NEET 2020 Biology Paper With Solutions September 13

NEET 2020 Question Paper with Solutions Biology Sept. 13th comprises 90 questions out of the total 180. At BYJU'S, students are provided with the best platform for learning, especially for competitive exam preparation. In view of this, the question paper of NEET is made available to students along with brief explanation for each question with solutions. This enables students to practice and comprehend every question easily, with just at a glance. Solutions for the Biology section of the NEET paper are provided with accurate answers and precise briefing for each question, equipping students with just the information required to crack that question.

NEET 2020 - Biology (Sep 13)

Question 1. Presence of which of the following conditions in urine are indicative of Diabetes Mellitus?

  1. a) Ketonuria and Glycosuria
  2. b) Renal calculi and Hyperglycemia
  3. c) Uremia and Ketonuria
  4. d) Uremia and Renal Calculi

Solution:

  1. Answer: a

    Diabetes mellitus leads to a complex disorder called prolonged hyperglycemia, which is associated with loss of glucose through urine known as glycosuria and when the cell are unable to utilize carbohydrates for energy instead they use fats & proteins, and degradation of these fats produces ketone bodies. The presence of these ketone bodies in urine is known as ketonuria


Question 2. Match the following columns and select the correct option

Column I

Column II

(a)

Placenta

(i)

Androgens

(b)

Zona pellucida

(ii)

Human Chorionic Gonadotropin hCG

(c)

Bulbo-urethral glands

(iii)

Layer of the ovum

(d)

Leydig cells

(iv)

Lubrication of the penis

  1. a) (iii) (ii) (iv) (i)
  2. b) (ii) (iii) (iv) (i)
  3. c) (iv) (iii) (i) (ii)
  4. d) (i) (iv) (ii) (iii)

Solution:

  1. Answer: b

    a) Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progestogens, etc.

    b) The secondary oocyte forms a new membrane called zonapellucida surrounding it.

    c) The male accessory glands include paired seminal vesicles, a prostate and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose, calcium and certain enzymes. He secretions of bulbourethral glands also helps in the lubrication of the penis.


Question 3. Match the following columns and select the correct option.

Column I

Column II

(a)

Bt cotton

(i)

Gene therapy

(b)

Adenosine deaminase deficiency

(ii)

Cellular defence

(c)

RNAi

(iii)

Detection of HIV infection

(d)

PCR

(iv)

Bacilus thuringiensis

  1. a) (ii) (iii) (iv) (i)
  2. b) (i) (ii) (iii) (iv)
  3. c) (iv) (i) (ii) (iii)
  4. d) (iii) (ii) (i) (iv)

Solution:

  1. Answer: c

    (a) Bt toxin is produced by a bacterium called Bacillus thuringiensis (bt for short). Bt toxin gene has been cloned from the bacteria and been expressed in plants to provide resistance to insects without the need for insecticides; in effect created a bio-pesticide. Examples are Bt cotton, Bt corn, rice, tomato, potato and soyabean etc.

    (b) The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency.

    (c) RNAi (RNA interference) is a biological process in which RNA molecules inhibit gene expression or translation, by neutralizing targeted mRNA molecules and takes place in all eukaryotic organisms as a method of cellular defense.

    (d) Polymerase chain reaction can be used in detection of HIV infection as it detects the genetic material of HIV i.e. its RNA


Question 4. The sequence that controls the copy number of the linked DNA in the vector, is termed

  1. a) Palindromic sequence
  2. b) Recognition site
  3. c) Selectable marker
  4. d) Ori site

Solution:

  1. Answer: d

    Origin of replication (ori): This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA.


Question 5. Match the following columns and select the correct option.

Column I

Column II

(a)

6-15 pairs of gill slits

(i)

Trygon

(b)

Heterocaercalcaudal fin

(ii)

Cyclostomes

(c) 

Air bladder

(iii)

Chondrichthyes

(d)

Poison sting

(iv)

Osteichthyes

  1. a) (iv) (ii) (iii) (i)
  2. b) (i) (iv) (iii) (ii)
  3. c) (ii) (iii) (iv) (i)
  4. d) (iii) (iv) (i) (ii)

Solution:

  1. Answer: c

    (a) Cyclostomata that belongs to agnatha which comprises the living jawless vertebrates.

    (b) Chondrichthyes consists of the cartilaginous fishes (eg. shark), a typical member of which has heterocercal (two unequal lobes) caudal fins.

    (c) Osteichthyes consists of the bony fishes in which air bladder is present which regulates buoyancy.

    (d) Chondrichthyes: poison sting (e.g., Trygon, whose common name is sting ray).


Question 6. In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?

  1. a) ICSI and ZIFT
  2. b) GIFT and ICSI
  3. c) ZIFT and IUT
  4. d) GIFT and ZIFT

Solution:

  1. Answer: c

    When females cannot conceive then embryo transfer is done by using ZIFT and IUT technique. In this method ova from female& sperms from male is collected and are induce in laboratory under simulated conditions to form zygote. The zygote with upto 8 blastomeres is transferred into fallopian tube through ZIFT or embryos with more than 8 blastomeres are transferred into uterus through IUT.


Question 7. Select the correct events that occur during inspiration.

(a) Contraction of diaphragm

(b) Contraction of external inter costal muscles

(c) Pulmonary volume decreases

(d) Intra pulmonary pressure increases

  1. a) (a), (b) and (d)
  2. b) Only (d)
  3. c) (a) and (b)
  4. d) (c) and (d)

Solution:

  1. Answer: c

    Inspiration is initiated by contraction of diaphragm which increases volume of thoracic chamber in antero-posterior axis and contraction of external inter –costal muscles which lifts up the ribs and sternum causing increases in volume of thoracic chamber in dorso-ventral axis.

    events that occur during inspiration


Question 8. The QRS complex in a standard ECG represents:

  1. a) Depolarization of ventricles
  2. b) Repolarization of ventricles
  3. c) Repolarization of auricles
  4. d) Depolarization of auricles

Solution:

  1. Answer: a

    The QRS complex represents the depolarization of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.


Question 9. The enzyme enterokinase helps in conversion of

  1. a) Caseinogen into casein
  2. b) Pepsinogen into pepsin
  3. c) Protein into polypeptides
  4. d) Trypsinogen into trypsin

Solution:

  1. Answer: d

    Trypsinogen is present in inactive form in pancreatic juice. So enterokinase enzyme converts inactive trypsinogen to active trypsin. This in turn activates the other enzymes in the pancreatic juice.


Question 10. Identify the correct statement with reference to human digestive system.

