Q1. In angiosperms, root hairs arise from which one of the following regions of the root?
Correct Answer: (d)
From the region of maturation, some epidermal cells form very fine and delicate, thread like structures called root hairs.
From the region of maturation, some epidermal cells form very fine and delicate, thread like structures called root hairs.
Q2. In which one of the following, the ovules are not enclosed by an ovary wall and remain exposed?
Correct Answer: (b)
The gymnosperms are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation. Pinus is a gymnosperm. Funaria is a moss. Selaginella is a pteridophyte.
The gymnosperms are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation. Pinus is a gymnosperm. Funaria is a moss. Selaginella is a pteridophyte.
Q3. In the lac operon, the z gene codes for
Correct Answer: (c)
In lac operon, i gene codes for regulator protein; z gene codes for beta-galactosidase; y gene codes for Permease; a gene codes for transacetylase.
In lac operon, i gene codes for regulator protein; z gene codes for beta-galactosidase; y gene codes for Permease; a gene codes for transacetylase.
Q4. Exploring molecular, genetic and species-level diversity for products of economic importance is called
Correct Answer: (c)
Exploring molecular, genetic and species-level diversity for products of economic importance is called bioprospecting.
Exploring molecular, genetic and species-level diversity for products of economic importance is called bioprospecting.
Q5. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Genetically modified organism | I. Agrobacterium tumefaciens |
| B. Thermostable DNA polymerase | II. Bt cotton |
| C. Ti plasmid | III. Thermus aquaticus |
| D. pBR322 | IV. Escherichia coli |
Correct Answer: (a)
Genetically modified organism - Bt cotton; Thermostable DNA polymerase - Thermus aquaticus; Ti plasmid - Agrobacterium tumefaciens; pBR322 - Escherichia coli.
Genetically modified organism - Bt cotton; Thermostable DNA polymerase - Thermus aquaticus; Ti plasmid - Agrobacterium tumefaciens; pBR322 - Escherichia coli.
Q6. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Productivity | I. Gross primary productivity minus respiration losses |
| B. Net primary productivity | II. Rate of formation of new organic matter by consumers |
| C. Gross primary productivity | III. Rate of biomass production |
| D. Secondary productivity | IV. Rate of production of organic matter during photosynthesis |
Correct Answer: (a)
Productivity is the rate of biomass production (III). Net primary productivity is GPP minus respiration losses (I). Gross primary productivity is the rate of production of organic matter during photosynthesis (IV). Secondary productivity is the rate of formation of new organic matter by consumers (II).
Productivity is the rate of biomass production (III). Net primary productivity is GPP minus respiration losses (I). Gross primary productivity is the rate of production of organic matter during photosynthesis (IV). Secondary productivity is the rate of formation of new organic matter by consumers (II).
Q7. Since the origin and diversification of life on Earth, there have been five episodes of mass extinction of species. How is the sixth extinction, which is in progress, different from the previous episodes?
Correct Answer: (c)
The current, sixth episode of mass extinction is estimated to be 100 to 1000 times faster than the ones in the pre-human times and our activities are responsible for the faster rates.
The current, sixth episode of mass extinction is estimated to be 100 to 1000 times faster than the ones in the pre-human times and our activities are responsible for the faster rates.
Q8. Alpha-helix is found in which level of protein structure?
Correct Answer: (a)
Alpha-helix is shown by secondary structure of protein as it has right handed helices. Primary structure is linear and tertiary structure is a hollow ball-like structure. Quaternary structures are formed by more than one polypeptide chains.
Alpha-helix is shown by secondary structure of protein as it has right handed helices. Primary structure is linear and tertiary structure is a hollow ball-like structure. Quaternary structures are formed by more than one polypeptide chains.
Q9. The main function of bulliform cells in grasses is :
Correct Answer: (d)
Bulliform cells are large empty colourless cells that lose water and become flaccid in water scarce condition. Hence they curl the leaf inwards to minimise water loss by reducing the exposed surface area.
Bulliform cells are large empty colourless cells that lose water and become flaccid in water scarce condition. Hence they curl the leaf inwards to minimise water loss by reducing the exposed surface area.
Q10. Identify the correct sequence of steps in each cycle of Polymerase Chain Reaction :
Correct Answer: (d)
The correct sequence of steps in PCR is Denaturation → Annealing → Extension.
The correct sequence of steps in PCR is Denaturation → Annealing → Extension.
Q11. Match List I with List II:
| List-I (Phase of cell cycle) | List-II (Activity) |
|---|---|
| A. G1 phase | I. Actual cell division occurs |
| B. S phase | II. Cell is metabolically active and continuously grows but does not replicate its DNA |
| C. G2 Phase | III. Synthesis of DNA occurs and the amount of DNA per cell doubles |
| D. M phase | IV. Proteins are synthesized while cell growth continues |
Correct Answer: (d)
During G₁ phase the cell is metabolically active and continuously grows but does not replicate its DNA (II). S-phase marks the phase during which DNA synthesis or replication takes place and amount of DNA per cell doubles (III). During the G₂ phase, proteins are synthesised in preparation for mitosis while cell growth continues (IV). During M phase the actual cell division takes place (I).
During G₁ phase the cell is metabolically active and continuously grows but does not replicate its DNA (II). S-phase marks the phase during which DNA synthesis or replication takes place and amount of DNA per cell doubles (III). During the G₂ phase, proteins are synthesised in preparation for mitosis while cell growth continues (IV). During M phase the actual cell division takes place (I).
Q12. Which of the following statements are correct?
A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss.
B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans.
C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake.
D. Water hyacinth is an invasive species.
E. When a species becomes extinct, the plant and animal species associated with it are not affected.
Choose the correct answer from the options given below:
A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss.
B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans.
C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake.
D. Water hyacinth is an invasive species.
E. When a species becomes extinct, the plant and animal species associated with it are not affected.
Choose the correct answer from the options given below:
Correct Answer: (b)
The Nile Perch introduced into Lake Victoria in East Africa led eventually to extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake. When a species become extinct, the plant and animal species associated with it in an obligatory way also become extinct. Statements A, B and D are correct.
The Nile Perch introduced into Lake Victoria in East Africa led eventually to extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake. When a species become extinct, the plant and animal species associated with it in an obligatory way also become extinct. Statements A, B and D are correct.
Q13. Which of the following statements are correct with reference to a transcription unit?
A. A transcription unit in DNA is defined primarily by three regions : promoter, structural gene and terminator.
B. The promoter is said to be located towards the 5′-end of the structural gene.
C. The promoter is a DNA sequence that provides binding site for RNA polymerase.
D. The promoter defines the template and coding strands.
E. The terminator is located towards the 3′-end of the coding strand and it defines the end of the process of transcription.
Choose the correct answer from the options given below:
A. A transcription unit in DNA is defined primarily by three regions : promoter, structural gene and terminator.
B. The promoter is said to be located towards the 5′-end of the structural gene.
C. The promoter is a DNA sequence that provides binding site for RNA polymerase.
D. The promoter defines the template and coding strands.
E. The terminator is located towards the 3′-end of the coding strand and it defines the end of the process of transcription.
Choose the correct answer from the options given below:
Correct Answer: (d)
The promoter and terminator flank the structural gene in a transcription unit. The promoter is said to be located towards 5′-end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase and it is the presence of promoter in a transcription unit that also defines the template and coding strands. The terminator is located towards 3′-end of the coding strand and it usually defines the end of transcription. All statements are correct.
The promoter and terminator flank the structural gene in a transcription unit. The promoter is said to be located towards 5′-end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase and it is the presence of promoter in a transcription unit that also defines the template and coding strands. The terminator is located towards 3′-end of the coding strand and it usually defines the end of transcription. All statements are correct.
Q14. Which one of the following statements is not true about the universal rules of binomial nomenclature?
Correct Answer: (d)
According to universal rules of nomenclature, the first word denoting the genus starts with a capital letter with the second components denotes the specific epithet and starts with small letter.
According to universal rules of nomenclature, the first word denoting the genus starts with a capital letter with the second components denotes the specific epithet and starts with small letter.
