NEET 2020 Biology Question Paper With Solutions [Official]

Here you can find the official NEET 2020 Biology question paper along with their solutions to each of the questions. Solving previous year question papers can give you a rough idea about the nature and difficulty level of questions which are going to be asked in the upcoming NEET exam.

I’m going to keep it short, simple and sweet.

Lets begin! Shall we?

NEET 2020 Biology Question Paper:

1. 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) Dicotyledonous root

(2) Monocotyledonous stem

(3) Monocotyledonous root

(4) Dicotyledonous stem

Answer: (2) Monocotyledonous stem

2. Which of the following would help in prevention of diuresis?

(1) Decrease in secretion of renin by JG cells

(2) More water reabsorption due to under secretion of ADH

(3) Reabsorption of Na+ and water from renal tubules due to aldosterone

(4) Atrial natriuretic factor causes vasoconstriction

Answer: (3) Reabsorption of Na+ and water from renal tubules due to aldosterone

Aldosterone acts mainly at the renal tubules and stimulates the reabsorption of Na+ and water and excretion of K+ and phosphate ions.Thus, aldosterone helps in prevention of diuresis.

3. Which of the following statements is not correct?

(1) Genetically engineered insulin is produced in E-Coli.

(2) In man insulin is synthesised as a proinsulin.

(3) The proinsulin has an extra peptide called C-peptide.

(4) The functional insulin has A and B chains linked together by hydrogen bonds.

Answer: (4) The functional insulin has A and B chains linked together by hydrogen bonds.

Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulphide bridges.

4. Embryological support for evolution was disapproved by:

(1) Oparin

(2) Karl Ernst von Baer

(3) Alfred Wallace

(4) Charles Darwin

Answer: (2) Karl Ernst von Baer

Embryological support for evolution was disapproved by Karl Ernst von Baer. He noted that embryos never pass through the adult stages of other animals.

5. Goblet cells of alimentary canal are modified from:

(1) Compound epithelial cells

(2) Squamous epithelial cells

(3) Columnar epithelial cells

(4) Chondrocytes

Answer: (3) Columnar epithelial cells

Certain cells of columnar epithelial cells contain mucus and are called goblet cells as they look like goblet. Such cells are present in the alimentary canal.

6. The QRS complex in a standard ECG represents:

(1) Repolarisation of ventricles

(2) Repolarisation of auricles

(3) Depolarisation of auricles

(4) Depolarisation of ventricles

Answer: (4) Depolarisation of ventricles

The QRS complex represents the depolarisation of the ventricles, that initiates the ventricular contraction.

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

(1) PS-I to ATP synthase

(2) PS-II to Cyt B6f complex

(3) Cyt B6f complex to PS-I

(4) PS-I to NADP+

Answer: (2) PS-II to Cyt B6f complex

After excitement, e is passed from PS-11(P680) to primary electron acceptor (Pheophytin). From primary e acceptor, e is passed to plastoquinone. Plastoquinone (PQ) in turn transfers its e to Cyt b, f complex. Therefore plastoquinone facilitates the transfer of electrons from PS Il to Cyt be f complex.

8. The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plants is/are:

(1) Ammonia and hydrogen

(2) Ammonia alone

(3) Nitrate alone

(4) Ammonia and oxygen

Answer: (1) Ammonia and hydrogen

The enzyme nitrogenase is a Mo – Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia. The reaction is as follows:

N2 + 8e + 8H+ + 16 ATP —> 2NH3 + 2H+ + 16ADP + 16Pi 

9. Match the following with respect to meiosis:

(a) Zygotene   (i) Terminalization

(b) Pachytene (ii) Chiasmata

(c) Diplotene  (iii) Crossing over

(d) Diakinesis (iv) Synapsis

Select the correct option from the following:

      (a) (b) (c) (d)

(1) (ii) (iv) (iii) (i)

(2) (iii) (iv) (i) (ii)

(3) (iv) (iii) (ii) (i)

(4) (i) (ii) (iv) (iii)

Answer: (3) (iv) (iii) (ii) (i)

10. Match the following columns and select the correct option.

            Column-I                 Column-II

(a) 6 -15 pairs of gill slits     (i) Trygon

(b) Heterocercal caudal fin (ii) Cyclostomes

(c) Air Bladder                    (iii) Chondrichthyes

(d) Poison sting                  (iv) Osteichthyes

     (a) (b) (c) (d)

(1) (i) (iv) (iii) (ii)

(2) (ii) (iii) (iv) (i)

(3) (iii) (iv) (i) (ii)

(4) (iv) (ii) (iii) (i)

Answer: (2) (ii) (iii) (iv) (i)

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

(1) Polysomes

(2) Endoplasmic reticulum

(3) Peroxisomes

(4) Golgi bodies

Answer: (4) Golgi bodies

Golgi apparatus is the important site of formation of glycoproteins and glycolipids. Protein synthesised by the rough endoplasmic reticulum and lipids synthesised by smooth endoplasmic reticulum reach the cisternae of the Golgi apparatus. Here, they combine with carbohydrates to form glycoproteins and glycolipids.