  1. a) Ileum is a highly coiled part
  2. b) Vermiform appendix arises from duodenum
  3. c) Ileum opens into small intestine
  4. d) Serosa is the innermost layer of the alimentary canal.

Solution:

  1. Answer: a

    (a) Small intestine is distinguishable into three regions, a ‘U’ shaped duodenum, a long coiled middle portion jejunum and a highly coiled ileum.

    (b) A narrow finger-like tubular projection, the vermiform appendix which is a vestigial organ, arises from the caecum.

    (c) Ileum opens into the large intestine.

    (d) The wall of alimentary canal possesses four layers from outer to inner namely serosa, muscularis, submucosa and mucosa. Serosa is the outermost layer and mucosa is the innermost layer.


Question 11. Ray florets have:

  1. a) Hypogynous ovary
  2. b) Half inferior ovary
  3. c) Inferior ovary
  4. d) Superior ovary

Solution:

  1. Answer: c

    Ray florets have inferior ovary and the reason is that the other parts of the flower are attached above the level of ovary. Example of such an ovary is ray florets of sunflower.


Question 12. Which of the following is put into anaerobic sludge digester for further sewage treatment?

  1. a) Effluents of primary treatment
  2. b) Activated sludge
  3. c) Primary sludge
  4. d) Floating debris

Solution:

  1. Answer: b

    Major portion of activated sludge is pumped into large tanks called anaerobic sludge digesters. So, here other kinds of bacteria which grow anaerobically digest the fungi and bacteria of the sludge.


Question 13. The number of substrate level phosphorylations in one turn of citric acid cycle is:

  1. a) Two
  2. b) Three
  3. c) Zero
  4. d) One

Solution:

  1. Answer: d

    During Krebs' or citric acid cycle, succinyl-CoA is acted upon by enzyme succinyl-CoA synthetase to form succinate (a 4C compound). The reaction releases sufficient energy to form ATP (in plants) or GTP (in animals) by substrate-level phosphorylation. GTP can then be used to form ATP.


Question 14. Identify the correct statement with regard to G1 phase (Gap I) of interphase

  1. a) Cell is metabolically active, grows but does not replicate its DNA
  2. b) Nuclear division takes place
  3. c) DNA synthesis or replication takes place
  4. d) Reorganization of all cell components takes place

Solution:

  1. Answer: a

    G1 Phase is metabolically active stage of cell cycle. Different type of amino acid RNA, Protein synthesis take place in G1 phase but DNA replication does not take place, (Note: DNA replication occur in S-Phase)


Question 15. Which of the following pairs is of unicellular algae?

  1. a) Anabaena and volvox
  2. b) Chlorella and spirulina
  3. c) Laminaria and Sargossum
  4. d) Gelidium and Gracilaria

Solution:

  1. Answer: b

    Chlorella and spirulina are unicellular algae, rich in proteins is used as food supplement even by space travelers. These algae also produce a nutritional biomass that astronauts could eat.


Question 16. Identify the wrong statement with reference to immunity:

  1. a) Active immunity is quick and gives full response
  2. b) Fetus receives some antibodies from mother, it is an example for passive immunity
  3. c) When exposed to antigen (living or dead) antibodies are produced in the host’s body. It is called “Active immunity”
  4. d) When readymade antibodies are directly given, it is called “Passive immunity”.

Solution:

  1. Answer: a

    Active immunity is slow and takes time to give its full effective response. Fetus receiving some antibodies from their mother, through the placenta during pregnancy is an example of passive immunity. Active immunity is the immunity in which when host is exposed to antigens either living or dead, antibodies produces in the host body. In passive immunity readymade antibodies are directly given to protect the body against foreign agents


Question 17. Match the following columns and select the correct option

Column I

Column II

(a)

Floating ribs

(i)

Located between second and seventh ribs

(b)

Acromion

(ii)

Head of the Humerus

(c) 

Scapula

(iii)

Clavicle

(d)

Glenoid cavity

(iv)

Do not connect with the sternum

  1. a) (iii) (ii) (iv) (i)
  2. b) (iv) (iii) (i) (ii)
  3. c) (ii) (iv) (i) (iii)
  4. d) (i) (iii) (ii) (iv)

Solution:

  1. Answer: b

    (a) There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. Last 2 pairs (11thand 12th) of ribs are not connected ventrally and are therefore, called floating ribs.

    (b) & (c)

    Scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs. The dorsal, flat, triangular body of scapula has a slightly elevated ridge called the spine which projects as a flat, expanded process called the acromion. The clavicle (collar bone) articulates with this.

    (d) Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.


Question 18. Identify the basic amino acid from the following

  1. a) Lysine
  2. b) Valine
  3. c) Tyrosine
  4. d) Glutamic acid

Solution:

  1. Answer: a

    Lysine is the basic amino acid. Valine is neutral amino acid. Tyrosine is aromatic amino acid. Glutamic acid is acidic amino acid.


Question 19. The plant parts which consist of two generations one within the other:

(a) Pollen grains inside the anther

(b) Germinated pollen grain with two male gametes

(c) Seed inside the fruit

(d) Embryo sac inside the ovule

  1. a) (c) and (d)
  2. b) (a) and (d)
  3. c) (a) only
  4. d) (b) only

Solution:

  1. Answer: b

    (a) Pollen grain inside the anther (2n).

    (d) Embryo sac inside ovule female gametophyte (2n).


Question 20. Identify the wrong statement with reference to transport of oxygen.

  1. a) Higher H+ conc. in alveoli favors the formation of oxyhemoglobin
  2. b) Low pCO2 in alveoli favors the formation of oxyhemoglobin
  3. c) Binding of oxygen with hemoglobin is mainly related to partial pressure of O2
  4. d) Partial pressure of CO2 can interfere with O2 binding with hemoglobin.

Solution:

  1. Answer: a

    The Oxygen dissociation curve is highly useful in studying the effect of factors like pCO2, H+ concentration, etc., on binding of O2 with hemoglobin. In the alveoli, where there is high pO2, low pCO2, lesser H+ concentration and lower temperature, the factors are all favorable for the formation of oxyhemoglobin. In the tissues, where low pO2, high pCO2, high H+ concentration and higher temperature exist, the conditions are favorable for dissociation of oxygen from the oxyhemoglobin.


Question 21. Match the following columns and select the correct option.

Column I

Column II

(a)

Organ of Corti

(i)

Connects middle ear and pharynx

(b)

Cochlea

(ii)

Coiled part of the labyrinth

(c) 

Eustachian tube

(iii)

Attached to the oval window

(d)

Stapes

(iv)

Located on the basilar membrane

  1. a) (iv) (ii) (i) (iii)
  2. b) (i) (ii) (iv) (iii)
  3. c) (ii) (iii) (i) (iv)
  4. d) (iii) (i) (iv) (ii)

Solution:

  1. Answer: a

    (a) Organ of corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors.

