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18351.

Question : Describe the steps involved in cytological techniques.

Answer»

Solution :There are different types of mounting based on the portion of a specimen to be observed.
1. Whole mount: The whole organism or smaller structure is mounted over a slide and observed.
2. Squash: Is a preparation where the material to be observed is CRUSHED/ squashed onto a slide so as to reveal their contents. example: Pollen grains, mitosis and meiosis in root TIPS and flower buds to observe chromosomes.
3. Smears: here the specimen is in the fluid (blood and microbial cultures etc) are scraped, brushed or aspirated from surface of organ. example: Epithelial cells.
4. Sections: Free hand sections from a specimen and thin sections are selected, strained and mounted on a slide. example: LEAF and stem of plants.
Discussion
18352.

Question : Write down the characteristic features of Rhodophyceae.

Answer»

Solution :1. Rhodophyceae commonly called as red algae.
2. Mostly marine habitats.
3. The thallus is multicellular, macroscopic, and may be filamentous, ribbon-like etc.
4. Chlorophyll 'a', r-phycoerythrin and r-phycocyanin are photosynthetic pigments.
5. ASEXUAL REPRODUCTION is by means of monospores, neutral spores and tetraspores
6. Floridean starch is the storage material
7. Sexual reproduction in oogamous.
8. Male sex organ is SPERMATANGIUM producing spermatium.
9. Female sex organ is carpogonium.
10. Spermatium is carried by water and FUSES with egg forming zygote.
11. Zygote undergoes meiosis forming carpospores.
12. Alternation of generation is seen. Example: Ceramium, GELIDIUM and Gigartina.
Discussion
18353.

Question : Describe the salient features of Phacophyeeae members.

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Solution :1. Phaeophyceae commonly called as Brown algae.
2. Majority are marine habitats. Pleurocladia is a fresh water form.
3. Thallus may be filamentous, frond-like or giant kelps.
4. Thallus is differentiated into photosynthetic part-frond, stalk-like structure-stipe and a holdfast for attachment.
5. Chlorophyll 'a' and 'c', carotenoids and XANTHOPHYLLS are photosynthetic pigments.
6. A golden brown fucoxanthin pigment GIVES olive green to brown colour.
7. Mannitol and Laminarin starch is the storage material.
8. Motile spores with unequal flagella (one whiplash and one tinsel) are present.
9. Oogamous is the major type of sexual reproduction. Isogamy is also SEEN.
10. Alternation of generation is seen. EXAMPLE: Sargassum, Fucus, Laminaria and Dictyota
Discussion
18354.

Question : Describe the salient features of Chlorophyceae members.

Answer»

Solution :1. Chlorophyceae commonly called as GREEN algae.
2. Mostly aquatic (fresh water or marine), few terrestrial.
3. Shape of chloroplast differs. It may be cup shaped (Chlamydomonas) or girdle-shaped or reticulate, or stellate etc.
4. Chlorophyll 'a' and 'b' are photosynthetic pigments.
5. Pyrenoids store starch & ALSO proteins.
6. Outer cell wall is MADE of PECTIN and inner is cellulose.
7. Vegetative reproduction is by fragmentation.
8. Asexual reproduction by zoospores, aplanospores and akinetes.
9. SEXUAL reproduction may be isogamous, anisogamous or oogamous.
E.g. Chlanydomonas, Volvox and Spirogyra.
Discussion
18355.

Question : Describe the role played by protein pumps during active transport in plants.

Answer»

SOLUTION :During active TRANSPORT, pumps are proteins that use energy to carry SUBSTANCES across the cell membrane. These pumps can transport substances from a low concentration to a HIGH concentration.
Discussion
18356.

Question : Describe the role of liver, lungs and skin in excretion.

Answer»

Solution :(i) Role of liver in excretion :
It helps in formation of bile pigments and urine pigment urochrome from Hb of dead RBCS.
It changes excess of anmino acids into urea by Ornithine cycle by deamination and detoxification.
(ii) Role of lungs : It helps in not only expelling foul air with more `CO_(2)` during expiration but ALSO to expel about 400 ml. of water per day.
(iii) Role of SKIN :
Sweat glands of skin expel water, inorganic salts like sodium chloride, lactic acid, some urea and `CO_(2)` in the form of sweat which helps in excretion and HOMEOTHERMY as evaporation of sweat gives cooling effect.
Oil glands of skin expel waxes, sterols, fatty acids and a number of HYDROCARBONS in the form of sebum.
Discussion
18357.

Question : Describe the role of haemoglobin in transport of respiratory gases.

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Solution :(i) TRANSPORT of oxygem - 97% of oxygen CARRIED as oxyhaemoglobin.
`Hb_4 + 4O_2 to Hb4O_8`
(ii) Transport of `CO_2` - 70% of `CO_2` in RBCs
`CO_2 + H_2O to H_2 CO_3 to HCO_3 + H^(+)`
23% of `CO_2` combines with amino group of haemoglobin - carbaminohemoglobin
Discussion
18358.

Question : Describe the role of abscisic acid in plant.

Answer»

Solution :Abscisic acid (ABA) was discovered for its role in regulating abscission and dormancy.
But LIKE other PGRs, it also has other wide ranging effects on PLANT growth and development.
It acts as a general plant growth inhibitor and an inhibitor of plant metabolism.
ABA inhibits seed germination.
ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.
Therefore, it is also called the stress hormone.
ABA plays an important role in seed development, maturation and dormancy. By inducing dormancy, ABA helps seeds to withstand desiccation and other FACTORS unfavourable for growth.
In most situations, ABA acts as an antagonist to GAS.
Discussion
18359.

Question : Describe the respiratorybalance sheet.

