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11451.

Question : Succulents are known their stomata closed during the day to chek transpiration . How dothey meet their photosynthetic CO_(2) requirements ?

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Solution :Xerophytic plants grow in area where - water is scarce. Stomata are closed during the day so exchanged of gases does not take PLACE.
These plants have a special arrangement. In that for fixation of `CO_(2)` is fixed in malic ACID during night `CO_(2). It has 4 carbons. During day `CO_(2)` is released and enters into PHOTOSYNTHETIC cellw which do work of PHOTOSYNTHESIS.
11452.

Question : Successions that occur on soils or areas which have recently lost their community are referred to as

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PRIMARY successions
secondary successions
lithoseres
priseres

Answer :B
11453.

Question : Succession stages that occur in salt marshes is called

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PSAMMOSERE
HALOSERE
Lithosere
Hydrosere

Answer :B
11454.

Question : Succelent performs CO_(2) fixation by

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CAM pathway
`C_(4)`pathway
`C_(3)`pathway
`C_(2)`pathway

Solution :Succulents PERFORM `CO_(2)` FIXATION by CAM pathway.
11455.

Question : Succession is

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Orderly PROCESS of community change till stability
Gradual, CONVERGENT DIRECTIONAL and continuous process
Series of biotic communities that appear GRADUALLY in a barren area
All of the above

Answer :D
11456.

Question : Succelentsare know to keep theirstomata closedduringthe dayto check transpiration . Howdo theymeet theirphotosyntheitcCO_(2) requirements ?

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Solution :Succulentsundergo a specialcarbon pathway called Crassulacean Acid Metabolism(CAM pathway). At nighttime , CAMplantsfor `CO_(2)`with thehelp of Phosphoenol PYRUVICACID ( PEP) and produe Oxalo AceticAcid (OAA) . Subsequently OAAis convertedinto malicacid like `C_(4)`cycleand produces `CO_(2)`. The `CO_(2)`thusformedentersCalvinscycleproduces carbohdrates .
11457.

Question : Substrate level phosphorylation occurs during which step of Krebs' cycle

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Succinly`CoA rarr` Succinic acid
Isocitric acid `rarr` Oxalosuccinic acid
Oxalosuccinic acid `rarr alpha-`ketoglutaric acid
Malic acid `rarr OA A`

SOLUTION :During Krebs' or citric acid cycle, succinyl-CoA is acted upon by enzyme sucinyl-CoA synthese to form succinate (a 4 C compound). The reaction releases sufficient energy to form ATP (in PLANTS) or GTP (in ANIMALS) by SUBSTRATE level phosphorylation GTP can form ATP through a coupled reaction
`"Succiny CoA + GDP/ADP +H"_(3)PO_(4) underset("synthetase")overset("Succinyl CoA")(hArr) "Succinate + CoA + GTP/ATP"`
11458.

Question :Subterranean cleistogamous and geophilous flowers occur in

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Viola
Anthocephalus
COMMELINA benghalensis
FICUS benghalensis

Answer :C
11459.

Question :Subterranean cleistogamous flowers are found in

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FICUS benghalensis
COMMELINA benghalensis
Adina cordifolia
Anthocephalus cadamba

ANSWER :B
11460.

Question : Substrate level phosphorylation occurs in TCA cycle during conversion of :

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Oxaloacetic ACID to CITRIC acid
Succinyl CoA to succinic acid
Succinic acid to fumaric acid
Fumaric acid to malic acid

ANSWER :B
11461.

Question : Substance whichbring about changesinallostericsites are called.

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activators
inhibitors
promoters
modulators

ANSWER :D
11462.

Question :Substance that accumulates in a fatigued muscle is ..............

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PYRUVIC ACID
LACTIC acid
`CO_(2)`
ADP

Solution :Lactic acid
11463.

Question : Subcellular organelle is :

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Peroxisome
Ribosome
Plastid
Lysosome

Solution :Ribosome is NAKED (without any UNIT membranous COVERING) so is CALLED subcellular organelle.
11464.