    (b) Cochlea is the coiled portion of the labyrinth (fluid filled inner ear)

    (c) Eustachian tube connects the middle ear cavity with the pharynx and helps in equalizing pressure of ear drums.

    (d) Stapes is one of the ossicle of middle ear and is attached to oval window of cochlea.


Question 22. Name the plant growth regulators which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.

  1. a) Ethylene
  2. b) Abscisic acid
  3. c) Cytokinin
  4. d) Gibberellin

Solution:

  1. Answer: d

    Gibberellin hormone regulates the growth, elongation of stem. It is able to promote the internode elongation because of which the size of sugarcane increases.


Question 23. The roots that originate from the base of the stem are:

  1. a) Prop roots
  2. b) Lateral roots
  3. c) Fibrous roots
  4. d) Primary roots

Solution:

  1. Answer: c

    Roots that originate from the base of stem constitute the fibrous root system as seen in the monocots example wheat plant.


Question 24. If the head of cockroach is removed, it may live for few days because:

  1. a) The head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.
  2. b) The head holds a 1/3rd of a nervous system while the rest is situated along the dorsal part of its body.
  3. c) The supra-oesophageal ganglia of the cockroach are situated in ventral part of abdomen.
  4. d) The cockroach does not have nervous system.

Solution:

  1. Answer: a

    Nervous system consists of nerve ring and ventral double nerve cord. Nerve ring is situated oesophegeal area. Head contain only branches of ganglia so there is no major part of nervous system in head region.


Question 25. Strobili or cones are found in:

  1. a) Marchantia
  2. b) Equisetum
  3. c) Salvinia
  4. d) Pteris

Solution:

  1. Answer: b

    Strobilus is structure present on many land plant species consisting of sporangia bearing structure densely aggregated along a stem. Equisetum produces strobili.


Question 26. Dissolution of the synaptonemal complex occurs during:

  1. a) Diplotene
  2. b) Leptotene
  3. c) Pachytene
  4. d) Zygotene

Solution:

  1. Answer: a

    Dissolution of the synaptonemal complex occurs at diplotene phase of prophase I in meiosis


Question 27. Match the following diseases with the causative organism and select the correct option.

Column-I

Column-II

(a) Typhoid

(i) Wuchereria

(b) Pneumonia

(ii) Plasmodium

(c) Filariasis

(iii) Salmonella

(d) Malaria

(iv) Haemophilus

  1. a) (ii) (i) (iii) (iv)
  2. b) (iv) (i) (ii) (iii)
  3. c) (i) (iii) (ii) (iv)
  4. d) (iii) (iv) (i) (ii)

Solution:

  1. Answer: d

    Salmonella typhi is a pathogenic bacterium which causes typhoid fever in human beings. Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air filled sacs) of the lungs. Wuchereria the filarial worms cause a slowly developing chronic inflammation Plasmodium are responsible for different types of malaria.


Question 28. The first phase of translation is:

  1. a) Aminoacylation of tRNA
  2. b) Recognition of an anti-codon
  3. c) Binding of mRNA to ribosome
  4. d) Recognition of DNA molecule

Solution:

  1. Answer: a

    In translation the first phase is activation of amino acids in the presence of ATP. The activated amino acids are then linked to their cognate tRNAs, a process commonly called as charging of tRNA or aminoacylation of tRNAs


Question 29. Match the following columns and select the correct option.

Column-I

Column-II

(a) Clostridium butylicum

(i) Cyclosporin-A

(b) Trichoderma polysporum

(ii) Butyric Acid

(c) Monascus purpureus

(iii) Citric Acid

(d) Aspergillus niger

(iv) Blood cholesterol lowering agent

  1. a) (i) (ii) (iv) (iii)
  2. b) (iv) (iii) (ii) (i)
  3. c) (iii) (iv) (ii) (i)
  4. d) (ii) (i) (iv) (iii)

Solution:

  1. Answer: d

    Clostridium butylicumis (a bacterium) of butyric acid. Cyclosporin A which is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum. Monascus purpureusis not a blood cholesterol lowering agent, statins produced by it are been commercialised as blood-cholesterol lowering agents. Aspergillus nigeris (a fungus) of citric acid.


Question 30. The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of:

  1. a) 1 molecule of 6-C compound
  2. b) 1 molecule of 4-C compound and 1 molecule of 2-C compound
  3. c) 2 molecules of 3-C compound
  4. d) 1 molecule of 3-C compound

Solution:

  1. Answer: d

    Photorespiration is the light dependent process. At high temperature, RuBP carboxylase functions as oxygenase and instead of fixing carbon dioxide (C3 cycle), oxidizes ribulose 1,5-biphosphate to produce a 3-carbon phosphoglyceric acid and a 2-carbonphosphoglycolate.


Question 31. Match the following concerning essential elements and their functions in plants:

Column-I

Column-II

(a) Iron

(i) Photolysis of water

(b) Zinc

(ii) Pollen germination

(c) Boron

(iii) Required for chlorophyll biosynthesis

(d) Manganese

(iv) IAA biosynthesis

Select the correct option:

  1. a) (iii) (iv) (ii) (i)
  2. b) (iv) (i) (ii) (iii)
  3. c) (ii) (i) (iv) (iii)
  4. d) (iv) (iii) (ii) (i)

Solution:

  1. Answer: a

    (a) Iron - (iii) Required for chlorophyll biosynthesis

    (b) Zinc - (iv) IAA (auxin) biosynthesis

    (c) Boron - (ii) Pollen germination

    (d) Manganese - (i) Photolysis of water


Question 32. Name the enzyme that facilitates opening of DNA helix during transcription.

  1. a) DNA polymerase
  2. b) RNA polymerase
  3. c) DNA ligase
  4. d) DNA helicase

Solution:

  1. Answer: b

    RNA polymerase holoenzyme binds to the promoter, unwinds DNA (open complex) and form phosphodiester links between the initiating nucleotides. DNA polymerase, DNA ligase & DNA helicase are involved in the process of replication and not transcription.


Question 33. From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask:

  1. a) CH4, H2, NH3 and water vapor at 600°C
  2. b) CH3, H2, NH3 and water vapor at 600°C
  3. c) CH4, H2, NH3 and water vapor at 800°C
  4. d) CH3, H2, NH4 and water vapor at 800°C

Solution:

  1. Answer: c

    S.L. Miller conducted an experiment to prove the theory of chemical origin. In their experiment, the conditions of primitive earth were created in the laboratory. The electric discharge was stimulated into a closed flask containing CH4, H2, NH3 and water vapor at 800°C. This proved that life originates from non-living components.