Q15. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Decomposition | I. Accumulation of dark coloured amorphous colloidal substance |
| B. Detritus | II. Release of inorganic nutrients by the activity of microbes in soil |
| C. Mineralisation | III. Breaking down of complex organic matter into inorganic substances |
| D. Humification | IV. Dead remains of plants and animals including fecal matter |
Correct Answer: (b)
Decomposition is the process of breaking down of complex organic matter into inorganic substance (III). Detritus includes dead remains of plants and animals including fecal matter and acts as the raw material for decomposition (IV). Degradation of humus by activity of microbes leading to release of inorganic nutrients is called Mineralisation (II). Accumulation of the dark coloured amorphous substance called humus, is called humification (I).
Decomposition is the process of breaking down of complex organic matter into inorganic substance (III). Detritus includes dead remains of plants and animals including fecal matter and acts as the raw material for decomposition (IV). Degradation of humus by activity of microbes leading to release of inorganic nutrients is called Mineralisation (II). Accumulation of the dark coloured amorphous substance called humus, is called humification (I).
Q16. Which one of the following is the site for active ribosomal RNA synthesis?
Correct Answer: (c)
Nucleolus is the site for active ribosomal RNA synthesis.
Nucleolus is the site for active ribosomal RNA synthesis.
Q17. 2(C51H98O6) + 145O2 → 102CO2 + 98H2O + energy
The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation would be :
The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation would be :
Correct Answer: (a)
R.Q. = Volume of CO₂ evolved / Volume of O₂ consumed. As per the given equation, R.Q. = 102/145 = 0.7. This value lies between 0.5 and 0.95.
R.Q. = Volume of CO₂ evolved / Volume of O₂ consumed. As per the given equation, R.Q. = 102/145 = 0.7. This value lies between 0.5 and 0.95.
Q18. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Incomplete dominance | I. Human skin colour |
| B. Co-dominance | II. Inheritance of flower colour in Antirrhinum sp. |
| C. Pleiotropy | III. Phenylketonuria disease in humans |
| D. Polygenic inheritance | IV. ABO blood groups |
Correct Answer: (a)
Inheritance of flower colour in snapdragon (Antirrhinum sp.) is an example of incomplete dominance (II). ABO blood groups exhibit codominance in case of individuals having AB blood group (IᴬIᴮ) (IV). Phenylketonuria in humans is an example of pleiotropy (III). Human skin colour is controlled by three pairs of non allelic genes, hence it is an example of polygenic inheritance (I).
Inheritance of flower colour in snapdragon (Antirrhinum sp.) is an example of incomplete dominance (II). ABO blood groups exhibit codominance in case of individuals having AB blood group (IᴬIᴮ) (IV). Phenylketonuria in humans is an example of pleiotropy (III). Human skin colour is controlled by three pairs of non allelic genes, hence it is an example of polygenic inheritance (I).
Q19. Arrange the following steps of DNA fingerprinting in a correct sequence.
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Hybridisation using a labelled VNTR probe.
C. Transferring of separated DNA fragments to synthetic membranes.
D. Detection of hybridised DNA fragments by autoradiography.
E. Separation of DNA fragments by electrophoresis.
Choose the correct answer from the options given below:
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Hybridisation using a labelled VNTR probe.
C. Transferring of separated DNA fragments to synthetic membranes.
D. Detection of hybridised DNA fragments by autoradiography.
E. Separation of DNA fragments by electrophoresis.
Choose the correct answer from the options given below:
Correct Answer: (c)
The correct sequence of steps of DNA fingerprinting is: Isolation of DNA and its digestion by restriction endonuclease (A) → Separation of DNA fragments by electrophoresis (E) → Transferring of separated DNA fragments to synthetic membranes (C) → Hybridisation using a labelled VNTR probe (B) → Detection of hybridised DNA fragments by autoradiography (D).
The correct sequence of steps of DNA fingerprinting is: Isolation of DNA and its digestion by restriction endonuclease (A) → Separation of DNA fragments by electrophoresis (E) → Transferring of separated DNA fragments to synthetic membranes (C) → Hybridisation using a labelled VNTR probe (B) → Detection of hybridised DNA fragments by autoradiography (D).
Q20. Which of the following statements are correct with reference to packaging of DNA helix?
A. Histones are organized to form a unit of eight molecules called histone octamer.
B. Histones are negatively charged basic proteins.
C. Histones are rich in the basic amino acid residues - lysine and arginine.
D. The positively charged DNA is wrapped around the histone octamer to form nucleosome.
E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins.
Choose the correct answer from the options given below:
A. Histones are organized to form a unit of eight molecules called histone octamer.
B. Histones are negatively charged basic proteins.
C. Histones are rich in the basic amino acid residues - lysine and arginine.
D. The positively charged DNA is wrapped around the histone octamer to form nucleosome.
E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins.
Choose the correct answer from the options given below:
Correct Answer: (a)
Eight Histones are combined to form a histone octamer (A). Histones are positively charged basic proteins that are rich in basic amino acids (lysine and arginine) (C). The negatively charged DNA is wrapped around the histone octamer to form nucleosome (D is incorrect as DNA is negatively charged). The packaging of chromatin at higher levels require NHC (non histone chromosomal) proteins (E).
Eight Histones are combined to form a histone octamer (A). Histones are positively charged basic proteins that are rich in basic amino acids (lysine and arginine) (C). The negatively charged DNA is wrapped around the histone octamer to form nucleosome (D is incorrect as DNA is negatively charged). The packaging of chromatin at higher levels require NHC (non histone chromosomal) proteins (E).
Q21. Find the incorrect statement(s) about photosynthesis from the following:
A. The water splitting complex is associated with PS I.
B. C4 plants use the C3 pathway of CO2 fixation as the main biosynthetic pathway.
C. In C4 plants, photorespiration does not occur.
D. C3 plants exhibit 'Kranz' anatomy.
E. ATP synthesis in chloroplast occurs through chemiosmosis.
Choose the answer from the options given below:
A. The water splitting complex is associated with PS I.
B. C4 plants use the C3 pathway of CO2 fixation as the main biosynthetic pathway.
C. In C4 plants, photorespiration does not occur.
D. C3 plants exhibit 'Kranz' anatomy.
E. ATP synthesis in chloroplast occurs through chemiosmosis.
Choose the answer from the options given below:
Correct Answer: (d)
The water splitting complex is associated with PS-II (A incorrect). C₃ pathway is the main biosynthetic pathway for CO₂ fixation in both C₃ and C₄ plants (B correct). C₃ plants do not exhibit 'kranz' anatomy (D incorrect). ATP synthesis in chloroplast occurs by chemiosmosis (E correct).
The water splitting complex is associated with PS-II (A incorrect). C₃ pathway is the main biosynthetic pathway for CO₂ fixation in both C₃ and C₄ plants (B correct). C₃ plants do not exhibit 'kranz' anatomy (D incorrect). ATP synthesis in chloroplast occurs by chemiosmosis (E correct).
Q22. Arrange the following steps of somatic hybridisation in a correct sequence.
A. Digestion of cell walls.
B. Isolation of naked protoplasts.
C. Fusion of protoplasts to get hybrid protoplast.
D. Isolation of single cells from two different varieties of plants.
E. Growing of hybrid protoplast to form a new plant.
Choose the correct answer from the options given below:
A. Digestion of cell walls.
B. Isolation of naked protoplasts.
C. Fusion of protoplasts to get hybrid protoplast.
D. Isolation of single cells from two different varieties of plants.
E. Growing of hybrid protoplast to form a new plant.
Choose the correct answer from the options given below:
Correct Answer: (a)
The correct sequence for somatic hybridisation is: Isolation of single cells from two different varieties of plants (D) → Digestion of cell walls (A) → Isolation of naked protoplasts (B) → Fusion of protoplasts to get hybrid protoplast (C) → Growing of hybrid protoplast to form a new plant (E).
The correct sequence for somatic hybridisation is: Isolation of single cells from two different varieties of plants (D) → Digestion of cell walls (A) → Isolation of naked protoplasts (B) → Fusion of protoplasts to get hybrid protoplast (C) → Growing of hybrid protoplast to form a new plant (E).
