1 - 5 CHAPTERS (JUNIOR BOTANY)

1. Match the following.

The correct matching is
I II III IV
A: D B E A

2. Read the following statements and find out incorrect one.
1) Lower the taxa, more are the characteristics that the members within the taxon share.
2) Taxa can indicate categories at different levels.
3) Increase in mass and increase in number of individuals are twin characters of growth.
4) Properties of tissues are present in the constituent cells and also arise as a result of interactions among the constituent cells.
A: 4 (Properties of tissues are present in the constituent cells and also arise as a result of interactions among the constituent cells.)

3. Number of taxons to which plants and animals belong in the taxonomic categories showing hierarchial arrangement
A: 7

4. Assertion (A): The deuteromycetes reproduce only by asexual conidio spores.
Reason (R): Once perfect stages of members of deuteromycetes were discovered they were often moved to other classes.
A: A and R are correct. R doesn't explain A.

5. Sex organs are absent in
A. Phycomycetes B. Bacteria C. Deuteromycetes
D. Cyanobacteria E. Basidiomycetes
A: B, C, D, E

6. Match the following.

The correct matching is
I II III IV
A: C A D E

7. Read the following statements. Find out correct one.
1) The study of geographical distribution of plants in the past, present and future is called phytogeography.
2) Lichenology is the study of algal and fungal members which show symbiosis with other plants.
3) The study of non-chlorophyllous, heterotrophic vascular plants is called Mycology.
4) Palaeobotany helps us in understanding the course of evolution in plants.
A: 4 (Palaeobotany helps us in understanding the course of evolution in plants)

8. Find the mis-match.
1) Biofertilisers - Avoids soil and water pollution
2) Bioremediation - Controls soil pollution
3) Sand binding plants - Prevention of soil pollution
4) Intensive tree plantation - Control of green house effect.
A: 3 (Sand binding plants - Prevention of soil pollution)

9. The following books or contributions were written or made in same period.
1) Micrographia - Description of sexual reproduction in plants
2) Anatomical study of plant tissues - Binomial nomenclature
3) Herbals - Vrikshayurveda
4) Discovery of genetic nature of RNA - Sexual system of classification
A: 1 (Micrographia - Description of sexual reproduction in plants)

10. The following are homosporous, vascular and heterosporous, seed bearing plants respectively.
1) Polytrichum - Cycas
2) Sphagnum - Lycopodium
3) Lycopodium - Salvinia
4) Lycopodium - Ginkgo
A: 4 (Lycopodium - Ginkgo)

11. The following plants show swollen leaf base and swollen petiole respectively.
1) Some leguminous plants, Eichornea
2) Tephrosia, Pistia
3) Eichornia, Crotalaria
4) Mangifera, Hydrilla
A: 1 (Some leguminous plants, Eichornea)

12. The stem modifications that are useful in perennation
A: Underground stem modifications

13. Unisexual and bisexual flowers are present in the inflorescence of
A: Tridax

14. Aggregate fruits are found in
A) Lotus B) Rose C) Helianthus D) Annona
A: A, B, D

15. The following plants are similar in producing adventitious roots with similar kind of growth.
1) Banyan tree, Taeniophyllum 2) Ficus, Monstera
3) Maize, Monstera 4) Monstera, Sugarcane
A: 2 (Ficus, Monstera)

16. Read the following table and select a pair of correct combinations.

A: II, III

Posted Date : 15-10-2020

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

VIRUS

Louis Pasteur proposed Germ theory of diseases & disproved theory of spontaneous generation.
Connecting link between livings & non-livings.
Virus is a Virus - Andre Lwoff.
Virus = Intracellular, obligatory, Pathogenic parasite.
Discovered by Iwanowski -- Proved that virus, is filterable.
Beijerinek called Contagium Vivum fluidum
Filterable agents - Virus.
Virus was first crystallised by Stanley (TMV)
Chemical analysis of TMV → Pierie & Bawden
Nucleic acid of TMV is infectious hereditary → Gierrer & Schramm.
The genetic material of TMV is RNA → Confirmed by Frankel - Conrat.
Mycophages - Virus that kills fungi
Cyanophage - Virus that kills cyanophyceae.
Zymophase - Virus that kills yeast.
Largest Virus = Vaccinia Virus. Longest Virus = TMV
Smallest Virus - f2 Bacteriophage.
Animal Rhabdo viruses are bullet shaped.
Plant Rhabdo viruses are rod shaped.
Nucleo proteins = Virus = Protein coat (Capside) + Core (RNA or DNA).
Never a virus contains both nucleic acids.
Symmetry is based on arrangement of capsomeres.
TMV - helical, Adeno virus - Cubical, Bacterio phage - Binal.
TMV structure - Franklin.
Lytic cycle is shown by Virulent Phages. e.g.: T2, T4, T6
Lysogenic Cycle - Coli Phage - λ. Temperate phage.
Virion = Virus out side the livings.
Viroid = Virus without protein → Potato spindle Tuber Virus, Citrus exocortosis virus.
Prion = Virus without Nucleic acid → Scrapie disease of sheep and 'Mad Cow disease' of cow.
Most of the animal viruses consist an additional outer envelope called Peplos. Its structural units are Peplomeres (lipids + carbohydrates + Proteins)
Lysozymes are produced at the time of penetration and they are also needed at the time of lysis.
The part of the genetic material belonging to virus, that is associated with the genetic material of Bacteria is called prophase Temperate Phage.
The protein coat that remains out side the body of the host is called Ghost.
Cytovirin, Thio uracil, Zinc sulphate, Malchite green - Chemotherapy
Vaccine for Small pox → Edward Jenner
Vaccine for Rabies - Louis Pasteur.

Posted Date : 10-08-2020

VIRUS

VIRUS

Posted Date : 17-04-2020

THE LIVING WORLD

1. Read the following statements.

A) Plants is a taxon
B) Non-living objects also grow
C) Growth, can not be taken as a defining property of living organisms
D) Mule do not reproduces
Find out the correct statements
A: All statements are correct

2. Darwin of 20th Century
A: Ernst Meyer

3. The twin characters of growth
A: Increase in mass and number of individuals

4. Match the following.

List - A	List - B
I. Fungi	A. Budding
II. Yeast	B. Mule
III. Protonema	C. Fragmentation
IV. No reproduction	D. Spore
	E. Non livings

The correct matching is
A: I II III IV
D A C E

5. Reproduction is synonymous with growth-
1) Yeast/ Amoeba 2) Bacteria 3) Chlamydomonas 4) All
A: 4(All)

6. Characteristic feature of living organism is
1) Growth 2) Reproduction 3) Metabolism 4) All
A: 4(All)

7. Technically complicated feature of all living organisms is
A: Irritability

8. The defining property of living organisms
A: Metabolism

9. Read the following statements and find out incorrect one.
1) Couplet is a pair of contrasting characters used in the identification and
nomenclature of plants and animals.
2) Biological museums are generally set up in educational institutions.
3) The largest herbarium is at RBG.
4) The order of Mango is Sapindales.
A: 1(Couplet is a pair of contrasting characters used in the identification and
nomenclature of plants and animals.)

10. The word systematics is derived from
A: Latin

11. Both the words in a biological name, when hand written, are separately underlined
to indicate that
1) They are important.
2) The former is a genus and the later is species.
3) To catch the attention of the reader.
4) None
A: 4(None)

12. Expand ICBN
A: International Code for Botanical Nomenclature

13. Match the following.

I. Dry plant specimens	A. Botanical Gardens
II. Preserved Animal specimens	B. Museum
III. Living specimens of plants	C. Flora
IV. Index of plant species	D. Herbarium

The correct matching is
A: I II III IV
D B A C

14. Match the following.

I. Habitat and distribution	A. RBG
II. Information on 1 taxon	B. Herbarium
III. Repository for future use in Identification	C. Monograph
IV. International Centre for Identification	D. Flora

The correct matching is
A: I II III IV
D C B A

15. Match the following.

I. IBG	A. International for plants nomenclature
II. RBG	B. Lucknow
III. NBRI	C. Howrah
IV. ICBN	D. London (International for plant Identification)

A: I II III IV
C B A D

16. Images of specimens are seen in
A: Digital Herbarium

17. Man belongs to the order
A: Primata

18. Arrange the following in sequence
A. Nomenclature B. Characterisation C. Identification
A: BCA(Characterisation, Identification, Nomenclature)

19. The defining properties of all organisms
A. Growth B. Metabolism C. Reproduction D. Irritability E. Self Consciousness
A: BD only(Metabolism, Irritability only)

20. Mango belongs to the family
A: Anacardiaceae

Posted Date : 15-10-2020

MICROBES IN HUMAN WELFARE

1. Which of the following is not a function of LAB (Lactic Acid Bacteria)
1) LAB produces acids that coagulate and partially digest milk proteins.
2) LAB improves nutritional quality by increasing vitamin B₁₂
3) It checks the disease causing microbes in our stomach
4) It is not a probiotic
A: 4 (It is not a probiotic)

2. The raising of dough in dosa and idli is due to
A. Respiration of bacteria
B. Fermentation caused by bacteria
C. Production of CO₂ gas
A: ABC

3. The dough used for making bread is fermented using
A: Saccharomyces cervisiae

4. Fermenting sap from palms yields
A: Toddy

5. Which of the following cannot be fermented by microbes?
1) Fish 2) Soya bean
3) Bamboo shoots 4) None of the above
A: 4 (None of the above)

6. Assertion (A): Swiss cheese has large holes.
Reason (R): Propionibacterium sharmanii produces large amount of oxygen during fermentation.
A: Only A is true, R is false

7. Roquefort cheese cannot be ripened without fermented with
A: Fungi

8. In which of the following is the yeast Saccharomyces cervisiae used
A. Production of beverages like wine, beer, whisky etc.
B. Bread making
C. Fermenting malted cereals and fruit juices
D. Production of ethanol
A: ABCD

9. Which of the following is produced without distillation?
1) Wine 2) Whisky 3) Brandy 4) Rum
A: 1 (Wine)

10. With reference to human beings antibiotics means
A: Pro life

11. Alexander Flemming discovered penicillin while working on
A: Staphylococci bacteria

12. The potential of penicillin to be used as an effective antibiotic was established by
A: Ernest Chain and Howard Florey

13. Citric acid is produced by which of the following micro organisms
1) Aspergillus niger 2) Acetobacter aceti 3) Clostridium butylicum 4) Lactobacillus
A: 1 (Aspergillus niger)

14. The enzymes used in detergent formulations to remove oily stains from laundry are
A: Lipases

15. The enzymes used to clarify fruit juices
A: Proteases and pectinases

16. Clots from blood vessels of patients who have undergone myocardial infection can be removed by using
A: Streptokinin

17. Which of the following is true about cyclosporin A
1) It is a bioactive molecule
2) It is used as an immunosuppressive agent in organ transplant patients
3) It is produced by the fungus Trichoderma polysporum
4) All the above
A: 4 (All the above)

18. The blood cholesterol lowering statins are produced by
A: Yeast Monascus purpureus

19. The following debris during primary treatment of sewage is removed by
1) Sedimentation 2) Sequential filtration 3) Mechanical agitation 4) Anaerobic digestion
A: 2 (Sequential filtration)

20. The grit (soil and small pebbles) during primary treatment of sewage is removed by
A: Sedimentation

21. Masses of bacteria associated with fungal filaments to form mesh like structure are called
A: Flocs

22. The amount of O₂ that would be consumed if all the organic matter in 1 litre of water were oxidised by bacteria is called
A: Biochemical oxygen demand

23. The BOD is
1) A measure of organic matter present in water
2) A measure of the rate of uptake of oxygen by microorganisms in a sample of water
3) BOD is directly proportional to the polluting potential of waste water
4) All the above
A: 4 (All the above)

24. Utilization of microbes to remove toxic substances such as oil or chemical spills released into the environment is called
A: Bio remediation

25. Bio gas contains a mixture of gases predominantly
A: Methane

26. Which of the following is true regarding methanogenic bacteria
A. They are found in anaerobic sludge during sewage treatment
B. They are present in rumen of cattle
C. They are present in human intestine
D. They help in breakdown of cellulose
A: ABD

27. The technology of Bio gas production was developed in India by
1) Indian Agricultural Research Institute
2) Khadi and Village Industries Commission
3) Indian Council of Agricultural Research
4) Both 1 and 2
A: 4 (Both 1 and 2)

28. Use of biological methods for controlling plant diseases and pests is called
A: Bio control

29. In view of organic farming the best way to control pests is
A: To keep them at manageable levels by a complex system of checks and balances

30. Red and black marked lady bird is used to get rid of
A: Aphids

31. The insects which are used to get rid of mosquitoes are
A: Dragon flies

32. The microbial bio control agent that can be introduced to control butterfly caterpillars is
A: Bacillus thuringiensis

33. An effective bio control agent of plant pathogens which lives in root ecosystems is
A: Trichoderma viridi

34. The best bio control agent to attack arthropod insects is
A: Baculo virus

35. Which of the following pairs are free living nitrogen fixers
1) Mycorrhizae and Azospirillum 2) Azospirillum and Azotobacter
3) Azotobacter and Rhizobium 4) Mycorrhizae and Azotobacter
A: 2 (Azospirillum and Azotobacter)

36. The Glomus fungus which is a best mycorrhizae former helps in absorption of
A: Phosphorus

37. Mycorrhizae are useful to the plant in the following ways
A. Absorption of phosphorus
B. Provide resistance to root borne pathogens
C. Provide tolerance to salinity and drought
D. Overall increase in plant growth and development
A: ABCD

38. Which of the following is not a nitrogen fixing cyano bacterium?
1) Anabaena 2) Nostoc 3) Oscillatoria 4) Chlorella
A: 4 (Chlorella)

39. The mutated pathogens are called
A: Super microbes

40. The use of or threat to use biological agents in order to spread fear or inflict disease and death upon large populations is called
A: Bio terrorism

41. Which of the following is not a re-emerging diseases presently
1) Cholera 2) TB 3) Dengue fever 4) SARS
A: 4 (SARS)

Posted Date : 30-11-2020

BIOMOLECULES

1. Match the following.

I) Starch	A) Cotton
II) Glycogen	B) Exoskeleton of Cockroach
III) Cellulose	C) Potato
IV) Chitin	D) Liver

I II III IV
A: C D A B

2. Glycine and Alanine are different in one of the 4 substituents of α - carbon and are similar in the rest. They differ in the following group.
1) H 2) COOH 3) R group 4) Amino group
A: 3 (R group)

3. Sugar of the DNA is linked with the following by Glycosidic bond.
1) Another sugar 2) Nitrogen base 3) Phosphate 4) Nitrogen base & Phosphate
A: 2 (Nitrogen base)

4. Match the followig.

I) Ester bond	A) Polysaccharide
II) Glycosidic bond	B) Amino acids of a protein
III) Peptide bond	C) Water
IV) Hydrogen bond	D) Fats

I II III IV
A: D A B C

5. Haemoglobin shows
A: Quarternary structure

6. The sequential position of an amino acid in a protein can be known through its
A: Primary structure

7. Assertion (A): Wasting paper is a loss of vegetation.
Reason (R): Paper is made of cellulose.
A: A & R are correct. R explains A.

8. Secondary metabolite differs from the primary metabolite in that it has no-
A: identifiable function

9. Assertion (A): All macrobiomolecules are polymers.
Reason (R): Lipids are also polymers.
A: A is correct. R is false.

10. Two adjacent molecules in a polymer are connected by Glycosidic bond in the following
1) Nucleic acids 2) Polysaccharides 3) Lipids 4) Proteins
A: Polysaccharides

11. Assertion (A): All amino acids are essential to plants and animals.
Reason (R): Plants do not have external supply of amino acids.
A: A is false. R is correct.

12. Lipids do fit to the following quality.
1) Macro molecules 2) Acid soluble

3) High molecular weight 4) Micro molecule
A: 1 (Macro molecules)

13. Metabolism of plants and animals involves energy conversions. The study of such
matter can be studied by a science called
A: Bio energytics

14. The most abundant protein in the biosphere
A: Rubisco

15. The most abundant protein in animals
A: Collagen

16. Proteins are not seen as
A: important Cell wall constituents of xylem

17. A polysaccharide present in the exoskeleton of arthropod
A: Chitin

18. A polysaccharide that occurs in the storage form of energy
A: Glycogen, starch

19. The following are hetero polymers
1) Starch, cellulose 2) Proteins, starch
3) Nucleic acids, proteins 4) Starch, nucleic acids
A: 3 (Nucleic acids, proteins)

20. Assertion (A): Lipids are biomacro molecules.
Reason (R): Lipids have molecular weight less than 1000 Da.
A: A & R are correct. R do not explains A.

21. Assertion (A): Strictly lipids are not biomacromolecules.
Reason (R): Their molecular weight is high.
A: A is correct. R is false.

22. Arrange the followig in ascending order.
A. Types of nucleotides in RNA.
B. Types of nucleotides in the histones of octamer.
C. Types of nucleotides.
D. Types of Pyrimidines.
E. Types of Purines.
F. Types of nucleotides energy currency.
A: BFEDAC

23. Read the following statements and find out incorrect one.
1) Living state is non - equilibrium steady - states
2) Living state and metabolism are synonyms.
3) Livings at equilibrium have least metabolism
4) ATP is a chemical link between anabolism and catabolism.
A: 3 (Livings at equilibrium have least metabolism)

24. Biomolecules are involved in-
A: Metabolic flux

25. Cholesterol is formed from-
A: Acetic Acid

26. Energy is stored in which part of Glucose-
A: Chemical bonds

27. A student described macro biomolecules as large sized, complex chemicals that have high molecular weight. Any other characteristic feature missing in it, if so it may be
A: Acid insoluble

28. Find mismatch.
1) Sucrose - Disaccharide 2) Starch - Polysaccharide
3) Glucose - Monosaccharide 4) Fructose - Disaccharide
A: 4 (Fructose - Disaccharide)

29. Glycosidic link is broken in the digestion of-
1) DNA 2) RNA 3) Starch 4) All
A: 4 (All)

30. Animal cells store food in the form of-
A: Glycogen

31. Cellulose is-
Ans: Hexose homopolymer

32. Most abundant water insoluble Polysaccharide is-
Ans: Cellulose

33. Sweetest sugar is-
Ans: Fructose

34. Assertion (A): Lipids are biomacromolecules.
Reason (R): They are water insoluble.
Ans: A & R are correct. R do not explains A.

35. One of the following is a cane sugar-
1) Glucose 2) Sucrose 3) Fructose 4) Lactose
Ans: 2 (Sucrose)

36. Inulin is-
Ans: Fructosan

37. CO₂ dissolving in water in living organisms
Ans: Physical process catalysed by enzyme

38. Human skeletal muscles show
Ans: Conversion of Glucose to Lactic acid

39. Turn over regarding biomolecules means-
Ans: Changing into some biomolecules and formation from some other biomolecules regularly

40. Dynamic state of body constituents means-
Ans: Flow of the metabolites in the metabolic pathways with definite rate in a definite route

41. Read the following statements and find out incorrect one.
1) Every chemical reaction in a living is a catalysed reaction
2) Catalysts are proteins
3) Majority of metabolic pathways are not linked
4) Metabolite flow in a cell has definite direction
Ans: 3 (Majority of metabolic pathways are not linked)

42. The following bonds are formed due to dehydration-
1) Peptide bond, Hydrogen bond 2) Ester bond, Glycosidic bond
3) Glycosidic bond, Peptide bond 4) Hydrogen bond, Ester bond
Ans: 3 (Glycosidic bond, Peptide bond)

43. The bond between 2 glucose molecules adjacent to each other in a polysaccharide is similar to the bond between-
Ans: Sugar and nitrogen base

44. One of the following is not related to DNA
1) Loop or Pitch measures 34 A°.
2) One step makes 36° with another step.
3) The bond between Uracil and Sugar is called glycosidic bond.
4) Adenine is paired with Thymine by 2 hydrogen bonds.
Ans: 3 (The bond between Uracil and Sugar is called glycosidic bond.)

45. Nitrogen base of Nucleic acids is
Ans: Heterocyclic

46. Match the following.

I) Ribose pentose	A) RNA only
II) 2 'deoxyribose pentose	B) DNA
III) G ≡ C	C) RNA
IV) A = U	D) DAN or RNA

I II III IV
Ans: C B D A

47. Acid nature of DNA or RNA is due to-
Ans: Phosphate

48. Assertion (A): Nucleic acids are macromolecules.
Reason (R): They are acid insoluble.
Ans: A & R are correct. R explains A.

49. Match the following.

I) Cellulose	A) Fruits/ Fructose
II) Starch	B) Paper
III) Inulin	C) Fungi
IV) Chitin	D) Potato/ Rice

I II III IV
Ans: B D A C

50. Cellulose is not found in-
Ans: Inulin

51. Inulin is a polymer of-
Ans: Fructose

52. Complex Polysaccharides are made of-
1) Amino sugars 2) Chemically modified sugars
3) Hexose & Pentose 4) 1 & 2
Ans: 4 (Amino sugars, Chemically modified sugars)

53. Starch gives blue colour with Iodine because-
Ans: it has complex helices and can hold I₂

54. Assertion (A): Polysaccharides are nutritive in function.
Reason (R): Cell wall is protective in function.
Ans: A & R are correct. R do not explains A.

55. One of the following is not a homopolymer-
1) Glycogen 2) Starch 3) Inulin 4) Cellulose
Ans: 1 ( Glycogen)

56. One of the following is a branched polysaccharide-
1) Cellulose 2) Starch 3) Glycogen 4) 1 & 2
Ans: 3 ( Glycogen)

57. If sweet corn is immersed in the boiling water for sometime, the water becomes sweet because-
Ans: Disaccharides is converted to monosaccharides

58. One of the following is not a protein-
1) Inter cellular ground substance of animals 2) Inulin 3) Insulin 4) GLUT - 4
Ans: 2 (Inulin)

59. The following protein helps in the transportation of Glucose into cells-
1) GLUT - 4 2) Pencillin 3) Insulin 4) Haemoglobin
Ans: 1 (GLUT - 4)

60. Arrange the following in ascending order (Composition of Cell)
A) Nucleic Acids B) Lipids C) Proteins D) Carbohydrates
Ans: BDAC

61. Match the following.

I) Consume energy	A) Catabolic
II) Release energy	B) Metabolic
III) Neither input nor output of energy	C) Anabolic
IV) Living	D) Equilibrium

I II III IV
Ans: C A D B

62. Esterbond is formed between-
Ans: Phosphate and OH of the sugar

63. The angle between back bone and nitrogen base
Ans: 90°

64. A nucleotide means-
Ans: Nucleoside attached to phosphate

65. Match the following.

I) Toxin	A) Codeine
II) Alkaloids	B) Rubber
III) Lectin	C) Abrin
IV) Polymer	D) Concanavalin

I II III IV
Ans: C A D B

66. Match the following.

I) Carotenoids	A) Cucumin
II) Essential oils	B) Carrot
III) Drug	C) Gums
IV) Polymer	D) Lemon grass oil

I II III IV
Ans: B D A C

67. One of the following is not a polymeric substance-
1) Gum 2) Starch 3) Nucleotide 4) Rubber
Ans: 3 (Nucleotide)

68. Match the following-

I) 800 - 1000 Da	A) Average molecular weight of micro molecules
II) > 1000 Da	B) Macro molecules
III) Blood concentration of Glucose	C) Micro biomolecules
IV) 8 - 800 Da	D) 4.5 to 5.0 mm

I II III IV
Ans: C B D A

69. The left hand side of polysaccharide and polypeptide consists respectively-
Ans: Non reducing, Amino

70. Read the following statements and find out incorrect one-
1) Nitrogen bases in DNA are projected inside to backbone
2) Types of DNA are more than a dozen
3) One full turn of helical strand consists minimum 20 nitrogen bases
4) Two strands of DNA run anti parallel
Ans: 3 (One full turn of helical strand consists minimum 20 nitrogen bases)

71. Assertion (A): DNA exists as a double helix.
Reason (R): It is a secondary structure.
Ans: A & R are correct. R do not explains A.

