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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
Which of the following is not required for chemiosmosis ?A. A thylaked membraneB. A proton gradientC. ATP synthese enzymeD. Carboxylase enzyme |
| Answer» Correct Answer - D | |
| 852. |
Identify the parts marked as A,B and C in the given figure showing ATP synthesis thorugh chemiosmosis. A. `{:(A,B,C),("Thylakoid lumen",CF_(0),CF_(1)):}`B. `{:(A,B,C),("Thylakoid lumen",CF_(1),CF_(0)):}`C. `{:(A,B,C),("Chloroplast lumen",CF_(0),CF_(1)):}`D. `{:(A,B,C),("Chloroplast lumen",CF_(1),CF_(0)):}` |
| Answer» Correct Answer - A | |
| 853. |
Breaddonw of proton gradient developed during chemiosmosis leads to the release ofA. oxygenB. waterC. energyD. protons. |
| Answer» Correct Answer - C | |
| 854. |
In `C_(4)` plants, the bundle sheath cellsA. Have thin walls to facilitate gaseous exchangeB. Have large intercellur spaceC. Have a high densily of chloroplastsD. Are rich in PEP carboxylase |
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Answer» Correct Answer - C Bundle sheath cells are characterized by: (i) Large number of chloroplasts (agranal) Thick walls impervious to gaseous exchange (ii) No intercellular space |
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| 855. |
Free radical has electronA. Unparied and extermelyB. Paried and extermelu inactiveC. Unapaired and extremely inactiveD. Parired and extremely reactive |
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Answer» Correct Answer - A Free radical has electron unpaired and extremely active. |
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| 856. |
In which cells of leaf, pyruvate is converted to PEP in C pathway?A. Epidermals cellsB. Mesophyll cellsC. Bundle sheath cellsD. Guard cells |
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Answer» Correct Answer - B In mesophyll cells pyruvate is converted to PEP in `C_(4)` pathway. |
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| 857. |
When two plants L and M were exposed to different light intensities and temperature they showed changes in their rates of photosynthesis. Which have been represented in the following graph. The graph indicates thatA. Plant L is a `C_(3)` plant for which the light saturation point is 100% of full sunlightB. Plant M is a `C_(4)` plant for which the optimum temperature s around `20^(@)C`C. Plant M is a `C_(3)` plant which is more affected at higher temperature and higher light intensity as compared to palnt LD. plant L is a `C_(4)` plant and cannot function at light intensities above the saturation point. |
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Answer» Correct Answer - C The light intensity at which a plant can achive maximum rate of photosynthesis is called as light saturation point. Its value is 50-70% full sunlight in `C_(3)` sun plants and upto 200% of full sunlight in `C_(4)` sun plants (e.g., sugarcane). Beyond satruation point (it is seldom relaised in nature in `C_(4)` sun plants). the rate of photosynthesis begins to decline. Optimum temperature at which the rate of photsynthesis is maximum is `10^(@)C-25^(@)C` for `C_(3)` plants and `30-45^(@)C` for `C_(4)` plants. Thus, it is clear that in the given graph, plant L is a `C_(4)` plant and plant M is a `C_(3)` plant. |
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| 858. |
In which cells of leaf, pyruvate is coverted to PEP in `C_(4)` pathwayA. Epidermal cellB. Mesophyll cellC. Bundle sheath cellD. Guard cell |
| Answer» Correct Answer - B | |
| 859. |
Refer to the given cross section of a `C_(4)` leaf and select the incorrect option. A. P are the chloroplasts in which thyiakoids are stacked together to form grans.B. P are the chloroplasts which can perform light reaction, evolve molecular `O_(2)` and produce assimilatory power.C. Q are the chloroplasts in which thylakoids occur as stroma lamellae.D. Q are the chloroplasts in which `CO_(2)` is fixed by phosphoenol pyruvic acid to form oxaloacetic acid. |
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Answer» Correct Answer - D The given croos ection of the leaf represents Kranz anatomy in a `C_(4)` plant. P are the chloroplasts of mesophyll cells, which are granal (i.e., thylakoids are stacked to from grana) and are specilasied to perform light reaction of photosynthesis, evolve `O_(2)` and produce assimilatory power (ATP+NADPH`_(2)`). Q are the chloroplasts of bundle sheath cells which are agranal, i.e., grana are absent and the thylakoids occur as stroma lamellae. In these chloroplasts. Final fixation of `CO_(2)` takes place through calvin cycle where `CO_(2)` is fixed by RuBP in the presence of enzyme RuBisCO. |
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| 860. |
Which metal ion is a constituent of chlorophyll?A. IronB. copperC. MagnesiumD. Zinc |
| Answer» Correct Answer - C | |
| 861. |
Number ofcarboxylation occur in Calvin cycle is |
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Answer» Correct Answer - B Calvin cycle is divided into three distinct phase but carboxylation occur only one time. |