Q23. Match List-I with List-II:
| List-I | List-II |
|---|---|
| A. Conjunctive tissue | I. Specialised cells in the vicinity of guard cells |
| B. Casparian strips | II. Endodermal cells rich in starch |
| C. Subsidiary cells | III. Tissue between xylem and phloem |
| D. Starch sheath | IV. Endodermal cells with suberin deposition |
Correct Answer: (c)
(A) Conjunctive tissue is the tissue between xylem and phloem (III). (B) Casparian strips are found in endodermal cells, they are suberin depositions in the cell wall (IV). (C) Subsidiary cells are specialised cells in the vicinity of guard cells (I). (D) Starch sheath is another name for endodermal cells rich in starch (II).
(A) Conjunctive tissue is the tissue between xylem and phloem (III). (B) Casparian strips are found in endodermal cells, they are suberin depositions in the cell wall (IV). (C) Subsidiary cells are specialised cells in the vicinity of guard cells (I). (D) Starch sheath is another name for endodermal cells rich in starch (II).
Q24. Which one of the following is not a characteristic of plant cells in the phase of elongation?
Correct Answer: (d)
Phase of elongation is characterised by new cell wall deposition, increased vacuolation and enlargement of cell. Presence of large conspicuous nuclei is a feature of cells in the meristematic phase.
Phase of elongation is characterised by new cell wall deposition, increased vacuolation and enlargement of cell. Presence of large conspicuous nuclei is a feature of cells in the meristematic phase.
Q25. Match List-I with List-II:
| List-I (Growth Regulator) | List-II (Function/Effect) |
|---|---|
| A. 2,4-D | I. Brewing industry |
| B. GA3 | II. Stimulation of stomatal closure |
| C. Kinetin | III. Herbicide |
| D. ABA | IV. Nutrient mobilisation |
Correct Answer: (a)
2,4-D (Auxin) is a herbicide (III). GA₃ (Gibberellic Acid) is used in brewing industry (I). Kinetin (Cytokinin) causes nutrient mobilisation (IV). ABA (Abscisic Acid) stimulates stomatal closure (II).
2,4-D (Auxin) is a herbicide (III). GA₃ (Gibberellic Acid) is used in brewing industry (I). Kinetin (Cytokinin) causes nutrient mobilisation (IV). ABA (Abscisic Acid) stimulates stomatal closure (II).
Q26. The enzyme required for carboxylation in the Calvin cycle is
Correct Answer: (c)
RuBisCO (RuBP carboxylase-oxygenase) is the enzyme required for carboxylation in the calvin cycle.
RuBisCO (RuBP carboxylase-oxygenase) is the enzyme required for carboxylation in the calvin cycle.
Q27. How many ATP and NADPH molecules are required to make one molecule of glucose through the Calvin pathway?
Correct Answer: (a)
Each turn of Calvin pathway utilizes 3 ATP and NADPH+H+ molecules for fixation of 1 CO₂ molecule. So for 1 Glucose 6 turns are required, hence 18 ATP and 12 NADPH+H+ are required for glucose synthesis.
Each turn of Calvin pathway utilizes 3 ATP and NADPH+H+ molecules for fixation of 1 CO₂ molecule. So for 1 Glucose 6 turns are required, hence 18 ATP and 12 NADPH+H+ are required for glucose synthesis.
Q28. Which of the following floral formula is the correct floral formula of Solanaceae family?
Correct Answer: (b)
The floral formula for Solanaceae is ⊕ ⚥ K(5) C(5) A5 G(2). It shows actinomorphic, bisexual, pentamerous flower with epipetalous condition. Generally in Solanaceae, calyx (K) and Corolla (C) shows fusion in sepals and petals respectively. Hence the floral formula must exhibit gamosepalous and gamopetalous condition.
The floral formula for Solanaceae is ⊕ ⚥ K(5) C(5) A5 G(2). It shows actinomorphic, bisexual, pentamerous flower with epipetalous condition. Generally in Solanaceae, calyx (K) and Corolla (C) shows fusion in sepals and petals respectively. Hence the floral formula must exhibit gamosepalous and gamopetalous condition.
Q29. Which of the following is an in situ conservation method?
Correct Answer: (a)
In-situ conservation is exemplified by sacred groves. Wildlife Safari parks, Botanical gardens, seed banks are examples of ex-situ conservation.
In-situ conservation is exemplified by sacred groves. Wildlife Safari parks, Botanical gardens, seed banks are examples of ex-situ conservation.
Q30. Which of the following statements are not true regarding restriction endonucleases?
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in E. coli.
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
Choose the answer from the options given below:
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in E. coli.
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
Choose the answer from the options given below:
Correct Answer: (d)
Statements C and D are incorrect. Restriction endonucleases usually cut the DNA slightly away from the centre of palindrome sites. They cannot remove nucleotides from the ends of the DNA fragment, which is function of restriction exonuclease.
Statements C and D are incorrect. Restriction endonucleases usually cut the DNA slightly away from the centre of palindrome sites. They cannot remove nucleotides from the ends of the DNA fragment, which is function of restriction exonuclease.
Q31. In racemose inflorescence,
Correct Answer: (d)
In racemose type of inflorescence, the main axis continues to grow and the flowers are borne laterally in an acropetal succession. On the contrary, in cymose type of inflorescence, the main axis terminates into a flower and hence, is limited in growth.
In racemose type of inflorescence, the main axis continues to grow and the flowers are borne laterally in an acropetal succession. On the contrary, in cymose type of inflorescence, the main axis terminates into a flower and hence, is limited in growth.
Q32. Arrange the following in the correct developmental sequence related to microsporogenesis :
A. Microspore tetrads
B. Sporogenous tissue
C. Pollen grains
D. Pollen mother cells
Choose the correct answer from the options given below :
A. Microspore tetrads
B. Sporogenous tissue
C. Pollen grains
D. Pollen mother cells
Choose the correct answer from the options given below :
Correct Answer: (b)
The correct developmental sequence related to microsporogenesis will be: Sporogenous tissue (B) → Pollen mother cell (D) → Microspore tetrads (A) → Pollen grains (C).
The correct developmental sequence related to microsporogenesis will be: Sporogenous tissue (B) → Pollen mother cell (D) → Microspore tetrads (A) → Pollen grains (C).
Q33. Identify the correct statements about biomolecules.
A. Lipids are generally water soluble.
B. Proteins are polypeptides.
C. Polysaccharides are long chains of sugars.
D. Adenine and guanine are substituted pyrimidines.
E. Almost all enzymes are proteins.
Choose the correct answer from the options given below :
A. Lipids are generally water soluble.
B. Proteins are polypeptides.
C. Polysaccharides are long chains of sugars.
D. Adenine and guanine are substituted pyrimidines.
E. Almost all enzymes are proteins.
Choose the correct answer from the options given below :
Correct Answer: (b)
Statements B, C and E are correct. Statements A and D are not true. Lipids are not water soluble. Adenine and guanine are substituted purines.
Statements B, C and E are correct. Statements A and D are not true. Lipids are not water soluble. Adenine and guanine are substituted purines.
Q34. Which of the following statements are true with reference to the sex-determination in honeybees?
A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker).
B. An unfertilized egg develops as a male by parthenogenesis.
C. A male has half the number of chromosomes than that of a female.
D. Males produce sperms by meiosis.
E. Honeybees have a haplodiploid sex-determination system.
Choose the correct answer from the options given below :
A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker).
B. An unfertilized egg develops as a male by parthenogenesis.
C. A male has half the number of chromosomes than that of a female.
D. Males produce sperms by meiosis.
E. Honeybees have a haplodiploid sex-determination system.
Choose the correct answer from the options given below :
Correct Answer: (a)
Statements A, B, C and E are correct. Males produce sperms by mitosis (not meiosis), so D is incorrect.
Statements A, B, C and E are correct. Males produce sperms by mitosis (not meiosis), so D is incorrect.
Q35. Heterophyllous development in response to environment is an example of which of the following phenomena?
Correct Answer: (d)
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g., heterophylly in cotton, coriander and larkspur.
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g., heterophylly in cotton, coriander and larkspur.
Q36. Which of the following statements are correct regarding amino acids?
A. They are substituted methanes.
B. Serine is an aromatic amino acid.
C. Valine is a neutral amino acid.
D. Lysine is an acidic amino acid.
Choose the correct answer from the options given below:
A. They are substituted methanes.
B. Serine is an aromatic amino acid.
C. Valine is a neutral amino acid.
D. Lysine is an acidic amino acid.
Choose the correct answer from the options given below:
Correct Answer: (c)
Statements A and C are correct while statements B and D are incorrect. Serine is an alcoholic amino acid and lysine is a basic amino acid.