72. Number of water molecules lost to form one full turn of helical strand-
Ans: 20

73. Number of glycosidic bonds in a loop-
Ans: 20

74. Types of sub units in Haemoglobin
Ans: 2

75. Number of sub units in Haemoglobin
Ans: 4

76. Match the following regarding the structure of protein-

I) Inorganic chemist	A) 2-D
II) Biologist	B) Quarternary structure
III) Organic Chemist	C) Molecular formula
IV) Physicist	D) 3-D

I II III IV
Ans: C B A D

77. Secondary structure of a protein shows-
Ans: Right handed helix

78. The shape of the protein when it shows tertiary structure-
Ans: Wooden ball

79. Lecithin is a-
Ans: Phospholipid

80. One of the following is different from the rest Identify it-
1) Adenosine 2) Thymidylic acid 3) Uridine 4) Guanosine
Ans: 2 (Thymidylic acid)

81. The bonds present in Adenylic acid
A. Hydrogen bonds B. Ester bond C. Glycosidic bond
Ans: B C

82. The genetic material is-
Ans: DNA & RNA

83. Backbone of DNA is-
A. Nitrogen base B. Sugar C. Phosphate
Ans: B C

84. Assertion (A): Oils have lower melting point.
Reason (R): Gingely oil is a fluid in summer.
Ans: A & R are correct. R do not explains A.

85. Neural tissues have-
Ans: Lipids with more complex structures

86. Simplest lipids are-
Ans: Fatty acids

87. Glycerol is called-
Ans: Trihydroxy propane

88. 'R' group of lipid consists-
Ans: upto 19 carbons

89. Number of carbons in Palmitic acid excluding carboxyl carbon-
Ans: 15

90. Number of carbons including carboxyl carbon in Palmitic acid and Arachidonic acid-
Ans: 16, 20

91. Assertion (A): Fatty acids are found esterified with glycerol.
Reason (R): Many lipids have both glycerol and fatty acids.
Ans: A & R are correct. R explains A.

92. Assertion (A): Biomolecules are called Building blocks of protoplasm.
Reason (R): Protoplasm is made of Biomolecules.
Ans: A & R are correct. R explains A.

93. Comparison of elements present in non-living matter and living matter is based on the work done by-
Ans: C.N.R.Rao

94. The least elemental composition of non living and living matter respectively are-
Ans: Nitrogen, Silicon

95. The highest elemental composition of non-living matter and living matter respectively are-
Ans: Oxygen, Oxygen

96. One form of inorganic matter of plant or animal is-
Ans: Ash

97. Simplest Amino acid is-
Ans: Glycine

98. If the 'R' group of proteinaceous amino acids is hydroxy methyl, it is-
Ans: Serine

99. Assertion (A): Amino acid is a zwitterion.
Reason (R): It can accept a H⁺ or donate a H⁺.
Ans: A & R are correct. R explains A.

100. Glutamic acid is-
Ans: Acidic

101. Lysine is-
Ans: Basic

102. One of the following is neutral-
1) Valine 2) Lysine 3) Glutamic acid 4) None
Ans: 1 (Valine)

103. Assertion (A): In solutions of different pH the structure of amino acid changes.
Reason (R): Amino acid has ionizable NH₂ and COOH groups.
Ans: A & R are correct. R explains A.

104. One of the following is considered as side chain of amino acid
1) H 2) COOH 3) NH₂ 4) R
Ans: 4 (R)

105. Number of Carbons in the simplest amino acid and simplest fatty acid respectively-
Ans: 2, 2

106. The simplest amino acid and fatty acid respectively are-
Ans: Glycine, Acetic acid

107. One of the following is not an aromatic amino acid-
1) Tryptophan 2) Serine 3) Tyrosine 4) Phenyl alanine
Ans: 2 (Serine)

108. Assertion (A): Amino acids resist any change in pH of the medium.
Reason (R): They are dipolar.
Ans: A & R are correct. R explains A.

109. Number of double bonds in the structure of saturated fatty acids-
Ans: 0

110. Number of double bonds in the unsaturated fatty acids-
Ans: Minimum 1

111. Assertion (A): Amino acids are substituted methanes.
Reason (R): An amimo acid has four substituent groups occupying the 4 valency positions.
Ans: A & R are correct. R explains A.

112. One of the following is not related to Lecithin-
1) It is synthesised by peroxisome
2) It is a phospholipid
3) It is a diglyceride with nitrogenous phosphorylated compound
4) It is present in the Cell wall
Ans: 4 (It is present in the Cell wall)

113. Steps in a loop are made of-
Ans: Nitrogen base pairs

114. Number of polynucleotide chains in DNA are-
Ans: 2

115. The 2 polynucleotide chains of DNA are-
1) Complimentary 2) Antiparallel 3) Right handed 4) All
Ans: 4 (All)

116. Number of C-N Rings in purines and pyrimidines respectively-
Ans: 2, 1

117. Read the following conversions. Separate them into anabolic and catabolic conversions respectively-
A. Starch Glucose
B. Glucose Lactic acid
C. Glucose Ethanol + CO₂
D. CO₂ + H₂O Glucose
E. Glycogen Glucose
F. Protein Aminoacids
G. Acetic acid Cholesterol
Ans: Anabolic - D G Catabolic - AB C E

118. The following are homopolymers-
1) Chitin 2) Glycogen 3) Cellulose 4) All
Ans: 4 (All)

119. Structural Polysaccharides are-
1) Cellulose 2) Chitin 3) Starch 4) 1 & 2
Ans: 4 (Cellulose, Chitin )

120. Storage Polysaccharides are-
1) Glycogen 2) Starch 3) 1 & 2 4) Glucose
Ans: 3 (Glycogen, Starch )

121. Inulin is a-
Ans: Storage Polysaccharide

122. Ratio between Nitrogen bases in a loop and C - N rings-
Ans: 2 : 3

Posted Date : 30-11-2020

THE UNIT OF LIFE

1. The most important difference between plant cell and animal cell is-
A: Cell wall

2. Next to Cell wall, the important difference between plant and animal cell is-
A: Plastids

3. The Cell wall was discovered by-
A: Robert Hooke

4. Living cell was first discovered by-
A: Leeuwenhoek

5. The unit of life is-
A: Cell

6. The physico chemical approach to study the physiological and behavioural processes and understand living organisms is called-
A: Reductionist biology

7. The concepts and techniques of the following sciences helped to understand biology-
1) Physics 2) Chemistry 3) Genetics 4) 1, 2
A: 4 (Physics, Chemistry)

8. Match the following.

I. Robert Hooke	A) Nucleus
II. Anton van Leeuwenhoek	B) Omnis cellula e cellula
III. Robert Brown	C) Dead cork plant cell
IV. Rudolf Virchow	D) Living cell

The correct matching
A: C D A B

9. All cells arise from pre existing cells. This was proposed by-
A: Virchow

10. The idea of cell division was first given by-
A: Rudolf Virchow

11. The Cells of animals are surrounded by a thin layer which is today called as plasma membrane was first proposed by-
A: Schwann

12. Presence of Cell wall is a unique character of the plant cells. It was proposed by the following based on the study of plant tissues-
1) Leeuwenhoek 2) Rudolf Virchow
3) German Botanist M.J.Schleiden 4) British zoologist Theodore Schwann
A: 4 (British zoologist Theodore Schwann)

13. Schwann and Schleiden did not explain one idea in their cell theory. This was explained by the following for the first time and a final shape came to cell theory-
A: Rudolf Virchow

14. Microbodies include-
A: Peroxisomes, Glyoxysomes

15. Microbodies are generally present in-
A: Plant cells

16. One of the following character is not related to microbodies.
1) They are single membrane bounded
2) They are the only the single membrane bounded organelles
3) They are usually present in animal cells
4) 2, 3
A: 4 (They are the only the single membrane bounded organelles, They are usually present in animal cells)

17. Phospholipid synthesis is done by-
A: Peroxisomes

18. Match the following.

I. Conversion of lipids to carbohydrates	A) Organelles concerned with Autolysis
II. Photorespiration in a group of green plants	B) A type of ER without the presence of ribosomes
III. Rich in hydrolytic enzymes/ Hydrolyses	C) A kind of microbodies generally found in fat rich germinating seeds
IV. Major site for synthesis of lipids	D) Organelles which play an important role in the synthesis of phospholipids

I II III IV
A: C D A B

19. Lipids are converted to carbohydrates in a metabolic pathway called-
A: Glyoxylate cycle

20. Cells are protected from the toxic effect of hydrogen peroxide by-
A: Peroxisomes

21. DNA has charge-
A: -ve

22. The enzyme required to convert H₂O₂ into non-toxic molecules-
A: Catalase

23. Catalase is present in-
A: Peroxisomes

24. The bead like structures in chromosome are called-
A: Nucleosome

25. Positively charged part of chromosome-
A: Histones

26. A typical nucleosome consists-
1) 8 Histones 2) 200 bp of DNA

3) 1, 2 4) Linker DNA
A: 3 (8 Histones, 200 bp of DNA)

27. Ratio between histones and base pairs of a typical nucleosome-
A: 1 : 8

28. Types of histones
A: 5

29. Charge of the Octamer-
A: +ve

30. Ratio between different histones of Octamer-
A: 1 : 1 : 1 : 1

31. The protein nature of the Octamer-
1) Basic, +vely charged 2) -vely charged, Acidic

3) Acidic, +vely charged 4) None
Ans: Basic, +vely charged

32. Two adjacent nucleosomes connected by-
Ans: Linker DNA

33. The parts of a chromosome which do not unite with similar part of other homologous chromosome-
Ans: Telomere

34. The term chromosome was coined by-
Ans: Waldeyer

35. The part of a chromosome whose division indicates the completion of chromosomal division-
Ans: Centromere

36. Arrangement of the chromosomes of a genome according to their height is called-
Ans: Karyotype

37. If all the pairs of chromosomes of an organism are represented by a single partner, it is called-
1) Genome 2) Set 3) Ploid 4) All
Ans: 4 (All)

38. Parts belonging to a chromosome are equal-
1) Kinetochores 2) Telomeres 3) Chromatids 4) All
Ans: 4 (All)

39. The material of the chromosome is-
Ans: Chromatin

40. Chromosomes are best seen during-
Ans: Metaphase

41. The part of the chromosome that helps in movements-
Ans: Centromere

42. Chromosome shapes are observed during-
Ans: Anaphase

43. The chromosome shape is decided by-
Ans: Position of Centromere

44. Number of DNA threads present in a chromosome-
Ans: 2

45. A metacentric chromosome in a metaphase shows
A) 2 Chromatids B) 2 Centromeres C) 4 Telomeres
D) 4 Polynucleotide Chains E) 4 Arms F) 2 Kineto chores
Ans: Except B

46. Arms are equal in-
Ans: Meta centric

47. Satellite is present-
Ans: Above the telomere

48. Arms are almost equal in-
Ans: Submetacentric

49. Match the following.

I. Chromatin	A) Organelle that divides intracellular space
II. Nucleus	B) Naked Organelle
III. KR Porter	C) Robert Brown
IV. Palade	D) Fleming

I II III IV
Ans: D C A B

50. Read the following statements and find out one unrelated to satellite-
1) Its location is constant 2) Few chromosomes in a genome have satellite
3) It takes deep stain 4) It is associated with secondary constriction
Ans: 3 (It takes deep stain)

51. Arrange the following in ascending order
A. Number of nuclei in a cell showing 3 free nuclear mitotic divisions
B. Number of nuclei in epidermal cells
C. Number of nuclei in megaspore tetrade
D. Number of nuclei in RBC
E. Number of nuclei in embryo sac
F. Number of nuclei in the pollen grain (male gametophyte) at the time liberation
G. Number of nuclei in a mature male gametophyte
Ans: DBFGCEA

52. Outer nuclear membrane characters are-
1) Separates perinuclear space and cytoplasm 2) Bears ribosomes on its outer side
3) It is continuous with ER 4) All
Ans: 4 (All)

53. Punctured organelle-
Ans: Nucleus

54. A nucleus can be seen under-
Ans: Compound Microscope

55. Coenocytic condition is seen in-
Ans: Rhizopus

56. Number of bp in one turn of DNA of nucleosome-
Ans: 100

57. The place of H₁ protein-
Ans: Seals entry and leaving point of DNA.

58. A chromosome with one extremely short and one very long arm, is called-
Ans: Acrocentric

59. RNA and Protein molecules pass in both directions between nucleus and cytoplasm through-
Ans: Nuclear Pores

60. Assertion (A): Nucleolus is naked organelle.
Reason (R): The contents of nucleolus are continuous with the rest of the nucleoplasm.
Ans: A & R are correct. R explains A.

61. Ribosomal RNA (rRNA) synthesis occurs in-
Ans: Nucleolus

62. 80 S Ribosomes are synthesised from-
Ans: Nucleolus

63. Centrosome is generally seen in-
Ans: Animals

64. Number of Centrioles and arrangement in centrosome-
Ans: 2, Parallel to each other

65. Shape of Centriole.
Ans: Cylindrical

66. Flagella and Cilia resemble differ from each other respectively-
A. Made of tubulin B. Useful in movement
C. Present in plant and animal cells D. Length
E. Location F. member
Ans: ABC, DEF

67. Read the following statements and find out incorrect one.
1) Cilia cause the movement of either the cell or the surrounding fluid.
2) Flagella are responsible for cell movement.
3) Flagella and Cilia are present in prokaryotes.
4) Flagella and Cilia acts as oars.
Ans: 4 (Flagella and Cilia acts as oars.)

68. Number of microtubules in a flagellum-
Ans: 20

69. 9 + 2 arrangement of microtubules is seen in-
Ans: Flagellum of eukaryotes

70. Cartwheel model is seen in the structure of-
Ans: Centrosome

71. Centriole duplication in animal cells occurs-
Ans: Parallel to DNA replication in nucleus.

72. Match the following.

I. Provides enzymes for protein synthesis.	A. Colourless part of photosynthetic organelle.
II. Provides enzymes for protein synthesis, photosynthesis.	B. Organelle that protects the cell from toxic effect of H₂O₂.
III. Enzymes for lipid synthesis.	C. Smooth endoplasmic Reticulum.
IV. Consists enzymes for the phospholipid synthesis.	D. Smallest primitive naked organelle.

I II III IV
Ans: D A C B

73. Number of radial spokes in flagellum-
Ans: 9

74. Centriole forms the following structures-
1) Blepharoplast of cilia 2) Blepharoplast of flagellum 3) Spindle fibres 4) All
Ans: 4 (All)

75. Number of fibrils in centriole-
Ans: 9

76. Fibril of centriole is made of-
Ans: 3 microtubules

77. Central part of centriole is made of & it is called-
Ans: Proteins, Hub

78. Flagellum is surrounded by-
Ans: Plasma membrane

79. Core of Flagellum is made of-
1) Plasma membrane 2) 9 pairs of duplets

3) 2 central microtubules 4) 2, 3
Ans: 4 (9 pairs of duplets, 2 central microtubule)

80. Match the following-

I. Hub	A) Centriole
II. 9 + 0	B) Connected with fibrils by radial spokes
III. 9 + 2	C) Protein
IV. Radial spokes	D) Flagellum

I II III IV
Ans: B A D C

81. Match the following-

I. ER	A) ER, Golgi, Lysosomes, Vacuole
II. Endomembrane system	B) Microfilaments, microtubules intermediate filaments
III. Cytoskeleton	C) Disc shaped sacs or cisternae
IV. Golgi	D) Network of tiny tubular structures

I II III IV
Ans: D A B C

82. Cytos Releton is not concerned with-
Ans: Secretion of enzymes

83. Assertion (A): Mitochondria, Chloroplast and peroxisomes are not considered as a part of Endomembrane system.
Reason (R): Their functions are not coordinated with those of endomembrane system.
Ans: A & R are correct. R explains A.

84. Assertion (A): ER, Golgi, Lysosomes and vacuoles constitute endomembrane system.
Reason (R): Their functions are coordinated.
Ans: A & R are correct. R explains A.

85. Extra luminal compartment means-
Ans: Cytoplasm

86. Intracellular space is divided into luminal and extraluminal compartments by the following-
1) Golgi 2) Nucleus 3) ER 4) Plastids
Ans: 3 (ER)

87. The function of RER-
1) Secretion 2) Protein synthesis
3) 1, 2 4) Mechanical support and protein synthesis
Ans: 3 (Secretion, Protein synthesis)

88. The function of SER in plant and animal cells respectively-
Ans: Lipid synthesis, Steroidal hormone synthesis

89. The organisms other than plants which consist plastids are-
Ans: Euglenoids

90. The division of plastids into 3 types is based on-
Ans: Pigments

91. Terpenoids of chloroplasts are-
Ans: Carotenoids

92. Protein synthesis requires enzymes. These are present in-
1) Cytoplasma of prokaryotes and eukaryotes 2) Mitochondria & Chloroplast
3) Nucleus 4) All
Ans: 4 (All)

93. Plastids concerned with storage are-
Ans: Leucoplasts

94. Leucoplasts with oils-
Ans: Elaioplasts

95. Carotenoid pigments are present in-
1) Chromoplasts 2) Chloroplasts 3) Elaioplasts 4) 1, 2
Ans: 4 (Chromoplasts, Chloroplasts )

96. No. of Chloroplasts in mesophyll cells-
Ans: 20 -40

97. Assertion (A): Chloroplasts consists enzymes useful in photosynthesis.
Reason (R): Ribosomes are present in chloroplasts.
Ans: A is false. R is correct.

98. Double membrane bounded organelles of plant and animal cells are-
Ans: 3, 2

99. Double membrane bounded organelles of plant cells are-
Ans: Chloroplasts, Mitochondria, Nucleus

100. Match the following-

I. Inner membrane is relatively less permeables	A) Organelle consists linear DNA
II. Inner membrane forms number of infoldings	B) Organelle that produces ATP without consuming food
III. Outer membrane and inner membranes are fused	C) Organelle that is concerned with polluting atmosphere
IV. Formation of glycoproteins here and there and glycolipids	D) Organelle involved in the synthesis of cell wall materials

I II III IV
Ans: B C A D

101. Functions seen in Chloroplasts are-
A. Photosynthesis B. Protein Synthesis C. DNA replication
Ans: A B C

102. Number of Chloroplasts in Chlamydomonas-
Ans: 1

103. Chloroplast resembles mitochondria in-
A. Produce ATP B. Evolve O₂ C. DNA synthesis
D. Protein synthesis E. 70 S Ribosomes F. Circular DNA G. Crystae
Ans: A B C D E F

104. DNA is present in-
1) Nucleus 2) Chloroplast 3) Mitochondria 4) All
Ans: 4 (All)

105. The organelle that divides by fission also performs-
Ans: Takes CO2 and gives out Oxygen

106. Match the following. (functions are given as a clue)

I. Cisternae	A) Synthesis of Cell wall materials
II. Cristae	B) Consists histones
III. Thylakoids	C) Power house of the cell
IV. Pores	D) Consists circular DNA

I II III IV
Ans: A C D B

107. Match the following-

I. Food vacuole	A) Aquatic bacteria
II. Vacuole	B) Protists
III. Gas vacuole	C) Amoeba
IV. Contractile vacuole	D) Anthocyanins

I II III IV
Ans: B D A C

108. Arrange the following events in a sequence-
A. Proteins are modified in the cisternae of forming face.
B. Protein sythesis occurs on the surface of ribosomes.
C. Proteins are released from transface.
D. Golgi plays an important role in the formation of cell plate.
The correct sequence is-
Ans: B A C D

109. Assertion (A): Golgi's function is packaging materials.
Reason (R): Vacuoles occupy 90% volume of the cell.
Ans: A & R are correct. R do not explains A.

110. One of the following functions is not related to Golgi apparatus-
1) Synthesis of Glycoproteins & Glycolipids
2) Synthesis of Phospholipids
3) Synthesis of Cell wall materials
4) Modifies proteins synthesised by ribosomes and releases into cytoplasm
Ans: 2 (Synthesis of Phospholipids)

111. Match the following-

I. Packaging Organelle	A) Lysosomes
II. Formed by packaging process	B) Trans
III. Modification of Proteins	C) Golgi apparatus
IV. Releasing of modified proteins	D) Cisternae

I II III IV
Ans: C A D B

112. The class of enzymes present in lysosomes-
Ans: 3

113. The following enzyme is not present in Lysosomes-
1) Lipases and Proteases 2) Nucleases
3) Carbohydrases 4) Transaminases
Ans: 4 (Transaminases)

114. Read the following statements and find out incorrect one-
1) Lysosomes cause the death of cells
2) Tonoplast facilitates the transport of ions and other materials against
concentration gradients in to the vacuole
3) Anthocyanins are present in the vacuole of all plants
4) Lysosomes are formed from Golgi apparatus
Ans: 3 (Anthocyanins are present in the vacuole of all plants)

115. Assertion (A): Concentration of cell sap significantly higher than the cytoplasm.
Reason (R): Tonoplast fecilitates the transport of a number of ions and other
meterials against concentration gradient in to the vacuole.
Ans: A & R are correct. R explains A.