12. 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 tumefaciens (iii) DNApolymerase

(d) Salmonella typhimurium      (iv) Cry proteins

Select the correct option from the following:

     (a) (b) (c) (d)

(1) (iii) (iv) (i) (ii)

(2) (ii) (iv) (iii) (i)

(3) (iv) (iii) (i) (ii)

(4) (iii) (ii) (iv) (i)

Answer: (3) (iv) (iii) (i) (ii)

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

(1) Morgan 

(2) Mendel

(3) Sutton 

(4) Boveri

Answer: (1) Morgan

Sutton and Boveri proposed the chromosomal theory of inheritance but its experimental verification was done by Thomas Hunt Morgan.

14. 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:

     (a) (b) (c) (d)

(1) (ii) (iii) (i) (iv)

(2) (ii) (iv) (iii) (i)

(3) (iii) (i) (iv) (ii)

(4) (iii) (iv) (i) (ii)

Answer: (2) (ii) (iv) (iii) (i)

15. Bilaterally symmetrical and acoelomate animals are exemplified by:

(1) Annelida

(2) Ctenophora

(3) Platyhelminthes

(4) Aschelminthes

Answer: (3) Platyhelminthes

Platyhelminthes are bilaterally symmetrical, triploblastic and acoelomate animals with organ level of organisation.

16. Floridean starch has structure similar to:

(1) Laminarin and cellulose

(2) Starch and cellulose

(3) Amylopectin and glycogen

(4) Mannitol and algin

Answer: (3) Amylopectin and glycogen

In Rhodophyceae, food is stored as floridean starch which is very similar to amylopectin and glycogen in structure.

17. Identify the correct  statement with regard to G1 phase (Gap 1) of interphase.

(1) Nuclear Division takes place.

(2) DNA synthesis or replication takes place.

(3) Reorganisation of all cell components takes place.

(4) Cell is metabolically active, grows but does not replicate its DNA.

Answer: (4) Cell is metabolically active, grows but does not replicate its DNA.

S or synthesis phase marks the period during which DNA synthesis takes place. Reorganisation of all cellular components takes place in M-phase. This phase also starts with nuclear division (Karyokinesis).

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

(1) the head holds a 1/3rd of a nervous system while the rest is situated along the dorsal part of its body.

(2) the supra-oesophageal ganglia of the cockroach are situated in the ventral part of the abdomen.

(3) the cockroach does not have a nervous system.

(4) the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.

Answer: (4) the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.

The head holds a bit of a nervous system of cockroach while the rest is situated along the ventral (belly-side) part of its body. So, if the head of a cockroach is cut off, it will still live for as long as one week.

19. The enzyme enterokinase helps in conversion of:

(1) pepsinogen into pepsin

(2) protein into polypeptides

(3) trypsinogen into trypsin

(4) caseinogen into casein

Answer: (3) trypsinogen into trypsin

Trypsinogen is activated by an enzyme, enterokinase, secreted by the internal mucosa into active trypsin.

20. 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

     (a) (b) (c) (d)

(1) (i) (ii) (iv) (iii)

(2) (ii) (iii) (i) (iv)

(3) (iii) (i) (iv) (ii)

(4) (iv) (ii) (i) (iii)

Answer: (4) (iv) (ii) (i) (iii)

21. Identify the wrong statement with reference to transport of oxygen.

(1) Low pCO2 in alveoli favours the formation of oxyhaemoglobin.

(2) Binding of oxygen with haemoglobin is mainly related to partial pressure of O2.

(3) Partial pressure of CO2 can interfere with O2 binding with haemoglobin.

(4) Higher H+ conc. in alveoli favours the formation of oxyhaemoglobin.

Answer: (4) Higher H+ conc. in alveoli favours the formation of oxyhaemoglobin.