Statements A and C are correct while statements B and D are incorrect. Serine is an alcoholic amino acid and lysine is a basic amino acid.
Q37. "The Evil Quartet" of biodiversity loss includes which of the following?
Correct Answer: (c)
'The Evil Quartet' is the Sobriquet used to describe the four major causes of biodiversity loss which includes Habitat loss and fragmentation, over-exploitation; Alien species invasions and Co-extinctions.
'The Evil Quartet' is the Sobriquet used to describe the four major causes of biodiversity loss which includes Habitat loss and fragmentation, over-exploitation; Alien species invasions and Co-extinctions.
Q38. Match List-I with List-II:
| List-I (Process) | List-II (Location) |
|---|---|
| A. Glycolysis | I. Inner mitochondrial membrane |
| B. ETS | II. Mitochondrial matrix |
| C. Accumulation of protons | III. Cytoplasm |
| D. Krebs' cycle | IV. Intermembrane space |
Correct Answer: (c)
The site of glycolysis is the cytoplasm (III). Electron transport system is localized in inner mitochondrial membrane (I). Accumulation of protons occur in intermembrane space (IV). Krebs's Cycle takes place in mitochondrial matrix (II).
The site of glycolysis is the cytoplasm (III). Electron transport system is localized in inner mitochondrial membrane (I). Accumulation of protons occur in intermembrane space (IV). Krebs's Cycle takes place in mitochondrial matrix (II).
Q39. Which one of the following is a triploid cell?
Correct Answer: (d)
Synergid is haploid, Zygote is diploid, Central cell initially contains two polar nuclei which fuse just before fertilization to form a secondary nucleus (2n). Primary endosperm cell (PEC) is triploid.
Synergid is haploid, Zygote is diploid, Central cell initially contains two polar nuclei which fuse just before fertilization to form a secondary nucleus (2n). Primary endosperm cell (PEC) is triploid.
Q40. Which one of the following types of pollination brings genetically different types of pollen grains to the stigma?
Correct Answer: (b)
Transfer of pollen grains from anther to stigma of a different plant is known as Xenogamy. This is the only type of pollination which brings genetically different types of pollen grains to stigma. Cleistogamy flowers are invariably autogamous.
Transfer of pollen grains from anther to stigma of a different plant is known as Xenogamy. This is the only type of pollination which brings genetically different types of pollen grains to stigma. Cleistogamy flowers are invariably autogamous.
Q41. Match List I with List II:
| List-I (Placentation) | List-II (Example) |
|---|---|
| A. Marginal | I. Mustard |
| B. Axile | II. Pea |
| C. Parietal | III. Marigold |
| D. Basal | IV. Lemon |
Correct Answer: (a)
Marginal placentation is found in pea (II), axile placentation is found in lemon (IV), parietal placentation is found in mustard (I) and basal placentation is found in marigold (III).
Marginal placentation is found in pea (II), axile placentation is found in lemon (IV), parietal placentation is found in mustard (I) and basal placentation is found in marigold (III).
Q42. The main criteria used for Five Kingdom Classification proposed By R.H. Whittaker (1969) included :
A. Cell structure
B. Body organization
C. Presence of flagellum
D. Reproduction
E. Phylogenetic relationships
Choose the correct answer from the options given below :
A. Cell structure
B. Body organization
C. Presence of flagellum
D. Reproduction
E. Phylogenetic relationships
Choose the correct answer from the options given below :
Correct Answer: (c)
The main criteria for five kingdom classification used by (R.H. Whittaker) includes cell structure, body organization, mode of nutrition, reproduction and phylogenetic relationships. Presence of flagellum is not a main criteria.
The main criteria for five kingdom classification used by (R.H. Whittaker) includes cell structure, body organization, mode of nutrition, reproduction and phylogenetic relationships. Presence of flagellum is not a main criteria.
Q43. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Trypsin | I. Intercellular ground substance |
| B. Morphine | II. Lectin |
| C. Concanavalin A | III. Enzyme |
| D. Collagen | IV. Alkaloid |
Correct Answer: (a)
Trypsin is a proteolytic enzyme (III). Morphine is a secondary metabolite that belongs to the category of alkaloid (IV). Concanavalin A is a lectin (II). Collagen acts as an intercellular ground substance (I).
Trypsin is a proteolytic enzyme (III). Morphine is a secondary metabolite that belongs to the category of alkaloid (IV). Concanavalin A is a lectin (II). Collagen acts as an intercellular ground substance (I).
Q44. Which of the following statements are correct with respect to DNA separation, isolation and visualization?
A. The cutting of DNA is done by molecular scissors.
B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis.
C. The separated DNA fragments can be seen without staining when exposed to UV light.
D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light.
Choose the correct answer from the options given below :
A. The cutting of DNA is done by molecular scissors.
B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis.
C. The separated DNA fragments can be seen without staining when exposed to UV light.
D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light.
Choose the correct answer from the options given below :
Correct Answer: (b)
The cutting of DNA is possible by the use of restriction enzymes (molecular scissors) that results in the fragments of DNA (A). These fragments of DNA can be separated by a technique known as gel electrophoresis (B). The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation (C and D incorrect).
The cutting of DNA is possible by the use of restriction enzymes (molecular scissors) that results in the fragments of DNA (A). These fragments of DNA can be separated by a technique known as gel electrophoresis (B). The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation (C and D incorrect).
Q45. Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule?
Correct Answer: (b)
Sickle cell anaemia is an autosome linked recessive trait that can be transmitted from parents to offsprings when both partners are carrier for the gene. The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at sixth position of beta globin chain of haemoglobin molecule.
Sickle cell anaemia is an autosome linked recessive trait that can be transmitted from parents to offsprings when both partners are carrier for the gene. The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at sixth position of beta globin chain of haemoglobin molecule.
Q46. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Cortisol | I. Stimulates the formation of alveoli in mammary glands |
| B. Aldosterone | II. Produces anti-inflammatory reactions |
| C. Cholecystokinin | III. Stimulates reabsorption of Na⁺ and water from renal tubule |
| D. Progesterone | IV. Stimulates secretion of pancreatic enzymes and bile juice |
Correct Answer: (c)
Cortisol produces anti-inflammatory reactions (II). Aldosterone stimulates reabsorption of Na⁺ and water from renal tubule (III). Cholecystokinin (CCK) stimulates secretion of pancreatic enzymes and bile juice (IV). Progesterone stimulates formation of alveoli in mammary glands (I).
Cortisol produces anti-inflammatory reactions (II). Aldosterone stimulates reabsorption of Na⁺ and water from renal tubule (III). Cholecystokinin (CCK) stimulates secretion of pancreatic enzymes and bile juice (IV). Progesterone stimulates formation of alveoli in mammary glands (I).
Q47. Arrange the following events occurring in Renin-Angiotensin mechanism in the correct order:
A. Increase in blood pressure and Glomerular filtration rate
B. Reabsorption of Na⁺ and water from distal parts of tubule due to Aldosterone
C. Fall in Glomerular filtration rate
D. Vasoconstriction by Angiotensin II and release of Aldosterone.
E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II.
Choose the correct answer from the options given below:
A. Increase in blood pressure and Glomerular filtration rate
B. Reabsorption of Na⁺ and water from distal parts of tubule due to Aldosterone
C. Fall in Glomerular filtration rate
D. Vasoconstriction by Angiotensin II and release of Aldosterone.
E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II.
Choose the correct answer from the options given below:
Correct Answer: (d)
The correct sequence is: Fall in GFR (C) → Renin converts Angiotensinogen to Angiotensin I and II (E) → Vasoconstriction and Aldosterone release (D) → Reabsorption of Na⁺ and water (B) → Increase in blood pressure and GFR (A).
The correct sequence is: Fall in GFR (C) → Renin converts Angiotensinogen to Angiotensin I and II (E) → Vasoconstriction and Aldosterone release (D) → Reabsorption of Na⁺ and water (B) → Increase in blood pressure and GFR (A).
Q48. In humans, respiration occurs in the following steps. Arrange these steps in the correct order.