Answer»

Solution :It is POSSIBLE to make calculation of the net gain of ATP for every glucose molecule oxidised but in realitythis can remain only a theoretical exericse. These calculations can be madeonly on certain assumptions that :
(1)Thereis a sequentialorderly PATHWAY functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETSpathway following one after another.
The NADH synthesised in glycolysisis trnasferred into the mitochondria and undergoesoxidative phosphorylation.
(3) None of the intermediates in the pathway are utilised to synthesize any othercompound.
(4) Only glucose is being respired-no other alternative substrates are entering in the pathway at any of the intermediary stages.
But this kind of assumptions are notreally valid in a living system .
All pathways work simultaneouslyand do not take PLACE one afteranother.
Substrates enter the pathways an arewithdrawn from it as and when necessary.
ATP is utilised as and when needed .
Enzymatic rates arecontrolled bymultiple means. Yet it is USEFUL to do this exercise to appreciatethe beautyand efficiencyof the living systemin extraction and storing energy.
Hence, there can be a net gain of 38 ATP moleculesduring aerobic respiration of one molecule of glucose.
Discussion
18360.

Question : Describe thereaction occuringin membrane of Cristaeof mitochondria .

Answer»

Solution :The following steps in the respiratory process are to releaseand utilisethe energy stored in `NADH + H^(+) and FADH_(2)`
Thisis accomplished when they are oxidised through the electron transport system and theelectrons are passed on to `O_(2)`resulting in the formation of `H_(2)O`
The metabolic pathwaythrough whichthe electron passes form one carrier to another is called the electron transportsystem .(ETS)
It is present in the inner mitochondrialmembrane .
Electronsfrom NADH produced in the mitochondrial matrix during citric acid cycleare oxidised by an NADHdehydrogenase (Complex-I).
Electronsare then transferred to ubiquinonc location within theinner membrane.
Ubiquinone also receives reducing equivalents via `FADH_(2)` (Complex -II)
That is generated during oxidation of succinate in the citric acidcycle.
The reduced ubiquinone is then oxidised with the TRANSFER of electronsto cytochrome c via cytochrome `bc_(1)` complex (Complex - III)
Cytochrome c is small PROTEIN attached to the other surface to the inner membrane and acts as a mobile carrierfor transfer of ELECTRONS between complexIII and IV .
Complex IV refers to cytochrome c oxidase complex containing cytochromes a and `a_(3)` and two copper centers.

ATP formation in ETS :
When the electrons PASS form one carrier to another via complex I to IV in the electron transport chain, they are coupledto ATP synthase (complex V)for the production of ATP from ADP and inorganic phosphate.
The number of ATP molecules synthesised depends on thenature of the electron donor.
Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of `FADH_(2)`produces 2 molecules of ATP.
Although the aerobic of respiration takes place only in the presences of oxygen .
The ROLE of oxygen is limited tot he terminal stage of the process. Yet the presence of oxygen is vital, since it drives the whole process byremoving hydrogen form the system .
Oxygen acts as the final hydrogen accepter . (For production of` H_(2)` O)
In respirationit is the energy of oxidation-reduction utilised for the same process . it isfor this reason that theprocess is called oxidative phosphorylation .
Discussion
18361.

Question : Describe the process of transformation.

Answer»

Solution :Transfer of DNA from one bacterium to another is called transformation. In 1928, the BACTERIOLOGIST Fredrick Griffith demonstrated transformation in mice using Diplococcus PNEUMONIAE. Two strains of this bacterium are present. One strain produces smooth colonies and are virulent in nature (S TYPE), in addition another strain produced rough colonies and are avirulent (R type). When S-type of cells were injected heat killed S-type cells into the MOUSE the mouse did not die. When the mixture of heat killed S-type cells and R - type cells were injected into the mouse. The mouse died. THe avirulent rough strain of Diplococcus had been transformed into S-type cells . The hereditary material of heat killed S-type cells had transfored R - type cell into virulent smooth strains. Thus the phenomenon of changing the character of one strain by transferring the DNA of another strain into the former is called transformation.
Discussion
18362.

Question : Describe the process of respiration occuring in cytoplasm.

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
18363.

Question : Describe the process of fermentation in detail.

Answer»

Solution :In fermentation by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvicacid is converted to `CO_(2)`and ethanol .
`C_(6) H_(12) O_(6) to 2CH_(3)CO*CO O H to 2C_(2) H_(5) OH to 2CO_(2)`
The enzymes, pyruvic acid, decarboxylase and alcohol dehydrogenasecatalyse thesereactions .
Other organismslike some bacteria produce lactic acid from pyruvic acid .
In animal cells also, like muscles during exercise, when oxygen is INADEQUATE for cellular respiration pyruvicacid is reduced to lactic acid by lactate dehydrogenase.
The REDUCING agent is `NADH + H^(+)`which is reoxidised to `NAD^(+)`in boththe processes.
In bothlactic acid and alcohol fermentation not much energy is RELEASE.
Less than 7% of the energy in glucose is released and not all of it is trapped as high energy bonds of ATP .
Also, the processes are hazardous.
Inthis synthesis 4 ATP are formed and 2 ATP are used meansbenefit of 2ATP occurs.
When the proportion of alcohol is more then 13% it becomes poisonous for yeast. It becomes the CAUSE of its death .
Ineukaryotes these steps take place within the mitochondriaand this requires `O_(2)` . Aerobic respiration is the process that leads to a complete oxidation of organicsubstances in the presenceof oxygen and releases `CO_(2)` water and a large amountof energy present in the substrate.
Discussion
18364.

Question : Describe the process of digestion seen in small intestine in detail.

