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18101.

Question : Write a not an DNA fingerprinting.

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Solution :The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening SEQUENCES (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers. DNA finger printing involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, asmall stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density GRADIENT centrifugation. The bulk DNA forms a major peak and the other SMALL peaks are referred to as satellite DNA. Depending on base composition (A: Trich or G: Crich), length of segment and number the of repetitive units, satellite DNA is classified into many sub - categories such as micro-satellites and mini- satellites, etc. These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of DNA POLYMORPHISM and form the basis of fingerprinting. DNA isolated from blood, hair, skin CELLS, or other at genetic evidences left the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.
Discussion
18102.

Question : Describe the post fertilization changes in a flower.

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Solution :Zygote is formed INSIDE the ovule and develops into embryo.
Ovule develops into seed.
Ovary develops into fruit, ovary wall become pericarp which enlcloses seeds.
NUCELLUS persistent in seeds of SOM especies and KNOWN as perisperm.
The sepals, petals and STAMENS wither and fall off.
Discussion
18103.

Question : Describe the post - fertilisation changes in a flower.

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Solution :Fertilisation is the PROCESS of the fusion of male and female gametes to form a DIPLOID zygote. After fertilisation, the following change occur in a FLOWER :
the zygote is FORMED inside the ovule and develops into the embryo.
The OVULES develop into the seed.
The ovary develops into the fruit.
The ovary wall is transformed into a protective layer called the pericarp.
Discussion
18104.

Question : Tabulate any 4 post fertilization changes in a flower.

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Solution :POST fertilization changes in a flower
After fertilization the sepals, petals and stamens of a flower wither and fall off. The pistil remains attached. The OVARY develops into FRUIT and ovule develops into the seed. The zygote develops into the EMBRYO. After DISPERSAL, seeds germinate under favourable conditions to produce new plants.
Discussion
18105.

Question : Describe the physiology or mechnismof urine formation.

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Solution :Physiology of urineformation /Mechnism of urine formation : Urineformationtakes place in three stages. These areultrafiltration , selective reabsorption and tubularsecretion.
(I) Ultrafiltration :
(1) This is a pysical process taking place in the Malpighian body. The diameter of afferentrenal arterioleis largerthan thediameterof efferentrenal arteriole and hencethere is hydrostaticpressuredevelopedon the contents inside the blood vessels in the GLOMERULUS. This pressure is alled effectivefiltration pressure or EFP.
(2) `99%` of the filtrate is reabsorbed and only 1 to 1.5 litre of urine is formed per day .
(3) Outer parietal layer and inner visceral layer of Bowman'scapsule enclose aspace called urinaryspace. The innr visceral layer contains podocytes which are specialized CELLS.
(4) Due to semiprmeable glomerular endothelial MEMBRANE, low molecular weight substances are pulled out of the blood and enter into urinary space. Thisfiltrate is blood plasma exceptproteinsand blood corpuscles.
(II) Selective reabsorption.
(1) In this process 99% of the filtrate is reabsorbed.
(2) There aretwo processes involvedin this step :Passive transport for osmosisand active transport. Active transport uses ATP molecules since it is against the concentratioin gradient.
(3)When the filtrate passes through the renal TUBULE there is exchange of substances betweenthe surroundingblood and renal tubular fluid.
(4) Highthreshold substances e.g.,glucose, amino acids, potassium and calciumions, etc. are completely reabsorbed. Urea anduric acid are low threshold substances. They are reabsorbednegligibly.
(5) Waterus reabsorbed everywhere except in the ascendingloopof Henle. This absorption occurs by osmosisand is known as obligatoryabsorptionof water.
(6) Amino acids and ions are pumped out from PCT. Glucoseis maximally reabsorbedin healthy persons sothat no glucosemolecules remainin the urine.
(III) Tubular secretion :
(1) Someabnormal and unwanted substanceswhich have passed into peritubular capillaries are secreted back in DCT,duringthis process.
(2) Nitrogenous substances like creatinine and ions like potassium and hydrogenare also takenbackin the lumenof uriniferous tubule through for peritubular capillaries.
(3) Secretion of hydrogen ions is in important homeostatic mechanism to keep the balance of the acidityof the blood.
(4) Sodium ion concentrationis maintainedby aldosterone while calcium ion concentration is maintaindby calcitonin and parathor-mone.
(5) If any toxicand abnormalconstituent is present in the body it is secreted back into urine, in this stage.
Discussion
18106.

Question : Describe the photoexcitation of chlorophyll-a in photosynthesis.

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Solution :(1) Chlorophyll-a is capable of converting the physical energy of light into chemical energy. It also acts as reaction centre of photosynthesis. Therefore chlorophyll-a is considered as an essential pigment.

(2) Initially, the chlorophyll-a lies at the ground level. But when it absorbs photons (solar energy) it gets activated and becomes energy RICH. This state of chlorophyll is called the excited state.
(3) In excited state, the chlorophyll-a expels an electron. The expelled electron is energy rich as it possesses extra amount of energy.
(4) With the loss of electron `(e^(-))` the chlorophyll-a becomes positively charged. This state of chlorophyll-a is called ionized state. The molecule of chlorophyll-a cannot remain in ionized state for more than 10 seconds.
(5) The energy rich expelled electron is TRANSFERRED through various acceptors and donors.
(6) During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy PRESENT in the electron is converted into chemical energy in the form of ATP.
(7) There are TWO forms of chlorophyll-a, viz. `P_(700)` and `P_(680)` which form the reaction centres of PS I and PS II respectively for the photochemical reaction.
Discussion
18107.

Question : Describe the pathway follway by phosphorus in an ecosystem.

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Solution :(1) The phosphorus cycle is the simplest of all the nutrient CYCLES, which is a type of sedimentary cycle.
(2) Since phosphorus is a heavy molecule it never goes into the atmosphere. Phosphorus remains in the bodies of ORGANISMS, dissolved in water or in the form of rock.
(3) Upon weathering of the rock due to ACTION of mildly acidic water, the phosphates in the rock go into the solution. Phosphorus is a major biological constituent of all the living organisms. It is found in DNA, RNA, ATP, etc. From this it is obvious that phosphorus is an essential element.
(4) Plants take up phosphorus in the form of phosphate. The roots of plants can absorb phosphate ions from the soil. Animals obtain phosphorus through food which they consume. Thus AUTOTROPHS supply phosphorus to heterotrophs.
(5) The requirement of phosphorus in animals is much more as it is the component of bones, teeth, shells, etc.
(6) Upon death of plant or animal and also through defecationof the animals, phosphate returns to the soil due to the action of decomposers. This in turn is used up by other growing plants.
Discussion
18108.

Question : Describe the parts of a biogas plant with a labelled diagram.

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Solution :
The biogas contain methane `60%` and `40%` CO,. Its production is as follows.
1) The biogas plant consist of 10 - 15 feet deep concrete tank, in which the slurry of the dung is FED.
2) A floating cover is placed over the slurry which keeps on rising as the gas is produced in the tank by the microbial activity. The biogas plant has an outlet that is connected to pipe, for removal of biogas and supply of it to the REQUIRED places.
4) There is also an outlet to remove the spent slurry which can be used as fertiliser.
Discussion
18109.