116. ATP producing centre of the cell
Ans: Mitochandria

117. Robert Hooke first observed except-
Ans: Vacuole

118. Cell wall was first observed by-
Ans: Robert Hooke

119. The Cell wall of the following plant was observed first-
1) Sequoeia 2) Cotton 3) Oak 4) Ficus
Ans: 3 (Oak)

120. Book written by Hooke -
Ans: Micrographia

121. Pits are present in the-
Ans: Secondary Cell wall

122. Pits are useful in-
Ans: Inter cellular transport

123. The following character is not related to primary Cell wall-
1) It shows growth
2) Its thickness gradually increases with maturity
3) It is not first formed Cell wall
4) It lies outside to secondary Cell wall
Ans: 2 (Its thickness gradually increases with maturity)

124. Cytoplasm of neighbouring cells is connected by-
Ans: Plasmodesmata

125. Middle lamellum is made of-
Ans: Calcium pectate

126. Middle Lamellum joins-
1) Different neighbouring cells 2) Similar adjacent cells
3) 1, 2 4) None
Ans: 3 (Different neighbouring cells, Similar adjacent cells)

127. Plasmodesmata pass through-
A. Middle Lamellum B. Primary Cell wall C. Secondary Cell wall
Ans: A B only

128. Cell wall of Bacteria and Fungi are made of respectively-
Ans: Mucopeptide, Chitin

129. Match the following Cell wall substances.

I. Fungi	A) Murein
II. Archaebacteria	B) Chitin
III. Bacteria	C) Mycolic acid
IV. Actinomycetes	D) Pseudomurein

I II III IV
Ans: B D A C

130. The Cell wall of Algae is made of-
1) Cellulose 2) Galactans and Mannans 3) Calcium carbonate 4) All
Ans: 4 (All)

131. The following new materials are not added to the cellulose interfibrillary spaces of secondary Cell wall-
1) Lignin 2) Chitin 3) Suberin 4) Cutin & Suberin
Ans: 2 (Chitin)

132. Cell wall is concerned with-
Ans: Barrier to desirable macromolecules

133. Common between plant cell and animal cell-
Ans: Plasma membrane

134. Vegetative plant cell is surrounded by the following layers in centripetal succession-
A) Middle Lamellum B) Primary cell wall
C) Secondary cell wall D) Plasma membrane
Ans: A B D

135. Plasma membrane can be seen under-
Ans: Electron microscope

136. Plasma membrane of Archaebacteria is made of-
Ans: branched chain lipids

137. Plasma membrane before flueid-mosaic model is said to be made of-
Ans: P-L-L-P

138. Common to all models of plasma membrane-
Ans: Bilayered lipids - head

140. Cell wall made of cellulosic plates is seen in-
Ans: Dinoflagellates

141. Plasma membrane of RBC is mainly composed of-
Ans: Phosphoglyceride

142. Which nature of the plasma membrane helps the cell in performing the functions like cell growth, formation of inter cellular junctions, secretion, endocytosis and cell division-
Ans: Fluid nature

143. Chemical composition of plasma membrane is-
Ans: 52% proteins 40% lipids

144. Position of proteins in the plasma membrane-
1) Integral 2) Peripheral 3) Partly in side 4) All
Ans: 4 (All)

145. Proteins show lateral movements within the bilayered lipid layers. It is because-
Ans: Quasi fluid nature of lipids

146. Plasma membrane in a plant cell is seen clearly in-
Ans: Flaccid cell

147. Match the following functions of plasma membrane.

I. Passive transport	A) Against concentration gradient
II. Active transport	B) Neutral solutes along concentration gradient
III. Diffusion	C) Diffusion of water across the membrane
IV. Osmosis	D) No requirement of energy

I II III IV
Ans: D A B C

148. Plasma membrane forms mesosomes in-
Ans: some Bacteria

149. One of the following process is an energy dependent process-
1) Passive transport 2) Osmosis 3) Active transport 4) Diffusion
Ans: 3 (Active transport)

150. The following are permeable and semipermeable respectively-
1) Primary Cell wall, Secondary Cell wall
2) Plasma membrane, Primary Cell wall
3) Primary Cell wall, Plasma membrane
4) Secondary Cell wall, Middle lamellum
Ans: 3 (Primary Cell wall, Plasma membrane)

151. Plasma membrane by nature-
Ans: Selectively permeable

152. Carrier proteins are required in the transportation of-
Ans: polar molecules

153. Assertion (A): Polar molecules can not pass through the non-polar lipid bilayer.
Reason (R): Carrier proteins are present in the plasma membrane.
Ans: A & R are correct. R do not explains A.

154. Read the following statements and findout correct one-
1) Carrier proteins of the Cell wall are required to facilitate the polar molecules
2) Non polar molecules require carrier proteins to pass through the lipid bilayer
3) Na⁺/ K⁺ pumps are associated with active transport
4) Cell membrane has polar lipid bilayer and carrier proteins
Ans: 3 (Na⁺/ K⁺ pumps are associated with active transport)

155. Assertion (A): Several ribosomes attach to a single mRNA and form polysomes.
Reason (R): These ribosomes are of only type i.e. 70 S type.
Ans: A & R are correct. R do not explains A.

156. The size of the 70 S ribosomes is-
Ans: 15 nm × 20 nm

157. In prokaryotes the ribosomes-
1) Present in cytoplasm only 2) Form Polysomes
3) Associated with plasma membrane 4) All
Ans: 4 (All)

158. Incusion bodies-
1) Present in Prokaryotes only 2) They are naked
3) They are reserve food materials 4) All
Ans: 4 (All)

159. The following are not present in prokaryotes-
1) Polysomes 2) Mesosomes
3) Plasmids 4) Membrane bounded/ Structures/ Organelles
Ans: 4 (Membrane bounded/ Structures/ Organelles)

160. The following structures do not play a role in motility-
1) Flagella, Cilia 2) Pili Cilia
3) Pili, Fimbriae 4) Fimbriae, Flagella
Ans: 3 (Pili Fimbriae)

161. The following are small bristle like fibres-
1) Fimbriae 2) Pili 3) Flagella 4) Cilia
Ans: 1 (Fimbriae)

162. Flagella in Bacteria are extensions of-
Ans: Cell wall

163. Mesosome consists-
Ans: Vesicles, Tubules, Lamellae

164. Chromatophores are the extensions of-
Ans: Plasma membrane

165. Which of the following is concerned with maximum functions-
1) Mesosomes 2) Cell wall 3) Gas vacuoles 4) Plasmid
Ans: 1 (Mesosomes)

166. Common between Prokaryotes and Eukaryotes
1) Cell wall 2) Nucleoid 3) Plasmid 4) None
Ans: 4 (None)

167. The following part of prokaryotes is structurally similar to that of eukaryotes-
1) Cell wall 2) Plasmid 3) Plasma membrane 4) Nucleoid
Ans: 3 (Plasma membrane)

168. Cell envelope of prokaryotic cells is-
Ans: 3- layered

169. The position of Cell wall in cell envelope is-
Ans: Between glycocalyx and Plasma membrane

170. The living part of cell envelope is-
Ans: Innermost

171. Slime layer and capsule are different types of-
Ans: Glycocalyx

172. Match the following-

I. Loose sheath	A) Staining
II. Thick, Tough	B) Cell wall
III. Shape	C) Slime layer
IV. Chemical Composition of cell envelope	D) Capsule

I II III IV
Ans: C D B A

173. The following structure prevents the bacterium from bursting or collapsing-
1) Cell wall 2) Cell envelope 3) Capsule 4) Slime sheath
Ans: 3 (Capsule)

174. Assertion (A): Glycocalyx differs in composition and thickness among different bacteria.
Reason (R): Glycocalyx is follwed by plasma membrane.
Ans: A is correct. R is false.

175. Three layers of cell envelope in Bacteria
1) Distinct from each other in structure 2) Distinct from each other in function
3) Act as single protective unit 4) All
Ans: 4 (All)

176. The following structure gives response to staining procedure-
1) Cell envelope 2) Cell wall
3) Protoplasm 4) Plasma membrane
Ans: 2 (Cell wall)

177. Read the folloiwng statements and find out correct one-
1) Cell envelope gives response to the gram's stain
2) Glycocalyx is similar in composition and thickness in different Bacteria
3) Plasma membrane in Bacteria differs structurally from that of eukaryotes
4) Mesosomes are the extensions of plasma membrane which are in the form of
vesicles, tubules and vacuoles
Ans: 1 (Cell envelope gives response to the gram's stain)

178. The extensions of the plasma membrane in Bacteria are-
A) Mesosomes B) Plasmids C) Chromatophores
Ans: A, C

179. The parts of bacterial flagellum are
A. Hook B. Filament C. Basal body
Arrange then in a sequence
Ans: C A B

180. The functions of fimbriae are-
A. Motility B. Attachment to the host C. Attachment to the rocks in streams
Ans: B C

181. The surface structures in Bacteria which help in the motility-
A. Flagella B. Pili C. Fimbriae
Ans: A only

182. One of the following is not a prokaryote-
1) Yeast 2) Bacteria, Cyano Bacteria 3) Mycoplasma 4) PPLO
Ans: 1 (Yeast)

183. Read the following statements-
1) The prokaryotes multiply rapidly than eukaryotes
2) Prokaryotes do not show much variations in size and shape as their organisation is simple
3) Prokaryotes show more variations in the organisation
4) All bacteria have plasmids
Which one among the above is correct?
Ans: 1 (The prokaryotes multiply rapidly than eukaryotes)

184. Assertion (A): Genomic DNA of bacteria is circular and naked.
Reason (R): Plasmid confers certain phenotype characters to the host Bacteria.
Ans: A & R are correct. R do not explains A.

185. Plasmid is a part of Genotype. It is-
Ans: completely incorrect

186. Resistance to antibiotics and monitoring Bacterial transformation with foreign DNA are concerned with respectively-
Ans: Plasmid is concerned with both

187. Which among the following characters do not applies to both prokaryotes and eukaryotes-
Ans: Vacuole

188. Match the following-

I. Bacteria	A) 7.0 μ m
II. Mycoplasma	B) 3 - 5 μ m
III. Human RBC	C) 0.02 - 0.2 μm
IV. Virus	D) 0.3 μ m

I II III IV
Ans: B D A C

189. The shape of the prokaryotic cell varies with-
Ans: function

190. Largest and smallest cells respectively are-
Ans: Ostrich egg, PPLO

191. Non-membrane bound organelles are-
Ans: Ribosomes, Centriole

192. Match the following-

I. No nucleus	A) Epithelial cell
II. In distinct nucleus	B) Nerve cell
III. Distinct nucleus	C) Tracheid
IV. Distinct nucleus in a branched cell	D) Prokaryote

I II III IV
Ans: C D A B

193. A typical eukaryotic cell measures-
Ans: 10 times more than a typical Bacteria

194. Ribosomes are seen in-
1) Cytoplasm 2) Mitochondria & Chloroplast
3) 1, 2 4) Vacuole also
Ans: 3 (Cytoplasm, Mitochondria & Chloroplast)

195. Triple helical structure refers to-
Ans: Collagen

196. Triple helical structure of collagen was discovered by-
Ans: G.N.Ramachandran

197. Who influenced G.N.Ramachandran to solve the structure of collagen-
Ans: Linus Pauling

198. G.N.Ramachandran published the following and got appreciation from Linus Pauling.
1) α - helix, β - sheet structures 2) a - helix, β - helix structures
3) a - helix, β - sheet structures 4) α - helix, β - helix structures
Ans: 3 (a - helix, β - sheet structures)

Posted Date : 30-11-2020

ECOLOGICAL ADAPTATION, SUCCESSION AND ECOLOGICAL SERVICES

1. Father of Ecology in India-
A: Ramdeo Mishra

2. Ramdeo Mishra started-
A: Postgraduate course in Ecology

3. Ecology is also called-
A: Environmental Biology

4. The plants which open the stomata during night are-
A: CAM Plants

5. The trees which loose leaves every year are seen in-
A: Deciduous forests

6. A distinct ecological, community of plants and animals living together in a particular climate is called-
A: Biome

7. The term Ecosystem was coined by-
A: A.G.Tansley

8. Flora and Fauna are found scanty in-
A: Desert

9. Ecology was started in which century?
A: 19th

10. Ecological studies began with the works of scientists-
A: Haeckel

11. Deserts, Rain forests and Tundra are-
A: Biomes

12. A desert is a-
A: Biome

13. The functional unit of ecology-
A: Ecosystem

14. Annual variations in the intensity and duration of temperature are caused by-
A: Rotation of earth around Sun and the tilt of its axis

15. Population means-
A: A group of similar individuals belonging to the same species found in an area

16. Community includes different-
A: Genera, species, same area

17. Adaptation helps the organism-
A: To survive and reproduce

18. The largest ecosystem is-
A: Biosphere

19. Structural and functional Earth is-
A: Gigantic ecosystem

20. Eugen Warming-
A: Botonist, ecologist

21. Eugen Warming-
A: Danish

22. Match the following-

I II III IV
A: D C B A

23. Find wrong match among the following-
1) Utricularia - Submerged root less
2) Typha - Limnophila
3) Pistia - Nymphaea
4) Lemna - Not rooted
A: 3 (Pistia - Nymphaea)

24. Assertion (A): Roots to hydrophytes are of secondary importance.
Reason (R): They have surplus of water in their habitat.
A: A & R are correct. R explains A.

25. Assertion (A): In submerged hydrophytes gas exchange takes place directly through the thin walls by diffusion.
Reason (R): Stomata are totally absent in them.
A: A & R are correct. R explains A.

26. Assertion (A): All hydrophytes contain aerenchyma.
Reason (R): Mechanical tissues and xylem are poor developed.
A: A & R are correct. R do not explains A.

27. We find extensively these plants on the surface of land-
1) Hydrophytes 2) Xerophytes 3) Mesophytes 4) 1 & 2
A: 3 (Mesophytes)

28. Adaptations in mesophytes-
1) Not necessary 2) Necessary
3) Necessary unless the habitat is specialised 4) 1 & 3
A: 4 ( Not necessary & Necessary unless the habitat is specialised)

29. Assertion (A): All ephemerals are annuals.
Reason (R): All annuals are annuals.
A: A is correct. R is false.

30. Ephemerals are found generally in-
A: Arid

31. Tribulus is-
A: ephemeral

32. Pore space in the habitat of mesophytes consists-
A: 50% air, 50% water

Posted Date : 15-10-2020

TAXONOMY OF FLOWERING PLANTS

1. The term taxonomy is coined by-
A: de Candolle

2. Taxonomy is also called-
A: New Systematics, Systematic Botany

3. Taxonomy - the word is derived from-
A: Greek

4. The branch of Botany that needs knowledge from various other branches is-
A: Taxonomy

5. Plant Systematics is other name for-
A: Systematic Botany

6. Taxonomy based on purely morphological description, it is called-
A: Alpha taxonomy

7. Match the following-

I) Chromosome Number	A) Numerical taxonomy
II) Phytochemical Data	B) Omega taxonomy
III) Mathematical Methods	C) Cytotaxonomy
IV) Information from other branches	D) Chemotaxonomy

A: I II III IV
C D A B

8. The ultimate aim of taxonomy is-
A: Classification

9. The first step in taxonomy-
A: Characterisation

10. Arrange the following steps in a sequence-
A) Identification B) Characterisation C) Classification D) Nomenclature
A: BADC

11. The earliest Classification of plants is based on
A: Economic uses

12. Identification of an unknown plant was made easier by the following system of classification-
1) Artificial System 2) Natural System

3) Sexual System 4) Phylogenetic System
A: 1(Artificial System)

13. Post Darwinian systems of classification are-
A: Phylogenetic system of classification

14. Match the following-

I) One of few characters	A) Natural system
II) Evolutionary trends	B) APG system
III) Latest system	C) Artificial system
IV) All the important morphological characters	D) Phylogenetic System

A: I II III IV
C D B A

15. Father of Botany in India-
A: Parasara

16. Principles of classification were proposed by-
A: Linnaeus

17. Floral characters were first used in the classification by-
A: Linnaeus

18. Father of taxonomy-
A: Linnaeus

19. Who first used binomial nomenclature-
A: Gaspard Bauhin

20. Nomenclature is essential for-
A: Communication

21. Rules for nomenclature are proposed by-
A: International Code of Botanical Nomenclature

22. The Knowledge of embryology, Phytochemistry and other branches is used in the following-
1) Alpha taxonomy 2) Omega taxonomy
3) Numerical taxonomy 4) Chemo taxonomy
A: 2(Omega taxonomy)

23. Length of the stamens was taken into consideration in classification by-
A: Linnaeus

24. Assertion (A) : The Classification of Linnaeus and Theophrastus are not accepted-
Reason (R) : They gave equal importance to vegetative and sexual charactres.
A: A & R are correct. R explains A.

25. Assertion (A) : Sexual or floral characters are given more importance in classification than vegetative characters.
Reason (R): Vegetative characters are affected more easily by environment.
A: A & R are correct. R explains A.

26. Assertion (A) : Floral characters are more stable than vegetative characters.
Reason (R): Vegetative characters are not affected by environment.
Ans: A is correct. R is false.

27. Match the following through which different types of classification were proposed by different taxonomists-

I) Genera Plantarum	A) Linnaeus
II) Species Plantarum	B) Hutchinson
III) Historia Plantarum	C) Bentham & Hooker
IV) Families of Flowering Plants	D) Theophrastus

A: I II III IV
C A D B

28. Assertion (A) : Floral characters are given more importance in classification.
Reason (R): They are more conserved.
A: A & R are correct. R explains A.

29. Taxonomy was established in which century?
A: 17

30. Who popularised binomial nomenclature-
A: Linnaeus

31. Emphasis on the following is given in plant systematics-
A: Phylogeny

32. Who classified plants into hearbs, shrubs, trees basing on habit-
A: Theophrastus

33. Linnaeus classified plants into how many groups in his system of classification-
A: 24

34. Bentham and Hooker had not considered the following characters in their natural system of classification-
A: Evolutionary characters

35. First phylogenetic system of Classification was proposed by-
A: Eichler

36. The book written by Engler and Prantl is-
A: Die Naturlichen Pflanzenfamilien

37. One of the following is not a taxonomist-
1) Endlicher 2) Charles Darwin 3) Eichler 4) Hutchinson
A: 2(Charles Darwin)

38. The following book is written in 20th century-
1) Die Naturlichen Pflanzenfamilien
2) Families of flowering plants
3) Genera plantarum
4) Micrographia
A: 2(Families of flowering plants)

39. Computer is essential in the following-
1) Numerical taxonomy 2) Chemo taxonomy
3) Omega taxonomy 4) Phylogenetic system
A: 1(Numerical taxonomy )

40. The unit of taxonomy-
A: Taxon

41. The term used by Bentham and Hooker for order and family respectively-
Ans: Cohort, Natural order

42. Number of families in gymnosperms dicots and monocots respectively in Bentham & Hookers classification-
Ans: 3, 165, 34

43. Ratio between Cohorts of Polypetalae and Gamopetalae-
Ans: 3 : 2

44. Ratio between the series of Polypetalae and Gamopetalae-
Ans: 1 : 1

45. Ratio between classes and sub classes in dicofyledonae-
Ans: 0 : 3

46. The taxonomic rank of polypetalae-
Ans: Subclass

47. Number of orders in monocots-
Ans: Zero

48. Common between monocots and monochlamydae
Ans: Series is directly divided into families

49. Number of families in Angiosperms-
Ans: 199

50. Cohort is a group of-
Ans: Families

51. Number of orders in Polypetalae-
Ans: 15

52. Number of series in Dicots-
Ans: 14

53. Number of series in dicots and monocots respectively-
Ans: 14, 7

Posted Date : 30-11-2020

SCIENCE OF PLANTS

1. Match the following.

(I) 19th Century	(A) Taxonomy
(II) 18th Century	(B) Cytology
(III) 20th Century	(C) Tissue Culture
(IV) 17th Century	(D) Genetics

The correct matching is
A: I II III IV
D A C B

2. Match the following.

(I) Chinese plant cultivation	(A) 1300 B.C.
(II) Adharvana veda	(B) 340 B.C.
(III) First Book on agriculture	(C) 2500 B.C.
(IV) de Historia Plantarum	(D) 2000 B.C.

The correct matching is
A: I II III IV
C D A B

3. Arrange the following scientists in ascending order.
A) Anton van Leeuwenhoek B) Gaspard Bauhin
C) Robert Hooke D) Camerarius
E) Nehemiah Grew and Marcello Malpighi
A: BCAED

4. Medicinal plants were first described in
A: Adharvana veda

5. Internal characters of plants were first described by
A: Theophrastus

6. Match the following.

(I) Hieroglyphics	(A) Camerarius
(II) Herbals	(B) Linnaeus
(III) Sexual reproduction	(C) Egyptians
(IV) Sexual system of classification	(D) Renaissance period

The correct matching is
A: I II III IV
C D A B

7. Match the following 2 lists belonging to the same century.

(I) Introduction of binomial nomenclature	(A) Conduction of water through xylem
(II) Popularisation of binomial nomenclature	(B) Discovery of Zymase
(III) Anatomical study of plant tissues	(C) Micrographia
(IV) Laws of inheritance	(D) Sexual reproduction in plants

The correct matching is
A: I II III IV
C A D B

8. The term Botany has the root word from
A: Greek

9. Arrange the following words in a sequence.
A) Bouskein B) Botany C) Bous D) Botane
A: C A D B

10. Recorded information related to crop plants and fruit trees in the form of pictures is called
A: Hieroglyphics

11. Internal characters of plants were first mentioned in
A: de Historia Plantarum

12. Match the following lists.

(I) Weeds	(A) Adharvana veda
(II) Forests	(B) de Historia Plantarum
(III) Medicinal plants	(C) Krishiparasharam
(IV) External & Internal characters of 500 plants	(D) Vrikshayurveda

A: I II III IV
C D A B

13. Books of Medicinal plants are called
A: Herbals

14. Taxonomy and Physiology were established in 17th Century by
A: Linnaeus & Priestley

15. Match the following.

(I) First discovery of cell	(A) Nehemiah Grew, Malpighi
(II) First discovery of living plant cell	(B) Robert Hooke
(III) Anatomical studies	(C) Stephen Hales
(IV) First Physiological studies	(D) Anton van Leeuwenhoek

A: I II III IV
B D A C

16. Match the following.

(I) Linnaeus	(A) Indian
(II) GJ Mendel	(B) Greek
(III) V.S. Ram Das	(C) Swedish
(IV) Aristotle	(D) Czekhoslovakia
	(E) American

A: I II III IV
C D A B

17. Rapid Development was apparent in different disciplines of Botany, in Particular Cell Biology during which century
A: 20

18. Find out mis-match.
1) Role of chromosomes in heredity - Sutton & Boveri
2) Genetic nature of RNA - Frankel Conrat
3) Genetic nature of DNA - Watson & Crick
4) Artificial Synthesis of Gene - H.G. Khorana
A: 3(Genetic nature of DNA - Watson & Crick)

19. The following scientists are not concerned with Tissue Culture
1) Hanning, Nietzsche 2) Shimakura, Maheshwari
3) White, Skoog 4) Went, Maheshwari
A: 4(Went, Maheshwari)

20. The scientists concerned with C₃, C₄ respectively are
A: Calvin, Slack

21. Urease was crystallised by
A: Sumner

22. One of the following is not concerned with light reaction
1) Ruben & Kamen 2) Hill 3) Arnon 4) Priestle
A: 4(Priestley)

23. Electron microscope was discovered by
A: Knoll & Rusko

24. TCA Cyle was discovered by
A: H.A. Krebs

25. Identify the taxonomist belonging to 19th century
A: Endlicher

26. Select the correct pair of plants as the source of SCP
1) Arnica, Aloe 2) Digitalis, Spirulina
3) Chlorella, Spirulina 4) Arnica, Chlorella
A: 3(Chlorella, Spirulina)

27. Biodiesel is produced from
A: Pongamia, Jatropa

28. Assertion (A): Jatropa and Pongamia are petro plants.
Reason (R): They are rich in hydrocarbons.
A: A & R are correct. R explains A.