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| 862. |
The family in which many plants are `C_(4)` typeA. MalvaceaeB. SolanaceaeC. CruciferaeD. Gramineae |
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Answer» Correct Answer - D This pathway was first reported in members of family gramineae (grasses) like sugarcane, maize etc. More than 300 species belong to dicots and the rest belong to monocots. There are no known `C_(4)` gymnosperms, bryophytes or algae. |
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| 863. |
Cyclic photophosphorylation results in the formation of :A. NADPHB. ATP and NADPHC. ATP, NADPH and `O_2`D. ATP |
| Answer» Correct Answer - D | |
| 864. |
Cyclic-photophosphorylation results in the formation ofA. NADPHB. ATP and NADPHC. ATP,NADPH and `O_(2)`D. ATP |
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Answer» Correct Answer - D Cyclic-photophosphorylation involves only pigment system-I and result in the formation of ATP only. When the photons activate PS-I, a pair of electrons are raised to a higher energy level. They are captured by primary acceptor which passes them on to ferredoxin, plastoquinone, cytochrome complex, plastocynin and finally back to reaction centre of PS-I, i.e., `P_(700)` At each step of electron transfer, the electrons lose potential energy. their trip down hill is caused by the transport chain to pump `H^(+)` across the thylokoid membrane. The proton gradient thus established is responsible for forming ATP (2 molecules). No reduction of NADP to NADPH+`H^(+)` |
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| 865. |
Cyclic photophosphorylation results in the formation of :A. Cyclic photophosphorylation results in the formation ofB. NADPHC. ATP and NADPHD. ATP , NADPH and `O_2` |
| Answer» Correct Answer - D | |
| 866. |
Name the scientist, who first pointed out that plants purify foul air by bell jar experiment .A. WillstatterB. Robert HookeC. PriestleyD. Iean Senebier |
| Answer» Correct Answer - C | |
| 867. |
Who demonstrated that green plansts purifty the foul air produced by breathing animals and burning candles?A. PriestleyB. IngenhouszC. SachsD. Engelmann |
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Answer» Correct Answer - A Joseph Priestley (1770) observed that a candle burning in a closed space a bell jar, soon gets extinguished. Similarly. A mouse kept in a closed space would soon get suffocated and die However, when he placed a mint plant and candle continued to burn. Priestley hypotesised that four air or phologiston produced during burning of candles or animal (mice) respiration could be converted into pure air or dephlogiston by plants (mint). In 1774, Priestley discovered oxygen. |
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| 868. |
Mathc column I with column II and selct the correct option from the given codes. `{:("Column I","Column II"),(C_(3)"Plants",(i)"Kalanchoe,Opuntia"),(C_(4)"plants",(ii)"Maize,sugarcane"),(C_(4)"plants",(iii)"Maize,sugarcane"):}`A. ii,iii,iB. i,ii,iiiC. iii,ii,iD. i,iii,ii |
| Answer» Correct Answer - C | |
| 869. |
Read the following statements and selct the correct ones. (i) PS I is involved in non-cyclic photophosphroylation only. (ii) PS II is involved in both cyclic and non-cyclic photophosphorylation. (iii) Stoma lamellae membranes possess PS I only whereas grana lamellae membranes possess both PS I and PS II.A. i onlyB. ii onlyC. iii onlyD. i,ii and iii |
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Answer» Correct Answer - C PSI is involved in both cyclic and non-cyclic photophosphorylation PS II is involved only in non-cyclic photophosphorylation PS II is present in the aprressed (inner) part of grana thylakoids, PS I is located in the non-appressed (outer) part of grana thylakoids as well as stroma thylakoids |
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| 870. |
Oxygen released in photosynthesis is formed duringA. oxidative phosporylationB. cyclic photophosphorylationC. non - cyclic photophosphorylationD. carbon assimilation during dark reactions |
| Answer» Correct Answer - C | |
| 871. |
Read the following statements:A. `F_(0)` Part if ATP ase is associated with breakdown of proton gradient .B. A H- Carrier contributes in creation of proton gradient.C. Movement of electrons in ETS is coupled to pumpling of protons into the lumen.D. |
| Answer» Correct Answer - C | |
| 872. |
Photorespiration is favoured byA. high oxygen and low carbon dioxideB. high carbon dioxide and low oxygenC. high temperature and low oxygenD. high humidity and temperature |
| Answer» Correct Answer - A | |
| 873. |
Photorespiration is favoured byA. Nigh `O_(2)` and low `CO_(2)`B. Low light and high `O_(2)`C. Low temperature and high `O_(2)`D. low `O_(2)` and high `CO_(2)` |
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Answer» Correct Answer - A Photorespiration is light induced oxidation of photosynthetic intermediates with the help of oxygen. It is stimulated by high `O_(2)` concentration or low `CO_(2)`. High light intensity, high temperature and ageing of leaf. |