Question : Describe the out breeding devices that prevent the auto-gamy.

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Solution :Continued self-pollination results in inbreeding depression, flowering plants have developed out-breeding devices to discourage self-pollination and to encourage cross-pollination.
Pollen grains are released before the stigma becomes RECEPTIVE or stigma becomes receptive before pollens are released.
2. In some species the anthers and stigma are PLACED in different position, so that pollen can not come in contact with stigma of the same flower.
3. Self-incompatibility, it is a genetie mechanism were pollens are PREVENTED from fertilizing the ovule by inhibiting pollen tube GERMINATION.
4. By-production of unisexual flowers.
Discussion
18110.

Question : Describe the number of species found on the earth.

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Solution :According to International Union of Conservation of Nature (IUCN) 2004, the total number of plants and animals species described so far is more than 1.5 million.
For many taxonomic groups, species inventories are more complete in temperate than in TROPICAL countries.
Assuming that a large number of species are undiscovered in the tropics, biologists make a statistical comparison of the temperate - tropical species RICHNESS of an exhaustively studied group of insects and extrapolate this ratio to other groups of animals and plants.
This leads to a gross estimate of the total number of species on earth ranges from 20 to 50 million.
A more conservative and scientifically sound estimate by Robert May places the global species diversity at about 7 million.
More than `70%` of all the species recorded are animals while plant comprise not more than `22%` of the total.
Insects are the most species rich taxonomic group in animal kingdom making more than `70%` of the total animals.
It means out of every 10 animals, 7 are insects on the earth.
Number of fungi species in the world is more than the total number of species of fishes, amphibians, reptiles and mammals.
But biologist are still not sure about the diversity of prokaryotic species like prostists, orchaea bacteria etc.
This is because of the following reasons :
(i) The conventional taxonomic methods are not sufficient for identifying these microbial species,
(ii) Most of the species can not be cultured under LABORATORY conditions.
(iii) If molecular and BIOCHEMICAL criteria are adopted for delineating the microbial species, this would put the diversity into MILLIONS.
Discussion
18111.

Question : Describethe morphology ofthyroidgland. Add a noteon functionsof thyroidhormonesin humans.

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SOLUTION :Morphology of thyroidgland :
(1) Thyroidis the largestendocrinegland in the body. Weighingto aboutto and measuresaboutin lengthand cm inwidthbut it canvaryin size asperage, sexand diet.
(2) It islocatedin the neckregionanteriorlyjust belowthe larynxand situated ventrolaterallyto thetrachea.
(3) Itis reddishbrown,H-shaped bilobed and highly vasculargland.
(4) THETWO lobesare joinedby connectivetissuecalled isthmuswhich islocatedat 2ndto 4th tracheal cartilage.

2. Functions of Thyroidhormones :
(1) Thyroxineis the mainmetabolic hormone in the bodythat maintains basalmetabolicrate (BMR)by increasing glucoseoxidation. It bringsabout calorigenic effectby energyproduction.
(2) Thephysicalgrowth, development of gonads anddevelopment of mentalfacultiesis under THECONTROL of thyroxine.
(3) Body weight, respirationrate,HEARTRATE,bloodpressure, temperature , digestionand proteinsynthesis , ETC. are regulated by thyroxine.
(4) Thyrocalcitoninregulatesbloodcalcium level.
Discussion
18112.

Question : Describe the molecular mechanism of RNA modification.

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Solution :The processing of pre-mRNA to mature mRNA/Molecular mechanism of RNA MODIFICATION.
In eukaryotes THREE major types of RNA, mRNA, tRNA and rRNA are produced from a precursor RNA MOLECULE termed as the primary transcript or preRNA. The RNA polymerase II transcribes the precursor of mRNA, which are also called the heterogenous nuclear RNA or hnRNA which are processed in the NUCLEUS before they are transported into the cytoplasm.
(i) Capping Modification at the 5' end of the primary RNA transcript (hn RNA) with methylguanosine triphosphate is called capping.
Purpose of capping
(a) Protects RNA from degradation.
(b) Capping plays an important ROLE in removal of first intron in pre mRNA.
(c) It regulates the mRNA export from the nucleus into the cytoplasm.
(d) It helps in binding of mRNA to the ribosome.
(ii) Tailing/Polyadenylation
The 3' end of hnRNA is cleaved by an endonuclease and a string of adenine nucleotides is added to the 3' end of hnRNA (pre mRNA) is known as Poly (A) tail : Polyadenylation. This process is called tailing or polyadenylation.
Purpose of tailing:
(a) Translation of RNA transcript is facilitated.
(b) Helps in the synthesis of polypeptides.
(c) It enhances the mRNA stability in the cytoplasm.
The protein coding regions are not continuous in eukaryotes. Exons are the coding sequences or expressed sequences contain biological informations in the matured processed mRNA. Introns are intervening sequences, which are non- coding sequences (non-amino acid- coding sequences) that should be removed from a gene before the mRNA product is made. These exons and introns are known as Split Genes.
(iii) RNA splicing in plants
(a) RNA splicing is a process which involves the cutting or removing out of introns and knitting of exons. This process takes place in spherical particles which is a multiprotein complex called SPLICISOMES. It is approximately 40 - 60 nm in diameter.
(b) A spliceosome removes the introns with an enzyme ribozyme. Now the mature mRNA comes away from the spliceosomes through the nuclear pore and is transported out from the nucleus into cytoplasm, and gets attached to ribosome to carry out translation.
Discussion
18113.

Question : Describe the method of artificial hybridisation

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Solution :In artificial HYBRIDISATION only the desired pollen grains are USED. This is ACHIEVED by emasculation and bagging techniques. If the female parent bears bisexual flowers, removal of anthers from the flower BUD before the anther dehisces using a pair of forceps is necessary. This tep is referred to as emasculation. Emasculated flowers have to be convered with a bag of suitable size, generally made up of butter paper to prevent contamination of its stigma iwht unwanted pollen. This process is called bagging. When the stigma of bagged flower attains receptivity mature pollen grain collected from anthers of the male parent are dusted on the stigma and the flowers are rebagged and the fruits allowed to develop. If the female parent produces unisexual flowers, there is no need for emasculation. The female flower buds are begged before the flowers open. When the stigma becomes receptive, POLLINATION is carried out using the desired pollen and the flower rebagged. This is one of the major approaches of crop improvement programme
Discussion
18114.

Question : Describe the mechanism of anaerobic respiration.