Question 34. Goblet cells of alimentary canal are modified from:

  1. a) Chondrocytes
  2. b) Compound epithelial cells
  3. c) Squamous epithelial cells
  4. d) Columnar epithelial cells

Solution:

  1. Answer: d

    Goblet cells are modified columnar epithelial cells. It is found in the lining of organs like respiratory tract and intestine


Question 35. Cuboidal epithelium with brush border of microvilli is found in:

  1. a) proximal convoluted tubule of nephron
  2. b) Eustachian tube
  3. c) lining of intestine
  4. d) ducts of salivary glands

Solution:

  1. Answer: a

    The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in ducts of glands and tubular parts of nephrons in kidneys and its main functions are secretion and absorption.


Question 36. In light reaction, plastoquinone facilitates the transfer of electrons from:

  1. a) PS-I to NADP+
  2. b) PS-I to ATP synthase
  3. c) PS-II to Cytb6f complex
  4. d) Cytb6complex to PS-I

Solution:

  1. Answer: c

    In non-cyclic photophosphorylation, Plastoquinone obtained from PS-II transfer to cyt b6-f complex.


Question 37. If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6 × 109dp, then the length of the DNA is approximately.

  1. a) 2.2 meters
  2. b) 2.7 meters
  3. c) 2.0 meters
  4. d) 2.5 meters

Solution:

  1. Answer: a

    The diploid content of human genome is 6.6 × 109 base pairs. The distance between two consecutive base pairs is 0.34 nm (0.34 × 10–9 m), so the length of DNA double helix in a typical mammalian cell is around 6.6 × 10–9bp×0.34 × 10–9m/bp = 2.2 metres.


Question 38. Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?

  1. a) Golgi bodies
  2. b) Polysomes
  3. c) Endoplasmic reticulum
  4. d) Peroxisomes

Solution:

  1. Answer: a

    Proteins and lipids are formed in the endoplasmic reticulum and some of them are modified to form glycoproteins and glycolipids in the Golgi apparatus.


Question 39. Which of the following statements is not correct?

  1. a) The functional insulin has A and B chains linked together by hydrogen bonds.
  2. b) Genetically engineered insulin is produced in E-Coli.
  3. c) In man insulin is synthesized as a pro-insulin
  4. d) The pro-insulin has an extra peptide called C-peptide.

Solution:

  1. Answer: a

    The functional insulin molecule has two chains A and chain B, which are linked together by disulphide bridges and not hydrogen bonds.


Question 40. Identify the incorrect statement.

  1. a) Sapwood is the innermost secondary xylem and is lighter in colour.
  2. b) Due to deposition of tannins, resins, oils etc., heart wood is dark in colour.
  3. c) Heart wood does not conduct water but gives mechanical support.
  4. d) Sapwood is involved in conduction of water and minerals from root to leaf.

Solution:

  1. Answer: a

    In a large tree, only the outer secondary xylem (sapwood) serves in water conduction, while the inner part (heartwood) is composed of dead but structurally strong secondary xylem and is darker in colour due to disposition of tannins, resins and oils.


Question 41. Floridean starch has structure similar to:

  1. a) Mannitol and algin
  2. b) Laminarin and cellulose
  3. c) Starch and cellulose
  4. d) Amylopectin and glycogen

Solution:

  1. Answer: d

    Floridean starch is the storage substance found in red algae (Rhodophyta). The structure of this compound is similar to amylopectin and glycogen.


Question 42. Match the following with respect to meiosis:

Column-I

Column-II

(a) Zygotene

(i) Terminalization

(b) Pachytene

(ii) Chiasmata

(c) Diplotene

(iii) Crossing over

(d) Diakinesis

(iv) Synapsis

Select the correct option from the following:

  1. a) (i) (ii) (iv) (iii)
  2. b) (ii) (iv) (iii) (i)
  3. c) (iii) (iv) (i) (ii)
  4. d) (iv) (iii) (ii) (i)

Solution:

  1. Answer: d

    Terminalization: It is the stage in which chromosomes condenses further. Sites of crossing over entangle together with effective overlapping which makes chiasmata clearly visible.

    Chiasmata: It is the region where crossing over occurs.

    Crossing over: Crossing over or chromosomal crossover occurs in the pachytene stage in which non sister chromatids of homologous chromosomes may exchange segments over regions of homology.

    Synapsis: Synapsis of homologous chromosomes takes place in the zygotene stage.


Question 43. Match the following columns and select the correct option.

Column-I

Column-II

(a) Eosinophils

(i) Immune response  

(b) Basophils

(ii) Phagocytosis 

(c) Neutrophils

(iii) Release histaminases, destructive Enzymes  

(d) Lymphocytes

(iv) Release granules containing histamine

  1. a) (i) (ii) (iv) (iii)
  2. b) (ii) (i) (iii) (iv)
  3. c) (iii) (iv) (ii) (i)
  4. d) (iv) (i) (ii) (iii)

Solution:

  1. Answer: c

    Eosinophils release mediators like histaminases during type I hypersensitivity or allergic

    reaction.

    Basophils are granulocytes i.e. they contain large cytoplasmic granules in the cell nucleus and it store histamine, a vasodilator and anticoagulant heparin.

    Neutrophils are phagocytic in nature and are able to engulf foreign substance (like bacteria)

    Lymphocytes have receptors on their surface and they produce antibodies so responsible for immune response.


Question 44. The process of growth is maximum during:

  1. a) senescence
  2. b) dormancy
  3. c) log phase
  4. d) lag phase

Solution:

  1. Answer: c

    Period of growth is characterized by the number of cells increasing at an exponential rate.


Question 45. Match the following:

(a) Inhibitor of catalytic Activity

(i) Ricin  

(b) Possess peptide bonds

(ii) malonate

(c) Cell wall material in fungi

(iii) Chitin  

(d) Secondary metabolite

(iv) Collagen

Choose the correct option from the following:

  1. a) (iii) (iv) (i) (ii)
  2. b) (ii) (iii) (i) (iv)
  3. c) (ii) (iv) (iii) (i)
  4. d) (iii) (i) (iv) (ii)

Solution:

  1. Answer: c

    Malonate resembles succinate in structure. Succinate is the substrate of the enzyme succinate dehydrogenase. Hence malonate acts as a competitive inhibitor of this enzyme. Collagen is a protein, and hence has peptide bonds. Cell wall of fungi is made of chitin.

    Ricin, a toxin is a secondary metabolite.


