InterviewSolution
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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.
| 1451. |
Find out the correct sequence of labelling of diagram given below A. A spike B racemie C dichsial cyme D monochaial cymeB. A raceme B spike C monochasial cyms D dichasial cymeC. A dichasial cyme B monochaisal cyme C raceme D spikeD. A spike B dichasial cyme C monocasial cyme D-racetne |
| Answer» Correct Answer - B | |
| 1452. |
Find out the correct sequence of labelling of diagram given below A. A Epicarp B seed C measocarp D endocarpB. A epicarp B mesocarp C endocarp D seedC. A Epicarp B seed C endocarp D mesocarpD. A seed B endocarp C mesaocarp D epicarp |
| Answer» Correct Answer - B | |
| 1453. |
Identify the correct combinationA. Neem the absence of buds in the aixls of leaflets pinnately comound leafB. sunflower oppsite phyllotaxy inferiro ovaryC. carrot flowers brought to same height due to varied lengths of pedicels involucre of bractsD. pistia discord stem lateral brachd with many internodes |
| Answer» Correct Answer - A | |
| 1454. |
Select correct statement w.r.t. hard walled berry.A. Multiseeded fruit developing from superior ovary.B. Edible part is juicy unicellular hairsC. Develops from `Goverline(3)`D. Develops from `Gunderline(3)` |
| Answer» Correct Answer - C | |
| 1455. |
Identify the incorrectly matched pair relating to the part of leaf modified into spine with the example of the plantA. margin of lamina - argemoneB. apex of lamina - acaciaC. stipules modifed into tendirl - lathyrusD. margin and apex of lamina - aloe |
| Answer» Correct Answer - C | |
| 1456. |
How would you differentiate between aggregate fruits and composite fruits ? Give one example of each. |
| Answer» Aggregate fruits develop from multicarpellary apocarpous ovary. These fruits form an aggregate of fruitlets. e.g., Calotropis. Composite fruits, on the other hand, develop from the flowers of complete io.florescence. e.g., pineapple. | |
| 1457. |
A modification of leaf isA. PhyllodeB. PhyllocladeC. CladodeD. Corn |
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Answer» Correct Answer - A Phyllode is a modification of leaf. |
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| 1458. |
Which of the following is not the modification of leaf?A. Tendril in AntigononB. Tendril in ClematisC. Tendril in GloriosaD. Tendril in Nepenthes |
| Answer» Correct Answer - A | |
| 1459. |
Select a correct set.A. `{:("Plant","Organ","Function"),("Vanda","Tap root","Moisture absorption"):}`B. `{:("Plant","Organ","Function"),("Jasmine","Offset","Photosynthesis"):}`C. `{:("Plant","Organ","Function"),("Pineapple","Sucker","Propagation"):}`D. `{:("Plant","Organ","Function"),("Nepenthes","Leaf tip","Photosynthesis"):}` |
| Answer» Correct Answer - C | |
| 1460. |
Inflorescence with thick , fleshy axis and large-colored bract isA. SpatheB. SpadixC. SpikeletD. Hypanthodium. |
| Answer» Correct Answer - B | |
| 1461. |
The correct sequence of types of corolla in the following figures is A. A - Caryophyllaceous, B - papilionaceous, C - bilabiate, D - tubular, E-bell-shapedB. A- papelionaceous, B-biliabiate, C-tubur, D-bell-shaped,E-caryophyllaceousC. A-bilabiate, B- papilionaceous,C-caryophyllaceous,D-bell-shaped, E-tubularD. A - caryophyllaceous, B - bilabiate, C - papilionaceous, D-tubular, E-bell-shaped |
| Answer» Correct Answer - A | |
| 1462. |
Multicostate parallel type of venation is found in the leavesA. Grasses and palmsB. Banana and cannaC. Castor and china roseD. Mango and peepal |
| Answer» Correct Answer - A | |
| 1463. |
`120^(@)` phyllotaxy is found inA. TristichousB. DistichousC. PentastichousD. Octastichous |
| Answer» Correct Answer - A | |
| 1464. |
Spiral phyllotaxy in which sixth leaf lies above the first one after completing two circles isA. DistinchousB. TristichousC. PentastichousD. Octastichous |
| Answer» Correct Answer - C | |
| 1465. |
In 3/8 alternate phyllotaxy (Called ostastichous)A. There are 8 leaves in 3 circlesB. 3 leaves in 8 circlesC. There are 3 rows of leavesD. There are 8 rows of leaves on three sided stem |
| Answer» Correct Answer - A | |
| 1466. |
Gynandours condition showsA. adhesion of stamens with petalsB. adhesion of stamenns with carpelC. stamens with carpelD. staments are united throiughout their whole length |
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Answer» Correct Answer - B In gynandroius statmens are fused with the carpel throughtoiut their whole length or by thei anthers only a.g asclepidaceae family |
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| 1467. |
Match list `I` with list `II` and select the correct option `{:(,"List I",,"List II"),("A.","Germmules",1.,"Agave"),("B.","Leaf-buds",2.,"Penicillium"),("C.","Bulbil",3.,"Water hyacinth"),("D.","Offset",4.,"Sponges"),("E.","Conidia",5.,"Bryophyllum"):}`.A. A-4,B-5,C-1,D-3,E-2B. A-4,B-3,C-2,D-1,E-5C. A-3,B-5,C-4,D-2,E-1D. A-4,B-1,C-5,D-3,E-2 |
| Answer» Correct Answer - A | |
| 1468. |
In 1/2 distichous phyllotaxyA. 2nd leaf lies on 1st leaf at `180^(@)` angleB. `3^(rd)` leaf on `1^(st)` leaf at `180^(@)` angleC. 1st leaf lies exactly below 2nd leaf and `120^(@)` angleD. None of these |