Answer»

Solution :Semi digested food chyme comes to intestine by pyrrolic stomach. Chyme comes into contact of three different secretions in small intestine.
Pancreatic juice, BILE juice and intestinal juice.
Various types of movements are generated by the muscularis layer of the small intestine. These movements help in a thorough mixing up of the food with various secretions in the intestine and there by facilitate DIGESTION.
Pancreatic juice possesses inactive enzymes trypsinogen, chymotrypsinogen and procarboxypeptidases and amylase, lipase and nuclease.
Trypsinogen is activated by an enterokinase, secreted by the intestinal mucosa into ACTIVE trypsin which in turn activates the other enzymes in the pancreatic juice.
Proteases + peptones `overset"Trypsin"to` polypeptide + Amino Acid
Trypsin convents chymotrypsinogen into chyme trypsin and procarboxy peptidases in to carboxyl peptidase.
proteoses + polypeptides + peptones `to`Chymotrypsin small polypeptide + Amino Acid.
Polypeptide `overset"Carboxypoptidose"to` Peptide, Amino acid
Amylase present in pancreatic, juice hydrolyses carbohydrates and disaccharide is formed. Polysaccharide (starch `overset"Amylase"to`Disaccharide Maltese).
The bile released into the duodenum contains bile pigments (bilirubin and BILIVERDIN), bile salts, cholesterol and phospholipids but no enzymes.
Bile helps in emulsification of fats i.e. breaking down of the fats into very small micelles.
Fats `overset"Lipases"to` Diglycerides `to` Monoglycerides
Nucleases of pancreatic juice reacts on nucleic acid and forms nucleotides and nucleosides.
Nucleic acids `overset"Nucleases"to` Nucleotides `to` Nucleosides
Digestion by intestinal juice in Jejunum and ileum: The enzymes of intestinal juice react on enzymes, FINAL products react on them and convert into absorbable simple form.
Dipeptide `underset"(Erepsin)"overset"Dipeptidases"to` Amino Acids
Maltose `overset"Maltase"to` Glucose + Glucose
Lactose `overset"Lactose"to` Glucose + Galactose
Sucrose `overset"Sucrose"to` Glucose + Fructose
Nucleotides `overset"Nucleotidases"to` Nucleosides `overset"Nucleosidases"to` Sugars+ `N_2` Bases
Di and Monoglycerides `overset"Lipases"to` Fatty acid+Glycerol
Simple components produced from bio mega molecules are absorbed into jejunum and ileum of small intestine.
Undigested and unabsorbed components are passed into large intestine.
Discussion
18365.

Question : Describe the process of digestion of protein in stomach.

Answer»

SOLUTION :When PROTEIN ENTERS the stomach HCl and two proteoses pepsin and rennin are secreted.This hydrolyses proteins into larger peptides known as PEPTONES.
Discussion
18366.

Question : Describe the process of digestion occurring in stomach.

Answer»

Solution :The mucous of stomach has gastric glands. Gastric glands liave three major types of cells namely:
(1) Mucous neck cells which secrete mucus
(2) Peptic of chief cells which secrete the proenzyme pepsinogen
(3) Parietal or Oxyntic cells which secrete HCl and intrinsic factor (factor essential) for absorption of vitamin `B_12`.
Food MATERIAL comes into stomach cavity at that TIME Gaskin is secreted into blood by pyloric stomach which stimulates secretion of gastric juice from the stomach glands.
Stomach stores the food for 4-5 hours. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme.
The proenzyme pepsinogen on exposure to hydrochloric ACID gets converted into the active enzyme pepsin. Pepsin converts proteins into proteoses and peptones.
Pepsinogen HCl `to` Pepsin
Proteins `to` Proteoses + Peptones
The mucus and bicarbonates present in the gastric juice PLAY an important role in lubrication and protection of mucosal epithelium from excoriation by the highly concentrated hydrochloric acid.
HCl provides the acidic pH optimal for pepsins. Rennin is a proteolytic enzyme found in gastric juice of INFANTS which helps in the digestion of milk proteins.
Small amounts of lipases are also secreted by gastric glands.
Discussion
18367.

Question : Describe the process of digestion occurring in oral cavity.

Answer»

Solution :There are two MAIN functions of buccal cavity, mastication of food and facillation of swallowing.
With the help of saliva, teeth and tongue masticate food and mix up the food thoroughly.
Mucus in saliva helps in lubricating and adhering the MASTICATED food particles into a bolus.
The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of MUSCULAR CONTRACTIONS called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the STOMACH.
The saliva secreted into the oral cavity contains electrolytes (`Na^(+), K^(+), Cl^(-) , HClO_3^(-)` ) and enzymes, salivary amylase and lysozyme.
The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of carbohydrate splitting enzyme, the salivary amylase.
About 30 percent of starch is hydrolysed here by this enzyme into disaccharide-maltose.
Starch `underset"pH 6.8"overset"ptylin"to` Maltose
Lysozyme present in saliva acts as an antibacterial agent that prevents infections.
Peristalsis - Bolus proceeds further in oesophagus by continuous wavy contraction of muscles.
Stomach - controls the food going to stomach by sphincter muscle.
Discussion
18368.

Question : Describe the process of digestion occurring in large intestine.

Answer»

SOLUTION :No important process of digestion OCCURS in large intestine.
Functions: (1) Absorption of some water, minerals and some medicines. (2) Secretion of mucus which helps in ADHERING the waste particles together and lubricating it for an easy passage.
The undigested, unabsorbed substances called faeces enters into the caecum of the large intestine through ileo-caecal valve which PREVENTS the back flow of the faecal matter. It is temporarily stored in the rectum TILL defaecation.
Discussion
18369.

Question : Differentiate cytokinesis in plant cells and animal cells.

Answer»

Solution :Cytokinesis in plant cell: DIVISION of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls-centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. the golgi vesicles contains carbohydrates such as pectin, HEMICELLULOSE which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall. the first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.
Cytokinesis in Animal Cells: It is a CONTRACTILE process. the contractile MECHANISM contained in contractile ring located inside the plasma membrane. the ring consists of a bundle of microfilaments assembled from actin and myosin. this fibril helps for the generation for a contractile force. this force DRAWS the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.
Discussion
18370.