A. Diffusion of O₂ and CO₂ between blood and tissues
B. Diffusion of O₂ and CO₂ across alveolar membrane
C. Pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out
D. Cellular respiration
E. Transport of gases by the blood
Choose the correct answer from the options given below
A. Diffusion of O₂ and CO₂ between blood and tissues
B. Diffusion of O₂ and CO₂ across alveolar membrane
C. Pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out
D. Cellular respiration
E. Transport of gases by the blood
Choose the correct answer from the options given below
Correct Answer: (c)
Respiration involves: (C) Breathing/pulmonary ventilation → (B) Diffusion of gases across alveolar membrane → (E) Transport of gases by the blood → (A) Diffusion of O₂ and CO₂ between blood and tissues → (D) Cellular respiration.
Respiration involves: (C) Breathing/pulmonary ventilation → (B) Diffusion of gases across alveolar membrane → (E) Transport of gases by the blood → (A) Diffusion of O₂ and CO₂ between blood and tissues → (D) Cellular respiration.
Q49. The following are the stages of life cycle of Plasmodium. Arrange the stages in the proper order.
A. The parasites reproduce asexually in RBCs, bursting the cells.
B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood.
C. Gametocytes develop in RBCs.
D. Sporozoites reach the liver through the blood.
E. Female mosquito injects sporozoites into humans during bite.
Choose the correct answer from the options given below:
A. The parasites reproduce asexually in RBCs, bursting the cells.
B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood.
C. Gametocytes develop in RBCs.
D. Sporozoites reach the liver through the blood.
E. Female mosquito injects sporozoites into humans during bite.
Choose the correct answer from the options given below:
Correct Answer: (a)
Plasmodium life cycle: Female mosquito injects sporozoites (E) → Sporozoites reach liver (D) → Asexual reproduction in liver cells, release into blood (B) → Asexual reproduction in RBCs (A) → Gametocytes develop in RBCs (C).
Plasmodium life cycle: Female mosquito injects sporozoites (E) → Sporozoites reach liver (D) → Asexual reproduction in liver cells, release into blood (B) → Asexual reproduction in RBCs (A) → Gametocytes develop in RBCs (C).
Q50. Insertion of a foreign DNA at BamHI site in an E.coli cloning vector pBR322 results in the loss of antibiotic resistance towards:
Correct Answer: (c)
If one ligates a foreign DNA at the BamHI site of tetracycline resistance gene in the vector pBR322, the recombinant plasmid will lose tetracycline resistance due to insertion of foreign DNA.
If one ligates a foreign DNA at the BamHI site of tetracycline resistance gene in the vector pBR322, the recombinant plasmid will lose tetracycline resistance due to insertion of foreign DNA.
Q51. The following reaction depicts the activity of a particular class of enzymes :
Identify the enzymes class 'E' from the following options :
Identify the enzymes class 'E' from the following options :
Correct Answer: (c)
Lyases are the enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds. Transferases catalyse transfer of a group between substrates. Isomerases catalyse inter-conversion of isomers. Ligases catalyse linking together of 2 compounds.
Lyases are the enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds. Transferases catalyse transfer of a group between substrates. Isomerases catalyse inter-conversion of isomers. Ligases catalyse linking together of 2 compounds.
Q52. The specific receptors for neurotransmitters in a synapse are present on
Correct Answer: (d)
The specific receptors for neurotransmitters in a synapse are present on post-synaptic membrane.
The specific receptors for neurotransmitters in a synapse are present on post-synaptic membrane.
Q53. What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively?
Correct Answer: (a)
Mother: Iᴬi (heterozygous A) × Father: Iᴮi (heterozygous B). Offspring: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O). Probability of O blood group = 1/4 = 25%.
Mother: Iᴬi (heterozygous A) × Father: Iᴮi (heterozygous B). Offspring: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O). Probability of O blood group = 1/4 = 25%.
Q54. Match List-I with List-II:
| List-I (Respiratory Volume) | List-II (Capacity in mL) |
|---|---|
| A. ERV (Expiratory Reserve Volume) | I. 2500 – 3000 mL |
| B. RV (Residual Volume) | II. 500 mL |
| C. IRV (Inspiratory Reserve Volume) | III. 1000 – 1100 mL |
| D. TV (Tidal Volume) | IV. 1100 – 1200 mL |
Correct Answer: (a)
ERV (Expiratory Reserve Volume) - 1000-1100 mL (III); RV (Residual Volume) - 1100-1200 mL (IV); IRV (Inspiratory Reserve Volume) - 2500-3000 mL (I); TV (Tidal Volume) - 500 mL (II).
ERV (Expiratory Reserve Volume) - 1000-1100 mL (III); RV (Residual Volume) - 1100-1200 mL (IV); IRV (Inspiratory Reserve Volume) - 2500-3000 mL (I); TV (Tidal Volume) - 500 mL (II).
Q55. Which of the following is not an example of convergent evolution?
Correct Answer: (c)
Fore limbs of whales and bats are examples of divergent evolution that show homology.
Fore limbs of whales and bats are examples of divergent evolution that show homology.
Q56. Male frogs can be distinguished from female frogs due to the presence of
A. Bulging eyes
B. Vocal sacs
C. Webbed digits in feet
D. Copulatory pad on first digit of fore limbs
E. Olive green-coloured skin with dark irregular spots
Choose the correct answer from the options given below
A. Bulging eyes
B. Vocal sacs
C. Webbed digits in feet
D. Copulatory pad on first digit of fore limbs
E. Olive green-coloured skin with dark irregular spots
Choose the correct answer from the options given below
Correct Answer: (d)
Male frogs can be distinguished from female frogs due to the presence of vocal sacs and copulatory pad on first digit of fore limbs. Bulging eyes, webbed digit in feet and olive green-coloured skin with dark irregular spots are common in both male and female frogs.
Male frogs can be distinguished from female frogs due to the presence of vocal sacs and copulatory pad on first digit of fore limbs. Bulging eyes, webbed digit in feet and olive green-coloured skin with dark irregular spots are common in both male and female frogs.
Q57. A group of researchers procured some fish like animals and upon investigation the following characters were observed:
A. Endoskeleton was made of cartilage.
B. Ectoparasitic; as they were found attached on fish skin with their circular sucking mouth.
C. Paired fins and scales were absent, but 7 pairs of gill slits were present.
Which of the following species of animals did they consider to fit best with these characters?
A. Endoskeleton was made of cartilage.
B. Ectoparasitic; as they were found attached on fish skin with their circular sucking mouth.
C. Paired fins and scales were absent, but 7 pairs of gill slits were present.
Which of the following species of animals did they consider to fit best with these characters?
Correct Answer: (b)
Petromyzon sp. have cartilaginous endoskeleton. They have circular sucking mouth. They are ectoparasites on some fishes. Their body is devoid of scales and paired fins. They have 7 pairs (6-15 pairs) of gill slits. Scoliodon and Exocoetus are not parasites.
Petromyzon sp. have cartilaginous endoskeleton. They have circular sucking mouth. They are ectoparasites on some fishes. Their body is devoid of scales and paired fins. They have 7 pairs (6-15 pairs) of gill slits. Scoliodon and Exocoetus are not parasites.
Q58. Match List I with List II with respect to chronology of evolution of life forms:
| List-I | List-II |
|---|---|
| A. About 65 mya | I. Jawless fish probably evolved |
| B. About 500 mya | II. The dinosaurs suddenly disappeared from the earth |
| C. About 350 mya | III. Seaweeds and few plants probably existed |
| D. About 320 mya | IV. Invertebrates were formed and became active |
Correct Answer: (d)
About 65 mya - Dinosaurs disappeared (II); About 500 mya - Invertebrates formed (IV); About 350 mya - Jawless fish evolved (I); About 320 mya - Seaweeds and few plants existed (III).
About 65 mya - Dinosaurs disappeared (II); About 500 mya - Invertebrates formed (IV); About 350 mya - Jawless fish evolved (I); About 320 mya - Seaweeds and few plants existed (III).
Q59. Match List I and List II:
| List-I | List-II |
|---|---|
| A. Progestasert | I. Barrier made of rubber used by females |
| B. Multiload 375 | II. Oral contraceptive |
| C. Diaphragm | III. Hormone releasing IUD |
| D. Saheli | IV. Copper releasing IUD |
Correct Answer: (a)
Progestasert - Hormone releasing IUD (III); Multiload 375 - Copper releasing IUD (IV); Diaphragm - Barrier made of rubber used by females (I); Saheli - Oral contraceptive (II).