| Answer» Correct Answer - B | |
| 1469. |
`120^(@)` phyllotaxy is found inA. distichous condit ionB. tristichous conditionC. monostichous conditionD. none of the above |
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Answer» Correct Answer - B The arrangemnt of leves on a stem or brach is called phyllotaxy the number of vertical rows in which leavs are arranged is called as rothositiches `120^(@)` phyllotaxy is found in tristichous condition |
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| 1470. |
In which of the following etiolar leaf tendril is found ?A. clematisB. citrusC. parkinsoniaD. trapa |
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Answer» Correct Answer - A In clematis petiolar leaf tendril is found in this petiole become thin sensitive nd helps is climbing |
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| 1471. |
Which one is example of subaerial modification of stemA. AsparagusB. PolyalthiaC. TridaxD. Oxalis |
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Answer» Correct Answer - D Oxalis is example of subaerial modification of stem. |
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| 1472. |
Subaerial stem modification with long internode isA. tuberB. phyllodeC. phylocladeD. runner |
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Answer» Correct Answer - D In subaseirl modification the stems are dilicated thin weak and unable to stand rerect runner grow prostrate in all direction s abvoe the soil level it has a creeping stem with long internodes on the lower side nodes bear adventitious roots |
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| 1473. |
Which one is example of subaerial modification of stemA. agaveB. oxalisC. asparagusD. tridax |
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Answer» Correct Answer - B Oxalis is an example of runner s which are the subaerial weak stem modification it bears scale leaves dn aillary buds on the upper side nd adventious roots on the lower side |
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| 1474. |
Distinguish between alternate and whorled phyllotaxy. |
| Answer» In case of alternate phyllotaxy, only one leaf is borne at each node. In case of whorled phyllotaxy, more than. two leaves arise at each node. | |
| 1475. |
Whorled type of phyllotaxy is found inA. mustardB. china roseC. guavaD. colotropis |
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Answer» Correct Answer - C In wholred or verticallate phyllotaxy three e.g nerium or more than three leaves are borne on a single node in a whole or circle the leaves of the wholr of one node generally alternate with the elaves of the wholr of adjacent nodes in orr to provide maximum exposure |
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| 1476. |
Identify in order, the plants showing alternate, opposite and whorled phyllotaxyA. China rose,Calotropis,NeriumB. China rose,Nerium,CalotropisC. Nerium,China rose,CalotropisD. Nerium,Calotropis,China rose |
| Answer» Correct Answer - A | |
| 1477. |
Assertion:The alternate type of phyllotaxy is the arrangement of leaves in which a single leaf arises at each node in alternate manner. Reason:The alternate type of phyllotaxy is seen in China rose and mustard plant .A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false. |
| Answer» Correct Answer - B | |
| 1478. |
A dicot leaf with parallel venation isA. ColocasiaB. AlocaciaC. EryngiumD. All correct. |
| Answer» Correct Answer - C | |
| 1479. |
Edible part in the fruit of hesperidum isA. EndocarpB. MesocarpC. Juicy hairsD. Pericarp |
| Answer» Correct Answer - C | |
| 1480. |
Geocarpic fruits are produced byA. WatermelonB. OnionC. CarrotD. Ground nut |
| Answer» Correct Answer - D | |
| 1481. |
Dried fruit used in making a musical instrument isA. Snake gourdB. Bitter gourdC. Bottle gourdD. All the above |
| Answer» Correct Answer - C | |
| 1482. |
The morphological nature of the edible part of coconut isA. perispermB. cotyledonC. endospermD. pericarp |
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Answer» Correct Answer - C The edible part of coconut is endosperrn. Coconut water is free nuclear endospenn and white kernel is the cellular endosperm. |
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| 1483. |
In Bougainvillea, thorns are the modifications ofA. Adventitious rootB. StemC. LeafD. Stipules |
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Answer» Correct Answer - B In Bougainvillea, thorns are the modifications of stem (axillary buds) |
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| 1484. |
In Bougainvillea, thorns are the modifications ofA. stipulesB. adventitius rootC. stemD. leaf |
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Answer» Correct Answer - C In Bougainvillea, thorns are the modifications of stem. They are stiff, sharp structures, which have lost their growing point and become hard. They reduce transpiration as well as browsing by animals. |
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| 1485. |
Plants, which produce characteristic pneumatophores and show vivpary belong toA. mesophytesB. halophytesC. pasmmophytesD. hyrophytes |