Question : Describe theprocess occurring in matrixof mitochondria.

Answer»

Solution :The TCA cycle start with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yieldcitric acid. The REACTION is catalysed by the enzyme citratesynthase and a molecule of CoAis released.
`OA A (4C) + "Acetyl "CoA + H_(2) O (2C) underset("Citrate synthatase")to` Citric acid 6C + CoA
Citrate is then isomerised to isocitrate.
It is followingby two successive steps ofdecarboxylation, leading to theformation of `alpha`-ketoglutaric acidand then succinyl-CoA
Citric acid `to`Iso Citrate `to alpha` - ketoglutaric acid
Remaining steps of citric acid cycle :
Succinyl-Coa is oxidised to OAAallowing the cycleto continue.
During the conversion of succinyl-CoAto succinic acid a molecule of GTP is synthesized .
This is a substrate levelphosphorylation.
Ina coupled reaction GTP is converted to GDPwith thesimultaneoussynthesis of ATPform ADP.
`GTP to GDP + Pi to ADP + Pi to ATP`
Also there are three points in the cyclewhere `NAD^(+)`is reduced to NADH + `H^(+)`and onepoint where `FAD^(+)`is reduced to `FADH_(2)` .The continued oxidation of acetyl CoAVia theTCA cyclerequiresthe continued replenishmentof oxaloaceticacid
It isthe first member of the cycle . INADDITION it alsorequires regeneration of `NAD^(+) and FAD^(+)`from NADH and `FADH_(2)`respectively
The summary equation for this phase of respiration may be written as follow :
Pyruvic acid `+ 4NAD^(+) + FAD^(+) + 2H_(2) O + ADP + Pi underset("Matrix")overset("Mitochondrial")to `
`3CO_(2) + 4NADH + 4H^(+) + FADH_(2) + ATP`
Glucose has been broken down to release `CO_(2)`and eightmolecules of `NADH + H^(+)` , two of `FADH_(2)` have been synthesised besidesjusttwo MOLECULES of ATP in TCAcycle .
Discussion
18371.

Question : Describe the procedure of Avena curvature test.

Answer»

Solution :When the Avena seedlings have attained a height of 15 to 30 mm, about 1mm of the coleoptile TIP is removed. This apical part is the source of natural auxin. The tip is now PLACED on AGAR blocks for few hours. During this period, the auxin diffuses out of these tips into the agar. The auxin containing agar block is now placed on one SIDE of the decapitated stump of Avena coleoptile. The auxin from the agar blocks diffuses down through coleoptile along the side to which the auxin agar block is placed. An agar block without auxin is placed on another decapitated coleoptile. Within an hour, the coleoptiles with auxin agar block bends on the opposite side where the agar block is placed. This curvature can be measured.
Discussion
18372.

Question : Describe the primary, secondary and tertiary structures of proteins

Answer»

SOLUTION :primary proteins – the sequence of amino acids i.e., the positional information in a protein.
Secondary proteins – a protein thread is folded in the form of helix.Tertiary proteins – a LONG protein CHAIN is folded upon itself LIKE a hollow woollen ball.
Discussion
18373.

Question : Describe the phase till formation of two molecules of pyruvic acid from Glucose takes place. ORDescirbe : C_(6) H_(12) O_(6) to 2CH_(3) COCO O H + 2ATP + 2NADH_(2)

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
18374.

Question : What is bioluminescence.

Answer»

Solution :Bioluminescence, the property of a living organism to emit light is well-marked in ctenophores.
Ctenophores, commonly known as sea walnuts or comb jellies are exclusively marine, radially symmetrical, DIPLOBLASTIC organisms with tissue level of organisation.
The body bears eight EXTERNAL rows of ciliated comb plates, which help in LOCOMOTION.
Digestion is both extracellular and intracellular.
SEXES are not separate.
REPRODUCTION takes place only by sexual means.
Fertilisation is external with indirect development.
Examples : Pleurobrachia and Ctenoplana.
Discussion
18375.

Question :Describe the pattern of leaf arrangement in mosaic leaf.

Answer»

Solution :In MOSAIC leaf, leaves tend to fit in with one another and ADJUST themselves in such a way that they MAY secure the maximum amount of sunlight with minimum amount of overlapping. The lower leaves have longer PETIOLES and successive upper leaves possess decreasing length petioles. eg., Acalypha, Begonia.
Discussion
18376.

Question : Describe the mechanism of photoperiodic induction of flowering.

Answer»

Solution : Photoperiodic induction
An appropriate photoperiod in 24 hours' cycle constitutes one INDUCTIVE cycle. Plants may require one or more inductive CYCLES for flowering. The phenomenon of conversion of leaf primordia into flower primordia under the influence of SUITABLE inductive cycles is called photoperiodic induction. EXAMPLE: XANTHIUM (SDP) -I inductive cycle and Plantago (LDP) 25 inductive cycles
Discussion
18377.

Question : Describe the mechanism of muscle contraction with diagram.

Answer»

Solution :Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.
This theory was postulated by A.F. Huxely and J.Jensen.
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motorneuron.
A Motor neuron alongwith the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction.
A neural signal reaching this junction release a neurotransmitter (Acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of `(Ca^(+2)"Calcium ions")` into the SARCOPLASM.
Increase in `Ca^(++)` level leads to the binding of `Ca^(++)` with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin.
Utilising the energy from ATP hydrolysis, the myosin head now binds to exposed active sites on actin to form a cross bridge.
This pulls the attached actin filaments towards the centre of .A. bands. The .Z. LINE attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction.
It is clea from the above steps, that during shortening of the muscle, (contraction, the .I. bands gets reduced where as the .A. bands retain the length.
The myosin, releasing the ADP and `P_(1)` goes back to its relaxed state. A new ATP binds and the cross-bridge is broken.
The ATP is again hydrolysed by the myosin head and the cycle of cross bridge formation and breakage is repeated causing further sliding. The process continues till the `Ca^(++)` ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments. This caused the retuen of .Z. lines back to their ORIGINAL position (i.e., relaxation).
The reaction time of the fibres can vary in different muscles, Repeated activation of the muscles can LEAD to the accumulation of LACTIC acid due to anarobic breakdown of glycogen in them, causing fatigue.
Discussion
18378.