Progestasert - Hormone releasing IUD (III); Multiload 375 - Copper releasing IUD (IV); Diaphragm - Barrier made of rubber used by females (I); Saheli - Oral contraceptive (II).
Q60. The WBC count of a person's blood sample is 8000/cu mm. How many eosinophils and lymphocytes would be in the same blood sample approximately?
Correct Answer: (b)
Eosinophils constitute 2–3% of total WBCs. 2-3% of 8000 = 160–240/cu mm. Lymphocytes constitute 20–25% of total WBCs. 20-25% of 8000 = 1600–2000/cu mm.
Eosinophils constitute 2–3% of total WBCs. 2-3% of 8000 = 160–240/cu mm. Lymphocytes constitute 20–25% of total WBCs. 20-25% of 8000 = 1600–2000/cu mm.
Q61. Match List I with List II:
| List-I (Drug) | List-II (Effect) |
|---|---|
| A. Nicotine | I. Causes sense of euphoria and increased energy |
| B. Morphine | II. Stimulates adrenal gland to release catecholamines into blood circulation |
| C. Heroin | III. Effective sedative and painkiller |
| D. Cocaine | IV. A depressant; slows down body function |
Correct Answer: (c)
Nicotine stimulates adrenal gland to release catecholamines (II). Morphine is an effective sedative and painkiller (III). Heroin is a depressant; slows down body function (IV). Cocaine causes euphoria and increased energy (I).
Nicotine stimulates adrenal gland to release catecholamines (II). Morphine is an effective sedative and painkiller (III). Heroin is a depressant; slows down body function (IV). Cocaine causes euphoria and increased energy (I).
Q62. The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of
Correct Answer: (a)
The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of emphysema. Transgenic models exist for the study of other diseases, such as Alzheimer's disease, rheumatoid arthritis and cystic fibrosis.
The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of emphysema. Transgenic models exist for the study of other diseases, such as Alzheimer's disease, rheumatoid arthritis and cystic fibrosis.
Q63. Select the set of fishes which belong to the class Osteichthyes:
Correct Answer: (d)
Flying fish (Exocoetus) is a marine bony fish. Angel fish (Pterophyllum) and fighting fish (Betta) are aquarium bony fishes. Devil fish (Octopus) and cuttlefish (Sepia) are molluscs; Hag fish (Myxine) is a cyclostome; Saw fish (Pristis) and dog fish (Scoliodon) are cartilaginous fishes; Star fish (Asterias) is an echinoderm.
Flying fish (Exocoetus) is a marine bony fish. Angel fish (Pterophyllum) and fighting fish (Betta) are aquarium bony fishes. Devil fish (Octopus) and cuttlefish (Sepia) are molluscs; Hag fish (Myxine) is a cyclostome; Saw fish (Pristis) and dog fish (Scoliodon) are cartilaginous fishes; Star fish (Asterias) is an echinoderm.
Q64. Select the incorrect statements from the following:
A. Digestive system in Platyhelminthes is incomplete.
B. Bilateral symmetry is a characteristic feature of adult Echinoderms.
C. Pseudocoelom is possessed by Aschelminthes.
D. Notochord is persistent throughout life in the class Chondrichthyes.
E. Members of class Reptilia maintain a constant body temperature.
Choose the answer from the options given below:
A. Digestive system in Platyhelminthes is incomplete.
B. Bilateral symmetry is a characteristic feature of adult Echinoderms.
C. Pseudocoelom is possessed by Aschelminthes.
D. Notochord is persistent throughout life in the class Chondrichthyes.
E. Members of class Reptilia maintain a constant body temperature.
Choose the answer from the options given below:
Correct Answer: (b)
(A) Correct: Platyhelminthes have incomplete digestive system. (B) Incorrect: Bilateral symmetry is in larvae; adults have radial symmetry. (C) Correct: Aschelminthes have pseudocoelom. (D) Correct: Notochord is persistent in Chondrichthyes. (E) Incorrect: Reptiles are cold-blooded. Incorrect statements are B and E.
(A) Correct: Platyhelminthes have incomplete digestive system. (B) Incorrect: Bilateral symmetry is in larvae; adults have radial symmetry. (C) Correct: Aschelminthes have pseudocoelom. (D) Correct: Notochord is persistent in Chondrichthyes. (E) Incorrect: Reptiles are cold-blooded. Incorrect statements are B and E.
Q65. Non-membrane bound cell organelles found in both prokaryotic and eukaryotic cells are
Correct Answer: (a)
Ribosome is a non-membrane bound cell organelle, found in both prokaryotic and eukaryotic cells.
Ribosome is a non-membrane bound cell organelle, found in both prokaryotic and eukaryotic cells.
Q66. Which of the following equations depicts Verhulst-Pearl logistic population growth?
Correct Answer: (b)
Verhulst-Pearl logistic population growth is depicted by the equation dN/dt = rN((K-N)/K). Where K = Carrying capacity, r = Intrinsic rate of natural increase, N = Population density at time 't'.
Verhulst-Pearl logistic population growth is depicted by the equation dN/dt = rN((K-N)/K). Where K = Carrying capacity, r = Intrinsic rate of natural increase, N = Population density at time 't'.
Q67. Select the incorrect statement with reference to Rh grouping.
A. Erythroblastosis foetalis is a condition observed having foetus with Rh-ve blood and mother with Rh+ve blood.
B. Rh antigen is observed on RBCs in the majority of human beings.
C. Before blood transfusion, Rh group should also be matched.
D. Rh incompatibility is observed when a pregnant mother is Rh-ve and the foetus is Rh+ve.
E. Erythroblastosis foetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child.
Choose the answer from the options given below:
A. Erythroblastosis foetalis is a condition observed having foetus with Rh-ve blood and mother with Rh+ve blood.
B. Rh antigen is observed on RBCs in the majority of human beings.
C. Before blood transfusion, Rh group should also be matched.
D. Rh incompatibility is observed when a pregnant mother is Rh-ve and the foetus is Rh+ve.
E. Erythroblastosis foetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child.
Choose the answer from the options given below:
Correct Answer: (c)
(A) Incorrect: Erythroblastosis foetalis occurs when mother is Rh-ve and foetus is Rh+ve. (B) Correct: Rh antigen is on RBCs of majority. (C) Correct: Rh grouping should be matched before transfusion. (D) Correct: Rh incompatibility when mother Rh-ve and foetus Rh+ve. (E) Incorrect: Anti-Rh antibodies are administered after delivery of first child, not second. Incorrect statements are A and E.
(A) Incorrect: Erythroblastosis foetalis occurs when mother is Rh-ve and foetus is Rh+ve. (B) Correct: Rh antigen is on RBCs of majority. (C) Correct: Rh grouping should be matched before transfusion. (D) Correct: Rh incompatibility when mother Rh-ve and foetus Rh+ve. (E) Incorrect: Anti-Rh antibodies are administered after delivery of first child, not second. Incorrect statements are A and E.
Q68. Match List I with List II:
| List-I (Bioactive molecules) | List-II (Importance) |
|---|---|
| A. Streptokinase | I. Immunosuppressive agent |
| B. Statins | II. Removal of clots from the blood vessels |
| C. Lipases | III. Blood cholesterol-lowering agent |
| D. Cyclosporin A | IV. Detergent formulations |
Correct Answer: (c)
Streptokinase → Removal of clots (II); Statins → Blood cholesterol-lowering agent (III); Lipases → Detergent formulations (IV); Cyclosporin A → Immunosuppressive agent (I).
Streptokinase → Removal of clots (II); Statins → Blood cholesterol-lowering agent (III); Lipases → Detergent formulations (IV); Cyclosporin A → Immunosuppressive agent (I).
Q69. Match List I with List II:
| List-I | List-II |
|---|---|
| A. Molluscs | I. Pulmonary respiration only |
| B. Reptiles | II. Branchial respiration |
| C. Adult amphibians | III. Cellular respiration |
| D. Amoeba | IV. Pulmonary and cutaneous respiration |
Correct Answer: (a)
Molluscs → Branchial respiration (II); Reptiles → Pulmonary respiration only (I); Adult amphibians → Pulmonary and cutaneous respiration (IV); Amoeba → Cellular respiration (III).