29. One of the following is not a Palynologist.
1) Maheshwari 2) P.K.K. Nair
3) C.G.K. Ramanujam 4) Wodehouse
A: 1( Maheshwari )

30. A SCP used by Astronauts
A: Chlorella

31. One of the following is a biofertiliser but not micro oranism
1) Azolla 2) Nostoc 3) Anabaena 4) Rhizobium
A: 1(Azolla )

32. Match the following.

(I) Soil erosion	(A) Saprophytes
(II) Soil pollution	(B) Biofertilisers
(III) Green house effect	(C) Bio-remediation
(IV) Recycling of nutrients	(D) Intensive tree plantation
(V) Soil pollution & Water pollution	(E) Sand binding plants

The correct matching is
A: I II III IV V
E C D A B

33. It is a fundamental requisite for classification of plants
A: Morphology

34. Anatomy is a branch of
A: Internal Morphology

35. The study of microspores of fruit yielding plants is called
A: Palynology

36. The following branch of Botany is built on the basis of information obtained from different fields of Botany
1) Physiology 2) Phytogeography 3) Systematic Botany 4) Embryology
A: 3(Systematic Botany)

37. Match the following lists.

(I) Autotrophic thallophytes	(A) Pteridology
(II) Heterotrophic thallophytes	(B) Phycology
(III) Amphibians of plant kingdom	(C) Mycology
(IV) First Vascular plants	(D) Bryology

A: I II III IV
B C D A

38. The branch of Botany that deals with the study of the geographic distribution of the plants in the past and present is
A: Phytogeography

39. The study of fossils is called
A: Paleo Botany

Posted Date : 30-11-2020

6 - 9 CHAPTERS (JUNIOR BOTANY)

1. A Rajasthani scientist, inspired by an American missionary teacher was the first to identify with..
A) Restriction Enzyme
B) Usage of Embryological characters in taxonomy.
C) Work on C₄ pathway
D) Test-tube fertilisation.
E) Propose nutritional medium for tissue culture
F) Intra-ovarian pollination.
A: B D F

2. Cell division is unequal in the following.
1) Budding in Yeast
2) First division of microspore in angiosperms.
3) Second division of pollen grain.
4) Budding in Yeast & First division of microspore in angiosperms.
A: 4 (Budding in Yeast. & First division of microspore in angiosperms.)

3. The following feature is responsible for meaningful DNA packaging inside the nucleus.
1) +Ve charge of Histones, +Ve charge of DNA
2) +Ve charge of Histones, −Ve charge of DNA
3) −Ve charge of Histones, +Ve charge of DNA
4) −Ve charge of Histones, −Ve charge of DNA
A: 2 (+Ve charge of Histones, −Ve charge of DNA)

4. Read the following statements and find out incorrect one.
1) Both fusing gametes may be motile in a few fungi and algae.
2) Gametes are haploid though the parent plant body from which they arise may be either haploid or diploid.
3) In plants, hormones are responsible for sexual reproduction.
4) Asexual reproduction is the common method of reproduction in all angiosperms.
A: 4 (Asexual reproduction is the common method of reproduction in all angiosperms.)

5. Types of cells in the embryosac basing on function.
A: 4

6. Assertion (A): One of the problems of hybrids is that hybrid seeds have to be produced every year.
Reason (R): If these hybrids are made into apomicts, there is no segregation of characters in the hybrid progeny.
A: A and R correct. R does not explain A.

7. Match the following.

Correct matching is
I II III IV
A: B D A C

8. Match the following.

The correct matching is
I II III IV
A: D A C B

9. Identify the wrong pair of statements.
I) Capsicum has number of locules equal to the number of carpels in Trigonella.
II) Number of natural orders in Gymnospermae of Bentham and Hooker's classification is not equal to the pairs of stamens in Allium.
III) Types of vascular tissues seen in the vascular bundles of Solanaceae is equal to the number of petals of similar size present at the anterior end of Tephrosia flower.
IV) Number of groups of plants in the sexual system of classification proposed by Linnaeus is equal to the floral appendages of Brassicaceae.
A: I, III

10. The position of Fabaceae in Bentham and Hooker's classification is as
A: Sub family Papilionoideae

11. Find out mis-match.
1) Aristotle - Historia plantarum
2) Linnaeus - Species plantarum
3) Hutchinson - Families of flowering plants
4) Bentham & Hooker - Genera plantarum
A: 1 (Aristotle - Historia plantarum)

12. Essential organs in Solanum species basing on maturity are similar to the maturity of sex organs in
A: Medow saffron

13. Glycoproteins and Lipoproteins are synthesised by the following respectively.
1) SER, RER 2) Peroxisomes, ER

3) Golgi apparatus, ER 4) ER, Peroxisomes
A: 3 (Golgi apparatus, ER)

14. Number of the following organelles in a cell depend on the physiological activity of the cell
1) Nucleus 2) Mitochondria

3) Flagella 4) Chromosomes
A: 2 (Mitochondria)

15. Cell shape is maintained by
A: Cytoskeleton

16. The following ensures continuity of species between sexually reproducing organisms of one generation and the next.
1) Fragments 2) Gametes 3) Spores 4) Zygote
A: 4 (Zygote)

17. Read the following table. Find out correct pair of combinations.

A: II, III

Posted Date : 15-10-2020

6 - 9 CHAPTERS (SENIOR BOTANY)

1. Assertion (A): Plants retain the capacity for unlimited growth through out their life.
Reason (R): Plants have meristems at certain locations in their body.
A: A and R are correct. R explains A.

2. Read the following statements and find out incorrect one.
1) There are 3 hierarchial levels only according to ICTV in the classification of viruses.
2) HIV is the name of the Genus that causes AIDS.
3) In nomenclature, genus and species both end with virus.
4) Polio virus is polyhedral.
A: 2 (HIV is the name of the Genus that causes AIDS.)

3. ABA is chemically identical to
A) Ethylene B) GA C) Dormin D) Auxins
E) Abscissin II F) Cytokinins G) Inhibitor B
A: CEG

4. The following hormone is responsible for respiratory climactic in ripened fruits.
1) The hormone that promotes bolting.
2) The hormone that induces flowering mango.
3) The hormone that induces parthenocarpy in tomato.
4) The hormone that helps in adventitious shoot formation.
A: 2 (The hormone that induces flowering mango.)

5. Read the following statements and find out correct statements.
A. Conjugation is facilitated by conjugative plasmids.
B. The discovery of DNA as the genetic material was based on the experiments of Zinder, Lederberg.
C. Some bacteria are used as Biocensors.
D. Louis Pasteur is considered as father of Bacteriology.
A: A

6. Chromosome mapping was done by using the frequency of recombination, between gene pairs of the same chromosome.
A: Sturtevant

7. Assertion (A): Genes contain the information that is required to express a particular trait.
Reason (R): For the expression of traits genes provide only the opportunity and environment provides potentiality.
A: A is correct. R is false.

8. Virus has?
1) RNA 2) DNA 3) 1 or 2 4) 1 & 2
A: 3 (RNA or DNA)

9. Assertion (A): Oncogenic viruses cause cancer.
Reason (R): Human Papilloma causes cancer in human beings only.
A: A is correct. R is false.

10. Cancer cells show the following mutations.
1) Deletions 2) Insertions
3) Frame shift mutations 4) Chromosomal aberrations
A: 4 (Chromosomal aberrations)

11. Match the following different parts of T₄.

The correct matching is
I II III IV
A: C D B A

12. Match the following.

The correct matching is
I II III IV
A: C D A B

13. Match the following the various parameters of growth.

The correct matching is
I II III IV
A: B D C A

14. Match the following.

The correct matching is
I II III IV
A: B D A C

Posted Date : 15-10-2020

BIOLOGICAL CLASSIFICATION

1. 50% to 70% of Biogas is-
A: Methane

2. Chitinous wall is present around-
A: Oospore

3. Assertion (A): Mycorrhizae are used as bio - fertilisers.
Reason (R): They help to absorb more phosphate.
A: A & R are correct. R explains A.

4. A colonial aggregate of immobile non-flagellated individual cells occuring in the life cycle of some flagellated green algae is called-
A: Palmella stage

5. The spores on germination produce gametophyte. The spores are produced-
A) By mitosis in higher plants B) By meiosis in higher plants
C) By mitosis in lower plants (Thallophytes) D) By meiosis in thallophytes
A: BCD

6. Small plants which float passively in water currents are called-
A: Phytoplanktons

7. The following are treated as orphans in the classification of plant kingdom-
1) Viruses 2) Prions 3) Viroids 4) All
A: 4 (All)

8. Lichens are included in-
1) 2 kingdom classification 2) 5 kingdom classification
3) 6 kingdom classification 4) None
A: 4 (None)

9. Cellular Organisation is seen in-
A: Lichens

10. The following relationship between Algae and Fungi is correct

A: (2)

11. The following consists neither DNA nor RNA-
1) Virus 2) Virion 3) Prion 4) Viroid
A: 3 (Prion)

12. Bacteriophage is a-
A: Virus

13. Find out mis-match
1) Virus - Obligatory parasite 2) Viroid - Potato spindle tuber disease
3) Prion - Scrapie disease of women 4) Woese - Three domains
A: 3 (Prion - Scrapie disease of women)

14. RNA is present in-
1) TMV 2) Viroid 3) HIV 4) All
A: 4 (All)

15. Assertion (A): The infectious agent of Potato spindle tuber disease is a viroid.
Reason (R): It has free RNA.
A: A & R are correct. R explains A.

16. Assertion (A): Lichens are very good pollution indicators.
Reason (R): They do not grow in polluted areas.
A: A & R are correct. R explains A.

17. DNA is present in-
A: Bacteriophage

18. The following are closely related-
1) Archaea & Eukarya 2) Prion & Viroid
3) Prion & Virion 4) Archaea & Bacteria
5) Eukarya & Bacteria
A: 1 (Archaea & Eukarya)

19. One of the following is a Kingdom as well as domain-
1) Bacteria 2) Archaea 3) Eukarya 4) Protista
A: 1 (Bacteria)

20. Assertion (A): Viruses do not find a place in classification.
Reason (R): Viruses are not truely living.
A: A & R are correct. R explains A.

21. The following are not non-cellular-
1) Lichens 2) Viruses 3) Viroids 4) Virions
A: 1 (Lichens)

22. The domains of Carl Woese-
(A) Archaebacteria (B) Bacteria
(C) Protista (D) Cyanobacteria
(E) Archaea (F) Monera (G) Eukarya
A: BEG

23. The following is used as a marker in the 6 Kingdom Classification-
1) 23s rRNA 2) 16s rRNA 3) 30s 4) 50s
A: 2 (16s rRNA)

24. Monera of Whittaker is divided into the following by Carl Woese-
1) Bacteria, Archaea 2) Archaebacteria and Eukarya
3) Protista & Eukarya 4) Archaebacteria and Archaea
A: 1 (Bacteria, Archaea)

25. One of the following is not a characteristic feature of Animalia-
1) Holozoic Nutrition 2) Elaborate sense organs
3) Definite growth & shape 4) Presence of Cell wall
A: 4 (Presence of Cell wall)

26. Match the following-

I. Absorptive mode of nutrition	A) Prions
II. Ingestion of food	B) Animals
III. Absence of genetic material	C) Viroids
IV. Absence of proteins	D) Animals
	E) Plants

The correct matching is
I II III IV
A: E D A C

27. Assertion (A): All the members under the Kingdom Plantae of Whittaker are eukaryotes and autotrophic.
Reason (R): All eukaryotic and chlorophyll containing plants are kept in Plantae of Whittaker.
A: A & R are correct. R explains A.

28. Match the following.

I. Sexual rep absent	A) Basidiomycetes
II. Asexual rep absent	B) Phycomycetes
III. Sexual rep is reduced	C) Deuteromycetes
IV. All 3 types of sex rep. present	D) Ascomycetes
	E) Basidiomycetes

The correct matching is
I II III IV
A: C E A B

29. Match the following.

I. Edible	A) Basidiomycetes
II. Biochemical & Genetic studies	B) Deuteromycetes
III. No asexual spores	C) Neurospora
IV. No sexual spores	D) Truffles

The correct matching is-
I II III IV
A: D C A B

30. Find out mis-match.
1) Polypores - Bracket fungus 2) Lycoperdon - Puff-ball
3) Trichoderma - Ascomycetes 4) Albugo - Brassica
A: 3 (Trichoderma - Ascomycetes)

31. Sexual spores of Ascomycetes are called-
A: Ascospores

32. One of the following is not the shape of fruiting body of Ascomycetes-
1) Club shape 2) Cup shape 3) Globose 4) Flask
A: 1 (Club shape)

33. Identify an odd pair-
1) Pencillium - Neurospora 2) Aspergillus - Claviceps
3) Yeast - Albugo 4) Neurospora - Aspergillus
A: 3 (Yeast - Albugo)

34. Exogenous asexual spores are found in-
1) Phycomycetes 2) Basidiomycetes
3) Deuteromycetes 4) Ascomycetes
A: 4 (Ascomycetes)

35. Dikaryotic mycelium is found in-
1) Deuteromycetes 2) Ascomycetes
3) Phycomycetes 4) Basidiomycetes
A: 4 (Basidiomycetes)

36. Karyogamy and Meiosis occur in a separate structure in-
1) Ascomycetes 2) Phycomycetes
3) Basidiomycetes 4) None
A: 4 (None)

37. Some of the members of the following class are aquatic-
1) Phycomycetes 2) Ascomycetes
3) Basidiomycetes 4) Deuteromycetes
A: 1 (Phycomycetes)

38. Oogamy occurs in-
A: Phycomycetes

39. Read the following statements and find out how many of them are not incorrect?
I. Aquatic fungi belong to Ascomycetes
II. Algal fungi are aseptate and coenocytic
III. Sac fungi do not have dikaryotic mycelium
IV. Dikaryotic mycelium produce basidia
V. Deuteromycetes produce flagellated bodies
VI. Zygospores are formed in algal fungi only
A: 2

40. A large number of the following help in mineral cycling-
1) Phycomycetes 2) Deuteromycetes
3) Ascomycetes 4) Basidiomycetes
A: 2 (Deuteromycetes)

41. Match the following-

I. Puff ball	A) Algal fungi
II. Bread mould	B) Sac fungi
III. Red rot	C) Club rungi
IV. Morels	D) Imperfect fungi

The correct matching is
I II III IV
Ans: C A D B

42. The effective fungal biocontrol agent is-
Ans: Trichoderma

43. One of the following character is not considered in the classification of Fungi-
1) Mode of spore formation 2) Fruiting body
3) Cell wall 4) Structure of Mycelium
Ans: 3 (Cell wall)

44. Phycomycetes is known for-
1) Zoospores 2) Zygospores
3) Coenocytic mycelium 4) All
Ans: 4 (All)

45. Sexual spores in sac fungi-
Ans: Exogenous ascospores

46. Fruiting body without opening is called-
Ans: Cleistothecium

47. Read the following statements and find out incorrect one-
1) Genus is Deuteromycetes indicates vegetative or asexual phase
2) Rusts and smuts belong to Club fungi
3) Dikaryotic stage is seen in ascomycetes & basidiomycetes
4) Karyogamy & Meiosis occur simultaneously in Ascomycetes & Basidiomycetes
Ans: 4 (Karyogamy & Meiosis occur simultaneously in Ascomycetes & Basidiomycetes)

48. In fungi sexual reproduction occurs by-
1) Oospores 2) Ascospores 3) Basidiospores 4) All
Ans: 4 (All)

49. Toad stools belong to-
Ans: Fungi

50. Match the following.

I. Bread & Beer	A) Morels & Truffles
II. Wheat Rust	B) Penicillium
III. Antibiotic	C) Yeast
IV. Delicacies	D) Puccinia

I II III IV
Ans: C D B A

51. The common character in the members of Plantae of 2 kingdom classification-
Ans: Presence of Cell wall

52. Match the following-

I) Cellulose	A) Fungi
II) Chitin	B) Blue green algae
III) Prokaryotic	C) Plants
IV) Multicellular Prokaryotic	D) Bacteria

The correct matching is
I II III IV
Ans: C A D B

53. The members of the first 3 kingdoms of Whittaker's classification belong to the following of Linnaeus-
Ans: Plantae & Animalia

54. Unicellular prokaryotic and eukaryotic micro organisms are kept under the following kingdoms of Whittaker's respectively-
1) Eubacteria, Protista 2) Arechaebacteria, Chrysophyta
3) Monera, Protista 4) Monera, Euglenoids
Ans: 3 (Monera, Protista)

55. White spots on Mustard leaves are caused by-
Ans: Albugo

56. Assertion (A): All fungi are filamentous.
Reason (R): Yeast is unicellular.
Ans: A is false. R is correct.

57. Assertion (A): Gamets are absent in Basidiomycetes.
Reason (R): Sexual reproduction occurs by basidiospores.
Ans: A, R are correct. R do not explains A.

58. Find out the odd one belonging to Fungi
1) Zoospores 2) Conidia
3) Sporangiospores 4) Ascospores
Ans: 4 (Ascospores)

59. Match the following-

I) Plasmodium	A) Thousands of cilia
II) Entamoeba	B) Several feet length
III) Paramecium	C) Sleeping sickness
IV) Trypanosoma	D) Parasite

The correct matching is
I II III IV
Ans: B D A C

60. Fruiting bodies are formed by-
Ans: Slime moulds

61. Match the following-

I) Silica shell	A) Plasmodium
II) Silicated cell wall	B) Amoeboid protozoans
III) Spore with true wall	C) Protozoans
IV) No cell wall	D) Diatoms
	E) Slime moulds

The correct matching is
I II III IV
Ans: B D E C

62. Protoplasm is surrounded by plasma membrane only but not Cell wall in-
1) Slime moulds 2) Protozoans
3) 1 & 2 4) Dinoflagellates
Ans: 3 (Slime moulds, Protozoans)

63. Cell wall is replaced by protein rich layer called pellicle in-
Ans: Euglenoids

64. Match the following-

I) Noctiluca	A) Euglena
II) Gonyaulax	B) Rejuvenatory spores
III) Auxospore	C) Diatoms
IV) Palmella	D) Bioluminescence
	E) Red tides in Mediterranean sea

The correct matching is
I II III IV
Ans: D E C A

65. Find out mis-match-
1) Red dinoflagellate - Gonyaulax
2) Marine dinoflagellate - Noctiluca
3) Chief producers in Ocean - Diatoms
4) Dinoflagellates - Histone rich chromosomes
Ans: 4 (Dinoflagellates - Histone rich chromosomes)

66. Assertion (A): Dinoflagellates are called whirling whips.
Reason (R): Their flagella produce spinning movements.
Ans: A, R are correct. R explains A.

67. Dinoflagellates differ from other protists in the following characters-
A. Cell wall has stiff cellulose plates B. Protein rich layer called pellicle is present
C. Nucleus is called mesokaryon D. Variously coloured
E. Cell walls are indestructible
Ans: A C D

68. Assertion (A): The nucleus of Dinoflagellates is called Mesokaryon.
Reason (R): Chromosomes show condensation even in interphase.
Ans: A, R are correct. R explains A.

69. Match the following-

I) No Cell wall but PellicleI) No Cell wall but Pellicle	A) Diatoms
II) Cell wall diveded into 2 shells	B) Marine amoeboid protozoa
III) Silica shell	C) Euglena
IV) Cell wall with cellulosic plates	D) Dinoflagellates
	E) Chryophytes

I II III IV
Ans: C A B D

70. Find out the correct statement-
1) Fruiting bodies in slime moulds enclose spores.
2) The life cycle of slime moulds never show Cell wall.
3) Euglena body is rigid due to pellicle.
4) All are incorrect
Ans: 4 (All are incorrect)

71. Desmids are also called-
Ans: Golden algae

72. Chromosomes without histones are present in-
Ans: Whirling whips

73. One of the following is toxic-
1) Noctiluca 2) Euglena
3) Gonyaulax 4) Diatoms
Ans: 3 (Gonyaulcax)

74. Match the following.

I) Radially symmetrical	A) Diatoms
II) Longitudinal binary fission	B) Pennales
III) Binary fission	C) Euglena
IV) Bilaterally symmetrical	D) Centrales

I II III IV
Ans: D C A B

75. Soap box like cells are present in-
Ans: Diatoms

76. The age of Kieselguhr is (in years)-
Ans: Billions

77. Smallest, obligatory anaerobic cells belong to
Ans: Mycoplasmas

78. One of the following is not a related disease to Mycoplasmas-
1) Cattle - Pleuropneumonia 2) Plants - Witches broom
3) Man - Mycoplasmal urethritis 4) Woman - Mycoplasmal urethritis
Ans: 2 (Plants - Witches broom)

79. Mycoplasmas were first called as-
Ans: PPLO

80. Match the following Cell wall substances-

I) Mycolic acid	A) Chrysophytes
II) Peptidoglycan	B) Archaebacteria
III) Pseudomurein	C) Dinoflagellates
IV) Cellulose plates	D) Bacteria
V) Silica	E) Actinomycetes

The correct matching is
I II III IV V
Ans: E D B C A

81. Genetic material is naked in-
A. Virus B. Bacteria C. Blue green algae
Ans: ABC

82. Most extensive metabolic diversity is observed in-
Ans: Bacteria

83. Assertion (A): All the bacteria are heterotrophic.
Reason (R): Some bacteria are saprophytes.
Ans: A is false. R is correct.

84. The following play a greater role in recycling nutrients like N, P & S-
1) Bacteria 2) Chemoautotrophs
3) Archaebacteria 4) Cyanobacteria
Ans: 2 (Chemoautotrophs)

85. Large number of Deuteromycetes are-
Ans: Decomposers

86. Majority of the bacteria are-
Ans: Decomposers

87. Assertion (A): Kieselguhr is useful in polishing, filtration of oils and syrups.
Reason (R): It is gritty.
Ans: A, R are correct. R explains A.