Question :Describe the mechanism of enzyme action

Answer»

SOLUTION :Each enzyme (E) has a substrate (S) binding site in its molecule substrate binds with it to form enzyme, substrate COMPLEX (ES). It dissociates soon into its product(s) P and unchanged enzyme with an intermediate FORMATION of (EP) Enzyme-product complex. The formation of the ES complex is essential for catalysis.
`E + S It can be explained in the following steps :
(1) First, the substrate binds to the active site of the enzyme, fitting into the active site.
(2) The binding of the substrate INDUCES the enzyme to alter its shape, fitting more tightly around the substrate.
(3) The active site of the enzyme, now in the close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
(4) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and RUN through the catalytic cycle once again.
Discussion
18379.

Question : Describe the mechanism by which the human heart beat is initiated and controlled.

Answer»

Solution :The heart in human is myogenic (cardiomyocytes can produce spontaneous rhythmic DEPOLARISATION that initiates contractions). The SEQUENCE of ELECTRICAL conduction of heart is shown below.

1. The CARDIAC cells with fastest rhythm are called the Pacemaker cells, which are located in the right sinuatrial (SA) node/ Pacemaker.
2. On the left side of the right atrium is a node called auriculo ventricular node (AV node).
3. Two special cardiac muscle fibres originate from the auriculo ventricular node and are called the bundle or llis which runs down into the interventricular septum and the fibres spread into the ventricles. These fibres are called the Purkinjle Abres.
4. Pacemaker cells produce excitatoion through depolaeisation of their call membrane, Early depolarisation is slow and takes place by sodium influx and reduction in potassium efflux.
5. Minimum potential is required to activate voltage paied calcium (Ca) channels that CAUSES rapid depolarisation which results in action potential. The pace maker cells repolarise slowly via K. efflux.
Discussion
18380.

Question : Write a short note on the life cycle of earthworm.

Answer»

Solution :Lampito mauritii begins its LIFE cycle, from the fertilized eggs. The eggs are held in a protective cocoon. These cocoons have an incubation PERIOD of about 14-18 days after which they hatch to release juveniles The juveniles undergo changes into non-clitellate forms in Phase-I after about 15 days, which then develops a clitellum, called the clitellate at the end of the growth phase - Il taking 15 - 17 days to complete. During the reproductive stage, EARTHWORMS copulate, and later shed their cocoons in the SOIL after about 10 days. The life cycle of Lampito mauritii takes about 60 days to complete.
Discussion
18381.

Question : Describe the life cycle of Bombyx mori. Life cycle of Bombyx mori:

Answer»

Solution :The adult of Bombyx mori is about `2.5 cm` in length and pale creamy white in colour. Due to heavy body and feeble wings. Flight is not possible by the female moth. This moth is unisexual in nature and does not feed during its very short life period of `2-3` days. Just after emergence, male moth copulates with female for about `2-3` hours and if not separated, they may die after few hours of copulating with female. Just after copulation, female starts egg laying which is completed in `1-24` hours. A single female moth lays 400 to 500 eggs depending upon the climatic conditions. Two types of eggs are generally found namely diapause type and non-diapause type. The diapause type is laid by silkworms inhabiting the temperate regions. whereas silkworms belonging to subtropical regions like India lay non-diapause type of eggs. The eggs after ten days of incubation hatch into larva called as caterpillar. The newly hatched caterpillar is about 3 mm in length and is pale. yellowish-white in colour. The caterpillars are provided with well developed mandibulate type of mouth-parts and adapted to feed easily on the mulberry LEAVES.
After `1^(st), 2^(nd), 3^(rd) and 4^(th)` moultings caterpillars get transormed into `2^(nd), 3^(rd), 4^(th) and 5^(th)` instars respectively. It takes about 21 to 25 datys after hatching. The fully frown caterpillar is `7.5` cm in length. It develops salivary glands, stops feeding and undergoes pupation. The caterpillars stop feeding and move towards the corne AMONG the leaves and secretes a sticky fluid through their silk gland. The secreted fluid comes out through spinneret (a narrow pore situated on the hypopharynx) and takes the form of long fine thread of silk which hardens on EXPOSURE to air and is the white coloured bed of the pupa whose OUTER threads are irregular while the inner threads are regular. The length of continuous thread secreted by a caterpillar for the formation of cocoon is about `100-1200` meters which requires 3 days to complete. The pupal period lasts for 10 to 12 days and pupae cut through the cocoon and emerge into adult moth.
Discussion
18382.

Question : Describe the K^(+) Transporttheory on transpiration .

Answer»

Solution :Theory of `K^(+)` transport theory was proposed by LEVIT (1974) and elaborated by Raschke (1975). According to this theory, the FOLLOWING steps are INVOLVED in the stomatal opening:

In light
i. In guard CELL, starch is converted into organic acid ( malic acid).
ii. Malic acid in guard cell dissociates to malate anion and proton `(H^(+))`.
iii. Protons are transportedthrough the membrane into nearby subsidiary cells with theexchange of `K^(+)` ( Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
iv.This ion exchange is an active process and consumes ATP for energy.
v. Increased`K^(+)` ions in the guard cell are balanced by `Cl^(-)` ions. INCERASE in solute concentration decreases the water potential in the guard cell.
vi. Guard cell becomes hypertonicand favours the entry of water from surrounding cells.
vii. Increased turgor pressure due to the entry of water opens the stomatal pore.
In Dark

Thoery of `K^(+)` transport Closing of stomata
i. In dark photosynthesisstops and respiration continues with accumulation of `CO_(2)` in the sub-stomatal cavity.
ii. Accumulation of `CO_(2)` in cell lowers the pH level.
iii. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
iv. ABA stops further entry of `K^(+)` ions and also induce `K^(+)` ions to leak out to subsidiary cell from guard cell.
v. Loss of water from guard cell reduces turgor pressure and causes closure of stomata.
Discussion
18383.