Molluscs → Branchial respiration (II); Reptiles → Pulmonary respiration only (I); Adult amphibians → Pulmonary and cutaneous respiration (IV); Amoeba → Cellular respiration (III).
Q70. The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ______.
Correct Answer: (a)
Sickle cell anaemia is caused by substitution of Glutamic acid by Valine at sixth position of beta globin chain. The substitution results due to single base substitution from GAG to GUG. Hence, GUG is responsible for the change in the shape of RBC.
Sickle cell anaemia is caused by substitution of Glutamic acid by Valine at sixth position of beta globin chain. The substitution results due to single base substitution from GAG to GUG. Hence, GUG is responsible for the change in the shape of RBC.
Q71. Choose the correct statements regarding muscle contraction.
A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre.
B. The neural signal generates an action potential which causes the release of Ca²⁺ into sarcoplasm.
C. Increase in Ca²⁺ inactivates the actin for breaking cross bridges.
D. Actin binds to the myosin head to form a cross bridge.
E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band.
Choose the correct answer from the options given below :
A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre.
B. The neural signal generates an action potential which causes the release of Ca²⁺ into sarcoplasm.
C. Increase in Ca²⁺ inactivates the actin for breaking cross bridges.
D. Actin binds to the myosin head to form a cross bridge.
E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band.
Choose the correct answer from the options given below :
Correct Answer: (d)
Statements A, B, D and E are correct. C is incorrect: Increase in Ca²⁺ activates actin for cross bridge formation, not inactivates.
Statements A, B, D and E are correct. C is incorrect: Increase in Ca²⁺ activates actin for cross bridge formation, not inactivates.
Q72. Which of the following statements are correct with reference to human endoskeleton?
A. Human skull is monoconvulic.
B. The joint between any two adjoining vertebrae is a cartilaginous joint.
C. In human beings, the number of cervical vertebrae is seven.
D. All ribs except the last 2 pairs are bicephalic.
E. The occipital bone of skull is articulated with atlas vertebrae.
Choose the correct answer from the options given below:
A. Human skull is monoconvulic.
B. The joint between any two adjoining vertebrae is a cartilaginous joint.
C. In human beings, the number of cervical vertebrae is seven.
D. All ribs except the last 2 pairs are bicephalic.
E. The occipital bone of skull is articulated with atlas vertebrae.
Choose the correct answer from the options given below:
Correct Answer: (b)
(A) Incorrect: Human skull is diconvulic (two occipital condyles). (B) Correct: Joint between vertebrae is cartilaginous. (C) Correct: 7 cervical vertebrae. (D) Incorrect: All ribs are bicephalic. (E) Correct: Occipital bone articulates with atlas.
(A) Incorrect: Human skull is diconvulic (two occipital condyles). (B) Correct: Joint between vertebrae is cartilaginous. (C) Correct: 7 cervical vertebrae. (D) Incorrect: All ribs are bicephalic. (E) Correct: Occipital bone articulates with atlas.
Q73. Spermatogonia undergo a series of cell divisions to produce sperms. Select the correct statements from the following :
A. Spermatogonia always undergo meiotic cell division.
B. Primary spermatocytes divide mitotically to produce secondary spermatocytes.
C. Secondary spermatocytes, through their second meiotic division, produce haploid spermatids.
D. Spermatids produce spermatozoa through mitosis.
E. Spermatids transform into spermatozoa by spermiogenesis.
Choose the correct answer from the options given below:
A. Spermatogonia always undergo meiotic cell division.
B. Primary spermatocytes divide mitotically to produce secondary spermatocytes.
C. Secondary spermatocytes, through their second meiotic division, produce haploid spermatids.
D. Spermatids produce spermatozoa through mitosis.
E. Spermatids transform into spermatozoa by spermiogenesis.
Choose the correct answer from the options given below:
Correct Answer: (b)
(A) Incorrect: Spermatogonia undergo mitosis. (B) Incorrect: Primary spermatocytes undergo first meiotic division. (C) Correct: Secondary spermatocytes undergo second meiotic division to form haploid spermatids. (D) Incorrect: Spermatids transform via spermiogenesis, not mitosis. (E) Correct: Spermatids transform into spermatozoa by spermiogenesis. Correct statements are C and E.
(A) Incorrect: Spermatogonia undergo mitosis. (B) Incorrect: Primary spermatocytes undergo first meiotic division. (C) Correct: Secondary spermatocytes undergo second meiotic division to form haploid spermatids. (D) Incorrect: Spermatids transform via spermiogenesis, not mitosis. (E) Correct: Spermatids transform into spermatozoa by spermiogenesis. Correct statements are C and E.
Q74. The JGA (Juxta Glomerular Apparatus) is a special sensitive region formed by cellular modifications in related to the same nephron.
Correct Answer: (d)
JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.
JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.
Q75. Which one of the following is an appropriate example of sexual deceit?
Correct Answer: (b)
Female wasp and fig - Mutualism; Ophrys and bumblebee - Sexual deceit; Sea anemone and clown fish - Commensalism; Cuckoo and crow - Brood Parasitism.
Female wasp and fig - Mutualism; Ophrys and bumblebee - Sexual deceit; Sea anemone and clown fish - Commensalism; Cuckoo and crow - Brood Parasitism.
Q76. Choose the correct statements regarding frog's anatomy:
A. Hepatic portal system is the special venous connection between liver and intestine.
B. There are twelve pairs of cranial nerves arising from the brain.
C. The ureters and oviducts open separately into the cloaca in female frogs.
D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes.
E. Sinus venosus joins the right atrium of heart.
Choose the correct answer from the options given below:
A. Hepatic portal system is the special venous connection between liver and intestine.
B. There are twelve pairs of cranial nerves arising from the brain.
C. The ureters and oviducts open separately into the cloaca in female frogs.
D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes.
E. Sinus venosus joins the right atrium of heart.
Choose the correct answer from the options given below:
Correct Answer: (d)
(A) Correct: Hepatic portal system is between liver and intestine. (B) Incorrect: Frogs have 10 pairs of cranial nerves. (C) Correct: Ureters and oviducts open separately into cloaca in females. (D) Incorrect: Hindbrain consists of cerebellum and medulla oblongata; optic lobes are in midbrain. (E) Correct: Sinus venosus joins right atrium. Correct statements are A, C and E.
(A) Correct: Hepatic portal system is between liver and intestine. (B) Incorrect: Frogs have 10 pairs of cranial nerves. (C) Correct: Ureters and oviducts open separately into cloaca in females. (D) Incorrect: Hindbrain consists of cerebellum and medulla oblongata; optic lobes are in midbrain. (E) Correct: Sinus venosus joins right atrium. Correct statements are A, C and E.
Q77. Match List I with List II related to embryonic development at various months of pregnancy:
| List-I | List-II |
|---|---|
| A. The foetus movement starts and hair appears on the head | I. 24 weeks of pregnancy |
| B. The foetus develops limbs and digits | II. 20 weeks of pregnancy |
| C. The foetus develops external genital organs | III. 8 weeks of pregnancy |
| D. The foetus body is covered with fine hair; eyelids separate and eyelashes are formed | IV. 12 weeks of pregnancy |
Correct Answer: (b)
8 weeks (III) - limbs and digits develop (B); 12 weeks (IV) - external genital organs develop (C); 20 weeks (II) - foetus movement starts and hair appears (A); 24 weeks (I) - body covered with fine hair, eyelids separate (D).
8 weeks (III) - limbs and digits develop (B); 12 weeks (IV) - external genital organs develop (C); 20 weeks (II) - foetus movement starts and hair appears (A); 24 weeks (I) - body covered with fine hair, eyelids separate (D).
Q78. In a population of a grasshopper species, the chromosome number of some members is 23 and some other members possess 24 chromosomes. The 23 and 24 chromosome-bearing members in this species are
Correct Answer: (c)
In grasshoppers, males have 23 chromosomes (XO) and females have 24 chromosomes (XX). So 23 chromosomes = males, 24 = females.
In grasshoppers, males have 23 chromosomes (XO) and females have 24 chromosomes (XX). So 23 chromosomes = males, 24 = females.
Q79. In which animal do haploid cells divide mitotically to produce gametes?