Binding of oxygen with haemoglobin is related to partial pressure of O2,partial pressure of CO2, hydrogen ion concentration and temperature .In the alveoli, high p02, low pCO2, lesser Ht concentration and lower temperature are factors favourable for the formation of oxyhaemoglobin, whereas in the tissues, low pO2, high pCO2, high H+ concentration and high temperature are favourable for dissociation of oxygen from the oxyhaemoglobin.

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

(1) insects and water

(2) insects or wind

(3) water currents only

(4) wind and water

Answer: (2) insects or wind

In many aquatic plants with emergent flowers, pollination occurs by wind or insects, e.g., lotus, water lily, water hyacinth.

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

(1) Insect predators

(2) Insect pests

(3) Fungal diseases

(4) Plant nematodes

Answer: (2) Insect pests

Bt cotton is resistance to cotton bollworm infestation. The genes cry I Ac and cry ||Ab control cotton bollworms, thus acts as bio-pesticide.

24. Select the correct statement.

(1) Insulin is associated with hyperglycemia.

(2) Glucocorticoids stimulate gluconeogenesis.

(3) Glucagon is associated with hypoglycemia.

(4) Insulin acts on pancreatic cells and adipocytes.

Answer: (2) Glucocorticoids stimulate gluconeogenesis.

Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilisation of amino acids. Insulin acts mainly on hepatocytes and adipocytes. Glucagon is associated with hyperglycemia while insulin is associated with hypoglycemia.

25. Identify the basic amino acid from the following.

(1) Valine 

(2) Tyrosine

(3) Glutamic Acid 

(4) Lysine

Answer: (4) Lysine

Glutamic acid, valine and tyrosine are acidic, neutral and aromatic amino acid respectively

26. Flippers of Penguins and Dolphins are examples of:

(1) Natural selection

(2) Adaptive radiation

(3) Convergent evolution

(4) Industrial melanism

Answer: (3) Convergent evolution

Analogous structures are the result of convergent evolution, i.e, different structures evolving for the same function and hence having similarity. For example, the eye of the octopus and of mammals, the flippers of penguins and dolphins.

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

(1) CH3, H2, NH3 and water vapor at 600°C

(2) CH4, H2, NH3 and water vapor at 800°C

(3) CH3, H2, NH4 and water vapor at 800°C

(4) CH4, H2, NH3 and water vapor at 600°C

Answer: (2) CH4, H2, NH3 and water vapor at 800°C

In 1953, S.L. Miller, an American scientist, created similar conditions on a laboratory scale. He created electric discharge in a closed flask containing CH4, H2, NH3 and water vapor at 800°C and observed formation of amino acids.

28. The specific palindromic sequence which is recognized by EcoRI is:

(1) 5′ – GGATCC – 3′

      3′ – CCTAGG – 5′

(2) 5′ – GAATTC – 3′

      3′ – CTTAAG – 5′

(3) 5′ – GGAACC – 3′

      3′ – CCTTGG – 5′

(4) 5′ – CTTAAG – 3′

      3′ – GAATTC – 5′

Answer: (2) 5′ – GAATTC – 3′

                    3′ – CTTAAG – 5′

The palindromes in DNA are base pair sequences that are the same when read forward (left to right) or backward (right to left) from a central axis of symmetry. Thus,GAATTC CTTAAG is a palindromic sequence which is recognised by EcoRI. 

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

(1) Effect on reproduction

(2) Nutritive value

(3) Growth response

(4) Defence action

Answer: (4) Defence action

Nicotine, strychnine and caffeine are examples of alkaloids. These are produced by plants and are used by them in their defense against herbivores and pathogens.

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

(1) Renal calculi and Hyperglycaemia

(2) Uremia and Ketonuria

(3) Uremia and Renal Calculi

(4) Ketonuria and Glycosuria

Answer: (4) Ketonuria and Glycosuria

Diabetes mellitus is associated with formation of harmful compounds known as ketone bodies i.e., Ketonuria and loss of glucose through urine i.e., Glycosuria.

31. 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) (b) and (c)

(2) (d) and (c)

(3) (c) and (a)

(4) (a) and (b)

Answer: (1) (b) and (c)

In Urochordata, notochord is present only in the larval tail. Phylum Chordata is divided into three subphyla: Urochordata or Tunicata, Cephalochordata and Vertebrata.