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Solution :1. Anaerobic respiration : Respiration in the absence of molecular oxygen is called anaerobic respiration.
2. Mechanism of anaerobic respiration : Anaerobic respiration occurs in the following three steps :
(1) Glycolysis : During glycolysis, GLUCOSE is broken down to two molecules of pyruvate, two molecules of `NADH_(2)` and two molecules of ATP ALONG EMP pathway as shown below: `C_(6)H_(12)O_(6) + 2NAD + 2ADP + 2iP rarr 2CH_(3)CO.COOH + 2NADH_(2) + 2ATP`
(2) Decarboxylatlon: The pyruvate undergoes decarboxylation to form acetaldehyde. This reaction occurs in the presence of pyruvate decarboxylase enzyme which REQUIRES TPP (THYMINE pyrophosphate) as coenzyme and `Zn^(++)` as cofactor as shown below :

(3) Reduction : The acetaldehyde is finally reduced to ethyl alcohol by accepting hydrogen atoms from NADH2 produced during glycolysis. This reaction occurs in the presence of the enzyme ethanol dehyd rogenase as shown below :

During this process, since ethyl alcohol is produced, as end product, it is used in fermentation industries for the production of alcohol from molasses. It is also used in bakeries. The overall equation of anaerobic respiration can be summarized :
`C_(6)H_(12)O_(6) + rarr 2C_(2)H_(5)OH + 2CO_(2) + 2ATP`
Discussion
18115.

Question : Describe the major STDs and their symptoms.

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Solution : Cervical cancer
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most COMMON symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include
1. Having multiple sexual partners
2. Prolonged use of contraceptive pills Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be USED to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.
Modern screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with VACCINATION. Primary prevention begins with HPV vaccination of girls aged 9 – 13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. HEALTHY diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.
Discussion
18116.

Question : Explain Biological control of pests and diseases.

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Solution : Biological control means life against life. It is a natural and eco-friendly concept. It employs the natural ORGANISMS to control the population of pathogens and pests in an ECOSYSTEM. Classical examples are Trichoderma which is antagonist against many soil borne plant pathogens. Similarly, Penicillium inhibits the growth of Staphylococcus and therefore has been SUCCESSFULLY used in the production of Penicillin antibiotic to control many HUMAN BACTERIAL pathogens.
Discussion
18117.

Question : Name the phases in the life cycle of Plasmodium .

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Solution :Plasmodium vivax is a digenic parasite,involving two hosts,man as the secondary host and female anopheles mosquito as the primary host.The life cycle of plasmodium involves three phases namely schizogony,gamogony and sporogony.The parasite first enters the human blood stream through the bite of an infected female the sporozoite within the blood stream immediately enters the hepatic cells of the liver.Further in the liver they undergo multiple asexual fission(Schizogony)and produce merozoites.After being released from liver cells,the merozoites penetrate the RBC.s.Inside the RBC ,the merozoite begins to develop as unicellular trophozoites.The trophozoite grows in size and a central vacuole develops pushing them to one side of cytoplasm and becomes the signet ring stage .The trophozoite nucleus then divides asexually to produce the schizont.The large schizont shows yellowish -brown pigmented granules called Schuffners granules .The schizont divides and produces mononucleated merozoites.Eventually the erythrocyte lyses,releasing the merozoites and haemozoin toxin into the blood stream to infect other erythrocytes.Lysis of red blood cells RESULTS in cycles of fever and other symptoms.This erythrocytic stage is cyclic and repeats itself aproximately every 48 to 72 hours or longer depending on the species of plasmosium involved .The sudden release of merozoites triggers an attack on the RBCs.
Occasionally ,merozoites DIFFERENTIATE into macroametocytes and microgametocytes.When these are ingested by mosquitoe,they develop into MALE and female gametes respectively.In the mosquito.s gut,the infected erythrocytes lyse and male and female gametes ferilize to form a diploid zygote called okkinete .The ookinete migrates to the mosquito.s wall and develop into an oocyte.The oocyte undergoes meiosis by a process called sporogony to form sporozoites.These sporozoites migrate to the salivary glands of the mosquito .The cycle is now complete and qhen the mosquito bites another human host,the sporozoites are injected and the cycle begins a new.
The PATHOLOGICAL chages caused by malaria ,affects not only the erythrocytes but also the spleen and other visceral organs.Incubation period of malaria is about 12 days.The early symptoms of malaria are headache,nausea and muscular pain.The classic symptoms first develop with the synchronized release of merozoites,haemozoin toxin and erythrocyte debris into the blood stream resulting in malarial paroxysms-shivering chills,high fever followed be sweating .Fever and chills are caused partly by malarial TOXINS that induce macrophages to release tumour necrosis factor (TNF-`alpha`) and interleukin.
Discussion
18118.

Question : Describethe internalstructureof human ear.

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Solution :(1)The EAR iscomposedof the outer ear, middle ear andinternal ear.
(2) The external earconsistsof the pinnaand externalauditorymeatus (canal). The pinnacollectssoundwaves comingfrom theenvironment.The externalauditory canal ISTHE circulartube leading insideup to the eardrum. (tympanic membrane).
(3) The ear drum isformedof connective tissueshavingouter skin cover andinner mucus membrane.
(4) Themiddle ear consists of chain of three ossiclescalled malleus,incus andstapes.The malleusis attached totympanicmembraneand the stapesis connectedto the oval window of theinternal ear.They helpin the transmission of soundwaves fromexternalauditorycanalto internalear.
(5) Connectingmiddle earwith the pharynx is aneustachiantube which HELPSIN equalizingthe AIR pressureon eitherside of thetympanicmembrane.
(6) The internal ear isfluidfilledstructurecalled bonylabyrinthand the membranouslabyrinth.
(7) The membranes consist of coiledcochlea,the reissner'smembraneand basilarmembranes.Thesemembranesdivide thesurroundingperilymphfilled bonylabyrinth into an upperscalavestibule and a lowerscalatympani.
(8)The space within COCHLEA which is knwon as scala media is filled with endolymph. The scala vestibule ends at the oval window at the base of cochlea.
(9) Thescalatympani terminates at the roundwindow whichopensto themiddle ear.The organ of thecorti is locatedonthe brasillarmembrane.It containsthe haircells whichact asauditory receptors.A thinelasticmembraneprojects above therows of thehairscellscalled tectorialmembrane.
(10) Abovethe cochlea. the internalear alsocontainsvestibularapparatus.It consistsof threesemicircularcanals whcihare in perilymph and the otolith organsformed of thesacculus andutriculus.
(11) Thebases of canalsare swollenand are calledampullae.
Whichcontain a projecting ridgeknow as crista ampullariswhichcontainhair calls.
(12)The sacculus and utriculus also haveprojectingridgecalledmacula. The cristaand maculaare the specificreceptorsof vestibular apparatus . They are responsiblefor maintenance of bodyposture and the balance.
.
Discussion
18119.

Question : Describe the inheritance of human skin colour.