Question : Sub oesophageal ganglion is called

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PRINCIPAL sensory CENTRE 
principal MIXED centre 
principal motor centre 
all the above 

Answer :C
11465.

Question : Sub neural blood vessel runs below

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a. nerve RING 
b. SEGMENTAL nerves 
C. nerve CORD 
d. DORSAL blood vessel 

Answer :C
11466.

Question : Study this structure of a root hair andanswer the following: Label parts a, b, c, d and e.

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Solution :a. CELL WALL b. cytoplasm c. VACUOLE d. NUCLEUS e. cell membrane
11467.

Question : Study this structure of a root hair andanswer the following: How many cells is a root hair made upof?

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SOLUTION :ONE CELL.
11468.

Question : Study this structure of a root hair andanswer the following: Give one important property of part'e'.

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SOLUTION :CELL MEMBRANE is SEMI PERMEABLE.
11469.

Question : Study this structure of a root hair andanswer the following: What is the function of root hair?

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SOLUTION :CELL WALLS are THIN and PERMEABLE.
11470.

Question : Study the set-up shown in the diagram and answer the following: What will happen if a boiled potato istaken?

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SOLUTION :OSMOSIS will not TAKE PLACE.
11471.

Question : Study the set-up shown in the diagram and answer the following: What change will you observe in part'B' of the setup after one hour? Givereason for this change.

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SOLUTION :the LEVEL of solution will RISE in PART .B.
11472.

Question : Study the set-up shown in the diagram and answer the following: What is the technical term for thisprocess?

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SOLUTION :OSMOSIS.
11473.

Question : Study the set-up given in the diagram and answer the following questions: Why is the membrane calledsemipermeable membrane?

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Solution :The membrane allows only water MOLECULES to pass and RESTRICTS the ENTRY of sugar molecules.
11474.

Question : Study the set-up given in the diagram and answer the following questions: What is this process known as?

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SOLUTION :OSMOSIS.
11475.

Question : Study the set-up given in the diagram and answer the following questions: Which limb of the set-up containsmore concentrated solution?

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SOLUTION :TUBE A CONTAINS more CONC. SOLUTIONS.
11476.

Question : Study the set-up given in the diagram and answer the following questions: What will be the level in tube A andtube B after 1 hour of set-up?

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Solution :LEVEL in tube B will RISE after 1 hour.
11477.

Question :Study the set-up given alongside andanswer the questions given below: Why is the surface of water coveredwith oil?

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Solution :Oil prevents EVAPORATION of water. It is to PREVENT evaporation from the TEST TUBE.
11478.

Question :Study the set-up given alongside andanswer the questions given below: Which physiological process does thissetup of experiment show?

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SOLUTION :ABSORPTION of WATER by ROOT.
11479.

Question :Study the set-up given alongside andanswer the questions given below: What will happen to the level of waterin test tubes A & B?

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Solution :In test tube A the PLANT absorbs WATER and transpires also, hence the water LEVEL FALLS, test tube B is a control, the water level. Does notefall SINCE there is no plant in it.
11480.

Question :Study the set-up given alongside andanswer the questions given below: What is the purpose of the setting uptest tube B?

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Solution :Test tube B ACTS as a control for SET up .A..
11481.

Question : Study the given figure in which two chambers A and B, containing solutions are separated by a semipermeable membrane : If one chamber has Psi of - 2000 kPa and the other-1000 k Pa, which is the chamber that has higher Psi?

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SOLUTION :`-1000 KP `
11482.

Question : Study thegivenfigurecarefullyandanswerthe followingquestions : (i)LabeltheParts maeked as(a),(b) ,( C) ,(d) and € (ii)giveonemajorfunctionof eachpart.

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SOLUTION :(i) (a)Sertoli cells
(b)Seminiferoustubule
( C) Spermatozoa
( D)SPERMATOGONIA
( e)Intersitial cells .
(II)FUNCTIONOF parts:
(a)sertolicellsprovidenutrition todevelopingspermatozoa .
( B)Spermatozoa areformedin theseminiferoustubules from spematoniathatarisefromgerminalepithelialcells . ( C)Theseare male gameteswhichfertilizethe ovum .
(d)Thespermatoeniccellsofgerminalepithliumliningtheseminiferoustubulesby mitotic divisionsproducespermatogonia , Thelatterundergomeioticdivisiontoform spermatozoa .
( E)Intersitialcellssecretesexhormones(Testosterone ).
11483.