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

(1) eustachian tube

(2) lining of intestine

(3) ducts of salivary glands

(4) proximal convoluted tubule of nephron

Answer: (4) proximal convoluted tubule of nephron

The cuboidal epithelium is composed of a single layer of cube-like cells which is commonly found in ducts of glands and tubular parts of nephrons in kidneys and its main functions are secretion and absorption. The epithelium of proximal convoluted tubules (PCT) of nephron in the kidney has microvilli.

33. 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

     (a) (b) (c) (d)

(1) (iv) (iii) (ii) (i)

(2) (iii) (iv) (ii) (i)

(3) (ii) (i) (iv) (iii)

(4) (i) (ii) (iv) (iii)

Answer: (3) (ii) (i) (iv) (iii)

34. Which of the following is correct about viroids?

(1) They have free DNA without a protein coat.

(2) They have RNA with a protein coat.

(3) They have free RNA without a protein coat.

(4) They have DNA with a protein coat.

Answer: (3) They have free RNA without a protein coat.

Viroids are free RNA particles that lack protein coat. They are infectious agents smaller than viruses.

35. The body of the ovule is fused within the funicle at:

(1) Chalaza 

(2) Hilum

(3) Micropyle 

(4) Nucellus

Answer: (2) Hilum

The hilum is a scar on the seed coat where the funicle and body of the ovule is attached.

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

(1) 1 molecule of 4-C compound and 1 molecule of 2-C compound.

(2) 2 molecules of 3-C compound

(3) 1 molecule of 3-C compound

(4) 1 molecule of 6-C compound

Answer: (3) 1 molecule of 3-C compound

During photorespiration in C3 plants, some O2 does bind to RuBisCO and RuBP instead of being converted to 2 molecules of PGA binds with 0, to form one molecule of phosphoglycerate (3 Carbon) and phosphoglycolate (2 Carbon).

37. 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 histaminase, destructive enzymes

(d) Lymphocytes (iv) Release granules containing histamine

     (a) (b) (c) (d)

(1) (ii) (i) (iii) (iv)

(2) (iii) (iv) (ii) (i)

(3) (iv) (i) (ii) (iii)

(4) (i) (ii) (iv) (iii)

Answer: (2) (iii) (iv) (ii) (i)

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

(1) Low concentration of FSH

(2) High concentration of Estrogen

(3) High concentration of Progesterone

(4) Low concentration of LH

Answer: (2) High concentration of Estrogen

 FSH, LH and estrogen are at peak during ovulation (release of ovum).

39. Select the correct events that occur during inspiration.

(a) Contraction of diaphragm

(b) Contraction of external intercostal muscles

(c) Pulmonary volume decreases

(d) Intra pulmonary pressure increases

(1) only (d)

(2) (a) and (b)

(3) (c) and (d)

(4) (a), (b) and (d)

Answer: (2) (a) and (b)

Inspiration is initiated by the contraction of diaphragm that increases the volume of the thoracic chamber in the antero-posterior axis. The contraction of external intercostal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intra pulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs.

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

(1) GIFT and ICSI

(2) ZIFT and IUT

(3) GIFT and ZIFT

(4) ICSI and ZIFT

Answer: (2) ZIFT and IUT

In vitro fertilisation i.e., fertilisation outside the body in almost similar conditions as that in the body, followed by embryo transfer (ET) method. In this method, ova from the wife or donor (female) and sperm from husband or donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory. The zygote or early embryos (with upto 8 blastomeres) could then be transferred into the fallopian tube (ZIFT – Zygote Intra Fallopian Transfer) and embryos with more than 8 blastomeres, into the uterus (IUT – Intra Uterine Transfer) to complete its further development.

41. The infectious stage of Plasmodium that enters the human body is:

(1) Male gametocytes 

(2) Trophozoites

(3) Sporozoites 

(4) Female gametocytes

Answer: (3) Sporozoites

Plasmodium enters the human body as sporozoites (infectious form) through the bite of an infected female Anopheles mosquito.

42. 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

     (a) (b) (c) (d)

(1) (ii) (iii) (iv) (i)

(2) (iv) (iii) (i) (ii)

(3) (i) (iv) (ii) (iii)

(4) (iii) (ii) (iv) (i)

Answer: (1) (ii) (iii) (iv) (i)

43. Select the correct match.

(1) Thalassemia – X linked

(2) Haemophilia – Y linked

(3) Phenylketonuria – Autosomal dominant trait

(4) Sickle cell anaemia – Autosomal recessive trait, chromosome-11

Answer: (4) Sickle cell anaemia – Autosomal recessive trait, chromosome-11

44. Which of the following statements is correct?

(1) Adenine does not pair with thymine

(2) Adenine pairs with thymine through two H-bonds

(3) Adenine pairs with thymine through one H-bond

(4) Adenine pairs with thymine through three H-bonds

Answer: (2) Adenine pairs with thymine through two H-bonds

Adenine pairs with thymine by forming two hydrogen bonds, A = T.