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Solution :(1) The inheritance of skincolour in humans is polygenic .
(2) DavenportandDavenportstudied the inheritance of skincolour of offspring after a marriagebetween a black and a white .
(3) The offsprings derived from this marriage show intermediate skin colour. Such individuals withintermediate skin colour are called mulattoes.
(4) When there is a marriage between two mulattoes, all shades of colour are observed in their offsprings in the ratio of
1 : 6: 15 : 20 : 15 :6 :1
(5) From the above results it is obvious that skin colour in humans is controlled by there pairs of genes, viz, Aa , Bb and Cc.
(6) The colour of skinin human beings is determined by the presence of pigment called melanin.
(7) The synthesis of melanin in the skin is controlled by dominant genes which are additive in their effect.
(8)Greater the number of dominant genes greater is the amountof melanin synthesized.
(9) The black has the genotype AABBCC and the albino has the genotype aabbcc.
(10) The offspring (mulatto)possesses the heterozygous geno TYPE AaBbCc.
(11) . Mulattoes produce eight different types of gametes and in `F_(2)`generation there are SIXTY four combinations. Owing to the cumnlativeof each dominant gene the results are as shown below :
(i) Pure black -1 /64(6 dominant genes)
(ii) Black ( LESS dark ) - 6/64(5 dominantgenes )
(iii)Lesser black or brown -15/64(4 dominantgenes)
(iv) Mulatto (Intermediate - Sanwla ) 20/64 (3 dominant genes)
(v) Fair 15/64 (2 dominant genes)
(vi)Very fair 6/64 (1 dominantgene)
(vii) Pure white(albino) 1/64(No dominantgene)
Discussion
18120.

Question : Describethe histology of human testis.Write a note on humansperm.

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Solution :I. Histological structure of tesits :
(1)The external covering oftestisis a fibrousconnectivetissuecalledtunicaalbuginea.
(2)Then thereis an incompleteperitoneal coveringcalledtunicavaginalis.
(3) Interiorto thisthere is a covering calledtunicavascularisformed bycapillaries .
(4)The testis is composedof manyseminiferous tubulesthat are linedby cuboidalgerminalepithelial cells.

(5) In the seminiferous tubules variousstages of developingspermsare seenas spermatogenesistakes place here.These stagesare spermatogonia, primaryand secondaryspermatocytes, spermatidsand sperms.
(6)Interruptedbetwengerminalepithelium arelarge,pyramidal subtentacularcells,nursecellsor Sertoli cells.Sperm BUNDLES remainattached to Sertoli cellswith theirheads.
(7) Seminiferoustubulesform spermswhereasSertoli cells provide nourishment to the sperm till maturation.
(8) In BETWEENTHE seminiferoustubulesthere areinterstitial cells of Leyding which areendocrine in nature. Theysecrete testosterone.
II.Structureof sperm :
The spermconsists of HEAD, neck, middle piece and tail.
(1) Head :
(i)Head isthe main part whichis flat and ovaland has a largenucleusand an anrosome.
(ii) Acrosome is formedfrom Golgi complex.It secretesenzymeshyaluronidase which helps in penetration of the egg during fertilization.
(2) Neck : Neck isshort regionhaving twocentrioles.
(i) The proximal centrioleplays a role in first cleavage of zygote.
(ii) The axialfilamentof the spermis formedby the distal centriolegivesrise to .
(3) Middle piece : ltbdgt (i) Middlepieceacts as a powerhouse for spermbecausemitochondria PRESENT here supplies energyto swimming sperm.
(ii) It bears manyspirally coiled mitochondria orNebenkernaroundthe axialfilament.
(4) Tail :
(i) The tail is formedof cytoplasm and is long, slenderand taperingstructure.
(ii) The axial filament is a finethread-like structurethat arisesfrom the distal centrioleand traverses the middlepieceand tail.It has 9+2 ARRANGEMENT.
(iii) Taillashes and helpsthe spermatozoa to swim.
Discussion
18121.

Question : Describe the technique which is used in the separation and isolation of DNA fragments to be used in recombinant DNA technology.

Answer»

Solution :DNA FRAGMENTS are negatively changed, they move towards the anode under the electric field through the medium and separate according to their size due to the SIEVING effect of AGAROSE gel.
The separated fragments can be visualised by staining the DNA with ethidium bromide followed by exposure to UV radiation.
Elution is the PROCESS in which the separated bands of DNA are cut out from the gel and extracted.
Discussion
18122.

Question : Shape ofheart in rabbit.

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Solution :(i) The heart is pear shaped and lies in the thoracic cavity in between the lungs. It is ENCLOSED b y pericardium, a double layered MEMBRANE.

(ii) The , heart is four chambered with two aurides and two ventricles. The right and LEFT auricles are separated by interauricular septum , similarly right and left ventricles are separated by interventricular septum.
(iii) The right auricle opens into the right ventricle by right auriculoventricular aperture , guarded by a trkuspid ,ALVE.
(iv) The left auricle opens into the left ventricle by left auriculoventricular aperture guarded by a bicuspid valve or mitral vahe.
(v) The opening of the pulmonary artery and aorta are guarded by three semilunar VALVES.
(vi) The right auricle receives deoxygenated blood through two precaval (superior vena cava) and one postca val (inferior vena cava) veins from all parts of the body.
(vii) The left auricle receives oxygenated blood from the pulmonary veins from the lungs.
(viii) From the right ventricle arises pulmonary trunk which carries the deoxygenated blood to the lungs and from the left ventricle arises the systemic arch (aorta) which supplies oxygenated blood to all parts of the body.
Discussion
18123.

Question : Describe the formation of first cell.

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Solution :The first non-cellular life forms must have been originated 3 billion years ago in the form ofgiant molecules like RNA, Protein and polysaccharides inside a SELF replicating metabolic capsule.
These capsules must have later reproduced their molecules.
The first cellular form of life originated about 2000 million years ago.
These might have been single cells formed in aquatic ENVIRONMENT only.
This theory of ABIOGENESIS Le. the formation of first form of life arose slowly through the evolutionary FORCES acting on non-living molecules was accepted by a great majority.
Therefore, once formed, these single cells must have been EVOLVED into the diverse complex biodiversity of today.
Discussion
18124.

Question : Describe the Frederick Griffith's experiment to show transformation in bacteria.

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Solution :In 1928 Frederich Griffith conducted some EXPERIMENTS to show that DNA is the molecule of inheritance `ul("Streptococcus") ul("pneumoniae")` has two strains virulent (or) smooth strain (S-TYPE) with gelatinous coat. The virulent is capable of producing the disease. The other strain is NON-virulent (or) rough type (R-type) without gelatinous coat, it is not capable of producing the disease.
1. When the virulent (S-type) pneumonia were injected into a mice, the mice died.
2. When the non virulent pneumonia were injected into a mice, the mice continued to live.
3. The virulent pneumonia was heated and killed and then it was injected into MOUSE,
4. The non virulent pneumonia and heat killed virulent pneumonia were together injected into the mouse, then the mouse died.
The heat killed virulent bacteria transferred the inheritance molecule into nonvirulent BACTERIAL, this process is called transferred, as a result mouse died of the disease.
Discussion
18125.

Question : Describe the experiment of Hershey and Chase to prove that DNA is the genetic material.