Question : Describe the largest phylum of Animalia.

Answer»

Solution :Arthropoda is the largest phylum of Animalia which includes insects.
Over two-thirds of all named species on EARTH are arthropods.
They have organ-system level of organisation.
They are bilaterally symmetrical, triploblastic, segmented and coelomate animals.
The body of arthropods is covered by chitinous exoskeleton.
The body consists of head, thorax and ABDOMEN.
They have jointed appendages (arthros-joint, poda-appendages).
Respiratory organs are gills, book gills, book lungs or tracheal system.
CIRCULATORY system is of open type.
Sensory organs like antennae, EYES (compound and simple), statocysts or balance organs are present.
Excretion takes place through malpighian tubules.
They are mostly dioecious.
Fertilisation is usually internal.
They are mostly oviparous.
Development may be direct or indirect.
Examples : ECONOMICALLY important insects -
Apis - (Honey bee)
Bombyx - (Silkworm)
Laccifer - (Lac insect)
Vectors : Anopheles, Culex and Aedes (Mosquitoes)
Gregarious pest - Locusta (Locust) Living fossil - Limulus (King crab).
Discussion
18384.

Question : Describe the internal structure of Heart in Frog.

Answer»

Solution :1. Blood vascular system consists of a heart with three CHAMBERS, blood vessels and blood.
2. Heart is covered by a double-walled membrane called pericardium.
h wnd Ogen Syswns in Animals
3. There are two thin walled ANTERIOR chambers called auricles (Atria) and a single thick walled posterior chamber called ventricle.
4. Sinus venosus is a large, thin walled, triangular chamber, which is PRESENT on the dorsal side of the heart.
5. Truncus arteriosus is a thick walled and cylindrical structure which is obliquely placed on the ventral surface of the heart.
6. It ARISES from the ventricle and divides into right and left aortic trunk, which is further divided into three aortic arches namely carotid, systemic and pulmo-cutaneous,
7. The Carotid trunk supplies blood to the anterior region of the body.
8. The Systemic trunk of each side is joined posteriorly to form the dorsal aorta.
9. They supply blood to the posterior part of the
10. Pulmo-cutaneous trunk supplies blood to the tungs and skin.
11. 11. Sinus venosus receives the deoxygenated blood from the body parts by two anterior precaval veins and one post caval vein.
12. It DELIVERS the blood to the right auricle, at the same time left auricle receives oxygenated bloodthrough the pulmonary vein.
13. Renal portal and hepatic portal systems are seen in frog.
Discussion
18385.

Question : Describe the important properties of enzymes

Answer»

Solution :Specificity: Enzymes are specific in their ACTION. i.e. a particular enzyme acts only on a particular substrate. EG. Lactase acts on lactose only.
Temperature: Enzyme shows its highest activity at optimum temperature, i.e. 25-`40^@` C.
pH: Enzyme shows its highest activity at optimum pH. the optimum pH is 6.5-7. Some enzymes like pepsin function only in strong acidic pH, i.e 1.5-2. Trypsin is proteolytic enzyme which functions only in a alkaline pH, i.e., 8.5.
Enzyme concentration: The greater the affinity of the enzyme for its substrate, the greater is its catalytic activity.(IV) Enzymes are not destroyed by poisonous SUBSTANCES like chloroform. (vii) Enzymes are more active than inorganic catalysts.
Discussion
18386.

Question :Describe the important properties of enzymes.

Answer»

Solution :Lowering the activation energy.
Lowering the energy of transition state.
PROVIDING an alternative pathways.
Reducing the reaction entropy changes.
Increase in temperatures SPEED up the reactions.
Unique ENZYME for unique substance.
Small quantity is enough to facilitate faster reaction.
Discussion
18387.

Question : Describe the general characters of solanaceae.

Answer»

Solution :1. Distribution: Family Solanaceae includes about 88 genera and about 2650 species, ofthese Solanum is the largest genus of the family with about 1500 species.
2. Hahit . Mostly annual herbs, shrubs, smalltrees (Solanum violaceum) lianas with prickles (Solanum trilobatum) many with stellate trichomes, rarely vines (Lycium sinensis).
3. Root : Branched tap root SYSTEM.
4. Stem · Herbaceous or woody, erect or twining, or creeping, sometimes modified into tubers (So fanum tuberosum) often with btcollateral vascular bundles.
5. Leaves : Alternate, simple, rarely pinnae Y compound (Solanum tuberosum and Lycopersicon esculentum) exstipulate, opposite or sub- opposite in upper part, urncostate reticulate venation.
6. Inflorescence : Generally axillary or terminal cymose (Solanum) or solitary flowers (Datura stramonium). Extra axillary scorpioid cyme called rhiphidium (Solanum americanum) solitary and axillary (Datura and Nicotiana) umbellate cyme (Withania somnifera).
7. Flowers Bracteate (Petunia), or ebracteate (Withania) pedicellate, bisexual, heterochlamydeous, actinomorphic or weakly zygomorphic DUE to oblique position of ovary pentamerous, hypogynous.
8. Calyx: Sepals 5, synsepalous, valvate, persistent rarely the sepals are 4 or 6. Often enlarging to envelop the fruit (Physalis, Withania).
9. Corolla · PETALS 5, sympetalous, rotate, tubular (Solanum) or bell- shaped (Atropa) or infundibuliform (Petunia) rarely bilipped and zygomorphic (Schizanthus) usually valvate, sometimes convolute (Datura).
10. Androecium: Stamens 5, epipetalous, filaments usually unequal in length, stamens only 2 in Schizanthus 4 and didynamous in (Salpiglossis) ANTHERS dithecous, dehisce longitudinally or poricidal.
11. Gynoecium: Bicarpellary, SYNCARPOUS obliquely placed, ovary superior, bilocular but looks tetralocular due to the formation of false septa, numerous ovules in each locule on axile placentation.
12. Fruit: A capsule or berry. In Lycopersicon esculentum, Capsicum, the fruit is a berry and in species of Datul.ll and Petunia, the fruit is a capsule.
13. Seed : Endospermous.
Discussion
18388.