45. Which of the following is the most abundant protein in the animals?

(1) Insulin 

(2) Haemoglobin

(3) Collagen 

(4) Lectin

Answer: (3) Collagen

Collagen is the most abundant protein in the animal world. RuBisCO is the most abundant protein in the world of the biosphere.

46. Which of the following pairs is of unicellular algae?

(1) Chlorella and Spirulina

(2) Laminaria and Sargassum

(3) Gelidium and Gracilaria

(4) Anabaena and Volvox

Answer: (1) Chlorella and Spirulina

Gelidium, Gracilaria, Laminaria and Sargassum are multicellular. Anabaena is filamentous blue green algae. Volvox is colonial.

47. 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) and (d) 

(2) (a) only

(3) (a), (b) and (c) 

(4) (c) and (d)

Answer: (1) (a) and (d)

48. Identify the incorrect statement.

(1) Due to deposition of tannins, resins, oils etc., heart wood is dark in colour

(2) Heart wood does not conduct water but gives mechanical support

(3) Sapwood is involved in conduction of water and minerals from root to leaf

(4) Sapwood is the innermost secondary xylem and is lighter in colour

Answer: (4) Sapwood is the innermost secondary xylem and is lighter in colour

Sapwood is the peripheral or outermost region of the secondary xylem and lighter in colour.

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

(1) Inbreeding

(2) Outcrossing

(3) Mutational breeding

(4) Cross breeding

Answer: (4) Cross breeding

Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams. In cross-breeding, superior males of one breed are mated with superior females of another breed.

50. 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) G2 phase 

(2) M phase

(3) G1 phase 

(4) S phase

Answer: (2) M phase / (3) G1 phase

51. Identify the correct statement with reference to the human digestive system.

(1) Vermiform appendix arises from duodenum

(2) Ileum opens into small intestine

(3) Serosa is the innermost layer of the alimentary canal

(4) Ileum is highly coiled part

Answer: (4) Ileum is highly coiled part

lleum opens into the large intestine. Serosa is the outermost layer of the alimentary canal. The vermiform appendix is a vestigial organ, arising from the caecum. 

52. Which of the following refer to the correct example(s) of organisms which have evolved due to changes in 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) Only (d) 

(2) Only (a)

(3) (a) and (c) 

(4) (b), (c) and (d)

Answer: (4) (b), (c) and (d)

Herbicide resistant weeds, drug resistant eukaryotes and man-created breeds of domesticated animals like dogs are examples of evolution by anthropogenic action. Darwin’s Finches of Galapagos islands are an example of natural selection, adaptive radiation and Founder’s effect.

53. 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) Addision’s disease

(a) (b) (c) (d)

(1) (ii) (i) (iv) (iii)

(2) (iv) (iii) (i) (ii)

(3) (iii) (ii) (i) (iv)

(4) (iii) (i) (iv) (ii)

Answer: (4) (iii) (i) (iv) (ii)

54. Select the option including all sexually transmitted diseases.

(1) Cancer, AIDS, Syphilis

(2) Gonorrhoea, Syphilis, Genital herpes

(3) Gonorrhoea, Malaria, Gential herpes

(4) AIDS, Malaria, Filaria

Answer: (2) Gonorrhoea, Syphilis, Genital herpes

Malaria, filaria and cancer are not sexually transmitted diseases (STDs).

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

(1) Three 

(2) Zero 

(3) One 

(4) Two

Answer: (3) One

In the citric acid cycle, during the conversion of succinyl CoA to succinic acid a molecule of GTP is synthesized which is called substrate level phosphorylation.

56. Montreal protocol was signed in 1987 for control of:

(1) Disposal of e-wastes

(2) Transport of Genetically modified organisms from one country to another

(3) Emission of ozone depleting substances

(4) Release of GreenHouse gases

Answer: (3) Emission of ozone depleting substances

Recognising the deleterious effects of ozone depletion, an international treaty, known as the Montreal Protocol was signed at Montreal (Canada)in 1987 (effective in 1989) to control the emission of ozone depleting substances.