88. Asexual reproduction in Bacteria occurs by-
Ans: Endospores

89. One of the following is not a member of Actinomycetes.
1) Mycobacterium 2) Streptomyces
3) E.Coli 4) Corynebacterium
Ans: 3 (E.Coli)

90. Find out the incorrect one-
1) Fragments of blue green algae are called Hormogonia.
2) Nitrogen fixing blue green algae have specialised cells called Heterocysts.
3) The filament in blue green algae are surrounded by gelatinous sheath & generally called as Trichomes.
4) Some members of blue green algae are branched.
Ans: 4 (Some members of blue green algae are branched.)

91. Blooms in polluted water bodies are formed by-
Ans: Cyanobacteria

92. Asexual reproduction in cyanobacteria occurs by-
Ans: Akinetes

93. The red colour of the red sea is due to
Ans: Trichodesmium

94. The most primitive and first oxygenic photosynthetic organisms are-
Ans: Blue green algae

95. Assertion (A): Archae bacteria live in most harsh habitats.
Reason (R): The cell membrane in them have branched chain pseudomurein.
Ans: A, R are correct. R explains A.

96. Find incorrect statement-
1) Some cyanobacteria are terrestrial
2) Chl-a is present in cyanobacteria
3) Cell membrane in Archaebacteria has branched chain lipids
4) Flagellated bodies are present in cyanobacteria
Ans: 4 (Flagellated bodies are present in cyanobacteria)

97. Scientific basis of classification was first attempted by-
Ans: Aristotle

98. The fruiting bodies of Ascomycetes can be identified basing on-
Ans: Shape

99. Match the following-

I) One flagellum longitudinal and one flagellum transverse.	A) Trypanosoma
II) One flagellum long and one flagellum short.	B) Chrysophytes
III) Thousands of cilia.	C) Dinoflagellate
IV) Neither flagella nor cilia.	D) Euglenoids
	E) Paramecium

The correct matching is
I II III IV
Ans: C D E B

100. Archaebacteria do not show following habitat-
1) Hot springs - Thermo acidophiles 2) Extremely salty areas - Halophiles
3) Salty areas - Thermohalophiles 4) Marshy areas - Methanogens
Ans: 3 (Salty areas - Thermohalophiles)

101. The members of Eubacteria are the following except
1) Bacteria, Cyanobacteria 2) Protozoans
3) Actinomycetes 4) Mycoplasmas
Ans: 2 (Protozoans)

Posted Date : 30-11-2020

ENZYMES

1. Assertion (A): Almost all enzymes are proteins.
Reason (R): Most of the enzymes are polymers of amino acids.
A: A and R are correct. R explains A.

2. First name of enzyme
A: Ferments

3. The term enzyme was coined by
A: Willy Kuhne

4. Match the following.

List - I	List - II
I) Enzyme	A) First discovered enzyme
II) Precursor of enzyme	B) Louis Pasteur
III) Zymase	C) Zymogen
IV) Ferments	D) Protein

Correct matching is
I II III IV
A: D C A B

5. The term ferments was coined by
A: Louis Pasteur

6. An enzyme has
1) Primary structure 2) Tertiary structure

3) Secondary structure 4) All
A: 4(All)

7. The similarity between enzymes and inorganic catalysts is that both are
A: Concerned with bringing change in the velocity of a chemical reaction without under
going any permanent change

8. Assertion (A): Enzymes are universal.
Reason (R): They are present in all organisms which show sexuality.
A: A and R are correct. R explains A.

9. The definition of enzyme was given by
A: Dixon, Web

10. The source of first enzyme discovered is
A: Yeast

11. Assertion (A): All enzymes are proteins.
Reason (R): Nucleic acid like 23s rRNA can act as enzyme.
A: A is false. R is correct.

12. Assertion (A): Enzyme is a polymer of amino acids.
Reason (R): Chemically enzymes are proteins.
A: A and R are correct. R explains A.

13. Find out mismatch.
1) First essay on enzyme - J.B.S. Haldane
2) First enzyme crystallised - Urease
3) First enzyme discovered - Zymase
4) First name of the enzyme - Zymogen
A: 4(First name of the enzyme - Zymogen)

14. Match the following.

List - I	List - II
I) Enzyme	A) Polymer of Amino acids
II) Ribozyme	B) Chromogenic substrate
III) Taq polymerase	C) RNA
IV) β − Galactosidase	D) Thermostable

Correct matching is
I II III IV
A: A C D B

15. A functional enzyme shows
A: Tertiary structure

16. Generally enzyme gets damaged at this temperature
A: Above 40^oC

17. Read the following statements and find out incorrect one.
1) Enzymes are hydrophilic colloidals.
2) Water becomes vapour. It is a physical reaction.
3) When bonds are broken and new bonds are formed, it is a chemical reaction.
4) None
A: 4(None)

18. Number of carbonic acid molecules formed from water and CO₂ in the presence and absence
of carbonic anhydrase respectively are
A: 21,60,000, 200 per hour

19. Substrate + Enzyme give
A: Product + Enzyme

20. Substrate occupies into the active site of enzyme by the following process.
A: Diffusion

21. Read the following statements and find out incorrect one.
1) Acleft or pocket in the 3-D structure of a catalytic protein into which substrate fits is
called as active site.
2) A protein is an enzyme.
3) Enzymes are macromolecules.
4) Enzymes are colloidal.
A: 2(A protein is an enzyme.)

22. Read the following statements.
1) Same substrate can undergo different types of reactions and each catalysed by a
separate enzyme.
2) The last step in enzyme action is the release of product from enzyme.
3) Between substrate and product unstable intermediate structural states are formed.
4) Enzyme reduce the energy for biochemical reaction.
A: 4(Enzyme reduce the energy for biochemical reaction.)

23. If E = Enzyme, S = Substrate and P = Product. Choose the following equation for action
of enzyme on the substrate to form a product
A: E + P → [ES] → [EP] → E + S

24. Formation of (ES) complex was first explained by
A: Emil Fisher

25. Geometry of the enzyme is altered in the hypothesis of
A: Daniel E. Koshland

26. Tertiary structure of enzyme is altered by the factors like (A) substrate concentration
(B) pH (C) temperature. The following graph is suitable to X = ?
A: BC

27. Assertion (A): High temperature destroys enzymatic activity.
Reason (R): Low temperature preserves the enzyme in a temporarily inactive state.
A: A and R are correct. R do not explains A.

28. Km value is
1)
2) Approximate inverse measures of affinity of the enzyme for a given substrate
3) Michaelis - Menten constant
4) All
A: 4(All)

29. Malonate is similar to
A: Succinic acid

30. Competitive inhibitors which occupy active site on enzyme surface
A: Competitive inhibitors

31. Physically enzymes are
A: Colloidals

32. Assertion (A): Enzymes are heat sensitive.
Reason (R): They are colloidals.
A: A and R are correct. R explains A.

33. Assertion (A): Enzymes are almost inactive at lower temperature.
Reason (R): They are denatured at lower temperature.
A: A is correct. R is false.

34. Protein nature of the enzyme was first established by
A: Sumner

35. Match the following.

List − I	List − II
I) Optimum temperature	A) 25 − 30^oC
II) Low temperature	B) Denatured
III) High temperature	C) Sensitive
IV) Heat	D) Less active

A B C D
A: I III IV II

36. Assertion (A): Enzymes speed up the rate of a reaction.
Reason (R): Enzymes change the equilibrium of a reaction.
A: A is correct. R is false.

37. Assertion (A): Food kept in refrigerator is not spoiled.
Reason (R): Very cold temperature keeps the enzyme in a permanently inactive state
A: A is correct. R is false.

38. Read the following statements and find out incorrect one.
1) Enzymes are specific in action.
2) Generally enzymes are polymers.
3) Most of the enzymes work at neutral pH.
4) All enzymes are reversible in their action.
A: 4(All enzymes are reversible in their action.)

39. Co-enzymes are
1) metals 2) small organic molecules
3) small inorganic molecules 4) large inorganic molecules
A: small organic molecules

40. The following are regenerated
A: Coenzyme

41. Enzymes are
A: hydrophilic

42. The cell free enzyme that participates in fermentation is
A: Zymase

43. Match the following.

List − I	List − II
I) Haem	A) Carboxylase
II) Niacin	B) Catalase
III) Zinc	C) IAA Oxidase
IV) Mn	D) NADP
	E) Peptidase

The correct matching is
I II III IV
A: B D A C

44. Match the following.

List − I	List − II
I) Peroxidase	A) Molybdenum
II) Carboxy peptidase	B) Haem
III) Urease	C) Nickel
IV) Nitrogenase	D) Zinc
	E) Copper

The correct matching is
I II III IV
A: B D C A

45. Metal ions of the enzymes are concerned with
A. To form co-ordination bonds with side chains at the active site.
B. To form one or more co-ordination bonds with the substrate.
1) Neither A nor B 2) B 3) A 4) A and B
A: A and B

46. Protein part of the holoenzyme is called
A: Apoenzyme

47. Assertion (A): Enzymes enhance the rate of the reaction.
Reason (R): Enzymes lower the activation energy.
A: A and R are correct. R do not explains A

48. Match the following.

List − I	List − II
I) Cofactor	A) Polymer of amino acids
II) Coenzyme	B) Non protein constituent of holoenzyme
III) Apoenzyme	C) Organic compound tightly bound to apoenzyme
IV) Prosthetic	D) Organic compounds transiently associated with Apoenzyme.

Correct matching is
I II III IV
A: B D A C

49. Assertion (A): Coenzymes are cofactors.
Reason (R): All cofactors are not prosthetic groups.
A: A and R are correct. R do not explains A.

50. Prosthetic group is a
A: Part of active site of holoenzyme

51. Turn Over Number (TON) indicates
Ans: Number of substrate molecules converted to products.

52. Turn Over Number indicates
Ans: nature of substrate

53. To calculate the turn over number of an enzyme the following data is not necessary
1) Number of products 2) Number of substrate molecules
3) Time taken 4) Moles of enzymes
Ans: Number of products

54. The following is a vitamin
1) NADP 2) FAD 3) NAD 4)Niacin
Ans: Niacin

55. The E.C. of Glucose-6-Phosphotransferase is
Ans: 2.7.1.2

56. The first number in E.C. must be
Ans: less than 7

57. The second number in E.C. is always
1) above 3 2) less than 13 3) less than 14 4) 1, 3
Ans: 4 (1, 3)

58. Enzyme code is
1) 4 digit number 2) Decided by I.U.B. 3) Permanent 4)All
Ans: 4 (All)

59. Read the following statements and find out incorrect one.
1) All the enzymes discovered so far are divided into 6 classes.
2) IUB means International Union of Biochemistry.
3) Enzyme classification is based on the type of chemical reactions they catalyse.
4) Competitive inhibitor closely resembles the substrate in its molecular structure.
Ans: 4 (All the enzymes discovered so far are divided into 6 classes.)

60. Malonate resembles
Ans: Succinate

61. Assertion (A): Catalytic activity is lost when cofactor is removed from the enzyme.
Reason (R): Cofactor is non-protein part of the enzyme.
Ans: A and R are correct. R do not explains A.

Posted Date : 30-11-2020

STRATEGIES FOR ENHANCEMENT IN FOOD PRODUCTION

1. Which of the following is true regarding M.S. Swaminathan?
A. He developed expertise in genetics and plant breeding.
B. He developed scented Basmati rice
C. He is the initiator of Lab-to-Land and Food Security programme
D. He has been honoured Bharat Ratna
A: ABC

2. M.S. Swaminathan is closely associated with which of the following institutes?
1) ICRISAT 2) NBRI 3) CFTRI 4) IRRI
A: 4 (IRRI)

3. The renowned person responsible for introducing crop cafeteria, crop scheduling and genetic improvement of yield and quality of crops in India is
A: M.S. Swaminathan

4. The recorded evidence of plant breeding dates back to
A: 9000-11000 years

5. The most important root for any breeding programme is
A: Genetic variability

6. The entire collection of plants or seeds having all the diverse alleles for all genes in a given crop is called
A: Germplasm collection

7. Arrange the following in proper order in breeding programme
A. Cross hybridization
B. Collection of variability
C. Commercialization of cultivars
D. Evaluation and selection of parents
A: BDAC

8. The percentage of agriculture in India’s Gross Domestic Product (GDP) is
A: 33%

9. Nearly 62% of Indian Population is employed in
A: Agriculture

10. During the period 1960-2000 wheat production increased from 11 million tonnes to
A: 75 million tonnes

11. During the period 1960-2000 Rice production increased from 35 million tonnes to
A: 889.5 million tonnes

12. The increase in wheat and rice production is mainly due to development of
A: Semi-dwarf varieties in wheat and rice

13. IRRI (International Rice Research Institute) is located at
A: Philippines

14. Sonalika and Kalyan Sona are high yielding varieties of
A: Wheat

15. International center for wheat and maize improvement is located at
A: Mexico

16. Which of the following varieties of rice is developed at IRRI?
1) IR-8 2) Sonalika
3) Kalyan Sona 4) Jaya and Sona
A: 1 (IR-8)

17. Taichung Native-1 of Rice is developed from
A: Taiwan

18. Assertion (A): Sachaarum officinarum has thin stems and high sugar content.
Reason (R): It can be successfully crossed with Sachaarum barberi which grows in North India.
A: A is false and R is true.

19. Which of the following are better yielding semi-dwarf varieties of rice developed in India?
1) IR-8, Sonalika 2) Jaya, Ratna
3) Sonalika, Kalyansona 4) Kalyansona, Jaya
A: 2 (Jaya, Ratna)

20. Which of the following statements regarding Sachaarum barberi is true?
A. It is grown in South India
B. It is high yielding
C. It has high sugar content
A: None

21. Which of these is not a fungal disease?
1) Red rot of sugar cane 2) Black rot of crucifer
3) Brown rust of wheat 4) Late blight of potato
A: 2 (Black rot of crucifer)

22. Late blight of potato is caused by a
A: Fungus

23. Match the following variety with its resistance to disease.

A: cabd

24. Pusa Sadabahar is a leaf curl resistant variety of
A: Chilli

25. Which of the following is a variety of wheat?
1) Pusa Swarnim 2) Pusa Subhra
3) Pusa Komal 4) None of these
A: 4 (None of these)

26. Leaf and stripe rust resistant wheat variety is
A: Himgiri

27. Parbhani Kranti is a variety of Bhendi (Abelmoschus esculentus) is resistant to
A: Yellow mosaic virus

28. Which of the following makes maize resistant to stem borer?
1) High aspartic acid 2) Low nitrogen content
3) High sugar content in maize 4) All of the above
A: 4 (All of the above)

29. Which of the following statements is/are false regarding Boll worms?
A. They are attracted by smooth leaves
B. They do not prefer nekton cotton varieties
C. They are attracted by solid stems
A: BC

30. Assertion (A): Sawfly does not feed on solid stems of wheat.
Reason (R): Insect resistance in crop plants may be due to morphological, bio-chemical or physiological characters.
A: Both A and R are true and R is the correct explanation for A

31. Which of the following are varieties of Okra (Bhendi)?
1) Pusa sem 2 and Pusa sem 3
2) Pusa Gaurau and Pusa sem 3
3) Pusa Sawani and Parbhani Kranti
4) Parbhani Kranti and Pusa Gaurau
A: 3 (Pusa Sawani and Parbhani Kranti)

32. Fill up the blanks.

A: Aphids, Pusa Sem 2, Okra

33. Breeding crops for high vitamin, mineral, protein or fat contents to improve public health is called
A: Bio fortification

34. Golden rice is rich in
A: Β-carotene

35. Which of the following IARI released crop is not enriched with vitamin A?
1) Carrot 2) Spinach 3) Bathua 4) Pumpkin
A: 3 (Bathua)

36. IARI released bitter gourd is enriched with
A: Vitamin C

37. Which of the following IARI varieties are enriched with Iron and calcium
1) Spinach and Bathua 2) Carrot and Pumpkin
3) Carrot and French bean 4) Spinach and Garden pea
A: 1 (Spinach and Bathua)

38. Approximately how much grain content is required to produce one kg of meat?
A: 3-10 kgs

39. Which of the following alga is most suitable for single cell protein production?
1) Chlamydomonas 2) Spirulina
3) Methylophilus 4) Chaetomium
A: 2 (Spirulina)

40. Which of the following fungi is not suitable for single cell production?
1) Mucor 2) Candida
3) Saccharomyces 4) Chaetomium
A: 1 (Mucor)

41. The conditions required for autoclaving tissue culture medium are
A: 15 min 121ºC 15 lbs.

42. The virus free part of a plant used for tissue culture is
A: Apical meristem

43. Pomato is produced by
A: Somatic hybridisation

44. Which of the following are true regarding somatic hybrids?
A. Two isolated protoplasts from different varieties can be fused.

B. It can be done with plants which show physical or chemical incompatibility under normal sexual crosses.
C. It facilitates direct transfer of cytoplasmic or nuclear genomes to plant cells.
D. Desired combination of characters is always possible with somatic hybrids.
A: ABC

Posted Date : 30-11-2020

MODES OF REPRODUCTION & SEXUAL REPRODUCTION IN FLOWERING PLANTS

1. One of the following method of reproduction is slow-
1) Vegetative 2) Asexual 3) Sexual 4) 2, 3
A: 3 (Sexual)

2. The significance of reproduction is-
A: Continuity of species

3. Read the following statements and find out incorrect one-
1) Reproduction always involves 2 plants or parents
2) Reproduction always involves gametic union
3) Reproduction always requires for the union of gamets
4) All are correct
A: 4 (All are correct)

4. Asexual reproduction is common in the following-
1) Organisms with complex organisation
2) Organisms must contain vascular tissues
3) Organisms with simple organisation
4) Highly advanced plants
A: 3 (Organisms with simple organisation)

5. Gamets are not involved in-
A: Asexual and vegetative reproduction

6. The method of reproduction depends on
A: Physiology of organism

7. One of the following is an ephimeral-
1) Wolffia 2) Palm 3) Banana 4) Pinus
A: 1 (Wolffia)

8. Offspring exactly resembles gentically and morphologically in asexual or vegetative reproduction because-
A: It involves mitosis only

9. Clones are formed as a result of-
1) Sexual reproduction 2) Asexual reproduction
3) Vegetative reproduction 4) 2, 3
A: 4 (Asexual reproduction, Vegetative reproduction)

10. Assertion (A): Vegetative propagules form clones.
Reason (R): They do not show meiosis.
A: A & R are correct. R explains A.

11. Spore formation is exogenous-
A: Pencillium

12. Arrange the following plants in ascending order basing on their span of life-
A) Banana B) Pinus C) Wolffia

D) Carrot E) Rose F) Rice
A: C F D A E B

13. Differentiate a spore from zygote-
A: Former undergoes mitosis, later shows either mitosis or meisosis

14. The benefit of formation of spores is-
A: These are adopted for dispersal and survive for extended periods of time in unfavourable conditions

15. Gametogenesis means
A: Formation of gametes from their precursor cells

16. Assertion (A): vegetative is also a type of asexual reproduction.
Reason (R): Vegetative reproction do not involves gametic union.
A: A & R are correct. R explains A.

17. Formation of seeds without fertilisations is called-
A: Apomixis

18. Formation of fruit without fertilisation from the ovary is called-
A: Parthenocarpy

19. Parthenocarpy do not requires-
1) Pollination 2) Microsporogenesis 3) Megasporogenesis 4) All
A: 4 (All)

20. Apomixis is-
A) Mimics sexual reproduction B) A kind of asexual reproduction
C) A kind of hybridisation D) A kind of vegetative reproduction
A: A B

21. Citrus shows-
A: Polyembryony

22. Match the following

I II III IV
A: C D A B

23. In Citrus or Mango additional embryos are formed in the-
A: embroyosac

24. Mango shows
A) True embryo in the embryosac
B) Many embryos in nucellus
C) Many embryos in embrysoac along with true embryo
D) Many embryos in the fruit
A: A C

25. Match the following.

I II III IV
A: D C A B

26. Assertion (A): Apomixis is an assured reprodcution.
Reason (R): It can take place in the absence of pollinators such as in extreme conditions.
A: A & R are correct. R explains A.

27. Arrange the steps in a sequence that takes place during the formation of embryo by apomixis. Discard the events not required
a) Zygote b) Megaspore mother cell c) Meiosis
d) Diploid egg e) No meiosis f) Mitosis
g) Male gametes h) Embryo
A: B E D F H

28. Assertion (A): The hybrid seeds have to produced every year.
Reason (R): If the seeds collected from hybrids are sown, the plants in the progeny will segregate and do not maintain hybrid characters.
A: A & R are correct. R explains A.

29. Assertion (A): If the hybrids are made into apomicts, the farmer does not have to buy hybrid seeds every year.
Reason (R): There is no segregation of characters in the apomict made hybrid progeny.
A: A & R are correct. R explains A.

30. Pollination and fertilisation are not required in
A) Apomixis B) Parthenocarpy C) Polyembryony
A: A B

31. Assertion (A): A tiny seed can produce a large biomass.
Reason (R): Ficus plant produced from a tiny seed can produce billions of seeds.
Ans: A & R are correct. R explains A.

32. A complete root parasite that produces a friut with thousands of tiny seeds.
Ans: Orobanche

33. A partial root parasite that produce a fruit with thousands of seeds.
Ans: Striga

34. Name the plant whose viable seeds were found during the archeological excavations at king Herod's palace near the Dead sea whose age is 2000 years.
Ans: Phoenix

35. Dormancy period of lupinus-
Ans: 10,000 years

36. One of the angiospermic plant is microscopic-
1) Lemna 2) Oryza 3) Hydrilla 4) Wolffia
Ans: 4 (Wolffia)

37. In mangrooves seeds germinate while still attached to the mother plant. It is known as-
Ans: Vivipary

38. A zygote undergoes the following events and produces an organism
a) Cell division b) Cell differentiation c) Modifications
Ans: A B C

39. Organisms of one genaration and the next are linked by the following for the continuity of species.
1) Meiosis 2) Mitosis 3) Gametes 4) Zygote
Ans: 4 (Zygote)

40. The fate of zygote depends up on-
Ans: Life cycle of organism

41. Match the following-

I II III IV
Ans: C A D B

42. Internal fertilisation involves the following characters
A) Male gametes when non motile are carried to female gamete by pollen tube
B) Large number of male gametes produced
C) Male gametes may be motile to reach egg cell
D) Significant reduction in the number of egg cells
E) Reduction in the number of female plants
Ans: A B C D

43. Every sexually reproducing organ begins life as-
Ans: Zygote

44. In Rockweed meiosis occurs in-
Ans: During the formation of gametes

45. Meiocytes are the gamete mother cells in-
Ans: Vascular plants

46. Pollengrain germinates on the-
Ans: Stigma

47. Parthenogenesis means-
Ans: Formation of embryo from egg cell without fertilisation

48. Assertion (A): Male gametes are produced in large numbers when compared to the female gametes.
Reason (R): A large number of male gametes fail to reach the female gametes.
Ans: A & R are correct. R explains A.