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Solution :(1) Hershey and Chase performed experiments on BACTERIOPHAGES which are VIRUSES that infect bacteria. Bacteriophages are composed of DNA and protein.
(2) When they infect the bacterial cell, they inject their genetic material into bacteria. For finding out what is the nature of genetic material. Hershey and Chase used radioactive phosphorus `(P^(32))` in the medium for some viruses and for other viruses they used radioactive sulphur `(S^(35))`.
(3) DNA has phosphorus but no sulphur whereas proteins have sulphur but no phosphorus. Thus some viruses were with radioactive DNA (those labelled with `S^(35)`).
(4) Both radioactive labelled viruses were used to infect E.coli bacteria. After some time of infection, the viruses were centrifuged and their viral coats were separated.
(5) Bacteria which were infected with viruses having radioactive DNA also show radioactive characters.
(6) Bacteria which were infected with viruses having radioactive protiens did not show radioactive characters. This shows that DNA enters the bacterial cells at the time of INFECTING and not the protien.
Discussion
18126.

Question : How did Hershey and Chase prove that DNA is the hereditary material?

Answer»

Solution :Hershey and chase worked with VIRUS that inflects bacteria called bacteriophage.
They grew some viruses on radio active phosphorous medium and some on radio sulphur. Virus grown in the presence of radio active phosphorous contained radioactive DNA & those grown in radioactive sulphur contained radio active PROTEIN.

Radio active phages were allowed to attach to the E.coli bacteria. As the INFECTION proceeded, the viral coats were removed from the bacteria by agitating them in a blender, The viral particles were separated from bacteria by spinning them in a centrifuge.
Bacteria which was infected with virus that had radio active phosphorous were radio active, indicating that DNA was the MATERIAL that has passed from virus to bacteria. Bacteria that was infected with radioactive sulphur were not radio active. There fore DNA is the genetic material that has passed from virus to bacteria.
Discussion
18127.

Question : Describe the evolutionary history of plant and vertebrates through geological period.

Answer»

Solution :About 2000 million years ago (mya) the first cellular forms of life appeared on earth.
The mechanism of how non-cellular aggregates of giant macromolecules could evolve into cells with membranous envelop is not known.
Some of these cells had the ability to release `O_2`.
The reaction could have been similar to the light reaction in photosynthesis where water is split with the help of solar energy captured and channelised by appropriate light harvesting pigments.
Slowly single - celled organisms became multi-cellular life forms.
By the time of 500 mya, invertebrates were formed and active.
Jawless fish probably evolved around 350 mya.
Sea weeds and few plants existed probably around 320 mya.
We are told that the first organisms that invaded land were plants.
They were widespread on land when animals invaded land.
Fish with stout and strong fins could move on land and go back to water.
This was about 350 mya.
In 1938, a fish caught in South Africa happened to be a Coelacanth which was thought to be extinct.
These animals CALLED lobefins evolved into the first amphibians that lived on both land and water.
There are no specimens of these left with us.
HOWEVER, these were ancestors of modern day frogs and salamanders.
The amphibians evolved into reptiles.

They lay thick shelled eggs which do not dry up in sun unlike those of amphibians.
Again we only see their modern day descendents, the turtles, tortoises and crocodiles.
In the next 200 millions years or so reptiles of different shapes and sizes dominated on earth.
Giant ferns (Pteridophytes) were present but they all fell to form. coal deposits slowly.
Some of these land reptiles went back into water to evolve into fish like reptiles probably 200 mya (eg. Ichthyosaurs).
The biggest of them i.e. 1)rrannosaurus rex was about 20 feet in height and had huge fearsome dagger like teeth.
About 65 mya the dinosaurs suddenly disappeared from the earth.
Some say climatic changes killed them. Some say most of them evolved into birds.
The truth may live in between.
Small sized reptiles of that era still exist today.
The first mammals were like shrews.
Their fossils are small sized.
Mammals wp.rp. viviparous and protected their unborn young inside the mother.s body.
Mammals were more intelligent in sensing and avoiding danger at least.
When reptiles came down mammals TOOK over this earth.
There were in South America mammals RESEMBLING horse, hippopotamus, bear, rabbit etc.
Due to continental drift, when South America joined North America, these animals were overridden by North American fauna.
Due to the same continental drift pouched mammals of Australia survived because of lack of competition from any other mammal.
A rough sketch of the evolution of the life forms their times on a geological scale are indicated in figure.
Discussion
18128.

Question : Describe the embryological support for evolution.

Answer»

Solution :Embryological support for evolution was also proposed by Ernest Heckel based upon the observation of certain features during embryonic STAGE COMMON to all vertebrates that are absent in ADULT.
For example the embryo of all vertebrates including human develop a row of vestigial gill slit just behind the head but it is a functional organ only in fish and not found in any other adult vertebrates.
However this proposal was disapproved on careful study performed hy KHrl Ernest Van BAER.
He noted that embryos never PASS through the adult stages of other animals.
Discussion
18129.

Question :Describe the double helix model of DNA with a labelled diagram

Answer»

Solution : The DOUBLE helix model of DNA was explained by Watson and Helix.
DNA consists of two polynucleotide strands.
The polynucleotide strands are right hand coiled to FORM double helix.
Sugars and phosphates are the back box of the two strands.
Nitrogenous base is attached at right angles to the strand.
The nitrogenous bases are paired with HYDROGEN bonds. This is called base pairing
Adenine always pairs with thymine by two hydrogen bonds. Guanine always pairs with cytosine with by 3 hydrogen bonds.
The number of purines is EQUAL to number of pyrimidines. This is called as Chargaff.s rule of base equivalence.
Discussion
18130.

Question : Describethe different types of photosyntheticpigmentsand explain their role .

Answer»

Solution :The differenttype of photosynthetic pigments and their roles are as follows :
(1) Chlorophylls : The chlorophylls are most abundant and important pigments of plants . They are green pigments SOLUBLE in organic solvent and insoluble in water.
The chlorophylls are of seven types, viz., chlorophyll-a, chklorophyll-b, chlorophyll-c, chlorophyll-d, chlorophyll-e , bacterio-chlorophyll and bacterioviridin. Of these chlorophyll -a, chlorophyll b. chlorophyll-c, chlorophyll-d and chlorophyll-e are most important.
(i) Chlorophyll -a : :Chlorophyll -a is blue green in colour. Its chemical formula is `C_(55)H_(72)N_(4)Mg.` With the exception of photosyn- THETIC becteria, chlorophyll-a is found in all photosynthetic organisms.
The pigment chlorophyll-a is considered as an essential pigment as it converts the light energy into chemical energy.
(ii ) Chloropohyll-b : Chlorophyll-b is yellow green in colour. Its chemical formula is `C_(55)H_(70)O_(6)N_(4)Mg`. It is found in higher plants such as green ALGAE, bryophytes and all VASCULAR plants. The pigment chlorophyll-b absorbs and transfers the light energy to chloropohyll-a. Hence it is called an accessory pigment.
(iii ) Chlorophyll-c, d and e. These are found in brown and red algae in association with chlorophyll-a.
(2) Carotenoids : Carotenoids are widely distributed in chloroplasts and chromoplasts. There are two types of carotenoids, viz., carotenes and xanthophylls. Carotenes are orange yellow in colour while xanthophylls are yellow in colour. The major carotene is `beta` carotene.
Carotenoids absorb blue violet light and transfer it to chlorophyll-a and as such function as accessory pigments. Besides, carotenoids protect chlorophyll-a from photoxidation.
(3) Phycobillins : Phycobillins are accessory pigments found in red algae and cyanobacteria. They are of two types, viz., phycocyanin and phycoerythrin.
Discussion
18131.