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

(a) Iron               (i) Photolysis of water

(b) Zinc              (ii) Pollen germination

(c) Boron           (iii) Required for chlorophyll biosynthesis

(d) Manganese  (iv) IAA biosynthesis

    (a) (b) (c) (d)

(1) (iv) (i) (ii) (iii)

(2) (ii) (i) (iv) (iii)

(3) (iv) (iii) (ii) (i)

(4) (iii) (iv) (ii) (i)

Answer: (4) (iii) (iv) (ii) (i)

58. Match the following columns and select the correct option.

                       Column-I          Column-II

(a) Gregarious, polyphagous   (i) Asterias pest

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

(c) Book lungs                         (iii) Ctenoplana

(d) Bioluminescence               (iv) Locusta

     (a) (b) (c) (d)

(1) (ii) (i) (iii) (iv)

(2) (i) (iii) (ii) (iv)

(3) (iv) (i) (ii) (iii)

(4) (iii) (ii) (i) (iv)

Answer: (3) (iv) (i) (ii) (iii)

59. According to Robert May, the global species diversity is about:

(1) 7 million 

(2) 1.5 million

(3) 20 million 

(4) 50 million

Answer: (1) 7 million

60. Ray florets have:

(1) Half inferior ovary

(2) Inferior ovary

(3) Superior ovary

(4) Hypogynous ovary

Answer: (2) Inferior ovary

The ovary is inferior in ray florets of sunflowers. It possess epigynous flower i.e., the margin of thalamus grows upward enclosing the ovary completely and getting fused with it, the other parts of flower arise above the ovary

61. 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 × 109 bp, then the length of the DNA is approximately:

(1) 2.7 meters 

(2) 2.0 meters

(3) 2.5 meters 

(4) 2.2 meters

Answer: (4) 2.2 meters

If the distance between two consecutive base pairs is 0.34 nm i.e., 0.3 x 10–9 m and the total number of base pairs of a DNA double helix in a typical mammalian cells is 6.6 x 10–9 bp then the length of DNA is calculated by multiplying the total number of base pair with distance between two consecutive base pair i.e., 6.6 x 10–9 bp x 0.34 nm = 2.2 m (approx.).

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

Column – I        Column – II

(a) Bt cotton    (i) Gene therapy

(b) Adenosine (ii) Cellular defence deaminase deficiency

(c) RNAi         (iii) Detection of HIVinfection

(d) PCR          (iv) Bacillus thuringiensis

     (a) (b) (c) (d)

(1) (i) (ii) (iii) (iv)

(2) (iv) (i) (ii) (iii)

(3) (iii) (ii) (i) (iv)

(4) (ii) (iii) (iv) (i)

Answer: (2) (iv) (i) (ii) (iii)

63. Match the trophic levels with their correct species examples in the 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:

     (a) (b) (c) (d)

(1) (i) (ii) (iii) (iv)

(2) (ii) (iii) (iv) (i)

(3) (iii) (ii) (i) (iv)

(4) (iv) (iii) (ii) (i)

Answer: (2) (ii) (iii) (iv) (i)

64. 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

     (a) (b) (c) (d)

(1) (iv) (i) (ii) (iii)

(2) (i) (iii) (ii) (iv)

(3) (iii) (iv) (i) (ii)

(4) (ii) (i) (iii) (iv)

Answer: (3) (iii) (iv) (i) (ii)

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

(1) Lateral roots 

(2) Fibrous roots

(3) Primary roots 

(4) Prop roots

Answer: (2) Fibrous roots

In monocotyledonous plants, the primary root is short lived and is replaced by a large number of roots. These roots originate from the base of the stem and constitute the fibrous root system e.g., wheat plant.

66. Meiotic division of the secondary oocyte is completed:

(1) At the time of fusion of a sperm with an ovum

(2) Prior to ovulation

(3) At the time of copulation

(4) After zygote formation

Answer: (1) At the time of fusion of a sperm with an ovum

In humans, the secondary oocyte is produced when the primary oocyte grows in size and completes its first meiotic division. The secondary oocyte will be arrested at this stage of metaphase of meiosis Il until fertilisation takes place. Thus, when a sperm cell fertilises the female sex cell, the secondary oocyte rapidly completes the remaining stages of meiosis II, giving rise to ovum, with which the sperm cell unites.

67.Identify the wrong  statement with regard to Restriction Enzymes.

(1) Sticky ends can be joined by using DNA ligases.