49. If a species produces male and female gametes it is called-
Ans: Heterogametic

50. A plant which produces gametes is
Ans: Haploid or diploid

51. Match the following-

I II III IV
Ans: C D B A

52. If male and female flowers are produced on different branches of the same plant it is called-
Ans: Monoecious

53. Find incorrect match-
1) Monoecious green alga - Chara 2) Dioecious kelp - Marchantia
3) Monoecious monocot - Maize 4) Dioecious angiosperm - Papaya
Ans: 2(Dioecious kelp - Marchantia)

54. In Chara-
A) Male and female sex organs differ in size, shape and position
B) Male sex organ is called antheredium and female sex organ is called archegonium
C) Male and female sex organs are side by side
D) Male sex organ is above and female sex organ is below
Ans: A

55. Homothallic fungi, monoecious plants denote-
Ans: Bisexual condition

56. Heterothallic fungi and dioecious (plants) are the terms used to indicate-
Ans: Unisexual condition

57. Match the following-

I. Homogametes	A) Ferns
II. Heterogametes	B) Rhodophyceae
III. Isogametes	C) Cladophora
IV. Spermatia	D) Chlamydomonas

I II III IV
Ans: D A C B

58. Heterogametes of Fucus are-
Ans: Male gamete is smaller with 2 unequal flagella on lateral side, female gamete is larger and spherical

59. Cladophora produces gametes of-
Ans: equal size, shape, number of flagella and their position

60. Read the following statements and find out correct one-
1) Century plant produces flowers only once in its life span of 10-30 years
2) Bamboo produces flowers only once in its life span of 50-100 years
3) Neelakuranji produces flowers once in its 10 years life span
4) The colour of the flowers of strobilanthus is blue
Ans: 3 (Neelakuranji produces flowers once in its 10 years life span)

61. An angiospermic stamen has
1) Bilobed anther 2) Each life has 2 pollen sacs
3) Pollen sac is also called microsporangium 4) All
Ans: 4 (All)

62. Young anther has 4-
Ans: microsporangia

63. A mature stamen at the time of dehiscence has-
Ans: 2 microsporangia, 2 stomia

64. The stalk of the stamen is called-
Ans: Filament

65. Number of meiotic divisions in a stamen and ovule respectively-
Ans: many, one

66. Multi nucleate condition is seen in-
Ans: Tapetum

67. Nutritive structure of Anther wall-
Ans: Tapetum

68. Protection to the microsporangium is given by-
1) Epidermis 2) Tapetum 3) Middle layers 4) All
Ans: 4 (All)

69. Dehiscence is helped-
Ans: The layers that give protection

70. Nutritive tissue is placed between-
Ans: Middle layers and microsporangium

71. The thin region in between the pollen sacs-
Ans: Stomium

72. The layer that helps in dehiscence by its capacity to loose water and contract-
Ans: Endothecium

73. Microsporogenesis involves-
Ans: Meiosis

74. Find a mismatch-
1) Stamen = Microsporophyll
2) Microspore mother cell = Pollen mother cell
3) Microspore = Pollen grain
4) Pollen grain = Male gametophyte
Ans: 4 (Pollen grain = Male gametophyte)

75. Pollen grain measures-
Ans: 25 - 50 micrometers

76. Assertion (A): Sporopollenin is one of the most resistant organic chemical materials known.
Reason (R): No enzyme that degrades sporopollenin is known.
Ans: A & R are correct. R explains A.

77. Sporopollenin is present in-
Ans: Outer layer of the wall

78. Wall of the pollengrain is made of-
1) Pectin 2) Cellulose 3) Sporopollenin 4) All
Ans: 4 (All)

79. Assertion (A): Pollen grains are preserved as fossils.
Reason (R): They consist sporopollenin.
Ans: A & R are correct. R explains A.

80. Match the following-

I II III IV
Ans: D C B A

81. Germ pores are present in-
Ans: Exine

82. Cytoplasm of the vegetative cell consists-
1) Irregularly shaped nucleus 2) Small generative cell
3) 1, 2 4) Pollen grain
Ans: 3 (Irregularly shaped nucleus, Small generative cell)

83. Characters of Generative cell-
1) Small 2) Fusiform
3) Floats in the cytoplasm of vegetative cell 4) All
Ans: 4 (All)

84. Pollen grains in majority of the plants are liberated at-
Ans: 2-celled stage

85. Parthenium came to India along with-
Ans: Wheat

86. Assertion (A): Pollen grains increase the performance of horses and athletes.
Reason (R): They cause severe allergies bronchial afflictions, chronic respiratory diseases and pollen allergy.
Ans: A & R are correct. R do not explains A.

87. Assertion (A): Pollen tablets are used as food supplements.
Reason (R): They are rich in nutrients
Ans: A & R are correct. R explains A.

88. Pollen bank means-
Ans: Pollen grains preserved in liquid nitrogen

89. Arrange the layers in a sequence from centre to periphery-
A. Tapetum B. Middle layers C. Endothecium D. Epidermis
Ans: A B C D

90. Multicarpellary syncarpous and apocarpous gynoecium are seen respectively in-
Ans: Papaver, Michelia

91. Arrange the following plants in ascending order basing on the number of carpels.
A. Allium B. Pea C. Michelia D. Solanum
Ans: B D A C

92. Precursor of the male gametophytes-
Ans: Microspore

93. Pollen grains germinate-
Ans: on the stigma

94. Number of ovules in the ovary of orchid-
Ans: Many

95. Cell division is unequal during the following kind of vegetative reproduction-
1) Fragementation 2) Gemmae 3) Budding 4) Tissue Culture
Ans: 3 (Budding)

96. Read the following statements. Select a right one-
1) In monocots radicle and root cap is enclosed by well differentiated sheath called Coleorhiza
2) In monocots epicotyl has shoot apex and a few leaf primordia enclosed in solid foliar structure, the coleoptile
3) Perisperm is a part of albumin
4) Hilum and micropyle are in one line
Ans: 4(Hilum and micropyle are in one line)

97. Larger part of Monocot embryo is-
Ans: Scutellum

98. Shape of Scutellum-
Ans: shield

99. Membranous seed coat is found in-
Ans: Cereals

100. As the seed matures one of the following character is not seen-
1) Weight gradually increases
2) Embryo enters in a state of inactivity
3) Water content is reduced (10-15% by mass)
4) seeds become relatively dry
Ans: 1(Weight gradually increases)

101. Find a wrong pair (regarding dispersal).
1) Coconuts - Float 2) Martynia - Animal fur 3) Gua - Birds 4) Fig - Wind
Ans: 4 (Fig - Wind)

102. Measurements of Orchid seeds-
Ans: 85 micrometers

103. Number of ovules is least in the ovaries of-
1) Caryopsis 2) Cypsela 3) Nut & Drupe 4) All
Ans: 4 (All)

104. An ovule is-
1) Integumented ovary 2) Integumented megasporangium
3) Integumented nucellus 4) 1, 2
Ans: 4 (Integumented ovary, Integumented megasporangium)

105. Match the following-

I II III IV
Ans: C D A B

106. Find incorrect match-
1) Unitegmic - Helianthus (Gamopetalae) 2) Ategmic - Loranthus
3) Number of embryosacs in ovule - One 4) Gamopetalae, Monocots - Unitegmic
Ans: 4 (Gamopetalae, Monocots - Unitegmic)

107. Micropyle, Chalaza and funiculus are in one line-
Ans: Vertical, Orthotropous, Polygonum

108. The angle formed by the inverted ovule along side of the funiculus-
Ans: 180°

109. Micropyle lies close to the funiculus in-
Ans: Anatropous ovule

110. Embryosac is curved in-
Ans: Campylotropous ovule

111. Assertion (A): The embryosac of Campylotropous ovule is curved though body of the ovule is placed at 90° to the funiculus.
Reason (R): Body of the ovule bends in such a way that micropyle comes towards the funiculus.
Ans: A & R are correct. R explains A.

112. Arrange the ovules in ascending order based on the distance between micropyle and funiculus-
A. Campylotropous ovule (Fabaceae)
B. Orthotropous ovule (Polygonium)
C. Anatropous ovule (Asteraceae)
Ans: C A B

113. The mother cell of female gametophyte is-
Ans: Megaspore

114. Assertion (A): Female gametophyte of angiosperms is also called embryosac.
Reason(R): Female gametophyte becomes haploid endosperm in Gymnosperms.
Ans: A & R are correct. R do not explains A.

115. Arrange the following events in ascending order based on the number mentioned.
A. Number of cells in the egg apparatus.
B. Number of free nuclear mitotic divisions in a megaspore.
C. Number of vegetative cells in male gametophyte.
D. Number of polar nuclei in the embryosac.
E. Number of nuclei in the female gametophyte.
F. Types of cells in the embryosac.
Ans: CDAFBE

116. Number of male gametes in a male gametophyte is equal to-
A. Number of Polar nuclei.
B. Types of cells in the egg appartus.
C. Number of synergids.
D. Types of cells in a fully developed male gametophyte.
E. Number of integuments in the ovules of monocots and polypetalae.
F. Number of mitotic divisions in the pollen grain.
Ans: ABCDEF

117. Assertion (A): A typical angiosperm embryosac, at maturity, though 8-nucleate is 7-celled.
Reason (R): Central cell has 2-polar nuclei.
Ans: A & R are correct. R explains A.

118. Egg apparatus is present towards-
Ans: Micropyle

119. Egg apparatus and Antipodal cells-
A. equal in number of cells. B. equal in number of genomes.
C. equal in function. D. equal in position.
Ans: A B

120. Filiform apparatus is possessed by-
Ans: synergids

121. Pollen tubes are guided into the synergid by-
Ans: filiform apparatus

122. Pollination is direct in-
Ans: Plants bearing naked seeds

123. Match the following-

I II III IV
Ans: D C A B

124. The following plants show cleistogamy
1) Oxalis 2) Viola 3) Commelina 4) All
Ans: 4 (All)

125. Seed-set is assured even in the absense of pollinators in the plants showing-
Ans: Cleistogamy

126. Assertion (A): Geitenogamy is genetically similar to Autogamy
Reason (R): Pollengrains belong to the same flower.
Ans: A is correct. R is false.

127. Majority of the plants use the following agents for pollination-
1) Animals 2) Wind 3) Water 4) None
Ans: 1(Animals)

128. Assertion (A): Plants showing pollination by wind and water produce enormous amount of pollen.
Reason (R): Pollengrains coming in contact with the stigma is a chance factor.
Ans: A & R are correct. R explains A.

129. Read the following statements and find out incorrect one-
1) Majority of the plants use biotic agents for cross pollination
2) Pollination by wind is more common amongst abiotic pollinations
3) Wind pollination is quite common in grasses
4) Plants showing pollination by water are mostly dicots
Ans: 4 (Plants showing pollination by water are mostly dicots)

130. Assertion (A): The distribution of Bryophytes and Pteridophytes is limited.
Reason (R): They need water for the transportation of male gametes to the egg cell.
Ans: A & R are correct. R explains A.

131. Water hyacinth and water lilly are pollinated by insects or wind because-
Ans: Flowers emerge above the level of water

132. A marine hydrophyte showing hydrophily-
Ans: Zostera

133. Assertion (A): Vallisneria shows epihydrophily.
Reason (R): It is a submerged hydrophyte
Ans: A & R are correct. R do not explains A.

134. Pollengrains are long and ribbon like in-
Ans: Zostera

135. Wind and water pollinated flowers are-
1) Not very colourful 2) Do not produce nectar
3) Big in size & sweet smelling 4) 1, 2
Ans: 4 (Not very colourful, Do not produce nectar)

136. Assertion (A): In most of the water pollinated plants pollen grains are protected from wetting.
Reason (R): They are covered by mucilagenous sheath.
Ans: A & R are correct. R explains A.

137. Match the following pollinating agents-

I II III IV
Ans: B D A C

138. Pollination by reptiles like snakes is called-
Ans: Ophiophily

139. Among the animals the dominant pollinating agents are-
Ans: Insects

140. Among the insects the dominant pollinating are-
Ans: Bees

141. Match the following-

I II III IV
Ans: C D A B

142. Match the following-

I II III IV
Ans: C A B D

143. The following 2 organisms, one plant and one animal can not complete their life cycle without each other-
1) Squirrel - Guava 2) Lemur - Apple
3) Moth - Yucca 4) Tegeticula - Amorphophallus
Ans: 3 (Moth - Yucca)

144. Silky hairs represent the style and stigma in the following wind pollinated plant-
1) Sorghum 2) Zea 3) Pea 4) Pennisetum
Ans: 2 (Zea)

145. Read the following statements and find out-
1) Wind pollinated flowers often have a single ovule in each ovary
2) Feathery stigma easily traps air-borne pollen
3) Wind pollinated plants often possess well-exposed stamens
4) Wind pollination requires light and sticky pollen grains
Ans: 4 (Wind pollination requires light and sticky pollen grains)

146. Many insects consume pollen or the nectar without bringing about pollination. They are called-
Ans: Pollen/ Nectar Robbers

147. Inbreeding depression is caused due to-
Ans: Continuous autogamy

148. Match the following-

I II III IV
Ans: C D A B

149. The following one is a genetic mechanism to prevent autogamy-
1) Heterostyly 2) Protandry
3) Self-sterility 4) Herkogamy
Ans: 4 (Herkogamy)

150. The following condition gives an assured condition for cross pollination-
1) Unisexual flowers 2) Bisexual flowers
3) Achlamydous flowers 4) Chasmogamous flowers
Ans: 1 (Unisexual flowers)

151. The following plants are monoecious-
1) Zea 2) Castor 3) Musa 4) All
Ans: 4 (All)

152. The following plants are dioecious-
1) Carica 2) Borassus 3) Date Palm 4) All
Ans: 4 (All)

153. Dioecious condition prevents-
A. Autogamy B. Allogamy C. Xenogamy D. Geitenogamy
Ans: A D

154. Arrange the following events in a sequence. They take place after pollination-
A. Interactive B. Acceptance or Rejection C.Recognition
Ans: A C B

155. The dialogue between pistil and pollen takes place. It is-
Ans: Chemical process

156. Pollengrain falls on the stigma at 2-celled stage in 60% of the plants. The generative cell divides and produces 2 male gametes while the pollen tube is-
Ans: in the stigma

157. The ratio between pollen grain and male gametes is-
Ans: 1 : 2

158. Ratio between mitotic divisions in the pollengrain and male gametes produced is-
Ans: 1 : 1

159. Number of pollengrains required to produce a seed-
Ans: 1

160. Number of pollengrains and male gametes required for double fertilisation in the angiosperms-
Ans: 1, 2

161. Number of meiotic divisions and mitotic divisions required from male parent to accomplish double fertilisation in female-
Ans: 1, 2

162. The male gametes in angiosperms reach the egg cell in a pollen tube. This phenomenon started actually in-
Ans: Gymnosperms

163. If the male gametes are carried to the egg cell through pollen tube, it is called-
Ans: Siphonogamy

164. One pollengrain mother cell can fertilise-
Ans: 4 Ovules

165. Pollen pistil interaction include the following events-
1) From pollination to the germination of the pollengrain on the stigma
2) Deposition of the pollengrains on the stigma, germenation of the pollen grains, travelling of the pollen tube through the style and enter into the ovary
3) Entry of the pollen tubes into the ovule 4) 2, 3
Ans: 4 (Deposition of the pollengrains on the stigma, germenation of the pollen grains, travelling of the pollen tube through the style and enter into the ovary, Entry of the pollen tubes into the ovule)

166. How many pollengrains fall on the stigma, how many can germinate, how many pass through the style, how many enter into the ovary, ovule and embryosac respectively-
Ans: many, many, many, many, many, 1

167. Match the following-

Entry of the pollen tube into the ovule through-

I II III IV
Ans: A C D B

168. Nearest way to reach the egg apparatus of the embryosac
Ans: Porogamy

169. Embryosac is surrounded by a cell wall belonging to-
Ans: Megaspore

170. Egg apparatus in the embryosac is present always towards-
Ans: Micropyle

171. Read the following statements and find out incorrect one-
1) Often flowers of animal pollinated plants are specifically adapted for a particular species of animal
2) Pollengrains generally are sticky in animal pollinated flowers
3) The cells of Endothecium absorb water and swell to help in the dehiscence of the anther to release pollengrains
4) Pollengrains of cereals have viability of 30 minutes while loose of Leguminaceae, Rosaceae and Solanaceae maintain viability for months
Ans: 3 (The cells of Endothecium absorb water and swell to help in the dehiscence of the anther to release pollengrains)

172. Embryosac is nothing but enlarged-
Ans: Megaspore

173. Pollen grains can be germinated in the laboratory and it requires-
A. 10 ml of sugar solution. B. A drop of sugar solution of about 10%.
C. 15 - 30 mts D. 10 - 15 mts E. 30 - 45 mts
Ans: B D

174. Mother cell of the female gametophyte/ embryosac is-
Ans: Megaspore

175. Megaspore shows how many mitotic divisions to produce female gametophyte-
Ans: 7

176. Female gametophyte at one stage shows 8 nuclei. It is due to-
Ans: 7 - free nuclear mitotic divisions

177. The life cycle of an angiosperm is-
Ans: Diplohaplontic

178. Maximum number of cells in the male and female gametophytes of angiosperms respectively-
Ans: 3, 7

179. Number of spindle apparatti formed from megaspore mother cell to embryosac-
Ans: 3 + 7

180. Number of spindle apparatti formed in a megaspore to produce embryosac-
Ans: 7

181. Embryosac consists nuclei belonging to how many generations-
Ans: 3

182. The set of 3 cells present in the embryosac towards chalaza are called-
Ans: Antipodal cells

183. Artificial hybridisation requires
Ans: 2 flowers belonging to 2 parent plants

184. Emasculation means-
Ans: Removal of stamens from bisexual flowers in bud stage

185. Emasculation is followed by-
1) cross pollination 2) collection of pollen
3) searching for a male parent 4) bagging
Ans: 4 (bagging)

186. Bagging is done to-
Ans: Prevent unwanted pollination

187. Rebagging is necessary after-
Ans: Dusting the pollen on the stigma

188. This step is not necessary if the female parent bears unisexual flowers-
Ans: Emasculation

189. Arrange the following steps in a sequence-
A. Rebagging B. Dusting of pollen on stigma
C. Selection of parents D. Bagging
E. Emasculation F. Collection of pollen from selected male parent
Ans: CEDFBA

190. Male gametes are seen first in the following-
Ans: cytoplasm of synergid

191. Syngamy means the union-
Ans: Between the nucleus of male gamete and nucleus of egg cell

192. Triple fusion means union between-
Ans: one male gamete and central cell

193. Significant features of angiosperms-
1) Double fertilisation 2) Formation of endosperm after fertilisation
3) Pollination by animals and water 4) All
Ans: 4 (All)

194. Number of genomes of female parent participating in the formation of a seed-
Ans: 3

195. Number of genomes involved in double fertilisation-
Ans: 5

196. Number of genomes of male and female parents involved in double fertilisation respectively-
Ans: 2 + 3

197. To produce 'n' number of seeds in angiosperms how many meiotic divisions are required-
Ans:

198. To know the chromosome number in endosperm of angiosperms the principle applied is-
Ans:

199. PEN of an angiospermic plant consists
Ans: 2 female genomes, one male genome

200. Zygote an angiosperm consists-
Ans: Nucleus of male gamete, nucleus and cytoplasm of female gamete

201. The result of triple fusion is-
Ans: PEC

202. The triploid body of the seed is-
Ans: endosperm

203. The following occurs first or formed first-
1) Triple fusion, endosperm 2) Syngamy, embryo
3) PEC, embryo 4) Syngamy, endosperm
Ans: 4 (Syngamy, endosperm)

204. Match the following-

I II III IV
Ans: B C D A

205. Multinucleate stage is seen once in the development of the following-
1) Embryosac 2) Albumin 3) 1, 2 4) Embryo
Ans: 3 (Embryosac, Albumin )

206. Assertion (A): Most zygotes divide only after certain amount of endosperm is formed.
Reason (R): Developing embryo requires nutrition.
Ans: A & R are correct. R explains A.

207. Read the following statements. Find out a correct one-
1) The early stages of embryogeny in monocots and dicots are different
2) Black Pepper and Beet show perisperm
3) All seeds are endospermic
4) First leaf of Dolichos is called Scutellum
Ans: 2 (Black Pepper and Beet show perisperm)

208. The following structures are seen in monocots only-
1) coleorhiza 2) scutellum
3) coleoptile 4) All
Ans: 4 (All)

209. Arrange the stages in a sequence found during the embryogeny of Dicots
A. Heart shaped embryo B. Globular embryo
C. Mature embryo D. Pro embryo
Ans: DBAC

210. Find a wrong match-
1) Scutellum = First leaf of monocots or grasses 2) Endosperm = Albumin
3) Plumule = First apical bud of shoot 4) Root cap = Coleorhiza
Ans: 4 (Root cap = Coleorhiza)

Posted Date : 30-11-2020

MORPHOLOGY OF FLOWERING PLANTS

1. We must consider the distinctive characters and the general nature of palnts ...
This is the opening sentence of
A: Enquiry into plants

2. In fibrous root system, the roots originate from
A: base of the stem

3. Main functions of the root
A) Absorption of water B) Absorption of food
C) Storage of reserve food D) Anchorage
E) Synthesis of some hormones
A: A, C, D, E

4. The cells in this region divide repeatedly
A: Proximal to the root cap region

5. Function of root hairs
A) Absorption of water B) Absorption of minerals
C) Anchorage D) Absorption of food
A: A only (Absorption of water)

6. Differentiation and maturation of the cells take palce in
A: region of maturation

7. The following phenomenon can be seen in the region of elongation
1) Cell maturation 2) Cell elongation
3) Cell differentiation 4) 1 & 3
A: 2 (Cell elongation)

8. Root hairs are produced from the following
1) Region of elongation 2) Region of maturation
3) Region of meristematic activity 4) Root cap
A: 2 (Region of maturation)

9. Root hair region is region of
A: maturation

10. Match the following regarding food storage

I. Carrot	A. Tap root
II. Turnip	B. Fibrous roots
III. Sweet Potato	C. Tap root
IV. Asparagus	D. Adventitious roots

I II III IV
A: C A D B

11. Find out mis-match.
1) Brace roots - Stilt roots
2) Adventitious roots - Fibrous roots
3) Prop roots - Pillar roots
4) Root hair region - Region of maturation
A: 2 (Adventitious roots - Fibrous roots)

12. Roots are green in
A: Taeniophyllum

13. Match the following.

I. Pneumatophores	A. Viscum
II. Velamen roots	B. Vanda
III. Nodular roots	C. Rhizophore
IV. Haustoria	D. Arachis

I II III IV
A: C B D A

14. The following plants have green leaves but still they are grouped as parasites
1) Viscum - Vanda 2) Striga - Rafflesia
3) Viscum - Striga 4) Viscum - Cuscuta
A: 3 (Viscum - Striga)

15. Epiphytes have the following adv-roots
1) Velamen roots 2) Fibrous roots
3) Photosynthetic roots 4) Pneumatophores
A: 1 (Velamen roots)

16. Roots of the following enter into the xylem and absorb water and minerals.
1) Cuscuta 2) Rafflesia 3) Vanda 4) Striga
A: 4 (Striga)

17. Function of nodular roots is
A: N₂fixation

18. The adventitious roots of Avicennia are
A: Aerial, respiratory

19. Assertion (A): Viscum is a partial parasite.
Reason (R): Its haustoria enter into phloem of the host
A: A is false. R is correct.