Question : Describe the different parts of human eyes.

Answer»

Solution :(1) Eyeball is sphericaland has three layers.
(2) Sclera isthe outerlayer ofdenseconnectivetissue.Its transparentanteriorportionistermed as a cornea.
(3) Choroidis the middler layer. The posteriortwo third REGION of this layeris thin, while the anteriorregion is thickwhich formsthe ciliarybody.
(4) Iris is the forward segmentof the ciliary body which is pigmented andopaque.
(5) Lens is a transparentcrystalline structure.It is presentanteriorlyinside the iris andis heldin positionby the ligaments ofciliarybody.
(6) The aperturesurroundedby the iris in fontof thelends is known as pupil.
(7) The INNERMOST layer of the eye is the RETINA which consists of three sub-layers FORMED by ganglion cells, bipolar cells and photoreceptor cells, which are sensitive to light.
(8) There are two types of photoreceptor cells viz. road and cones containing photogments, rhodopsin which is a derivative of vitamin A (in roads) and iodopsin (in cones).
(9) The cones are responsiblefor daylightor photopic visionand colourvision.The rods functionin dim light givingscotopic vision.
(10) The opticnerveleavesthe eye at a point slightlyaway fromthe medianposteriorpole of theeyeball . In thisregion, the rods andconesare absentthereforethis regionis knownas blindspot.Maculalutea, a yellowishpigmentedspot ispresent lateral to theblindspot.Forvea is a centralpit presentbeside it.
(11)A spacebetween thecorneaand the lensis calledaqueous chamber, containingaqueoushumor.
(12) Vitreouschamberis the largerspacebetween thelens andthe retinawhichisfilled with vitreous homor.
Discussion
18132.

Question : Describe different steps involved in tissue culture technique.

Answer»

Solution :Steps involved in tissue culture TECHNIQUE : The VARIOUS steps that are involved in tissue culture are as follows :
(1) Explant culture : Explants are removed from the selected plants. These are in the form of cell, tissue or some other parenchymatous plant matter. Meristem is the most suitable tissue as an explant. Explants are sterilized and grown on solid culture medium where they start cell division.
(2) Callus formation and culture: Callus is unorganized mass of cells possessing thin wall parenchymatous cells. Explants proliferate and produce callus. All the cells of callus are identical as they are produced by mitosis.
(3) Organogenesis: Plant hormones such as auxins and cytokinins are given to the growing cells. Formation of organs is thus induced in callus. Rhizogenesis (growth of roots) takes PLACE if auxins are provided in more quantity. Cauliogenesis (growth of shoot system) takes place if cytokinins are provided in large amount.
(4) PREPARATION of suspension culture : For the preparation of suspension culture, callus is transferred to a liquid nutrient medium and is agitated. Cells of callus are separated due to agitation. Constant agitation at 100-250 rpm serves the purpose of aeration, mixing and prevention of aggregation. This suspension culture is subcultured every WEEK. Suspension cultures grow rapidly than the callus culture. They need to be subcultured every week.
Discussion
18133.

Question : Describe the development of monrospric embryo sac.

Answer»

Solution :The functional megaspore is the EMBRYO sac or female gametophyte . The megaspore elongates along micropylar - chalazal axis. The NUCLEUS undergoes a mitotic division. Wall formation does not follow the nuclear division . A large central vacuole now appears between the two draughts nuclei. The vacuole EXPANDS and pushes the nuclei towards the opposite poles of the embryo sac. Both the nuclei divide twice mitotically , forming four nuclei at each pole . At this stage all the eight nuclei are present in a common cytoplam (free nuclear division) . After the last nuclear division the cell undergoes appreciable elongation , assuming a sac - like appearance . This is followed by cellular organization of the embryo sac. Of the four nuclei at the micropylar end of the embryo sac, THREE organize into an egg apparatus, the fourth one is left free in the cytoplasm of the central cell as the upper polar nucleus. Three nuclei of the chalazal end form three antipodal cells whereas the fourth one functions as the lower polar nucleus. Depending on the plant the 2 polar nuclei may remainfree or may fuse to form a secondary nucleus (central cell) . The egg apparatus is made up of a central egg cell and two synergids , one on each side of the egg cell . Synergids secrete chemotropic substance that help to attract the POLLEN tube. The special cellular thickening called filiform apparatus of synergids help in the absorption , conduction of nutrients from hte nucellus to embryo sac. It also guides the pollen tube into the egg . Thus , a 7 called with 8 nucleated embryo sac is formed.
Discussion
18134.

Question : Describe the cross between a homozygous tall round seeded pea plant and a dwarf wrinkled seeded pea plant, what will be types of propency in the F_(2)generation of thiscrossand in what proportion will it be ? Name and state the law which is explained by this example.

Answer»

Solution :If we represent the GENE for tall habit of pea plant with T and roundseed by R,then the genotype of tall round seeded pea plant will be TTRR.Similarly the genotypeof dwarfwrinkledseeded peaplantwill the ttrr.


`F_(2)`progeny : Tall round = 9
Tall wrinkled = 3
Dwarfround = 3
Dwarf wrinkled = 1
Phenotypic ratio = 9: 3: 3: 1
NAME of law = Law of independent ASSORTMENT
Statementof law = The law of independentassortment
Statesthat when two parentsdifferingfromeachotherin twoor moreinheritance of one pairof charactersis independentof the other pairof CHARACTERS.
Discussion
18135.

Question : Describe the cross between a homozygous tall round- seeded pea plant and a dwarf, wrinkled-seeded pea plant. What will be the types of progeny in the F_(2) generation of this cross and in what proportion will it be? Name and state the law which is explained by this example.

Answer»

SOLUTION :The LAW EXPLAINED by this EXAMPLE is known as Law of Independent Assortment.
Discussion
18136.

Question : Describe the development of embryo in angiosperms.

Answer»

Solution :Development of embryo (dicot) in angiosperm :
(1) The development of embrlyo from a zygote is called embryogensis .
(2) The fusion of male gameteand an egg cell duringfertilization results in the formation of a diploidzygote.
(3) The zygote developsa wall around it and is CONVERTED into oospore.
(4) The oosporeundergoes a transverse DIVISION to from a large basal cell towards the microphyleand a small apical cell towards the interior of the embryo sac. This two celled structure is called proembryo.
(5) The basal cell undergoes repeated transverse divisions to form a multicellular structure called suspensor.
(6) The suspensor pushes the embryo towards the endosperm to draw its nutrition .
(7) The apical cell of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
(8) The cells of actant now undergo an unequal periclinal division to form an outer layer of eitht smaller cells and an inner layer or eight LARGER cells.
(9) The cells of the outer layer give rise to the dermatogen. The dermatogen in turn gives rise to the epidermis.