Question 46. Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G0). This process occurs at the end of:

  1. a) S phase
  2. b) G2 Phase
  3. c) M phase
  4. d) G1 phase

Solution:

  1. Answer: c

    In M-Phase, some cells do not divide further exist G1 phase to enter an inactive stage called quiescent stages (G0) of the cell cycle.


Question 47. Which of the following would help in prevention of diuresis?

  1. a) Atrial natriuretic factor causes vasoconstriction
  2. b) Decrease in secretion of renin by JG cells
  3. c) More water reabsorption due to undersecretion of ADH
  4. d) Reabsorption of Na+ and water from renal tubules due to aldosterone

Solution:

  1. Answer: d

    • ANF acts on the kidney to increase Na+ excretion and GFR also inhibit rennin secretion.
    • Due to decrease in secretion of renin, it reduced concentration of angiotensin I & II.
    • ADH stimulates water reabsorption by stimulating insertion of water channels or aquaporins into the membranes of kidney tubules.
    • Reabsorption of Na+ and water from renal tubules due to aldosterone help in prevention of diuresis.

Question 48. Which of the following is correct about viroids?

  1. a) They have DNA with protein coat.
  2. b) They have free DNA without protein coat.
  3. c) They have RNA with protein coat.
  4. d) They have free RNA without protein coat.

Solution:

  1. Answer: d

    Viroids have free RNA without protein coat, found in viruses named viroid. Potato spindle tuber disease is a disease caused by the viroids.


Question 49. The infectious stage of plasmodium that enters the human body is:

  1. a) Female gametocytes
  2. b) Male gametocytes
  3. c) Trophozoites
  4. d) Sporozoites

Solution:

  1. Answer: d

    The life cycle of plasmodium that infect humans follows three stages: (i) infection of a human with sporozoites; (ii) asexual reproduction and (iii) sexual reproduction. The two first takes place exclusively into the human body, while the third one starts in the human body and is completed into the mosquito organism. The human infection begins when an infected female anopheles mosquito bites a person and injects infected with sporozites saliva into the blood circulation i.e., the first life stage of plasmodium (stage of infection). 


Question 50. Which of the following statements is correct?

  1. a) Adenine pairs with thymine through three H-bonds.
  2. b) Adenine does not pair with thymine.
  3. c) Adenine pairs with thymine through two H-bonds.
  4. d) Adenine pairs with thymine through one H-bond.

Solution:

  1. Answer: c

    Adenine pairs with thymine in DNA through two hydrogen bonds.


Question 51. Flippers of Penguins and Dolphins are examples of:

  1. a) Industrial melanism
  2. b) Natural selection
  3. c) Adaptive radiation
  4. d) Convergent evolution

Solution:

  1. Answer: d

    Analogous organs have similar functions but different origins. Flippers of penguin and dolphin have a similar function (helps in swimming). However, Penguin and dolphin are not closely related to each other, and hence their flippers have different origins and are called analogous organs. This phenomenon is called convergent evolution.


Question 52. Montreal protocol was signed in 1987 for control of :

  1. a)Release of Green House gases
  2. b)Disposal of e-wastes
  3. c)Transport of Genetically modified organisms from one country to another
  4. d)Emission of ozone depleting substances

Solution:

  1. Answer: d

    Montreal Protocol is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion.


Question 53. Identify the wrong statement with regard to restriction enzymes.

  1. a) They are useful in genetic engineering.
  2. b) Sticky ends can be joined by using DNA ligases.
  3. c) Each restriction enzyme functions by inspecting the length of a DNA sequence.
  4. d) They cut the strand of DNA at palindromic sites.

Solution:

  1. Answer: b

    (a) Genetic engineering or recombinant DNA technology can be accomplished only if we have the key tools, i.e., restriction enzymes, polymerase enzymes, ligases, vectors and the host organism.

    (b) The sticky ends in DNA that result from the action of restriction endonucleases cannot be joined by the restriction enzymes, but by DNA ligases. So, this statement does not apply to restriction enzymes and hence is the correct option for the question.

    (c) Restriction enzymes begin their action by first scanning the length of the DNA sequence and then recognizing a specific sequence. This specific base sequence is known as the recognition sequence.

    (d) Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA.


Question 54. By which method was a new breed ‘Hisardale’ of sheep formed by using Bikaneri ewes and Marino rams?

  1. a)Cross breeding
  2. b)Inbreeding
  3. c)Out crossing
  4. d)Mutational breeding

Solution:

  1. Answer: a

    Cross-breeding is the process where the breeding between the two individuals of different species takes place. “Hisardale" is a new breed of sheep developed by crossing Bikaneri ewes and Marino rams in Punjab.


Question 55. Which of the following refer to correct example(s) of organisms which have evolved due to changes of environment brought about by anthropogenic action?

(a) Darwin’s Finches of Galapagos islands.

(b) Herbicide resistant weeds

(c) Drug resistant eukaryotes.

(d) Man-created breeds of domesticated animals like dogs.

  1. a) (b), (c) & (d)
  2. b) only (d)
  3. c) only (a)
  4. d) (a) & (c)

Solution:

  1. Answer: a

    Evolution by Anthropogenenic action is because of the interference by human beings. Anthropogenenic shows herbicide resistant weeds, drug resistant eukaryotes and man-created breeds of domesticated animals like dogs and Darwin’s Finches of Galapagos islands is an example of natural selection.


Question 56. Meiotic division of the secondary oocyte is completed:

  1. a)After zygote formation
  2. b)At the time of fusion of a sperm with an ovum
  3. c)Prior to ovulation
  4. d)At the time of copulation

Solution:

  1. Answer: b

    When sperm enters into the secondary oocyte, it provides the anaphase promoting factor that induces the completion of meiosis in the secondary oocyte.


Question 57. In relation to Gross primary productivity and Net primary productivity of an ecosystem. Which one of the following statements is correct?

  1. a) Gross primary productivity and Net primary productivity are one and same.
  2. b) There is no relationship between Gross primary productivity and Net primary productivity
  3. c) Gross primary productivity is always less than Net primary productivity.
  4. d) Gross primary productivity is always more than Net primary productivity.

Solution:

  1. Answer: d

    The gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Some of this organic matter is lost because of the respiration of plants. The remaining primary productivity is the net primary productivity.


Question 58.  Identify the wrong statement with reference to the gene ‘I’ that controls ABO blood groups.

  1. a) When IA and IB are present together, they express same type of sugar.
  2. b) Allele ‘i’ does not produce any sugar.
  3. c) The gene (I) has three alleles.
  4. d) A person will have only two of the three alleles.