20. Parasitic roots are called
A: Haustoria

21. Plants stealing water and minerals from other plants are called
A: Parasites

22. Pneumatophores are seen in
A: Mangrooves

23. Assertion (A): Rafflesia is a total parasite.
Reason (R): Its haustoria enter into the xylem and phloem of the host.
A: A, R are correct. R explains A.

24. Prop roots of Banyan and pneumatophores of Avicennia are
A: Hanging and erect

25. The following roots arise from lower nodes of the stem
1) Prop roots 2) Brace roots

3) Assimilatory roots 4) Velamen roots
A: 2 (Brace roots)

26. The plants which grow in swampy areas are called
A: Mangrooves

27. The following are not adventitious roots
1) Velamen roots 2) Brace roots

3) Prop roots 4) Nodular roots
A: 4 (Nodular roots)

28. The micro organism present in the modular roots
A: Only one species of Rhizobium

29. Stem is formed from
A: Plumule

30. Root may bear the following
1) nodes 2) leaves 3) flowers 4) None
A: 4 (None)

31. Nodes are separated from each other by
A: inter nodes

32. Assertion (A): Ginger is a stem.
Reason (R): It stores food.
A: A, R are correct. R do not explains A.

33. Assertion (A): Zaminkand is a underground stem.
Reason (R): It bears nodes, buds and adventitious roots.
A: A, R are correct. R explains A.

34. Assertion (A): Colocasia is an organ of perennation.
Reason (R): It is useful in vegetative reproduction.
A: A, R are correct. R do not explains A.

35. Assertion (A): Potato is an organ of perennation.
Reason (R): It can tide over conditions unfavourable for growth.
A: A, R are correct. R explains A.

36. Match the following.

I. Turmeric	A. Eyes
II. Potato	B. Corm
III. Onion	C. Rhizome
IV. Zaminkand	D. Bulb

I II III IV
A: C A D B

37. Potato bears the following distinct characters by which it can be differentiated from others under ground stem modifications
A) Eyes B) Scale leaves
C) Absense of adventitious roots D) Presense of adventitious roots
E) storage of food
A: AC

38. Fleshy scale leaves are present in
A: Bulb

39. Adventitious roots are absent in
A: Tuber

40. Stem do not stores food in
A: Bulb

41. Scape is present in
Ans: Bulb

42. The distinct characters of Bulb
1) Tunic 2) Scape 3) Fleshy scale leaves 4) All
Ans: 4 (All)

43. Apical bud and axillary bud are side by side in
Ans: Bulb

44. Food is stored in the following parts of bulb
1) Stem 2) Scale leaf 3) Branch 4) Root
Ans: Scale leaf

45. One plant shows many under ground stems storing food in
1) Solanum tuberosum 2) Zingiber 3) Allium cepa 4) Amorphophallus
Ans: Solanum tuberosum

46. The tendril in cucurbita is the modification of
Ans: stem

47. Axillary bud is modified into tendril in
Ans: Cucumber

48. All bud modifications are modifications of
Ans: stem

49. Axillary bud and terminal bud are modified into tendrils respectively in
Ans: Cucumber, Grape vine

50. Aerial stem modifications meant for photosynthesis
A) Tendrils B) Phylloclade C) Thorns D) Cladode or Cladophyll
Ans: B D

51. Match the following modification

I. Needle like	A. Euphorbia
II. Flattened	B. Asparagus
III. Cylindrical	C. Casurina
IV. Branch of Limitted growth	D. Opuntia

I II III IV
Ans: C D A B

52. Thorns are present in
1) Citrus 2) Bougainvillea 3) Opuntia 4) 1 & 2
Ans: 4 (Citrus & Bougainvillea)

53. Phylloclades are seen in the plants growing in-
Ans: Arid regions

54. Match the following parts useful in veg reproduction

I. Agave	A. Underground stem
II. Diascorea	B. Floral bud
III. Grasses	C. Lateral Slender branch
IV. Nerium	D. Vegetative bud

I II III IV
Ans: B D A C

55. Stem is discoid in
1) Allium 2) Pistia 3) Eichornea 4) All
Ans: 4 (All)

56. Find out correct pair regarding stem modifications
1) Strawberry - Jasmine 2) Grasses - Nerium
3) Pistia - Oxalis 4) Grasses - Strawberry
Ans: 4 (Grasses - Strawberry)

57. Find out incorrect match regarding stem modifications for vegetative reproduction
1) Oxalis, Grasses, Strawberry 2) Pineapple, Chrysanthemum, Banana
3) Pistia, Eichornea, Oxalis 4) Nerium, Jasmine
Ans: 3 (Pistia, Eichornea, Oxalis)

58. Rosette leaves are present in
Ans: Eichornea, Pistia

59. Underground stems are useful in vegetaive reproduction when older parts die
Ans: Grasses, Strawberry

60. Balancing roots are present in
Ans: Offset

61. A lateral branch of one internode length and useful in vegetative reproduction is called
Ans: Offset

62. A lateral branch grows obliquely down words and upwards respectively in the following pair of plants and useful in vegetative reproduction
1) Strawberry, Nerium 2) Jasmine, Banana
3) Pineapple, Jasmine 4) Pineapple, Banana
Ans: 2 (Jasmine, Banana)

63. Tap root system and adventitious root system are present in respectively
Ans: Nerium, Chrysanthemum

64. Stem is spongy, petiole is swollen and balancing roots with root pocket are seen in
Ans: Eichornea

65. Axillary bud is modified into bulbil in
Ans: Diascorea

66. Lateral organ of the plant
Ans: leaf

67. Petiole is swollen in
Ans: Eichornea

68. Pulvinus leaf base is seen in
Ans: Legumes

69. The incisions of the lamina are called
1) Leaf 2) Lobes 3) Leaflets 4) 2, 3
Ans: 4 (Lobes, Leaflets)

70. Common character between simple and compound leaf is the presence of
Ans: Axillary bud

71. Rachis is absent in
1) Simple leaf 2) Palmately compound leaf
3) Pinnately compound leaf 4) 1 & 2
Ans: 4 (Simple leaf & Palmately compound leaf)

72. If the leaflets are attached at one common point the leaf is-
Ans: Palmately Compound

73. If the incisions of the lamina donot touch mid rib, the leaf is
Ans: Simple lobed

74. The type of leaf in Bombax ceiba is
Ans: Palmate Compound

75. Match the following Phyllotaxy with suitable example-

I. Alstonia	A. Neither Alternate nor opposite
II. Hibiscus	B. as in Guava
III. Calotropis	C. Whorled
IV. Nerium	D. Alternate

I II III IV
Ans: C D B A

76. Number of leaves present at each node in Alstonia.
Ans: More than 2

77. A Modified leaf in Nepenthes is concerned with the following functions
A) Photosynthesis B) Climbing C) Fulfils N₂ requirement
Ans: ABC

78. Citrus shows
Ans: spines

79. The idea behind the formation of spines in Cacti-
Ans: 1 & 2(Protection, To reduce the rate of respiration)

80. Match the following leaf modifications

I. Photosynthesis	A. Bryophyllum
II. Food storage	B. Cacti
III. Defence	C. Acacia (Australian)
IV. Vegetative reproduction	D. Onion

Ans: I II III IV
C D B A

81. The following plants do not perform protein synthesis if the area in which they grow is free from insects-
(A) Nepenthes (B) Venus flytrap (C) Dionaea
Ans: ABC

82. Tendrils in Pea are the modifications of the following-
1) Leaf 2) Petiole 3) Terminal leaflets 4) Basal leaflets
Ans: 3 (Terminal leaflets)

83. Phyllode is seen in-
1) Australian Acacia 2) Nepenthes 3) 1 & 2 4) Opuntia & Nepenthes
Ans: 1 & 2(Australion Acacia, Nepenthes)

84. One of the following is not concerned with Nepenthes-
1) Lamina-Pitcher 2) Upper petiole-Tendril
3) Lower petiole-tendril 4) Lower petiole-phyllode
Ans: 3(Lower petiole-tendril)

85. The leaf of Bryophyllum develop epiphyllous buds-
Ans: from notches of the leaf margin

86. Find the mis-match-
1) Nepenthes - Pitcher plant
2) Dionaea - Venus flytrap
3) Bombax - Silk cotton
4) Poaceae - Fabaceae
Ans: 4( Poaceae - Fabaceae)

87. Inflorescence deals with arrangement of-
Ans: flowers

88. A flower is a modified-
Ans: shoot

89. Racemose inflorescence has-
Ans: Only lateral flowers

90. Assertion (A): The inflorescence in Cassia and Cauliflower is corymb.
Reason (R): The flowers are brought to the same height in them.
Ans: A & R are correct. R explains A.

91. Match the following:

I. Flowes appear to arise from the same point on peduncle	A. Sun hemp
II. Flowers are produced in acropetal succession	B. Achyranthes
III. Flowers are brought to the same height	C. Carrot
IV. Sessile flowers are produced in acropetal succession	D. Cauliflower

Ans: I II III IV
C A D B

92. Find out mismatch-
1) Simple raceme - Fabaceae 2) Spike - Poaceae
3) Head - Asteraceae 4) Umbel - Apiaceae
Ans: 4(Umbel - Apiaceae)

93. Find out mis-match-
1) Graminae - Poaceae 2) Umbelliferae - Apiaceae
3) Asteraceae - Compositae 4) None
Ans: 4(None)

94. Simple Raceme and spike differ in-
Ans: Flowers in the later are sessile and former are pedicellate.

95. Characters of Spike-
1) Bisexual, acropetal, sessile 2) Unisexual, acropetal, pedicellate
3) Bisexual, acropetal, pedicellate 4) Bisexual, spathe, acropetal
Ans: 1(Bisexual, acropetal, sessile)

96. Spathe is the modification of-
Ans: Bract

97. Inflorescence with spathe-
Ans: Spadix

98. Neuter flowers are present in-
Ans: Spadix

99. Match the following.

I. Unisexual flowers	A. Head
II. Bisexual flowers	B. Musa
III. Unisexual and bisexual flowers	C. Spadix
IV. Unisexual and neuter flowers	D. Carrot

Ans: I II III IV
B D A C

100. Immature flowers in umbel are present-
Ans: in the centre

101. Find out mismatch-
1) Compound raceme - Panicle 2) Head - Capitulum
3) Cyathium - Hypanthodium 4) Simple Cyme - Cymule
Ans: 3(Cyathium - Hypanthodium)

102. Minimum number of flowers in Cymule and dichasial Cyme respectively-
Ans: 3, 7

103. Peduncle is condensed in-
Ans: Capitulum

104. Flowers are produced in centripetal succession in-
Ans: Head

105. Head and Umbel are similar in that both have
Ans: Centripetal succession

106. Involucre forms a cup like structure in-
Ans: Cyathium

107. Gall flowers are present only in-
Ans: Hypanthodium

108. Blastophaga lays its eggs in-
Ans: Gall flowers

109. One of the following entire inflorescence becomes fruit-
1) Head 2) Hypanthodium 3) Verticellaster 4) Cyathium
Ans: 2(Hypanthodium)

110. False whorl of flowers are formed in-
Ans: Verticellaster

111. Flowers are enclosed by fleshy Cup like peduncle in-
Ans: Hypanthodium

112. Arrange Cyathium (A), Hypanthodium (B) and Verticellaster (C) in ascending order basing on the number of types of flowers in them-
Ans: C, A, B(Verticellaster, Cyathium, Hypanthodium)

113. Arrangement of the folowers is irregular in-
Ans: Hypanthodium

114. Position of the verticellaster is-
Ans: at the node

115. Arrange the following in ascending order-
A) Types of flowers in Hypanthodium
B) Minimum number of flowers in dichasial cyme
C) Number of flowers in the inflorescence of Datura
D) Types of cymose inflorescence in verticellaster
Ans: C, D, A, B

116. The flower is opposite to the bract in-
Ans: Monochasial Cyme

117. Inflorescence without monochasial Cyme is seen in-
A) Hamelia B) Euphorbiaceae C) Ficus D) Lamiaceae E) Solanum F) Ipomoea
Ans: C, F(Ficus, Ipomoea)

118. The following characters are seen only in Hypanthodium-
1) Fleshy deep cup like peduncle 2) Irregular arrangement of the flowers
3) Presence of Gall flowers 4) All
Ans: 4(All)

119. Match the following-

I. Irregular arrangement	A. Capitulum
II. Acropetal succession	B. Main axis terminates into a flower
III. Centripetal succession	C. Fleshy deep cup like peduncle
IV. Basipetal succession	D. Continuous growth of the main axis

Ans: I II III IV
C D A B

120. Inflorescence showing Polygamous condition is seen in-
Ans: Mangifera

121. Varied lengths of pedicels are seen in-
Ans: Corymb

122. Match the following.

I. Only one bisexual flowers	A. Lamiaceae
II. Only one Flower	B. Spadix
III. Only one type of flowers	C. Datura
IV. Neuter flowers	D. Cyathium

I II III IV
Ans: C D A B

123. Sterile female flowers are called-
Ans: Gall flowers

124. Male and female flowers produced not in the same succession-
Ans: Cyathium

125. Number of sporophylls in a Cyathium is 20. The male and female flowers in its are-
Ans: 17, 1

126. Number of stamens in the male flowers of Cyathium and Ficus respectively-
Ans: 1, 3

127. A cyathium has 16 flowers. The number of sporophylls in it are-
Ans: 18

128. Ratio between the microsporophylls of a male flower and megasporophylls of a female flower in cyathium-
Ans: 1 : 3

129. Special inflorescence with sessile flowers-
Ans: Hypanthodium

130. Special inflorescence with bisexual flowers-
Ans: Verticellaster

131. Match the following-

I. Umbel	A. Moraceae
II. Hypanthodium	B. Apiaceae
III. Verticellaster	C. Euphorbiaceae
IV. Cyathium	D. Lamiaceae

Ans: I II III IV
B A D C

132. Arrange the following whorls from outside to inside-
1) P G A 2) C K A G 3) K C A G 4) G A K C
Ans: 3(K C A G)

133. Read the following statements. Choose the Correct one-
1) Largest flower is present in a saprophytic angiosperm
2) Smallest flower is present in a smallest angiosperm
3) Accessory organs are always distinct
4) A unisexual flower has neither androecium nor gynoecium
Ans: 2(Smallest flower is present in a smallest angiosperm)

134. Maximum Number of whorls on the thalamus in an unisexual flower-
Ans: 3

135. The flower of Gulmohur can be cut into 2 similar halves passing through the centre in-
Ans: vertical plane

136. Spathe is present in-
1) Cocos 2) Colocasia 3) Banana 4) All
Ans: 4(All)

137. Arrangement of floral parts in a hypogynous flower from top to the bottom of receptacle-
1) K C A G 2) G A C K 3) A C G K 4) G A K C
Ans: 2(G A C K)

138. Position of gynoecium is not observed in-
Ans:

139. If gynoecium occupies highest position-
Ans: Ovary is superior

140. If other parts of flower are arranged on rim of thalamus and gynoecium is in the centre in the same level-
Ans: Flower is perigynous

141. Thalamus fuses with ovary completely in-
Ans: Epigynous flower

142. Match the following-

I. Plum	A. Ovary occupies lowest position
II. Brinjal	B. Ovary ½ superior
III. Rose	C. Ovary not protected by thalamus
IV. Sun flower	D. Perigynous

Ans: I II III IV
D C B A

143. If other parts of the flower arise above the ovary the flower is-

Ans: Epigynous

144. Find out mismatch-
1) Guava - Superior ovary
2) Peach - Ovary ½ inferior
3) Mustard - Superior ovary
4) Cucumber - Epigynous
Ans: 1 (Guava - Superior ovary)

145. Find out the unrelated basing on position of ovary-
1) Plum 2) Rose 3) Rayflorets 4) Peach
Ans: 3(Rayflorets)

146. Ovary is protected by-
Ans: Thalamus

147. Match the following shapes & bundles of stamens

I. Datura	A. Bi-lipped
II. Ocimum	B. Citrus
III. Diadelphous	C. Pea
IV. Polyadelphous	D. Funnel shaped

Ans: I II III IV
D A C B

148. Shape of the bisexual flowers in Sunflower-
Ans: Tubular

149. Wheel shaped flower is seen in-
1) Datura 2) Ocimum 3) Rayflorets 4) None
Ans: 4(None)

150. Aestivation means the arrangement of
Ans: Perianth in the bud

151. The perianth members in a whorl are free or just touch without overlapping the aestivation is-
Ans: Valvate

152. Aestivation is twisted in-
Ans: Lady’s finger, Cotton, Chinarose

153. If the margins of sepals or petals overlap one another but not in any particular direction, the aestivation is-
Ans: Inbricate

154. The flower, a pentamerous one has 3 types of petals based on size has the following aestivation.
1) Imbricate 2) Valvate 3) Vexillary 4) Twisted
Ans: 3(Vexillary)

155. Vexillary aestivation is also called-
Ans: Papilionaceous

156. Number of chambers in the mature and immature anther of a stamen-
Ans: 2, 4

157. If the stamens of a flower are free, the condition is-
Ans: Polyandrous

158. Match the following-

I. Epiphyllous	A. Solanaceae
II. Polyadelphous	B. Fabaceae
III. Epipetalous	C. Citrus
IV. Diadelphous	D. Liliaceae

Ans: I II III IV
D C A B

159. Stamens show variation in the length of filments in-
Ans: Salvia, Brassica

160. Rose shows or Lotus shows
Ans: Perigynous, apocarpous

161. Match the following-

I. Parietal	A. Placenta along ventral suture
II. Axile	B. Inner wall of the ovary or peripheral part
III. Marginal	C. Central axis, Unilocular ovary
IV. Free central	D. Placenta is axial

Ans: I II III IV
B D A C

162. Ovary is unilocular showing the following placentation-
1) Marginal 2) Basal 3) Free central 4) All
Ans: 4(All)

163. Ovary without septum is seen in the following aestivation-
1) Marginal 2) Basal 3) Free central 4) All
Ans: 4(All)

164. Find mismatch-
1) Dianthus, Primrose - Free Central 2) Sun flower, Marigold - Basal
3) Mustard, Lemon - Parietal 4) Pea, Bean - Marginal
Ans: 3(Mustard, Lemon - Parietal)

165. Ovules are arranged on
Ans: Cushion like structure

166. One ovule, one locule and no septum are seen in
Ans: Basal

167. Number of nodes and internodes respectively on the thalamus of a bisexual and complete flower.
Ans: 4, 3

168. One of the following pairs have similar placentation-
1) Pea, Marigold 2) Mustard, Argemone
3) Dianthus, Sunflower 4) Citrus, Sunflower
Ans: 2(Mustard, Argemone)

169. Assertion (A): Banana is a parthenocarpic fruit.
Reason (R): It is seedless.
Ans: A & R are correct. R explains A.

170. If a fruit is formed from ovary and other parts like thalamus and pedicel, it is called-
Ans: False fruit

171. A runner that produces false fruit
Ans: Strawberry

172. Fleshy fruits are differentiated from dry fruits basing on-
Ans: Nature of pericarp

173. Match the following developments

I. Drupe	A. Tricarpellary, unilocular inferior ovary
II. Berry	B. Monocarpellary superior ovary
III. Pome	C. Bi or multicarpellary superior
IV. Pepo	D. Bi or Multicarpellary inferior

Ans: I II III IV
B C D A

174. Match the following-

I. Endocarp stony	A. Mango
II. Mesocarp fibrous	B. Drupe
III. Mesocarp fleshy	C. Cocos
IV. One seeded	D. Mango

Ans: I II III IV
B C A D

175. Match the following-

I. Pulp	A. Hesperidium
II. Endocarp smooth	B. Pome
III. Cartilagenous endocarp	C. Pepo
IV. Mesocarp papery	D. Berry

Ans: I II III IV
D C B A

176. Match the following-

I. Thalamus fleshy	A. Hesperidium
II. Epicarp like a rind	B. Pome
III. Glandular epicarp	C. Citrus
IV. Juicy hairs on endocarp	D. Pepo

Ans: I II III IV
B D A C

177. Inferior berry-
Ans: Guava

178. Match the following-

I. Endocarp smooth	A. Berry
II. Endocarp Cartilagenous	B. Pepo
III. Endocarp hard	C. Pome
IV. Seeds hard	D. Drupe

Ans: I II III IV
B C D A

179. Mesocarp and endocarp are fused to form pulp in-
Ans: Berry

180. Match the following-

I. Pappus	A. Nut
II. Mericarps	B. Legume
III. Stony Pericarp	C. Cypsela
IV. Bursts dorsiventrally	D. Acacia

Ans: I II III IV
C D A B

181. Fleshy fruit and dry indehiscent fruit are externally seen together in-
Ans: Anacardium

182. The following fruits are all one seeded.
1) Fleshy 2) Indehiscent 3) Dehiscent 4) Schizocarpic
Ans: 2 (Indehiscent)

183. Pappus is formed from-
Ans: Calyx

184. Find out mismatch.
1) Rice - caryopsis 2) Cashew - nut
3) Pea - capsule 4) Tridax - cypsela
Ans: 3 (Pea - capsule)

185. The fruit in Cotton and Datura is-
Ans: Capsule

186. Husk is found in-
Ans: Caryopsis

187. Mericarps are released by
Ans: Schizocarpic

188. Entire inflorescence becomes a fruit in-
Ans: Composite

189. Match the following common names-

I. Jack fruit	A. Pyrus malus
II. Pine apple	B. Artocarpus integrifolia
III. Apple	C. Annona Squamosa
IV. Custard apple	D. Ananas sativus

Ans: I II III IV
B D A C

190. Find out incorrect pair-
1) Sweet Orange - Citrus sinensis 2) Cashew - Anacardiaceae
3) Pine apple - Bromeliaceae 4) Ficus - Euphorbiaceae
Ans: 4 (Ficus - Euphorbiaceae)

191. Read the following statements-
A. Pineapple is a sorosis
B. Jack fruit is a sorosis
C. Fiscus is a syconus
D. Ficus and jack fruit belong to the same family
E. Cashew and Mango belong to the same family
Find the incorrect statements.
Ans: None

192. Assertion (A): The fruit of Fiscus is a compound one.
Reason (R): It is formed from entire inflorescence.
Ans: A & R are correct. R explains A.