Question : Study the given data and answer the questions that follow. A sample of an enzyme called lactase was isolated from the intestinal lining of a calf. Assays were undertaken to evaluate the activity of the enzyme sample. The substrate of lactase is the disaccharide lactose. Lactase breaks a lactose molecule in two, producing a glucose molecule and a galactose molecule. Two assays were carried out. {:("Lactose concentration (% w/v)",15,15,15,15,15,15),("Concentration of enzyme smaple (%v/v)",0,5,10,15,20,25),("Reaction rate "mu"mole glucose "sec^(-1)mL^(-1),0,25,50,75,100,125):} {:("Lactose concentration (% w/v)",0,5,10,15,25,30),("Concentration of enzyme smaple (%v/v)",5,5,5,5,5,5),("Reaction rate "mu"mole glucose "sec^(-1)mL^(-1),0,15,25,35,40,40):} Which of the following statements can be concluded from the two assays ?

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The reaction rate of the LACTASE assay is always proportional to the amount of ENZYME present.
The amount of lactose in an assay has no effect on the rate of the reaction.
The reaction rate of the lactase assay is proportional to the amount of lactose present.
The reaction rate is proportional to the amount of enzyme present at a lactose concentration of 15%w/v.

Solution :The reaction rate is proportional to the amount of enzyme present at a lactose concentration of `15% w//v`. The rate of reaction is proportional with the SUBSTRATE concentration only to a limit, after which any increase in substrate concentration MAKES no difference in rate of reaction, the rate of reaction then remains constant.
11484.

Question : Study the four statements (i-iv) given below and select the two correct ones out of them. i) Alion eating a deer and a sparrow feeding on grains are ecologically similar in being consumers. ii) Predator star fish pisaster helps in maintaining species diversity of some invertebrates. iii) Predators ultimately lead to the extinction of prey species iv)Production of chemicals such as nicotine, strychnine by the plants are metabolic disorders. The two correct statements are

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a and d
a and B
b and C
c and d

ANSWER :B
11485.

Question : Study the following statements regarding inbreedind and select the incorrect ones. i) The inbreeding strategics allow the desirable qualities of two different breeds to be combined ii) It increases homozygosity. iii) It also helps in elimination of less disirable genes. iv) Continued inbreeding increases fertility and productivily.

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i and ii
ii and iii
iii and iv
i and iv

Answer :D
11486.

Question : Study the following statements carefully and give the answer A) Belladonna and ashwagandha are the medicinal plants that belong to Solanaceae family B) Tulips produce very beautiful flowers and belong to Liliaceae family C) A flower is a modified shoot, meant for sexualreproduction D) After fertilisation ovary and ovule convertinto fruit and seed respectively in gymnospermic plants angiosperme, How many statements are correct from these?

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Two
Three
Four
Five

ANSWER :B
11487.

Question : Study the following statements and select the incorrect option.

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Tapetum NOURISHES the developing pollen GRAINS
Hilum represents the junction between OVULE and funicle
In aquatic plants such as water hyacinth and water LILY, pollination occurs by water
The primary endosperm nucleus is triploid

Answer :A
11488.

Question : Study the following statements and select the incorrect ones. (i) Pyramids of energy can never be inverted, since this would violate the laws of thermodynamics. (ii) Pyramids of standing crop and numbers can be inverted since the number of organisms at a time does not indicate the amount of energy flowing through the system. (iii) There are certain limitations of ecological pyramids such as they do not take into account the same species belonging to two or more trophic levels. (iv) Saprophytes are not given any place in ecological pyramids even though they play a vital role in the ecosystem.

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(i) and (III)
(iii) and (IV)
(II) and (iii)
NONE of these

Answer :D
11489.

Question : Study the following statements and select the incorrect one.