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Answer» Correct Answer - B Plants that produce pneumatophores, i.e. negatively geotropic roots and show vivipary, i.e. germination of seeds inside the fruits are halophytes. These plants are adapted to grow in highly saline areas such as mangroves. Pneumatophores help these plants in respiration as they do not get sufficient oxygen from the soil. On the other hand vivipary aids in perennation. |
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| 1486. |
Leafless stem of onion which produces cluster of terminal flowers is calledA. PeduncleB. Floral axisC. ScapeD. Rachis |
| Answer» Correct Answer - C | |
| 1487. |
The primitive type of stamens are found in the flowers familyA. LiliaceaeB. MalvaceaeC. Gramineae/PoaceaeD. Degeneriacae/Magnoliaceae |
| Answer» Correct Answer - D | |
| 1488. |
Oblique septum and swollen placenta is characteristic feature ofA. Gloriosa superbaB. Capsium frutescenceC. Althea roseaD. Dalbergia sissoo |
| Answer» Correct Answer - A | |
| 1489. |
Go through the following matches (i) Cassia - lmbricate aestivation (ii) Lady finger - Twisted aestivation (iii) Calotropis - Vexillary aestivation (iv) Lily - Epipetalous stamens (v) Alstonia - Whorled phyllotaxy (vi) Silk cotton - Pinnately compound leaf Find out the correct matchesA. (i), (iii), (v), (vi)B. (ii), (iv), (v), (vi)C. (i), (ii) & (v)D. (iii), (iv), (v), (vi) |
| Answer» Correct Answer - C | |
| 1490. |
Go through the following matches (i) China rose - Twisted aestivation (ii) Calotropis - Valvate aestivation (iii) Gulmohur - Ascending imbricate aestivation (iv) Pea - Descending imbricate aestivation Which of these are correct ?A. (i), (ii) and (iii)B. (ii), (iii) and (iv)C. (i), (iii) and (iv)D. All are correct. |
| Answer» Correct Answer - D | |
| 1491. |
Find out the pairs which are correctly, matched with respect to aestivation of petals Igt Valvate-Calotropis II. Twisted-Bean III. Imbricate-Cassia IV. Vexillary-China roseA. II and IVB. I and IIC. I and IIID. III and IV |
| Answer» Correct Answer - C | |
| 1492. |
Read the following matches Datura- Actinomorphic flower (ii) Canna - Zygomorphic flower (iii) Gulmohur - Zygomorphic flower (iv) Cassia - Asymmetric flower Which of these are correct ?A. (i), (ii) and (iii)B. (i) and (iii)C. (ii) and (iv)D. (ii), (iii) and (iv) |
| Answer» Correct Answer - B | |
| 1493. |
Go through the following mathces (i) Cotton - Open aestivation (ii) Lady finger - Quincuncial aestivation (iii) Cassia - Ascending imbricate aestivation (iv) Bean - Descending imbricate aestivation Which of these are correct ?A. (i), (iii) & (iv)B. (iii) and (iv)C. (ii), (iii) and (iv)D. All are correct. |
| Answer» Correct Answer - B | |
| 1494. |
Select the wrong matchA. Castor oil seed - Dicot, endospermic and perispermicB. Bean seed - Dicot and nonendospermicC. Maize - Monocot and endospermicD. Mustard - Dicot and endospermic |
| Answer» Correct Answer - D | |
| 1495. |
Butterfly shaped flower with one stranded, two wing-like and two keeled petal belong toA. CompositaeB. RubiaceaeC. MalvaceaeD. Papionaceae |
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Answer» Correct Answer - D Descending imbricate aestivation in petals are found in Papilionaceae family. |
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| 1496. |
Assertion : There are two alae in Pisum sativum flower. Reason : Both alae are covered by largest petal .A. If both Assertion and Reason are true and the Rea-son is the correct explanation of the Assertion.B. If both Assertion and Reason are true and the Rea-son is not the correct explanation of the Assertion.C. If Assertion is true, but Reason is false.D. If both Assertion and Reason is false. |
| Answer» Correct Answer - B | |
| 1497. |
Angionsperm to which the largest flower belong isA. Total root parasiteB. Partial root parasiteC. Total stem parasiteD. Partial stem parasite |
| Answer» Correct Answer - A | |
| 1498. |
Angionsperm to which the largest flower belong isA. total stem parasiteB. partial stem parasiteC. total root parasiteD. partial root parasite |
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Answer» Correct Answer - C Rafflesia, a total root parasite, produces largest flower in the world. |
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| 1499. |
Clinging roots occure inA. TrapaB. orchidC. screw pineD. podostemon |
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Answer» Correct Answer - B Orchids have clinging roots that are modified adventitious roots meant for providing mechanical support. These arise from the axils of leaves or nodes of the stem and pierce the substratum plant to facilitate fixation. |
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| 1500. |
The embryo in sunflower hasA. one cotyledonB. two cotyledonsC. many cotyledonsD. no cotyledon |
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Answer» Correct Answer - B Sunflower is a dicotyledonous plant, so the number of cotyledons in sunflower is two. Monocotyledons contain only one cotyledon in their embryo. It belong to order-Asterales and family-Asteraceae. |
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