193. Cotyledons are edible in-
A. Bean B. Pea C. Groundnut D. Cashew
Ans: A, B, C, D(Bean, Pea, Groundnut, Cashew)

194. Match the following-

I. Pine apple	A. Perianth
II. Custard apple	B. Peduncle, juicy bracts
III. Jack fruit	C. Endosperm
IV. Cocos	D. Mesocarp and endocarp of each fruitlet

Ans: I II III IV
B D A C

195. The edible part in cucumber-
A) Epicarp B) Mesocarp C) Rind D) Endocarp E) Pericarp F) Placenta
Ans: B, D, F(Mesocarp, Endocarp, Placenta)

196. Peduncle is edible in-
Ans: Ficus & Pineapple

197. Read the following synonyms and find out an incorrect one.
1) Keel - carina 2) Wings - alae
3) Compound fruit - composite fruit 4) Vexillum - standard 5) None
Ans: 5(None)

198. One seeded bits of schizocarpic fruits are called-
Ans: Mericarp

199. Function of involucre-
Ans: Protection to inflorescence

200. Under grond stem that grows erect is called-
Ans: Corm

201. Inflorescence with naked flowers-
Ans: Cyathium

202. Function of the bract-
Ans: Protects the flower in bud condition

203. Bracteole arises from-
Ans: pedicel

204. The fruit of Rice is covered by-
Ans: Husk

205. The stem is sub-aerial and useful as a runner-
Ans: Oxalis

206. Inflorescence with single bisexual flower-
Ans: Datura

Posted Date : 30-11-2020

PLANT KINGDOM

1. Match the following-

I) Unicellular flagellated	A) Spirogyra
II) Unicellular non flagellated	B) Chara
III) Filamentous	C) Chlorella
IV) Filamentous, Branched	D) Chlamydomonas
	E) Ulothrix

I II III IV
A: D C A E

2. All kinds of sexual reproduction are present in-
A: Chlorophyceae

3. Match the following.

I) Isogamy (not flagellated gamets)	A) Chlamydomonas
II) Anisogamy	B) Volvox
III) Isogamy (Flagellated gamets)	C) Phaeophyceae
IV) Oogamy	D) Spirogyra

I II III IV
A: D C A B

4. Match the following.

I) Chl - a, Chl - b	A) Rhodophyceae
II) Chl - a	B) Phaeophyceae
III) Chl - a, Chl - c	C) Cyanophyceae
IV) Chl - a, Chl - d	D) Chlorophyceae

I II III IV
A: D C B A

5. Identifiy incorrect pair-
1) Chloroplasts variously shaped - Chlorophyceae
2) Fucoxanthin - Reserve food
3) Floridean starch - Rhodophyceae
4) Mannitol - Reserve food
A: 2 (Fucoxanthin - Reserve food)

6. The following reserve food is similar to Glycogen of Fungi-
1) Starch 2) Mannitol 3) Floridean starch 4) Laminarin
A: 3 (Floridean starch )

7. Kelps belong to-
A: Phaeophyceae

8. The algal plant used by Astronauts-
A: Chlorella

9. Phyrenoids are present in-
A: Chloroplasts of Chlorophyceae

10. The algae which grow to a height of 100 metres are called and they belong to-
A: Kelps, Phaeophyceae

11. Pear shaped gametes with laterally attached flagella are seen only in-
A: Phaeophyceae

12. Flagellated bodies are absent in-
A: Rhodophyceae

13. The following algae are olive green to various shades of brown-
1) Chlorophyceae 2) Cyanophyceae

3) Chrysophytes 4) None
A: 4 (None)

14. Identify incorrect pair-
1) No flagella - Rhodophyceae 2) No flagella - Cyanobacteria
3) Lateral flagella - Phaeophyceae 4) Colonial form - Chlamydomonas
A: 4 (Colonial form - Chlamydomonas)

15. The outer layer in the Cell wall of Chlorophyceae is made of-
A: Pectose

16. Different shades of Phaeophyceae members is due to the presence of amount of the following pigment-
1) Carotenoids 2) Xanthophylls

3) Fucoxanthin 4) Chl - a
A: 3 (Fucoxanthin)

17. Outer layer of the Cell wall of Phaeophyceae is made of-
A: Algin

18. Leaf like Photosynthetic organ in Phaeophyceae is-
A: Frond

19. Identify incorrect statement
1) Zoospores in Phaeophyceae are pear shaped and biflagellated laterally
2) Gamets in Phaeophyceae are pear shaped and biflagellated laterally
3) The flagella in Phaeophyceae are unequal
4) Minimum number of flagella in Chlorophyceae is 4
A: 4 (Minimum number of flagella in Chlorophyceae is 4)

20. The Cell wall of Rhodophyceae members is made of-
1) Cellulose 2) Pectin

3) Polysulphate esters 4) All
A: 4 (All)

21. One of the following is not a set members of Phaeophyceae
1) Ectocarpus, Dictyota, Laminaria 2) Laminaria, Sargassum, Fucus
3) Dictyota, Porphyra, Sargassum 4) Fucus, Ectocarpus, Laminaria
A: 3 (Dictyota, Porphyra, Sargassum)

22. Sexual reproduction occurs by Oogamy only in-
A: Red algae

23. The Characteristic feature of Rhodophyceae
A. Chl - d
B. Phycoerythrin
C. Non-motile spores and gamets
D. Absence of Isogamy and Anisogamy
E. Flask shaped female sex organ called carpogonium
F. Spermatium
G. Post fertilisation developements
H. All
1) A B C 2) D E F G H 3) 1 & 2 4) A B D F G H
A: 3 (1 & 2)

24. Air Bladder is present and frond is absent in-
A: Fucus, Polysiphonia

25. Sexual reproduction is Oogamous and internal in -
A: Phaeophyceae

26. Find out incorrect set members of Rhodophyceae-
1) Polysiphonia, Porphyra 2) Gelidium, Polysiphonia
3) Gracilaria, Gelidium 4) Fucus, Gelidium
A: 4 (Fucus, Gelidium)

27. Marine speices of Algae used as food-
1) Sargassum 2) Laminaria 3) Porphyra 4) All
A: 4 (All)

28. Rhodophyceae member used as food-
A: Laminaria

29. Phaeophyceae members used as food-
A: Sargassum, Laminaria

30. Iodine is obtained from the following phaeophyceae edible member-
1) Laminaria 2) Sargassum 3) Porphyra 4) Ectocarp
A: 1 (Laminaria)

31. Air bladder is present in-
A: Fucus

32. Rhodophyceae member with frond-
A: Porphyra

33. The hydrocolloid of Rhodophyceae is-
A: Carrageen

34. The hydrocolloid of Phaeophyceae is-
A: Algin

35. Algae is divided into classes based on-
A. Cell wall B. Pigments

C. Reserve food D. Sexual reproduction
A: A B C

36. Source of Agar Agar-
A: Gelidium, Gracilaria

37. The female sex organ in Rhodophyceae-
A: Club shaped, Carpogonium

38. Life cycle of Rock weed is-
A: Diplontic

39. Rock, weed is a member of-
A: Phaeophyceae

40. If the haploid stage is restricted to the gamets the life cycle of that organism is called-
A: Diplontic

41. Match the following life cycles-

I) Haplontic	A) Rockweed or Fucus
II) Diplontic	B) Vascular plants
III) Haplodiplontic	C) Many Algae
IV) Diplobiontic	D) Bryophytes
	E) Polysiphonia

I II III IV
Ans: C A D E

42. The life cycles of Bryophytes and Vascular plants respectively are-
Ans: Haplodiplontic, Diplohaplontic

43. The life cycle of Bryophytes is similar to-
A. Ectocarpus B. Polysiphonia C. Laminaria D. Rockweed
Ans: A C

44. Match the following events with life cycles

I. Diploid stage is restricted to Zygote only	A. Diplontic
II. Haploid stage is restricted to gametes only	B. Haplontic
III. Leading gametophyte, dependent sporophyte	C. Diplohaplontic
IV. Dependent gametophyte, leading sporophyte	D. Haplodiplontic

I II III IV
Ans: B A D C

45. Pteridophytes, Gymnosperms and Angiosperms resemble each other in the following characters-
1) Life cycle type 2) Sex organs 3) Seed 4) Flower
Ans: 1 (Life cycle type)

46. Life cycle of the following shows an intermediate condition-
1) Algae 2) Fungi 3) Bryophytes 4) Rockweed
Ans: 3 (Bryophytes)

47. Vascular, non-archegoniates-
Ans: Angiosperms

48. Highly Advanced archegoniates-
Ans: Gymnosperms

49. Highly advanced non vascular plants-
Ans: Bryophyta

50. Smallest and tallest angiosperms-
Ans: Wolfia, Eucalyptus

51. Giant redwood tree is also called-
Ans: Sequoiadendron

52. Long shoot, dwarf shoot and seeds are seen in-
Ans: Ginkgo

53. Leaves are needle like in-
Ans: Pinus

54. The following plant is considered as living fossil-
1) Cycas 2) Cedrus 3) Gnetum 4) Ginkgo
Ans: 4 (Ginkgo)

55. Find out incorrect matching-
1) Smallest angiosperm - Wolfia 2) Tallest angiosperm - Eucalyptus
3) Tallest gymnosperm - Sequoia 4) Compound leaves - Ginkgo
Ans: 4 (Compound leaves - Ginkgo)

56. Roots are specialised in-
Ans: Cycas, Pinus

57. Stem is branched in-
Ans: Cedrus, Pinus

58. Assertion (A): The leaves in Gymnosperms are well adapted to reduce water loss.
Reason (R): Flowers in Gymnosperms are produced in the form of strobili.
Ans: A & R are correct. R do not explains A.

59. Vascular bundles in stems of Gymnosperm resemble those of-
Ans: Only dicots

60. Pick out the odd group basing on homosporous and heterosporous condition-
1) Bryophytes 2) Pteridophytes 3) Gymnosperms 4) Angiosperms
Ans: 2 (Pteridophytes)

61. Male and Female strobilus are produced in the same and different plants respectively in-
Ans: Pinus, Cycas

62. Assertion (A): Seeds are formed in Gymnosperms for the first time.
Reason (R): Female gametophytes is retained within the megasporium.
Ans: A & R are correct. R explains A.

63. Gametophytes have no independent existence in-
Ans: Phanerogams

64. Oogamy is siphonogamous in all the plants of-
Ans: Angiosperms

65. Match the following-

I) Siphonogamy	A) Rhodophyceae
II) Zoidogamy & Siphonogamy	B) Pteridophyta
III) Zoidogamous Oogamy	C) Cycas
IV) Oogamy	D) Angiosperms

I II III IV
Ans: D C B A

66. Assertion (A): Seeds are naked in Gymnosperms.
Reason (R): Seeds are not covered by Seed Coat.
Ans: A is correct. R is false.

67. Multiciliated male gametes are seen in-
Ans: Cycas

68. Fern Characters in Cycas-
1) Circinate Vernation 2) Ramenta, Archegonia 3) Multiciliated sperms 4) All
Ans: 4 (All)

69. Arrange the following in ascending order basing-
A) Maximum number of structures useful in movements in the gametes of Bryophytes.
B) Number of flagella present on the sides of gametes in Phaeophyceae.
C) Number of flagella on the anterior end of gametes in Rhodophyceae.
D) Types of sexual reproductions in Phaeophyceae.
E) Maximum number of flagella in members of Chlorophyceae.
F) Number of cilia in the sperms of ferns.
Ans: CBADEF

70. Assertion (A): Bryophytes are also called Amphibians of plant kingdom.
Reason (R): They are primitive land plants.
Ans: A & R are correct. R do not explains A.

71. Find out incorrect statement.
1) Bryophytes plants play an important role in hydrarch succession
2) Bryophytes are dependent on water to live on moist soils
3) Male sex organs are called antheredia for the first time in Bryophytes

4) Flask shaped female sex organ is found in Bryophytes for the first time
5) All are correct
Ans: 5 (All are correct)

72. Assertion(A): Mosses are highly advanced Bryophytes.
Reason(R): They have vegetative structures analogous to those of angiosperms or phanerogams.
Ans: A & R are correct. R explains A.

73. Assertion (A): Bryophytes are gametophytes.
Reason (R): The main plant body produces gametes.
Ans: A & R are correct. R explains A.

74. Multicellular, jacketed, stalked sex organs are present only in-
A) Bryophytes B) Thallophytes C) Pteridophytes
1) A, B, C 2) A, C 3) A, B 4) A only
Ans: 4 (A only)

75. Biflagellated antherozoids and archegonia are seen in-
Ans: Bryophytes, Pteridophytes

76. Zygote shows meiosis in-
1) Algae 2) Fungi

3) Non Vascular Cryptogams 4) 1 & 2
Ans: 4 (Algae, Fungi)

77. Read the following statements and find out incorrect one.
1) Bryophytes are homosporous and show haplodiplontic life cycle
2) Some pteridophytes are heterosporous and show diplohaplontic life cycle
3) Some pteridophytes are homosporous and show haplodiplontic life cycle
4) All phanerogams are heterosporous and show diplohaplontic life cyle
Ans: 3 (Some pteridophytes are homosporous and show haplodiplontic life cycle)

78. Common between Bryophytes and Pteridophytes-
1) Structure of sex organs 2) Type of division in Zygote
3) Zoidogamous oogamy 4) 2 & 3 5) 1, 2 & 3
Ans: 4 (Type of division in Zygote, Zoidogamous oogamy )

79. Assertion (A): Sphagnum is used in the transshipment of living material.
Reason (R): Peat has capacity to hold water.
Ans: A & R are correct. R explains A.

80. Assertion (A): Mosses have great ecological importance.
Reason (R): They are the first organisms along with lichens to colonise on rocks.
Ans: A & R are correct. R explains A.

81. Assertion (A): Some mosses reduce the impact of falling rain and prevent soil erosion.
Reason (R): They form dense mats on the soil.
Ans: A & R are correct. R explains A.

82. Match the following-

I) Protonema	A) Fungi
II) Pseudoelaters	B) Marchantia
III) Gemmae	C) Funaria
IV) Gemmae and elaters	D) Hornworts

The correct matching is
I II III IV
Ans: C D A B

83. Find out mis-match.
1) Anthoceros - Hornwort 2) Polytrichum - Haircap Moss
3) Funaria - Cord Moss 4) Sphagnum - Peatwort
Ans: 4 (Sphagnum - Peatwort)

84. Match the following-

I) Second stage in Mosses	A) Protonema
II) Dominant stage in Bryophytes	B) Moss
III) Juvenile stage	C) Gametophyte
IV) Paraphyses	D) Gametophore

I II III IV
Ans: D C A B

85. Leefy bryophytes are seen in
A) Hepaticopsida B) Anthocerotopsida C) Bryopsida
1) A, B 2) A, C 3) B, C 1) A, B, C
Ans: 2 (A, C)

86. The sporophyte is divided into foot, seta and capsule in-
A) Hepaticopsida B) Anthocerotopsida C) Bryopsida
Ans: C

87. Intercalary meristamatic zone is seen in and in the place of-
Ans: Hornworts, seta

88. Elaters are useful in the spore dispersal of-
Ans: Marchantia

89. Horn like structure in Anthocerotopsida represents-
Ans: Sporophyte

90. Number of Classes in Algae, Fungi and Bryophyta in Whittaker's classification-
1) 3, 3, 4 2) 3, 4, 3 3) 4, 4, 3 4) 4, 4, 4
Ans: 2 (3, 4, 3)

91. Marchantia is-
Ans: Gametophytic, homosporous, dioecious

92. Horn like structure in Anthocerotopsida represents-
Ans: erect, diploid structure on the gametophyte

93. Assertion (A): Bryophytes show heteromorphic alternation of generations.
Reason (R): Their dominant sporophyte body is conspicuously different from gametophyte.
Ans: A is correct. R is false.

94. The following plants grow as dense mats on the soil-
1) Blue green algae 2) Hornworts 3) Mosses 4) Liverworts
Ans: 3 (Mosses)

95. Match the following.

I) Club Moss	A) Polytrichum
II) Haircap Moss	B) Funaria
III) Peat Moss	C) Pteridophyte
IV) Cord Moss	D) Sphagnum

I II III IV
Ans: C A D B

96. Match the following.

I) First terrestrial plants	A) Gymnosperms
II) First Vascular plants	B) Angiosperms
III) First Seed plants	C) Bryophytes
IV) First Vascular, with out archegonia	D) Pteridophyta

I II III IV
Ans: C D A B

97. First tracheophytic cryptogams-
Ans: Pteridophytes

98. Pteridophytes are-
A) Highly advanced cryptogams B) Primitive vascular plants
C) First terrestrial tracheophytes D) First vascular embryophytes
E) First sporphyte dominant plants
Ans: ABCDE

99. Macrophyll bearing pteridophytes belong to-
Ans: Pteropsida

100. Match the following.

I) Central core of xylem surrounded by phloem	A) Siphonostele
II) Medullated protostele	B) Dictyostele
III) Siphonostele with scattered leaf gaps	C) Protostele
IV) Dissected siphonostele with overlapping leaf gaps	D) Solenostele

I II III IV
Ans: C A D B

101. Medullated protostele is called-
Ans: Siphonostele

102. Leaf gaps are overlapping in-
Ans: Dictyostele

103. Scattered leaf gaps are the characteristic feature of-
Ans: Solenostele

104. Leaf gaps are not seen in-
Ans: Protostele and Siphonostele

105. Majority of the Pteridophytes resemble the following group regarding the types of spores the plants produce-
1) Bryophytes 2) Gymnosperms 3) Phenerogams 4) Angiosperms
Ans: 1 (Bryophytes)

106. The following set of pteridophytes are heterospore-
1) Psilotum, Equisetum 2) Selaginella, Salvinia
3) Salvia and Selaginella 4) Equisetum, Lycopodium
Ans: 2 (Selaginella, Salvinia)

107. The gametophyte of pteridophytes is-
1) Prothallus, indepdenent 2) Thallus, independent, non-vascular
3) Non vascular 4) 1 & 3
Ans: 4 (Prothallus, indepdenent & Non vascular)

108. Assertion (A): The spread of pteridophytes is limited and restricted to narrow geographical regions.
Reason (R): Their gametophytes grow in cool, damp, shady places and need water for fertilisation.
Ans: A & R are correct. R explains A.

109. Female gametophyte is partly retained on the sporophyte for the first time in-
Ans: Pteridophytes

110. Female gametophyte is completely retained on the sporophyte for the first time in-
Ans: Gymnosperm

111. Precursor to the seed habit-
Ans: Development of Zygote into embryo within the female gametophyte while in megasporangium.

112. If the young leaves in pteridophytes coil from the tip to the base, it is called-
Ans: Vernation

113. Match the following-

I) Indusium	A) Pteris
II) False indusium	B) Dryopteris
III) Ramenta	C) Venation
IV) Dichotomous	D) Petiole

I II III IV
Ans: B A D C

114. Sporophylls produce sporangia on-
Ans: Ventral side

115. Identify wrong matching-
1) Sorus - Sporangia 2) Sporophylls - Strobilus/ cone
3) Multiciliate - Antherozoids 4) Indusium - Archegonium
Ans: 4 (Indusium - Archegonium)

116. Multiciliated antherozoids are present in-
Ans: Dryopteris

117. Sorus in Pteris is protected by-
Ans: Reflexed margin of the sporophyll

118. Match the following.

I) Psilotum	A) Lycopsida
II) Adiantum	B) Sphenopsida
III) Selaginella	C) Pteropsida
IV) Equisetum	D) Psilopsida

I II III IV
Ans: D C B A

119. One of the following is not a fern character-
1) Indusium 2) Furcate venation 3) Microphylls 4) Ramenta
Ans: 3 (Microphylls)

120. Match the following.

I) Oogamy	A) Pteridophytes
II) Zoidogamy	B) Angiosperms
III) Siphonogamy	C) Cycas
IV) Siphonogamy & Zoidogamy	D) Chlorophyceae

I II III IV
Ans: C A B D

121. Common character in tracheophytes is-
Ans: Complex tissue

122. Match the following-

I) Highly reduced archegonia	A) Plants with post fertilized endosperm
II) Highly developed archegonia	B) First true land plants
III) No Archegonia	C) First flowering plants
IV) Sessile Embedded archegonia	D) First embryo bearing plants

I II III IV
Ans: C D A B

123. Uncommon between archegoniates and embryophytes
Ans: Angiosperms

124. Uncommon between tracheata and archegoniates-
Ans: Angiosperms

Posted Date : 30-11-2020

study-material

Previous Papers

TS EAPCET: తెలంగాణ ఈఏపీసెట్‌ హాల్‌టికేట్లు డౌన్‌లోడ్‌ చేసుకోండి!

1 - 5 CHAPTERS (JUNIOR BOTANY)

Tags :

VIRUS

Tags :

VIRUS

Tags :

THE LIVING WORLD

Tags :

MICROBES IN HUMAN WELFARE

Tags :

BIOMOLECULES

Tags :

THE UNIT OF LIFE

Tags :

ECOLOGICAL ADAPTATION, SUCCESSION AND ECOLOGICAL SERVICES

Tags :

TAXONOMY OF FLOWERING PLANTS

Tags :

SCIENCE OF PLANTS

Tags :

6 - 9 CHAPTERS (JUNIOR BOTANY)

Tags :

6 - 9 CHAPTERS (SENIOR BOTANY)

Tags :

BIOLOGICAL CLASSIFICATION

Tags :

ENZYMES

Tags :

STRATEGIES FOR ENHANCEMENT IN FOOD PRODUCTION

Tags :

MODES OF REPRODUCTION & SEXUAL REPRODUCTION IN FLOWERING PLANTS

Tags :

MORPHOLOGY OF FLOWERING PLANTS

Tags :

PLANT KINGDOM

Tags :

study-material

Previous Papers

విద్యా ఉద్యోగ సమాచారం

Model Papers

లేటెస్ట్ నోటిఫికేష‌న్స్‌

Connect with Us

Quick links

© 2024 Ushodaya Enterprises Private Limited. Powered by Margadarsi Computers