Question : Describe the function of fruit.

Answer»

SOLUTION :Functions of Fruit:
1. EDIBLE part of the fruit is a source OFFOOD, energy for animals.
2. They are source of many chemicals like sugar, pectin, organic acids, viamins and minerals.
3. The fruit protects the seed from unfavourable CLIMATIC conditions and animals.
4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
5. In certain cases, fruit may provide nutrition to the developing seedling.
6. Fruits provide source of medicine to humans.
Discussion
18389.

Question : Describe the general characters of Liliaceae.

Answer»

Solution :1.Distrbution :Liliaceae are fairly large family COMPRISING about 15 genera and 550 species.
2. Mah,t Mostly perennial herbs persisting by means of a sympodial rhizome (Polygonatum), by a bulb (Lilium) CORM (Colchicum) shrnbby or tree like ( Yucca and Dracaena). WOODY climbers, climbing with the help of stipular tendrils in SMILAX. Trees in Xanthorrhoea· succulents in Aloe.
3. Root: Adventitious and fibrous, and typically CONTRACTILE.
4. Stem : Sterns usually bulbous, rhizomatous in some, aerial, erect (Dracaena) or climbing (Smilax) in Ruseus the ultimate branches are modified into phylloclades, In Asparagus stem is modified into cladodes and the leaves are reduced to scales.
5. Leaf : Radical (Lilium) or cauline (Dracaena) usually alternate, opposite (Gloriosa), reduced to scales (Ruscus and Asparagus). The ventation is parallel but in species of Smilax it is reticulate. Leaves are usually exstipulate.
6. Infkorescence : Flowers are usually borne in simple or branched racemes (Asphodelus) spikes in Aloe, huge terminal panicle in Yucca, sohtary flowers are also seen.
7.Flowers : Flowers are often showy, pedicellate, bracteate, usually ebracteolate bisexual, actinomorphic, trimerous, hypogynous, rarely unisexual (Smilax) and are dioecious, slightly zygomorphic (lilium) and hypogynous.
8. erianth· Tepals 6 biseriate arranged in two whorls of 3 each, apotepalous or rarely syntepalous as in Aloe. Usually petaloid or sometimes sepaloid, odd tepal of the outer whorl is anterior in position, valvate or imbncate, tepals more than six in Paris quadrifolia.
9. Androecium: Stamens 6, arranged in 2 whorls of 3 each. Number of stamens vary from 4 to 12 in different species. Apostamenous, opposite to the tepals, sometimes epiphyllous, filaments distinct or connate, anthers dithecous, basifixed or versatile, extrose, or introse, dehiscing usually by vertical slit and sometimes by terminal pores, rarely synstamenous (Ruscus).
10. Gynoecium: Tricarpellary, syncarpous, the odd carpel usually anterior, ovary superior, trilocular, with 2 rows of numerous ovules on axile placextation, rarely unilocular with parietal iacentation, style usually one, stigmas 1 or 3, rarely the ovary is inferior (Haernodorum), nectarsecreting septal glands are present in the ovary.
11. Fruit : Fruit usually a septicidal or loculicidal capsule or a berry as in Asparagus & Smilax.
Discussion
18390.

Question : Describe the following: Bivalent Draw a diagram to illustrate your answer.

Answer»

Solution :BIVALENT. In zygotene, the synaptonemal complex formed by a pair of synapsed HOMOLOGOUS CHROMOSOMES is called a bivalent (TETRAD).
Discussion
18391.

Question : Which of the following is a true nut?

Answer»

Solution : Chiasmata. During diplotene af meiotic I, the homologous chromosomes are HELD TOGETHER only at certain points, and these points of contact are KNOWN as chiasmata. They represent the PLACES of CROSSING over
Discussion
18392.

Question : Describe the following: Synapsis Draw a diagram to illustrate your answer.

Answer»

SOLUTION : (a) Synapsis. During zygotene HOMOLOGOUS chromosomes start pairing TOGETHER and this process of association is called synapsis.
Discussion
18393.

Question :Describe the following: (a) synapsis (b) bivalent (c) chiasmata Draw a diagram to illustrate your answer

Answer»

Solution :(a) During Zygotene of Prophase-I of Meiosis homologous chromosomes start pairing, This association is called synapsis.
(b) The COMPLEX formed by a PAIR of synapsed homologous chromosomes in Zygotene stage is called a bivalent.
(c) During DIPLOTENE stage paired chromosomes form X shaped structure is called chaismata.
Discussion
18394.

Question : Draw the female reproductive system cockroach?