Solution:

  1. Answer: a

    When IA and IB are present together they both express their own types of sugars, because of co-dominance. Hence red blood cells have both A and B types of sugars.


Question 59. Match the following columns and select the correct option.

Column-I

Column-II

(a) Pituitary gland

(i) Grave’s disease  

(b) Thyroid gland

(ii) Diabetes mellitus 

(c) Adrenal gland

(iii) Diabetes insipidus

(d) Pancreas

(iv) Addison’s disease

  1. a)(iii) (i) (iv) (ii)
  2. b)(ii) (i) (iv) (iii)
  3. c)(iv) (iii) (i) (ii)
  4. d)(iii) (ii) (i) (iv)

Solution:

  1. Answer: a

    The antidiuretic hormone released by the pituitary gland stimulates reabsorption of water by the kidneys. The deficiency of this hormone causes increased urine production, a condition called diabetes insipidus. Graves’s disease is an immune system disorder that results in the overproduction of thyroid hormones. Addison's disease is caused by a deficiency of the hormones secreted by the adrenal cortex. Pancreas is a composite gland which acts as both exocrine and endocrine gland. The deficiency of insulin, a hormone secreted by the pancreas, causes a complex disorder called diabetes mellitus which is associated with loss of glucose through urine and formation of harmful compounds known as ketone bodies.


Question 60. According to Robert May, the global species diversity is about

  1. a) 50 million
  2. b) 7 million
  3. c) 1.5 million
  4. d) 20 million

Solution:

  1. Answer: b

    Robert May was a theoretical ecologist the established, who field of theoretical ecology and population biology. According to him the global species diversity is about 7 million.


Question 61. The body of the ovule is fused within the funicle at

  1. a) Nucellus
  2. b) Chalaza
  3. c) Hilum
  4. d) Micropyle

Solution:

  1. Answer: c

    The body of the ovule fuses with the funicle in the region called the hilum. The funicle is a stalk through which the ovule is attached to the placenta.


Question 62. Match the following columns and select the correct option –

Column - I

Column - II

(a) Gregarious polyphagous pest

(i) Asterias

(b) Adult with radial symmetry and larva with bilateral symmetry

(ii) Scorpion

(c) Book lungs

(iii) Ctenoplana

(d) Bioluminescence

(iv) Locusta

  1. a)(iii) (ii) (i) (iv)
  2. b)(ii) (i) (iii) (iv)
  3. c)(i) (iii) (ii) (iv)
  4. d)(iv) (i) (ii) (iii)

Solution:

  1. Answer: d

    (a) Locust is a gregarious pest

    (b) In Echinodermata, adult are radially symmetrical and larva are bilaterally symmetrical. For example: Asterias.

    (c) In Arthopoda, scorpion respires through book lungs.

    (d) Bioluminescence (the property of a living organism to emit light) is well marked in ctenophores. For example: Ctenoplana.


Question 63. Embryological support for evolution was disapproved by:

  1. a)Charles Darwin
  2. b)Oparin
  3. c)Karl Ernst von Baer
  4. d)Alfred Wallace

Solution:

  1. Answer: c

    Karl Ernst and von Baer proposed four rules to explain the observed pattern of embryonic development in different species.


Question 64. Match the organism with its use in biotechnology.

(a) Bacillus Thuringiensis

(i) Cloning vector

(b) Thermus aquaticus

(ii) Construction of first rDNA Molecule

(c) Agrobacterium tunefaciens

(iii) DNA polymerase

(d) Salmonella typhimurium

(iv) Cry proteins

Select the correct option from the following

  1. a)(iii) (ii) (iv) (i)
  2. b)(iii) (iv) (i) (ii)
  3. c)(ii) (iv) (iii) (i)
  4. d)(iv) (iii) (i) (ii)

Solution:

  1. Answer: d

    (a) Bacillus thuringiensis produces the cry proteins, which are Bt toxins and have insecticidal properties.

    (b) The highly thermostable DNA polymerase from Thermus aquaticus is ideal for both manual and automated DNA sequencing because it is fast & highly processive and remain active during the high temperature induced denaturation of double stranded DNA.

    (c) The tumor-inducing (Ti) plasmid of Agrobacterium tumifaciens has been modified into a cloning vector. It is used to deliver required genes into plants.

    (d) Salmonella typhimurium is the first recombinant DNA that cut the piece of DNA from a plasmid carrying antibiotic resistance gene in the bacterium and linked it to the plasmid of E.coli.


Question 65. Which of the following is not an inhibitory substance governing seed dormancy?

  1. a)Phenolic acid
  2. b)Para - ascorbic acid
  3. c)Gibberellic acid
  4. d)Abscisic acid

Solution:

  1. Answer: c

    Gibberellins are involved in the natural process of breaking dormancy and other aspects of germination. Giberellic acid, one of the gibberellins, stimulates the cells of germinating seeds to produce mRNA molecules that code for hydrolytic enzymes.


Question 66. Which of the following statements about inclusion bodies is incorrect?

  1. a)They lie free in the cytoplasm
  2. b)These represent reserve material in cytoplasm.
  3. c)They are not bound by any membrane.
  4. d)These are involved in ingestion of food particles.

Solution:

  1. Answer: d

    Inclusions bodies are distinct granules that may occupy a substantial part of the cytoplasm. Inclusion, granules are usually reserve material of some sort. and for food particles ingestion the cell membrane helps in this process. Hence option 4 is wrong.


Question 67. The ovary is half inferior in:

  1. a)Sunflower
  2. b)Plum
  3. c)Brinjal
  4. d)Mustard

Solution:

  1. Answer: b

    Plum/peach belongs to the family Rosaceae and it shows half-inferior ovary. Hence the flowers are perigynous. Sunflower have inferior ovary. Plum/peach flowers have a half-inferior ovary. Brinjal (Solanaceae) has a superior ovary. Mustard have superior ovary.


Question 68. Match the trophic levels with their correct species examples in grassland ecosystem.

(a) Fourth trophic level

(i) Crow

(b) Second trophic level

(ii) Vulture

(c) First trophic level

(iii) Rabbit

(d) Third trophic level

(iv) Grass

Select the correct option:

  1. a)(iv) (iii) (ii) (i)
  2. b)(i) (ii) (iii) (iv)
  3. c)(ii) (iii) (iv) (i)
  4. d)(iii) (ii) (i) (iv)

Solution:

  1. Answer: c

    (a) Fourth trophic level - (ii) Vulture

    (b) Second trophic level - (iii) Rabbit

    (c) First trophic level - (iv) Grass

    (d) Third trophic level - (i) Crow


Question 69. The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is:

  1. a)Imbibition
  2. b)Plasmolysis
  3. c)Transpiration
  4. d)Root Pressure

Solution:

  1. Answer: d

    Root pressure (positive pressure) can be responsible for pushing up water to small heights in the stem. It also observable at night and early morning when evaporation is low, and excess water collects in the form of droplets around special openings of veins near the tip of grass blades, and leaves of many herbaceous parts. Such water loss in its liquid phase is known as guttation. Imbibition is the adsorption leading to absorption of water by hydrophillic substances. Plasmolysis is the shrinking of the cell membrane and cytoplasm when a cell is undergoing exosmosis. The loss of water in the form of water vapor from the aerial parts of the plant is called transpiration.