(2) Each restriction enzyme functions by inspecting the length of a DNA sequence.

(3) They cut the strand of DNA at palindromic sites.

(4) They are useful in genetic engineering.

Answer: (1) Sticky ends can be joined by using DNA ligases.

68. In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct?

(1) There is no relationship between Gross primary productivity and Net primary productivity.

(2) Gross primary productivity is always less than net primary productivity.

(3) Gross primary productivity is always more than net primary productivity.

(4) Gross primary productivity and Net primary productivity are one and same.

Answer: (3) Gross primary productivity is always more than net primary productivity.

Gross primary productivity (GPP) is the rate of production of organic matter during photosynthesis. A considerable amount of GPP (Gross Primary Productivity) is utilised by plants in respiration. GPP minus Respiration Losses (R) is the Net Primary Productivity (NPP). GPP – R = NPP. Therefore, GPP is always more than NPP.

69. The process of growth is maximum during:

(1) Dormancy 

(2) Log phase

(3) Lag phase 

(4) Senescence

Answer: (2) Log phase

In most systems, the initial growth is slow i.e., lag phase, and it increases rapidly thereafter – atan exponential rate i.e., log or exponential phase.

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

(1) Recognition site

(2) Selectable marker

(3) Ori site

(4) Palindromic sequence

Answer: (3) Ori site

Ori site 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 in the vector. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin supports a high copy number.

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

(1) RNA polymerase

(2) DNA ligase

(3) DNA helicase

(4) DNA polymerase

Answer: (1) RNA polymerase

RNA polymerase facilitates the opening of the DNA helix and continues elongation.

72. Snow-blindness in Antarctic region is due to:

(1) Damage to retina caused by infra-red rays

(2) Freezing of fluids in the eye by low temperature

(3) Inflammation of cornea due to high dose of UV-B radiation

(4) High reflection of light from snow

In the human eye, cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of cornea. This leads to a disorder called snow-blindness cataract. It leads to diminishing of eye sight, photo burning and later permanent damage to cornea that results in actual cataract

Answer: (3) Inflammation of cornea due to high dose of UV-B radiation

73. Strobili or cones are found in:

(1) Equisetum 

(2) Salvinia

(3) Pteris 

(4) Marchantia

Answer: (1) Equisetum

The sporophytes of pteridophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases sporophylls may form distinct compact structures called strobili or cones, e.g., Selaginella, Equisetum.

74. 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

(a) (b) (c) (d)

(1) (iv) (iii) (i) (ii)

(2) (ii) (iv) (i) (iii)

(3) (i) (iii) (ii) (iv)

(4) (iii) (ii) (iv) (i)

Answer: (1) (iv) (iii) (i) (ii)

75. Which of the following is put into an Anaerobic sludge digester for further sewage treatment?

(1) Activated sludge

(2) Primary sludge

(3) Floating debris

(4) Effluents of primary treatment

Answer: (1) Activated sludge

The sediment in a settling tank is called activated sludge. A small part of it is pumped back into the aeration tank to serve as the inoculum. While the remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters.

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

(1) Allele ‘i’ does not produce any sugar.

(2) The gene (I) has three alleles.

(3) A person will have only two of the three alleles.

(4) When IA and IB are present together, they express the same type of sugar.

Answer: (4) When IA and IB are present together, they express the same type of sugar.

ABO blood groups are controlled by the gene I. The gene I has three alleles IA, IB and i. The alleles IA and IB produce a slightly different form of the sugar while allele i does not produce any sugar. Because humans are diploid organisms, each person possesses any two of the three I gene alleles. When IA and IB are present together they both express their own types of sugars, because of co-dominance.

77. The ovary is half inferior in:

(1) Plum 

(2) Brinjal

(3) Mustard 

(4) Sunflower

Answer: (1) Plum

The ovary is superior in brinjal and mustard while it is inferior in sunflower

78. The first phase of translation is:

(1) Recognition of an anticodon

(2) Binding of mRNA to ribosome

(3) Recognition of DNA molecule

(4) Aminoacylation of tRNA

Answer: (4) Aminoacylation of tRNA

Translation is the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. The amino acids are joined by a bond that is known as a peptide bond. Formation of a peptide bond requires energy. So, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA-a process commonly called as charging of tRNA or aminoacylation of tRNA.