(10) The cells of the inner layer gives rise to an embryonal mass which on differentiation form PROCAMBIUM and ground meristem.
(11) The procambium and ground meristem undergo differentiation andform an embryonal axis consisting of a PLUMULE, a radical and two cotyledons.
Discussion
18137.

Question : Describe the development of endosperm in angiosperms.

Answer»

Solution :(1) In angiosperms, the ENDOSPERM formation takes place soon after the fusion of one of the two male gametes with the secondary nucleus.
(2) The fusion of the male gamete with the secondary nucleus is CALLED triple fusion or second fertilization.
(3) After the second fertilization, the diploid secondary nucleus gets converted into a triploid primary endosperm nucleus (PEN).
(4) The triploid primary endosperm nucleus develops into the endosperm of the seed.
(5) The triploid endosperm nucleus UNDERGOES a number of free nuclear mitotic divisions to form a group of triploid cells.
(6) The triploid endosperm acts as a NUTRITIVE tissue as it supplies food materials to the developing embryo.
Discussion
18138.

Question : Describe the consumptive use value of biodiversity as food, drugs and medicines, fuel and fiber with suitable examples.

Answer»

Solution :The direct use values where the biodiversity products can be harvested and consumed directly are called consumptive use value of biodiversity.
Humans derive countless direct economic BENEFITS from nature-food (cereals, pulses, FRUITS), firewood, fibre, construction material, industrial products (tannins, lubricants, dyes, RESINS, perfumes) and products of medicinal importance.
More than 25 per cent of the drugs currently sold in the market worldwide are derived from plants and 25,000 species of plants contribute to the traditional medicines used by native PEOPLES around the WORLD.
Discussion
18139.

Question : The connecting link between glycolysis and Krebs cycle is

Answer»

Solution :(1) Glycolysis is connected to KREBS CYCLE by a link reaction called acetylation.
(2) Pyruvic acid or pyruvate formed during glycolysis undergoes oxidative decarboxylation before entering the Krebs cycle.
(3) This reaction is catalysed by the enzyme pyruvic dehydrogenase. The coenzyme NAD and the coenzyme A are required for this reaction.
(4) Pyruvic acid (pyruvate) undergoes decarboxylation followed by oxidation in which the `H_(2)` which is removed is TAKEN up by NAO leading to the formation of `NADH_(2)`
(5) The acetyl fraction (2C) which remains after decarboxylation is taken up by the coenzyme A resulting in the formation of acetyl Co-A. This reaction is known as acetylation of pyruvate.
(6) Glycolysis terminates with the formation of pyruvate and food enters the Krebs cycle in the form of acetyl Co-A. THEREFORE acetylatlon of pyruvate or formation of acetyl Co-A is known as connecting link between glycolysis and Krebs cycle or link reaction.
(7) The link reaction can be summarized as follows:
Discussion
18140.

Question : Describe the concept of energy flow.

Answer»

Solution :(1) For considering the energy flow in an ecosystem the following aspects have to be taken into account :
(i) The efficiency of the producers in the absorption and conversion of the solar energy.
(II) The quantity of converted energy is used by the consumers.
(iii) The total input of energy in the form of food and its efficiency of assimilation by the consumers.
(iv) The amount of energy is lost through respiration, heat, excretion,etc. and the ultimate GNP or gross net production.
(2) The energy captured by autotrophs never returns back to the solar energy. The energy obtained by the berbivores will never go back to autotrophs. Hence energy flow is always unidirectional.
(3) The energy flow through different trophic levels is PROGRESSIVE and hence previous trophic level cannot GET this back.
(4) The amount of energy keeps on decreasing as it TRAVELS to further trophic level. This loss of energy is due to dissipation at heat formed during various metabolic activities of the organism. The energy loss is measured as respiration coupled with unutilized energy.
(5) If the food chain is shorter there is greater amount of available food energy. When the length of food chain increases, there is corresponding more loss of energy.
Discussion
18141.

Question : Describe the components of an ecosystem.

Answer»

Solution :The components of ecosystem may be divided into two main types biotic component comprising the various kinds of ORGANISMS and abiotic component consisting of environmental factors.
a.Biotic components : The living organisms present in an ecosystem FORM the biotic components.They are interconnected through food.Based on the mode of obtaining food, the organisms occurring in an ecosystem are classified into three main categories. They are PRODUCER, consumers and decomposers.
i. Producers : They produce food materials. These are of two types photoautotrophs and chemoautotrophs.
Photoautotrophs : These are green plants that having chlorophylls. They trap the light energy and change it into chemical energy.
Chemoautotrophs : Some bacteria such as sulphur bacteria, iron bacteria etc. capture energy released during chemical reactions and prepare organic food.
ii. Consumers : These are mainly animals which are unable to prepare their own food. So they utilize the materials and energy stored by the producers. Consumers are of 3 types. They are primary consumers, secondary consumers, tertiary consumers etc.
iii. Decomposers : These are the organisms which decompose the dead organic BODIES of producers and consumers. These are mainly bacteria and fungi. They are also known as saprotrophs.
b. Abiotic factors : These include the non - living, physicochemical factors of the ENVIRONMENT. These components affect the distribution, structure, behaviour and relationship of organisms. Abiotic factors include inorganic substances, organic compounds and climatic factors.
Discussion
18142.

Question : Describe the classification of species as given by IUCN.

Answer»

Solution :IUCN has classified the species into four categories, viz.endangered, vulnerable, rare and indeterminate species.
(1) Endangered species: In endangered species the number ofindividuals is greatly reduced to a critical LEVEL due to unfavourable environmental conditions or human factors. If the conditions in the habitat REMAIN the same, the endangered species become extinct soon. E.g. Asiatiet WILD ass, Psilotum nudum, OSMUNDA regalis.
(2) Vulnerable species : In vulnerable species the number of individuals is greatly reduced in recent times and keeps on decreasing. Some of the vulnerable species are abundant in the PRESENT times but are under threat of depletion due to their exploitation. E.g. Clouded leopard, musk deer (Antelope), Ophioglossum pendulum.
(3) Rare species : In rare species, the number of members is few. They live in small geographical area or in unusual environment. They are at risk in near future. E.g. Great Indian Bustard, Hawaiian Monk, seal.
(4) Indeterminate species: Due to reasons that are unRnown, these species are in danger of extinction. The information about them is not enough to determine their true condition. E.g. Three-banded Armadillo of Brazil, short-eared rabbit of Sumatra, Rhinoceros, etc.
Discussion
18143.

Question :Describe the biogas plant with a neat labelled diagram

Answer»

Solution :STRUCTURE of biogas plant with a neat labelled diagram.
a) The biogas plant consists of a concrete tank (10-15 ft deep) in which bio-wastes are collected and slurry of dung is fed.
b) A floating COVER is PLACED over the slurry, which keeps on rising as the gas is produced in the tank due to MICROBIAL (methanogens) activity.
C) The biogas plant has an outlet, which is connected to a pipe to supply biogas to nearby houses.
d) Used slurry is removed through another outlet and may be used as fertiliser.
e) The biogas thus produced is used for cooking and lighting.
Discussion
18144.