Question 70. Choose the correct pair from the following

  1. a)Nucleases - Separate the two strands of DNA
  2. b)Exonucleases - Make cuts at specific positions within DNA
  3. c)Ligases - Join the two DNA molecules
  4. d)Polymerases - Break the DNA into fragments

Solution:

  1. Answer: c

    DNA ligase is an enzyme which can connect two strands of DNA together by forming a bond between the phosphate group of one strand and the deoxyribose group on another. Nucleases cleave the phosphodiester bonds of nucleic acids and may be endo or exo, DNases or RNases, topoisomerases, recombinases, ribozymes, or RNA splicing enzymes. Exonucleases make cuts at the ends of the DNA strand. Polymerases help in the polymerization of a DNA or RNA molecule. DNA polymerases and RNA polymerases are the enzymes that perform this function.


Question 71. The transverse section of a plant shows following anatomical features:

a) Large number of scattered vascular bundles surrounded by bundle sheath.

b) Large conspicuous parenchymatous ground tissue.

c) Vascular bundles conjoint and closed.

d) Phloem parenchyma absent.

Identify the category of plant and its part:

  1. a)Dicotyledonous stem
  2. b)Dicotyledonous root
  3. c)Monocotyledonous stem
  4. d)Monocotyledonous root

Solution:

  1. Answer: c

    All anatomical features showing that plant is monocotyledonous stem so option 3 is correct. The monocot stem has vascular bundles near the outside edge of stem. Vascular bundles are scattered in parenchymatous ground tissue. There is no pith region in monocots. The vascular bundles are closed as they do not have cambium in it.


Question 72. Experimental verification of the chromosomal theory of inheritance was done by:

  1. a)Boveri
  2. b)Morgan
  3. c)Mendel
  4. d)Sutton

Solution:

  1. Answer: b

    Thomas Hunt Morgan, who studied fruit flies, provided the first strong confirmation of the chromosome theory. Morgan discovered a mutation that affected fly eye color. The chromosome theory of inheritance states that genes are found at specific locations on chromosomes, and that the behavior of chromosomes during meiosis can explain Mendel's laws of inheritance.


Question 73. Bt cotton variety that was developed by the introduction of toxin gene of Bacillus thuringiensis (Bt) is resistant to:

  1. a)Plant nematodes
  2. b)Insect predators
  3. c)Insect pests
  4. d)Fungal diseases

Solution:

  1. Answer: c

    B. thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. Bt cotton is an insect-resistant transgenic crop designed to combat the bollworm.


Question 74. Select the correct statement.

  1. a)Insulin acts on pancreatic cells and adipocytes.
  2. b)Insulin is associated with hyperglycemia.
  3. c)Glucocorticoids stimulate gluconeogenesis.
  4. d)Glucagon is associated with hypoglycemia.

Solution:

  1. Answer:  c

    (a) Insulin acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilization.

    (b) Glucagon acts mainly on the liver cells (hepatocytes) and stimulates glycogenolysis resulting in an increased blood sugar (hyperglycemia).

    (c) Glucocorticoids stimulate, gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilization of amino acids.

    (d) Glucagon hormone stimulates the process of gluconeogenesis which also contributes to hyperglycemia.


Question 75. The specific palindromic sequence which is recognized by EcoRI is:

  1. a)5’- CTTAAG -3’, 3’GAATTC - 5’
  2. b)5’- GGATCC - 3’, 3’- CCTAGG - 5’
  3. c)5’- GAATTC - 3’, 3’ -CTTAAG - 5’
  4. d)5’ - GGAACC - 3’, 3’ - CCTTGG - 5’

Solution:

  1. Answer: c

    Ecor I is the restriction enzyme which recognizes 6 base pair palindromic sequence and cuts both the strands of DNA at

    ecori


Question 76. Identify the substances having glycosidic bond and peptide bond, respectively in their structure.

  1. a)Cellulose, lecithin
  2. b)Inulin, Insulin
  3. c)Chitin, cholesterol
  4. d)Glycerol, trypsin

Solution:

  1. Answer: b

    Inulin is a polysaccharide have glycosidic bond, insulin is a polypeptide which is composed of two peptide chains.


Question 77. The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plant is/are:

  1. a)Ammonia and oxygen
  2. b)Ammonia and hydrogen
  3. c)Ammonia alone
  4. d)Nitrate alone

Solution:

  1. Answer: b

    Nitrogenase enzyme are responsible for the reduction of N2 to ammonia the product of reaction catalised by nitrogenase in root nodule of leguminous plants are ammonia and hydrogen.


Question 78. Which of the following hormone levels will cause release of ovum (ovulation) from the graafian follicle?

  1. a)Low concentration of LH
  2. b)Low concentration of FSH
  3. c)High concentration of Estrogen
  4. d)High concentration of Progesterone

Solution:

  1. Answer: c

    LH (Leutenizing hormone) is produced by pituitary gland in the brain. It triggers ovulation & promotes the development of corpus luteum by rupture of graafian follicle. Which high concentration of estrogen is released by mature graafian  follicle

    Leutenizing hormone


Question 79. Which of the following statements are true for the phylum - chordata?

(a) In urochordata notochord extends from head to tail and it is present throughout their life.

(b) In Vertebrata notochord is present during the embryonic period only.

(c) Central nervous system is dorsal and hollow.

(d) Chordata is divided into 3 subphyla: Hemichordata, Tunicata and Cephalochordata.

  1. a)(a) and (b)
  2. b)(b) and (c)
  3. c)(d) and (c)
  4. d)(c) and (a)

Solution:

  1. Answer: b

    (a) In Urochordata, notochord is present only in larval tail, while in Cephalochordata, it extends from head to tail region and is persistent throughout their life.

    (b) Vertebrata possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.

    (c)difference between chordates and non-chordates

    (d) Phylum Chordata is divided into three subphyla: Urochordata or Tunicata, Cephalochordata and Vertebrata.