Answer»

Solution :The FEMALE reproductive system of cockroach consists of a pair of ovaries, vagina, genital pouch, colleterial GLANDS, spermathecae and the external genitalia.
2. A pair of ovaries lies laterally in the 2nd and 6th abdominal segnment.
3. Each ovary is formed of a group of eight ovarian tubules or ovarioles, containing a chain of developing ova.
3. The lateral oviducts of each ovary unite into a broad median common oviduct known as vagina, which opens into the genital chamber.
5. The vertical opening of the vagina is the female genital pore.
6. A pair of spermathecae is present in the 6th segment, which opens by a median aperture in the dorsal wall of the genital pouch.
7. During COPULATION, the ova descend to the genital chamber, where they are fertilized by the sperms.
8. A pair of white and branched colleterial glands present behind the ovaries forms a HARD egg case called Ootheca around the eggs. Genital pouch is formed by the 7th 8th and 9th abdominal sterna.
9. The genital pouch has two chambers, a genital chamber into which the vagina opens and an oothecal chamber where oothecae are formed,
10. Three pairs of plate LIKE chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into the ootheca as ovipositors.
11. Ootheca is a dark reddish to blackish brown capsule contains nearly 16 fertilized eggs and dropped or glued to a suitable surface, of high relative humidity near a food source.
Discussion
18395.

Question : Describethe experiment of Priestley.

Answer»

Solution :Joseph Priestly (1733 - 1804) in 1770 performed series of experiments that reveated the essential role of air in the growth of green plants
Priestley, discorvered oxygen in 1774.

Priestley observed that a candle burning in a closed space - a BELL jar, soon gets extinguished (Figure a, b, c ,d)
However a MOUSE would soon suffocatein a closed space.
On the basis of these observations he CONCLUDED that a burning candle or an animal that breath the air, both somehow damage the air.
Supplies deficit of air.
But when he placed a mint plant in the same bell jar, he found that the mouse stayed alive and the candle continued to burm.
Priestley hypothesised as follows :
Plants restore to the air whatever BREATHING animals and burning candles remove.
Discussion
18396.

Question : Describe the events taking place during interphase.

Answer»

Solution :The interphase, though called the RESTING PHASE, is the time during which the cell is PREPARING for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is DIVIDED into `G_1` phase, S phase and `G_2` phase. For the events taking place during interphase, please see TEXT
Discussion
18397.

Question : Describe the enzymatic action on fats in the duodenum.

Answer»

Solution :The fat is DIGESTED by the enzyme lipase with HELP of bile. Fats are converted into di and monoglyceride and after that into fatty ACID and GLYCEROL.
Fat `overset"Lipases"to` Diglycerides `to` Monoglycerides
Di/Monoglycerides `overset"Lipases"to` Fatty ACIDS + Glycerol.
Discussion
18398.

Question : Describe the effect of substrate concerntration on enzyme activity.

Answer»

Solution : With the INCREASE in substrate concentration the velocity of the enzymatic reaction rises at first. This reaction ultimately REACHES a MAX velocity (Vmax) which is not exceeded any further because the enzyme molecules are FEWER than the substrate molecule so after SATURATION, no free enzyme molecules to bind with additional substrate.
Discussion
18399.

Question : Ifyour friendis addictedto cigarette smokingasa biology studentshow wouldyou explain theill-effectsof smoking ?

Answer»

Solution :The digestive system of the earthworms consists of the alimentary canal and the digestive glands.The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus. The mouth opens into the buccal cavity which occupies the `I^(st)` and `2^(nd)` segments.The buccal cavity leads into a thick muscular pharynx, which occupies the `3^(rd)` and `4^(th)` segments is surrounded by the pharyngeal glands.A SMALL narrow tube , oesophagus lies in the `5^(th)` segments and continues into a muscular gizzard in the `6^(th)` segment. The gizzard helps in the grinding of the soil particles and decaying leaves.Intestine starst from the `7^(th)` segment and continues till the last segment.Thedorsal wall of the intestine is folded into the cavityas the typhlosole.This fold contains blood vessels and increases the absorptive area of the intestine.The inner epithelium consists of columnar cells andglandular cells.The alimentary canal opens to the exterior through the anus.
The ingested organic rich soilpasses through the digestive system tract where digestive enzymes breakdown complex food into smaller absorbable units. The simpler molecules are absorbed through the intestinal membrane and are utilized.The undigested particles along with earth are passed out through the anus, as worm castings or vermicasts.The pharyngeal or salivary gland cells and the glandular cells of the intestine are supposed to be the digestive glands which secrete digestive enzymes for digestion of food.

[OR] Today due to curiosity, excitement or adventure youngsters start to smoke and later get ADDICTED to smoking. Research says about 80% of the lung cancer is due to cigarette smoking.Smoking is INHALING the smoke from burning tobacco. There are thousands of known chemicals which INCLUDES nicotine, tar. carbon monoxide, ammonia, sulphur-dioxide andeven small quantities of arsenic.Carbon monoxide and nictoine damage the cardiovascular system and tar damages the gaseous exchange system.Nictonine is the chemical that causes addiction is a stimulant which makes the heart beat faster and the narrowing of blood vessels results in raised blood pressue and coronary heart diseases. Presence of carbon monoxide reduces oxygen supply.Lung cancer, mouth cancer and larynex is more common in smokers than non-smokers.Smoking also causes cancer of the stomach, pancreas and bladder and lowers sperm count in men.
Smoking can cause lung diseases by damaging the airways and ALVEOLI and results in emphysema and chronic bronchitis.Thses two diseases with asthma are often refered as Chronic Obstructive Pulmonary Disease(COPD).When a person smokes, nearly 85% ofthe smoke released is inhaled by the smoker himself and others in the vicinity, called passive smokers are also affected.Guidance or counselling should be done in such users to withdraw this habit.
Discussion
18400.

Question : Describe the digestive role of chymotrypsin.Which two other digestive enzymes of the same category are secreted by its source gland?

Answer»

Solution :CHYMOTRYPSIN hydrolyses proteins into PEPTIDES and peptones.it also hydrolyses the MILK protein CASEIN to paracasein and whey proteins. The other two ENZYMES are trypsin and carboxypeptidase.
Discussion