Question : Describe the acetylation of respiration.

Answer»

Solution :(1) Glycolysis is connected to KREBS cycle by a link reaction called acetylation.
(2) Pyruvic acid or pyruvate formed during glycolysis undergoes oxidative decarboxylation before ENTERING the Krebs cycle.
(3) This reaction is catalysed by the enzyme pyruvic dehydrogenase. The coenzyme NAD and the coenzyme A are required for this reaction.
(4) Pyruvic acid (pyruvate) undergoes decarboxylation followed by oxidation in which the `H_(2)` which is removed is taken up by NAO leading to the formation of `NADH_(2)`
(5) The acetyl fraction (2C) which remains after decarboxylation is taken up by the coenzyme A resulting in the formation of acetyl Co-A. This reaction is known as acetylation of pyruvate.
(6) Glycolysis terminates with the formation of pyruvate and food enters the Krebs cycle in the FORM of acetyl Co-A. Therefore acetylatlon of pyruvate or formation of acetyl Co-A is known as connecting link between glycolysis and Krebs cycle or link reaction.
(7) The link reaction can be summarized as FOLLOWS:
Discussion
18145.

Question : Describe test cross.

Answer»

Solution :Though the genotypic ratios can be calculated using mathematical probability but it is not POSSIBLE to know the genotypic composition by looking at the phenotype of dominant trait.
= That is for example whether a tall plant from `F_1 or F_22` has TT or Tt composition cannot be predicted.
=Therefore, to solve this problem Mendel devised test cross.
=Back cross is a cross of `F_1` PROGENY back to one of their parents.
= A special back cross to the recessive parent is known as test cross.
= This METHOD was devised by Mendel to determine whether the dominant phenotype is homozygous or heterozygous.
=e.g. In a monohybrid cross, between violet COLOUR flower (W) and white colour flower (w). the `F_1` hybrid was a violet colour flower.
= If all the `F_1` - progeny are of violet colour, then the dominant flower is homozygous and if the progenies are in 1:1 RATIO, then the dominant flower is heterozygous.
=In case of double heterozygous i.e. heterozygous violet and axial flower (WwAa) crossed with double recessive i.e. recessive white and terminal flower [wwaa] the ratio will be 1 : 1:1:1.
Discussion
18146.

Question : Describe species area relationships.

Answer»

Solution :The great German naturalist and geographer Alexander Von Humboldt during his pioneering and extensive explorations in the wilderness of South American jungles observed that within a region species richness increased with increasing explored area but only up to a limit.
In fact the relation between species richness and area for a wide VARIETY of taxa [angiosperm plants, BIRDS, bats, fresh water fishes] turns out to be a rectangular hyperbola.

On a logarithmic scale, the relationship is a straight line described by the equation
`log S = log C+Z log A`
Where
S=Species richness
V = Area
Z = Slope of the line [regression co-efficient]
C = Y-intercept
Ecologists have discovered that the value of Z lines in the range of 0.1 to 0.2, regardless of the taxomomic GROUP or the region [whether it is the plants in Britain, birds in California or molluscs in New York state, the slopes of the regression line are amazingly similar.]
But if we analyse the species area relationship among very LARGE areas like the entire continents we will find that the slope of the line to be much steeper [Z values in the range of 0.6 to 1.2.]
For example for frugivorous [fruit eating] birds and mammals in the tropical forests of different continents, the slope is found to be 1.15.
Discussion
18147.

Question : Decribe steady population with the help of a pyramind.

Answer»

Solution :(1) The relative proportion of the INDIVIDUALS of various age groups in the population is called the age structure of that population.
(2) There are THREE age groups, viz. young or pre-reproductive age group (0 to 14 years), adults or reproductive group (15 to 59 years) and old or post-reproductive group (60 years and above).
(3) The age structure of a population decides the trend of a population.
(4) Based on the DISTRIBUTION of the age groups the populations can be either, growing population, steady population or declining population.
(i) Growing population : Population having larger number of individuals of the pre-reproductive age groups grows at a very rapid RATE. Therefore such a population is called a growing population.
(ii) Steady population: When the pre-reproductive age group is larger, the population remains steady.
(iii) Declining population : Larger number of old people and make the number of young people make the population decline.
Discussion
18148.

Question : Describe sterilization method with diagram.

Answer»

SOLUTION :Surgical method is also called sterilization. These are generally advised for the male / female partner as a terminal method to prevent any more pregnancies.

Surgical intervention blocks GAMETE transport and thereby prevent conception.
(a) Vasectomy : Sterilization procedure in the male is called vasectomy. In vasectomy, a small part of the vas deferens is removed or tied up through a small incision on the SCROTUM.
(b) Tubectomy : In tubectomy a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through VAGINA.
These techniques are highly effective but their reversibility is very poor.
Discussion
18149.

Question : Describe semiconservative replication of DNA step by step.

Answer»

Solution :It is proven that DNA replicates semiconservatively.
It was shown first in Escherichia coli and subsequently in higher organisms, such as plants and human cells.
Mathew Meselson and Franklin Stahl performed the following experiment in 1958 :
(i) They grew E. coli in a medium containing SNH,CI (15N is the heavy isotope of nitrogen) as the only nitrogen source for many generations.
The result was that 15N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds).
This heavy DNA molecule could be DISTINGUISHED from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

(ii) Then they transferred the cells into a medium with normal `""^(14)NH_(4)Cl` and took SAMPLES at various definite time INTERVALS as the cells multiplied and extracted the DNA that remained as double stranded helices.
The various samples were separated independently on CsCl gradients to measure the densities of DNA.
The results are shown in figure.
(iii) Thus, the DNA that was extracted from the culture one generation after the transfer from `""^(15)N` to `""^(14)N` medium (that is after 20 minutes, E. coli divides in 20 minutes) had a hybrid or intermediate density.
DNA extracted from the culture after another generation (that is after 40 minutes. II generation) was composed of equal amounts of this hybrid DNA and of .light. DNA.
Very similar experiments involving use of radioactive thymidine to detect distribution of newly synthesised DNA in the chromosomes was perfomed on Vicia faba (faba beans) by TAYLOR and Colleagues in 1958.
The experiments proved that the DNA in chromosome also replicate semiconservatively.
Discussion
18150.

Question : Describe sacred groves.

Answer»

Solution :(d) Sacred groves : These are small group of FORESTS with special religious importance in a particular cultures and are also of mythological importance.
These are undisturbed forests without any HUMAN interventions and include a number of rare, endangered and endemic species.
Sacred groves are found in Khasi and Jaintia Hills in Meghalaya.
Aravalli Hills of Rajasthan, Western GHATS regions of Karnataka and Maharashtra and the Sarguja, Chanda and Bastar areas of Madhya Pradesh.
These are PROTECTED by native people as a part of CULTURAL traditions.
In Meghalaya, the sacred groves are the last refuges for a large number of rare and threatened plants.
Discussion