Question 80. Bilaterally symmetrical and acoelomate animals are exemplified by:

  1. a)Aschelminthes
  2. b)Annelida
  3. c)Ctenophora
  4. d)Platyhelminthes

Solution:

  1. Answer: d

    Flatworms are bilaterally symmetrical, triploblastic and acoelomate animals with organ level of organisation. Aschelminthes is triploblastic, bilaterally symmetrical and pseudocoelomate. Annelida is triploblastic, bilaterally symmetrical and coelomate. Ctenophorais radially symmetrical, diploblastic.


Question 81. Which of the following regions of the globe exhibits highest species diversity?

  1. a)Himalayas
  2. b)Amazon forests
  3. c)Western Ghats of India
  4. d)Madagascar

Solution:

  1. Answer: b

    Species diversity is the number of different species that are represented in a given community. Amazon forest is the richest ecosystem on earth. It containing millions of species of plants, insects, birds and other forms of life, many still unrecorded by science.


Question 82. Select the correct match

  1. a)Sickle cell anaemia - Autosomal recessive trait, chromosome - 11
  2. b)Thalassemia - X linked
  3. c)Haemophilia - Y linked
  4. d)Phenylketonuria - Autosomal dominant trait

Solution:

  1. Answer: a

    This is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele, HbA and HbS on chromosome -11,

    The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the hemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG. Thalassemia is caused by mutations in the DNA of cells that make hemoglobin — the substance in red blood cells that carries oxygen throughout your body. This is an autosomal inheritance. Hemophilia is caused by a mutation or change, in one of the genes, that provides instructions for making the clotting factor proteins needed to form a blood clot. This change or mutation can prevent the clotting protein from working properly or to be missing altogether. These genes are located on the X chromosome.  Phenylketonuria or PKU is inherited in families in an autosomal recessive pattern. A birth defect that causes an amino acid called phenylalanine to build up in the body.


Question 83. Which one of the following is the most abundant protein in the animals?

  1. a)Lectin
  2. b)Insulin
  3. c)Hemoglobin
  4. d)Collagen

Solution:

  1. Answer: d

    Collagen is the most abundant protein in animal world and Ribulose bisphosphate Carboxylase-Oxygenase (RUBISCO) is the most abundant protein in the whole of the biosphere. Insulin is the hormone that helps in controlling the level of sugar (glucose) in the blood. Hemoglobin is the iron bound protein molecule (pigment) in the blood that helps to transport oxygen.


Question 84. Select the option including all sexually transmitted diseases.

  1. a)AIDS, Malaria, filaria
  2. b)Cancer, AIDS, syphilis
  3. c)Gonorrhoea, Syphilis, Genital Herpes
  4. d)Gonorrhoea, Syphilis, Genital Herpes

Solution:

  1. Answer: c

    Diseases or infections which are transmitted through sexual intercourse are collectively called sexually transmitted diseases (STD) or venereal diseases (VD) or reproductive tract infections (RTI). Gonorrhoea, syphilis, genital herpes, chlamydiasis, genital warts, trichomoniasis, hepatitis-B. Malaria, Filariare caused by the transmission of the parasite through the mosquitoes which act as a vector. AIDS is a disease caused by HIV virus and can be transmitted by several methods (through bodily fluids). Cancer is the uncontrollable cell division which leads to the formation of tumors and the causes can be several factors such as physical, chemical and biological factors called carcinogens.


Question 85. In water hyacinth and water lily, pollination takes place by:

  1. a)wind and water
  2. b)insects and water
  3. c)insects or wind
  4. d)water currents only

Solution:

  1. Answer: c

    In a majority of aquatic plants such as water hyacinth and water lily, the flowers emerge above the level of water and are pollinated by insects or wind as in most of the land plants. Their stem part which is above the thalamus is not in the water. The pollen grains are in the upper part of thalamus so pollination can't done by water. That's y it is done by insects and wind.


Question 86. In gel electrophoresis, separated DNA fragments can be visualized with the help of:

  1. a)Acetocarmine in UV radiation
  2. b)Ethidium bromide in infrared radiation
  3. c)Acetocarmine in bright blue light
  4. d)Ethidium bromide in UV radiation

Solution:

  1. Answer: d

    The separated DNA fragments can be visualized only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation (you cannot see pure DNA fragments in the visible light and without staining). You can see bright orange coloured bands of DNA in a ethidium bromide stained gel exposed to UV light.


Question 87. Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their

  1. a)defense action
  2. b)effect on reproduction
  3. c)nutritive value
  4. d)growth response

Solution:

  1. Answer: a

    Secondary plant metabolites are several chemical compounds produced by the plant cell through metabolic pathways obtained from the primary metabolic pathways. Primary metabolites include small molecules such as sugars, amino acids, tricarboxylic acids, or Krebs cycle intermediates, proteins, nucleic acids and polysaccharides. Many secondary metabolites are toxic or repellent to herbivores and microbes and help defend plants producing them.

    Secondary metabolism produces a large number of specialized compounds that do not aid in the growth and development of plants but are required for the plant to survive in its environment.


Question 88. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?

  1. a)14
  2. b)8
  3. c)4
  4. d)2

Solution:

  1. Answer: a

    A true breeding line refers to the plant that has been self-pollinated continuously and produced generations that were stable in the inheritance of the character.

    Mendel selected 14 true-breeding pea plant varieties, as pairs which were similar except for one character with contrasting traits.


Question 89. Which of the following is not an attribute of a population?

  1. a)Mortality
  2. b)Species interaction
  3. c)Sex ratio
  4. d)Natality

Solution:

  1. Answer: b

    Species interaction refers to direct and indirect interrelationship between different organisms, while on the other hand population attributes include population size, population density, population spacing, and age structure. Mortality is the ratio of deaths in an area to the population of that area; expressed per 1000 per year. 

    Sex ratio is the ratio of males to females in a population. 

    Natality is defined as the birth rate which is the total number of live births per 1000 in a given population during a given time period or a year.


Question 90. Snow - blindness in Antarctic region is due to:

  1. a)High reflection of light from snow
  2. b)Damage of retina caused by infra -red rays
  3. c)Freezing of fluids in the eye by low temperature
  4. d)Inflammation of cornea due to high dose of UV - B radiation

Solution:

  1. Answer: d

    In human eye, cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of cornea, called snow-blindness It is a painful eye condition caused by overexposure to ultraviolet (UV) light. Cornea sunburn results due to the effects of UV light on the cornea.


Video Lessons - NEET 2020 - Biology (Sep 13)

NEET 2020 Biology Paper With Solutions September 13

NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13
NEET 2020 Biology Paper With Solutions September 13