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

(1) Ethidium bromide in infrared radiation

(2) Acetocarmine in bright blue light

(3) Ethidium bromide in UV radiation

(4) Acetocarmine in UV radiation

Answer: (3) Ethidium bromide in UV radiation

In gel electrophoresis, separated DNA fragments can be visualised only after staining the DNA with a compound i.e., ethidium bromide and followed by exposure to UV radiation as bright orange coloured bands.

80. Dissolution of the synaptonemal complex occurs during:

(1) Leptotene

(2) Pachytene

(3) Zygotene

(4) Diplotene

Answer: (4) Diplotene

During diplotene in prophase I of meiosis I dissolution of the nucleoprotein synaptonemal complex occurs.

81. Identify the substances having glycosidic bond and peptide bond, respectively in their structure:

(1) Inulin, insulin 

(2) Chitin, Cholesterol

(3) Glycerol, trypsin 

(4) Cellulose, lecithin

Answer: (1) Inulin, insulin

Inulin is a polymer of fructose (polysaccharide). In a polysaccharide the individual monosaccharides are linked by a glycosidic bond. While insulin is a polymer of amino acids linked by a peptide bond.

82. Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.

(1) Abscisic acid 

(2) Cytokinin

(3) Gibberellin

(4) Ethylene

Answer: (3) Gibberellin

Spraying sugarcane crops with gibberellins increases the length of the stem and thus increasing the yield by as much as 20 tonnes per acre.

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

(1) These represent reserve material in cytoplasm.

(2) They are not bound by any membrane.

(3) These are involved in ingestion of food particles.

(4) They lie free in the cytoplasm.

Answer: (3) These are involved in ingestion of food particles.

Reserve material in prokaryotic cells are stored in the cytoplasm in the form of inclusion bodies. These are not bound by any membrane system and lie free in the cytoplasm, e.g., phosphate granules, cyanophycean granules and glycogen granules.

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

(1) Amazon forests

(2) Western Ghats of India

(3) Madagascar

(4) Himalayas

Answer: (1) Amazon forests

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

(1) 8

(2) 4

(3) 2 

(4) 14

Answer: (4) 14

Mendel conducted artificial pollination or cross pollination experiments using several true-breeding pea lines. He selected 14 true-breeding pea plant varieties, as pairs which were similar except for one character with contrasting traits. Some of the contrasting traits selected were smooth or wrinkled seeds, yellow or green seeds, inflated (full) or constricted green or yellow pods and tall or dwarf plants.

86. Identify the wrong statement with reference to immunity.

(1) Foetus receives some antibodies from mother,it is an example for passive immunity.

(2) When exposed to antigen (living or dead) antibodies are produced in the host’s body. It is called “Active immunity”.

(3) When ready-made antibodies are directly given, it is called “Passive immunity”.

(4) Active immunity is quick and gives full response.

Answer: (4) Active immunity is quick and gives full response.

Active immunity is slow and takes time to give its full effective response.

87. Which of the following is not an attribute of a Population?

(1) Species interaction

(2) Sex ratio

(3) Natality

(4) Mortality

Answer: (1) Species interaction

88. Choose the correct pair from the following:

(1) Exonucleases: Make cuts at specific positions within DNA

(2) Ligases: Join the two DNA molecules

(3) Polymerases: Break the DNA into fragments

(4) Nucleases: Separate the two strands of DNA

Answer: (2) Ligases: Join the two DNA molecules

89. 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) Plasmolysis 

(2) Transpiration

(3) Root pressure 

(4) Imbibition

Answer: (3) Root pressure

Transportation of ions from the soil into the vascular tissues of the roots, increases the pressure inside the xylem, known as root pressure. Effects of root pressure is 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.

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

(1) Para-ascorbic acid

(2) Gibberellic acid

(3) Abscisic acid

(4) Phenolic acid

Answer: (2) Gibberellic acid

Gibberellic acid is involved in growth promoting activities, such as cell division, cell enlargement, pattern formation, tropic growth, flowering, fruiting and seed formation and is called plant growth promoter. It breaks seed dormancy which is antagonistic to abscisic acid.


If you’ve made it till here, good job! Proud of you 🙂

To boost your preparation, I’d advise you to go through my detailed analysis of NEET 2019 & 2020 where in I’ve carefully examined the difficulty level and the trend of distribution of marks over the two years.

And if you intend to download a pdf of the NEET 2020 Biology question paper with solutions, visit here.

Hope you found the article helpful. Would love to know your thoughts/suggestions in the comments below!

Good luck Medicoholics! Until next time.

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