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| 874. |
Protochlorophyll differs from chlorophyll in lackingA. 2 hydrogen atoms in two of its pyrrole ringsB. 2 hydrogen atoms in two of its pyrrole ringsC. 4 hydrogen atoms in one of its pyrrole ringsD. 4 hydrogen atoms in two of its pyrrole rings |
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Answer» Correct Answer - A Protochlorophyll differs from chlorophyll in lacking 2 hydrogen atoms in one of its pyrrole rings. |
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| 875. |
NADPH is generated throughA. Anaerobic respirationB. Cyclic photophorylationC. Non-cyclic photophosphorylationD. Glycolysis |
| Answer» Correct Answer - C | |
| 876. |
Excitation of chlorophyll by light isA. Exergonic reactionB. Anabolic reactionC. Photochemical reactionD. Photooxidation reaction |
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Answer» Correct Answer - C Excitation of chlorophyll by light is photochemical reaction. |
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| 877. |
NADPH is generated throughA. photosystem-IB. photosystem-IIC. anaerobic respirationD. glycolysis |
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Answer» Correct Answer - B NADPH is generated through photosystem-II in non-cyclic photophosphorylation (which involves both PS-I and II) protons released from photolysis and electrons emitted from `P_(700)` are ultimately passed on to `NADP^(+)` resulting in the formation of NADPH. In cyclic photophosphorylation (which involves only PS-I) electrons flow in a cyclic manner but there is not net formation of NADPH and `O_(2)`. |
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| 878. |
Which of the following is present in Calvin Cycle.A. Reductive carboxylationB. Oxidative carboxylationC. PhotophosphorylationD. Oxidative phosphorylation |
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Answer» Correct Answer - A In dark phase or Calvin cycle, carbon dioxide is assimilated with the helpof assimilatory power (ATP and `NADPH_(2)`) to produce organic acid. The cycle involves reduction of carbon involving carboxylation, glycolytic reversal and regeneration of RuBP. `C_(3)` cycle is also known as reductive pentose pathway or photosynthetic carbon Reduction (PCR). |
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| 879. |
The law of limiting factors was given by ___ in the year___A. LeibigB. BlackmanC. CalvinD. Arnon |
| Answer» Correct Answer - B | |
| 880. |
The principle of limiting factors was proposed by:-A. BlackmannB. HillC. ArnonD. Liebig |
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Answer» Correct Answer - A The principle of limiting of factors was given by blackmann, a british plant physiologist in 1905, according to him, light intesity, carbon dioxide concentration and temperature are the limiting factors in photosynthesis. When a process is conditioned as to its rapidity by a number of seprate factors, the rate of the process is limited by the pace of the slowest factor. |
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| 881. |
Which one of the following is wrong in relation to photoespiration ?A. Ist oxidation in chloroplastB. It occurs in day time onlyC. It is a characteristic of `C_(4)` plantsD. It is a characteristic of ` C_(3)` plants |
| Answer» Correct Answer - C | |
| 882. |
Maximum absorption of light occures in the region (PAR) ofA. 400-700 nmB. 700-900 nmC. 1000-1200 nmD. 1500-2000 nm |
| Answer» Correct Answer - A | |
| 883. |
The principle of limiting factors was proposed by:-A. LeibigB. Hactch and slackC. BlackmannD. Arnon |
| Answer» Correct Answer - C | |
| 884. |
How many additional ATP are used during synthesis of two molecules of hexose sugar in maxize than tmato ?A. 12B. 36C. 24D. 8 |
| Answer» Correct Answer - C | |
| 885. |
`C_(4)` Plants have higher net photosynthesis rate as they haveA. No photorespirationsB. PEP as `CO_(2)` acceptorC. Kranz anatomyD. Photosynthesis even at low light intensity |
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Answer» Correct Answer - A `C_(4)` plants have higher net photosynthetic rate as they have no photorespiration. |
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| 886. |
Maximum absorption of light occures in the region (PAR) ofA. 1000-1200 nmB. 1500-2000 nmC. 400-700 nmD. 700-900 nm |
| Answer» Correct Answer - C | |
| 887. |
In photorespiration , glycolate and glycine synthesis occurs respecitvely inA. Chloroplast and mitochondriaB. peroxisome and chloroplastC. chloroplast and peroxisomeD. peroxisome and mithochondria |
| Answer» Correct Answer - C | |
| 888. |
the rate of photosynthesis is higher inA. Very high lightB. Continuous lightC. Red lightD. Green light |
| Answer» Correct Answer - C | |
| 889. |
Hill reaction occurs inA. High altitude plantsB. Total darknessC. Absence of waterD. Presence of ferredoxin |
| Answer» Correct Answer - B | |
| 890. |
Plants adapted to low light intensity haveA. larger photosynthetic unit size than the sun plantsB. higher rate of `CO_(2)` fixation that the sun plantsC. more extended root systemD. leaves modified to spines |