Correct Answer: (a)
Male honeybees are haploid and form gametes by mitosis. Female honeybees form gametes by meiosis.
Male honeybees are haploid and form gametes by mitosis. Female honeybees form gametes by meiosis.
Q80. Arrange the following cell layers/structures around the female gamete, from outer to inner side :
A. Zona pellucida
B. Perivitelline space
C. Corona radiata
D. Plasma membrane of ovum
Choose the correct answer from the options given below :
A. Zona pellucida
B. Perivitelline space
C. Corona radiata
D. Plasma membrane of ovum
Choose the correct answer from the options given below :
Correct Answer: (c)
From outer to inner side: Corona radiata → Zona pellucida → Perivitelline space → Plasma membrane of ovum.
From outer to inner side: Corona radiata → Zona pellucida → Perivitelline space → Plasma membrane of ovum.
Q81. What is the reason behind production of large holes in 'Swiss Cheese'?
Correct Answer: (a)
The large holes in swiss cheese are due to production of large amount of CO₂ by the bacterium Propionibacterium sharmanii.
The large holes in swiss cheese are due to production of large amount of CO₂ by the bacterium Propionibacterium sharmanii.
Q82. The toxin proteins isolated from Bacillus thuringiensis, coded by which of the following genes would control cotton bollworms and corn borer, respectively?
Correct Answer: (a)
cryIAc controls cotton bollworms and cryIAb controls corn borer.
cryIAc controls cotton bollworms and cryIAb controls corn borer.
Q83. Ecological pyramids represent the relationship between the organisms at different trophic levels and they are generally inverted for:
Correct Answer: (d)
Pyramid of number in grassland is upright. Pyramid of energy in pond is upright. Pyramid of biomass in grassland is upright. Pyramid of biomass in sea is inverted.
Pyramid of number in grassland is upright. Pyramid of energy in pond is upright. Pyramid of biomass in grassland is upright. Pyramid of biomass in sea is inverted.
Q84. Choose the correct statement regarding GIFT to overcome infertility.
Correct Answer: (d)
GIFT (Gamete Intra Fallopian Transfer) involves transfer of an ovum from a donor into the fallopian tube of a female who cannot produce ovum but can provide suitable environment for fertilization and development.
GIFT (Gamete Intra Fallopian Transfer) involves transfer of an ovum from a donor into the fallopian tube of a female who cannot produce ovum but can provide suitable environment for fertilization and development.
Q85. Choose the correct statements regarding cell organelles and their inclusions.
A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria.
B. Rough endoplasmic reticulum bears ribosomes on its surface.
C. Both mitochondria and plastids have circular DNA.
D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton.
E. Mitochondrion is a single membrane-bound structure.
Choose the correct answer from the options given below :
A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria.
B. Rough endoplasmic reticulum bears ribosomes on its surface.
C. Both mitochondria and plastids have circular DNA.
D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton.
E. Mitochondrion is a single membrane-bound structure.
Choose the correct answer from the options given below :
Correct Answer: (d)
Endomembrane system does not include mitochondria (A incorrect). RER bears ribosomes (B correct). Mitochondria and plastids have circular DNA (C correct). Cytoskeleton is network of microtubules, microfilaments and intermediate filaments (D correct). Mitochondrion is double membrane-bound (E incorrect). Correct statements are B, C and D.
Endomembrane system does not include mitochondria (A incorrect). RER bears ribosomes (B correct). Mitochondria and plastids have circular DNA (C correct). Cytoskeleton is network of microtubules, microfilaments and intermediate filaments (D correct). Mitochondrion is double membrane-bound (E incorrect). Correct statements are B, C and D.
Q86. Select the correct statements regarding cell membrane in eukaryotic cell.
A. Membrane of human RBCs has approximately 52% protein.
B. Major phospholipids are arranged in a bilayer.
C. Extensions of the plasma membrane into the cell form mesosomes.
D. Tails towards the inner part of lipids are hydrophobic and thus protected from aqueous medium.
E. Glycocalyx is present on the outer surface of the plasma membrane.
Choose the correct answer from the options given below:
A. Membrane of human RBCs has approximately 52% protein.
B. Major phospholipids are arranged in a bilayer.
C. Extensions of the plasma membrane into the cell form mesosomes.
D. Tails towards the inner part of lipids are hydrophobic and thus protected from aqueous medium.
E. Glycocalyx is present on the outer surface of the plasma membrane.
Choose the correct answer from the options given below:
Correct Answer: (c)
In prokaryotes, extensions of plasma membrane form mesosomes. Eukaryotes lack such structure (C incorrect). Glycocalyx is present on outer surface of plasma membrane in prokaryotes, not eukaryotes (E incorrect). Statements A, B and D are correct.
In prokaryotes, extensions of plasma membrane form mesosomes. Eukaryotes lack such structure (C incorrect). Glycocalyx is present on outer surface of plasma membrane in prokaryotes, not eukaryotes (E incorrect). Statements A, B and D are correct.
Q87. Match List I with List II related to muscular/skeletal system:
| List-I | List-II |
|---|---|
| A. Tetany | I. Inflammation of joints |
| B. Arthritis | II. Autoimmune disorder affecting neuromuscular junction |
| C. Myasthenia gravis | III. Wild contraction in muscle due to low Ca²⁺ in body fluid |
| D. Muscular dystrophy | IV. Progressive degeneration of skeletal muscle |
Correct Answer: (a)
Tetany - wild contraction due to low Ca²⁺ (III); Arthritis - inflammation of joints (I); Myasthenia gravis - autoimmune disorder affecting NMJ (II); Muscular dystrophy - progressive degeneration of skeletal muscle (IV).
Tetany - wild contraction due to low Ca²⁺ (III); Arthritis - inflammation of joints (I); Myasthenia gravis - autoimmune disorder affecting NMJ (II); Muscular dystrophy - progressive degeneration of skeletal muscle (IV).
Q88. Evolution of human appears parallel to the progressive development of brain and language skills. As such, the evolution of individual species in the sequence of their appearance is:
Correct Answer: (a)
The correct chronological order: Ramapithecus → Australopithecines → Homo habilis → Homo erectus → Neanderthalensis → Homo sapiens.
The correct chronological order: Ramapithecus → Australopithecines → Homo habilis → Homo erectus → Neanderthalensis → Homo sapiens.
Q89. The flightless bird with forelimbs modified as paddle-like structures suited for swimming is known as:
Correct Answer: (b)
Neophron is vulture and Psittacula is a parrot. Both perform flight. Struthio is ostrich and Apetnodytes is penguin. Both are flightless birds. In penguins, forelimbs are modified into flippers (paddle like structure) and are used for swimming. In ostrichs, forelimbs are small and used for balance while running, not for swimming.
Neophron is vulture and Psittacula is a parrot. Both perform flight. Struthio is ostrich and Apetnodytes is penguin. Both are flightless birds. In penguins, forelimbs are modified into flippers (paddle like structure) and are used for swimming. In ostrichs, forelimbs are small and used for balance while running, not for swimming.
Q90. Choose the correct statements regarding population interactions between two species.
A. In both parasitism and commensalism, only one species benefits and the other species is harmed.
B. Both species benefit in mutualism.
C. Both species benefit in commensalism.
D. In parasitism, only one species benefits and the other species is harmed.
E. In amensalism, one species is harmed and the other is unaffected.
Choose the correct answer from the options given below:
A. In both parasitism and commensalism, only one species benefits and the other species is harmed.
B. Both species benefit in mutualism.
C. Both species benefit in commensalism.
D. In parasitism, only one species benefits and the other species is harmed.
E. In amensalism, one species is harmed and the other is unaffected.
Choose the correct answer from the options given below:
Correct Answer: (c)
In parasitism, one species is benefitted and the other is harmed (D correct). In commensalism, one species gets benefitted and the other remains unaffected (A incorrect). Both species benefit in mutualism (B correct). In amensalism, one species is harmed and the other is unaffected (E correct). Correct statements are B, D and E.
In parasitism, one species is benefitted and the other is harmed (D correct). In commensalism, one species gets benefitted and the other remains unaffected (A incorrect). Both species benefit in mutualism (B correct). In amensalism, one species is harmed and the other is unaffected (E correct). Correct statements are B, D and E.