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SHORTER food chains PROVIDE more energy as compared to longer food chains.
Ecological factors connected with PHYSICAL geography of earth are CALLED topographic factors
The pyramid of biomass is upright in a grassland ecosystem and the pyramid of NUMBERS is upright in a parasitic food chain.
None of these

Answer :C
11490.

Question : Study the following plants, I. Hibiscus II Brassica III. Papaver IV. Annona V. Michelia VI. Rosa VII. Lotus. In how many plants the gynoecium is apocarpous and multicarpellary.

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3
4
5
2

Answer :B
11491.

Question :Study the following in nitrogen cycle and identify A and B NH_3 overset((A))to NO_2^(-) overset(B)to NO_3^(-)

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A-Nitrosomonas, B-Nitrobactor
A-Nitrobactor, B-Nitrobactor
A-Nitrobactor, B-Nitrosomonas
A-Thiobacillus, B-Nitrobacter

Answer :A
11492.

Question : Study the following flowchart and fill up the blanks by selecting the correct option.

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`{:("(A)","(B)","(C)",(D)),("THORACIC skeleton","Limbs","Skull","Ribs"):}`
`{:("(A)","(B)","(C)",(D)),("APPENDICULAR skeleton","Skull","Ribs","Limbs"):}`
`{:("(A)","(B)","(C)",(D)),("Appendicular skeleton","Limbs","Ribs","Skull"):}`
`{:("(A)","(B)","(C)",(D)),("LUMBAR skeleton","Limbs","Skull","Ribs"):}`

Solution :N//A
11493.

Question : Study the following equation and name the reaction A and B. Glycogen in liverunderset(B) overset(A)hArr Glucose in blood

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SOLUTION :REACTION A is GLYCOGENOLYSIS. Reaction B is GLYCOGENESIS.
11494.

Question : Studythe following figure representingthe life cycle of a typical cnidarian and choose the correct option .

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`{:(A,B),("SEXUAL REPRODUCTION ","ASEXUAL reproduction"):}`
`{:(A,B),("Asexual reproduction"," Sexual reproduction"):}`
`{:(A,B),("Asexual reproduction"," ASexual reproduction"):}`
`{:(A,B),("Sexual reproduction"," Sexual reproduction"):}`

ANSWER :B
11495.

Question : Study the following characters: Bicarpellary ovary, syncarpous ovary, bilocular ovary, Superior ovary, unequal lengthed stamens, polypetalous corolla, six stamens, dry fruit, ovules attached to false septum, tetramerous flower, ebracteate flower, bisexual flower, hypogynous flowers. How many characters from the above list observed in the taxon "Brassica"

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10
6
12
13

Answer :D
11496.

Question : Study the following given table and select the correct option that represents the letters.

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`{:("CHONDRICHTHYES","Osteichthyes"),("(1)A, C & E","B ,D ,F & G "),((2)"A ,D ,E & G ","B ,C & F "),((3)"A,D & E","B,C ,F &G"),((4)"B,D,E & F","A ,C,& G"):}`
`{:("Chondrichthyes","Osteichthyes"),("(1)A,D,E& G","B ,C& F "):}`
`{:("Chondrichthyes","Osteichthyes"),("A,D & E","B,C,F & G"):}`
`{:("Chondrichthyes","Osteichthyes"),("B ,D E & F","A,C & G"):}`

Answer :C
11497.

Question : Study the figures given below and identify the option which represents correct grouping of the labelled figures A,B,C and D.

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ANSWER :C
11498.

Question : Study the diagram given alongsideshowing a plant cell kept in a strong sugar solution. What will happen to this cell if it is kept in water?

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SOLUTION :ENDOSMOSIS LEADING to TURGIDITY
11499.

Question : Study the diagram given alongsideshowing a plant cell kept in a strong sugar solution. Name the process the cell hasundergone. What is the reason of the processnamed?

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SOLUTION :EXOSMOSIS
11500.

Question : Study the diagram given alongsideshowing a plant cell kept in a strong sugar solution. Name the process the cell hasundergone.

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SOLUTION :PLASMOLYSIS