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Answer» Correct Answer - A Shade tolerant plants have lower photosynteic rates and hence, lower growth rates. O the other hand, these plants have larger photosynthetic unit size than the sun plants. |
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| 891. |
Plants adapted to low light intensity haveA. More extended root systemB. Leaves modified to spinesC. Larger photsysthetic unit size than the sun plantsD. Higher rate of `CO_(2)` fixation than the sun plants |
| Answer» Correct Answer - D | |
| 892. |
Plants adapted to low light intensity haveA. Large photosynthetic unit size than the sun plantsB. Higher rate of ` CO_(2)` fixation than the sun plantsC. More extended root systemD. Leaves modified to spines |
| Answer» Correct Answer - A | |
| 893. |
Plants adapted to low light intensity haveA. Leaves modified to spinesB. Large photosynthetic unit size than the sun plantsC. Higher rate of `CO_(2)` fixation than the sun plantsD. More extended root system |
| Answer» Correct Answer - B | |
| 894. |
As compared to sun plants, plants adapted to low light intensity possessA. High rate of `CO_(2)` fixationB. Larger photosynthetic unitsC. More extended root systemD. Spiny leaves |
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Answer» Correct Answer - B As compared to sun plants, plants adapted to low light intensity possess larger photosynthetic unit. |
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| 895. |
Four electrons produced during photolysis of water will enterA. PS IB. PQC. PS IID. PC |
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Answer» Correct Answer - C Four electrons produced during photolysis of water will enter PS II. |
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| 896. |
Which of the following is produced during the light phase of photosynthesis?A. ATPB. NADPH`_(2)`C. Both ATP and `NADPH_(2)`D. Carbohydrates |
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Answer» Correct Answer - C Photochemical phase, also called light of hill reaction occurs inside the thylakoids, espeically upon light. The function of this phase is to produce assimilatory power consisting of reduced coenzyme NADPH and energy ATP molecules. |
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| 897. |
`O_(2)` evotution is directly associated with Or Which of the following does not participates when the light reaction synthesizes only ATP or performs the cyclic flow of electronsA. PS-IB. PS-IIC. PhytochromeD. Phycocyanin |
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Answer» Correct Answer - B The photosystem-II (reaction centre of _680) extracts an `e^(-)` from water returning to its unexcited state. The removal of four `e^(-)` from two molecules of water requires 4 quanta of light to fall on PS-II and leads to the production of `4H^(+)` ions and one molecules of `O_(2)`. |
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| 898. |
Photosynthetic pigmetns such as chl a,chl b,xanthophyll and carotene can be separated by which of the following techniques?A. paper chromatographyB. Gel ElectophoresisC. X-ray diffusionD. ELISA test |
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Answer» Correct Answer - A Paper chromatography is a tehnique used to separte pigments accoring to their relative solubility and molecular weight. A chromatographic separation of the leaf pigments shows that the color that we see in leaves is not due to a single pigment but due to four pigments-Chlorophyll (appears bright or blue green in the chromatogram). chlrophyll b(yellow green). xanthopylls (yellow) and carotenoids (yellow to yellow-orange) |
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| 899. |
In PS I, the reaction centre chl a has absorption maxima at____whereas in PS II, the reaction centre Chl a has absorption maxima at ___A. 700nm ,680nmB. 680nm,700nmC. 400nm,500nmD. 700nm,800nm |
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Answer» Correct Answer - A |
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| 900. |
During non-cyclic photophosphorylation in which of the following , `4e^(-)` produced through photolysis will enterA. PS-IIB. PCC. PQD. PS-I |
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Answer» Correct Answer - A Non cyclic photophosphorylation involves both PS-I and PS-II. The process begins with the absorption of ilght energy by PS_II. As light energy is absorbed , `4e^(-)` become excited from chlorophyll `-a` at the reaction centre. The `4e^(-)` released by these molecules are accepted by an electron-acceptor-substance. the other effect of this event is that photolysis of water is induced. the chlorophyll-a molecules n PS-II act as strong oxidising agent As described earlier, 4 molecules of `H_(2)O` are thus decomposed through light-induced energy. the `4H^(+)` ions become associated with 2 NADP. The `4OH^(-)` ions are associated with the release of oxygen. the four electrons `(4e^(-))` released from 4 `OH^(-)` are received by chlorophyll-a molecules at reaction-centre of PS-II |
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