Annuals

Answer :C
Discussion
18534.

Question : Dendrites transmits impulses away/towards the cell body.

Answer»

SOLUTION :TOWARDS
Discussion
18535.

Question : Dendrites : Bring impulse to cyton :: axon :

Answer»

SOLUTION :CARRY IMPULSE from CYTON
Discussion
18536.

Question : Demonstration of normal body temperature in persons with black bile

Answer»

supported the 'humor' hypothesis of health
INDICATED that they BELONG to HOT personality
disproved the 'humor' hypothesis of health
indicated the significance of 'humors'.

ANSWER :C
Discussion
18537.

Question : Demonstrate an experiment to prove osmosis.

Answer»

Solution :Thistle funnel EXPERIMENT :
(i) Mouth of a thistle funnel is tied with goat bladder. It acts as a semipermeable membrane.
(ii) Pour CONCENTRATED SUGAR solution in the thistle funnel and mark the level of solution.
(iii) PLACE this in a beaker of water. After some time water level in the funnel RISES up steadily.
(iv ) This is due to the inward diffusion of water molecules through the semipermeable membrane .
Discussion
18538.

Question : Explain Kuhne's fermentation experiment.

Answer»

Solution :Demonstration of ALCOHOLIC fermentation
Take a Kuhne's fermentation TUBE which consists of an upright glass tuve with side bulb. Pour 10% sugar solution mixed with baker's yeast into the fermentation tube the side tube is filled plug the mouth with lid. After some time, the GLUCOSE solution will be FERMENTED .The solution will give out an alcoholic smell and level of solution in glass column will fall due to the accumulation of `CO_2` gas. It is due to teh presence of zymase enzyme in yeast which CONVERTS the glucose solution into alcohol and `CO_2`. Now introduce a pellet of KOH into the tube, the KOH will absorb `CO_2` and the level of solution will rise in upright tube.
Discussion
18539.

Question : Explain Kuhne's experiment and the action of enzyme zymase.

Answer»

Solution :DEMONSTRATION of alcoholic fermentation :
Take a Kuhne'sfermentation tubewhichconsistsof anuprightglass tubewith sidebulb .Pour10 % sugarsolutionmixedwithbaker'syeastinto thefermentationtubethe sidetube is filledplugthe mouthwithlid . After some time ,theglucose colution will befermented. The solution willgiveout an alcoholic smell and levelof solution in glass column will FALL due to the accumulationof `CO_(2)`GAS . It is dueto the presenceof zymasa enzyme in yeast whichconverts the GLUCOSE solution into ALCOHOL and `CO_(2)`. Nowintroduce a pellet of of KOHintothe tube ,the KOH will absorb`CO_(2)` and the levelof solution will risein upright tube .
Discussion
18540.

Question : Demineralisation is caused due to .....

Answer»

SOLUTION :(HYPERPARATHYROIDISM)
Discussion
18541.

Question : Degradation of detritus into simple inorganic substances by the enzymatic action of bacteria and fungi is called

Answer»

Fragmentation
Leaching
Catabolism
Humfication

Answer :C
Discussion
18542.

Question :Degeneration and destruction of liver cells resulting in abnormal blood vessel and bile duct leading to the formation of fibrosis is known as ………..

Answer»

GALL stones
jaundice
HERNIA
LIVER CIRRHOSIS

Solution :N/A
Discussion
18543.

Question : Definnitions / Explanation : Thrombus

Answer»

Solution :Fatty substances accumulate in injured part of ARTERY and FORMS PLAQUE. Nthis plaque if remains permanent then it is called THROMBUS.
Discussion
18544.

Question : Definnitions / Explanation : Embolus :

Answer»

Solution :Plaque formed on the WALL of ARTERY when LEAVES its PLACE and circuates with BLOOD then it is called embolous.
Discussion
18545.

Question : Definitions Root Nodules

Answer»

Solution :SMALL nodules located on the roots of leguminous PLANTS are CALLED ROOT nodules.
Discussion
18546.

Question : Definitions Nitrogen fixation

Answer»

Solution :The PROCESS of conversion of nitrogen `(N_2)` to AMMONIA is termed as nitrogen FIXATION.
Discussion
18547.

Question : Definitions Necrosis

Answer»

Solution :The DEATH of tissues of leaves is CALLED NECROSIS.
Discussion
18548.

Question : Definitions Mobilised elements

Answer»

Solution :The elements that are actively MOBILISED within the PLANTS and EXPORTED to young DEVELOPING tissues. These are CALLED mobilised elements.
Discussion
18549.

Question : Definitions Micronutrient

Answer»

Solution :The elements which are NEEDED in very small amoungs class then `10M MOL e kg^-1` of dry water are called micronutrients.
Discussion
18550.

Question : DefinitionsMineral nutrition:

Answer»

Solution :The ABSORPTION of minerals , DISTRIBUTION and metabolism of plants is CALLED mineral nutrition.
Discussion