Explore topic-wise MCQs in NEET.

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9651.

Which is correct about energy changes in living cells                         [AMU 2001]

A. First energy transfer, then energy transformation
B. First energy transformation, then energy transfer
C. Both occur discontinuously
D. Both occur continuously
Answer» E.
9652.

Choose the correct description depicted by floral diagram                                          [KCET 2002]

A. United valvate sepals, free twisted petals, free stamens, unilocular ovary with marginal placenta
B. United valvate sepale, free imbricate petals, free stamens, unilocular ovary with axile placenta
C. United valvate sepals, free imbricate petals, epipetalous stamens, unilocular ovary with marginal placenta
D. United valvate sepals, free imbricate petals, free stamens, unilocular ovary with marginal placentation
Answer» E.
9653.

The group that does not fit into this category [MP PMT 1993]

A. Amphibia
B. Reptiles
C. Aves
D. Mammals
Answer» B. Reptiles
9654.

Match the names of items listed under column I with the items given under column II; choose the answer which gives the correct combination of the alphabets of the two columns Column I Column II A         Ketone p         has 5 carbon sugar, a phospphate group and a nitrogenous base  B         Fatty acid q         has long chain of carbon and hydrogen atoms and a carboxyl group at one end C         Aldehyde r          has a carbonyl group at the end of the carbon chain D        Nucleotide s          has a carbonyl group within the carbon chain

A. A = s, B = q, C = r, D = p
B. A = q, B = r, C = s, D = p               
C. A = s, B = r, C =q, D = p                
D. A = r, B = q, C = s, D = p
Answer» B. A = q, B = r, C = s, D = p               
9655.

Chief features of family Brassicaceae/Cruciferae is presence of                                  [APMEE 1994; BHU 2002]

A. Latex
B. Pectin
C. Alkaloids
D. Myrosin enzyme
Answer» E.
9656.

Haemolytic jaundice is caused due to a dominant gene but only 10% of the people actually develop the disease. A heterozygous man marries a homozygous normal woman; what proportion of the children would be expected to develop the haemolytic disease [AIIMS 1982]

A. 44317
B. 44470
C. 42005
D. 43831
Answer» E.
9657.

Match the names of biomolecules listed under column I with the characteristics given under column II; choose the answer which gives the correct combination of the alphabets of the two columns Column I (Biomolecule) Column II (Characteristics) A     ATP p       One or more sugar monomers B     DNA                              gf q       The main source of energy C     Protein r        Two strands of nucleotides D     Carbohydrate s        Long sequence of amino acids

A. A = s, B = q, C = r, D = p
B. A = q, B = r, C = s, D = p
C. A = s, B = r, C = q, D = p
D. A = r, B = q, C = s, D = p
Answer» C. A = s, B = r, C = q, D = p
9658.

A monocot showing reticulate venation is     [APMEE 1995]

A. Bombusa
B. Smilax
C. Callophyllum
D. Ginkgo
Answer» C. Callophyllum
9659.

Which of them is in tracheate group [MP PMT 1995]

A. Crab-Centipede-Cockroach
B. King crab-Scorpion-Housefly
C. Spider-Peripatus-Mosquito
D. Bedbug-Sandfly-Silkworm
Answer» E.
9660.

If a human mother has ?O' blood group, the foetus would die if the blood of foetus is [NCERT 1982]

A. A
B. B
C. AB
D. Would remain unaffected by blood group whether it is A, B or AB
Answer» E.
9661.

The floral formula for Malvaceae is    [RPMT 1991, 97]

A. Å      Epi(3?7) K(5) C(5) A(¥) \[\underline{{{G}_{(5)}}}\]         
B. Å      Epi(3?7) K(5) C5 A5 \[\underline{{{G}_{(5)}}}\]
C. Å      Epi(3-7) K(5) C5 A(¥) \[\underline{{{G}_{(5)}}}\]
D. Å      Epi(3-7) K(5) C(5) A(¥) \[\underline{{{G}_{(3-\infty )}}}\]
Answer» D. Å      Epi(3-7) K(5) C(5) A(¥) \[\underline{{{G}_{(3-\infty )}}}\]
9662.

A myriapoda has [NCERT 1983]

A. Chitinous exoskeleton, dorsal nerve cord, three body  segments and one pair of antennae
B. Chitinous exoskeleton, ventral nerve cord, three body segments and two pair of antennae
C. Soft body, ventral nerve cord, numerous body segments and two pair of antennae
D. Chitinous exoskeleton, ventral nerve cord, numerous body segments and one pair of antennae
Answer» E.
9663.

A pair of contrasting characters is termed as [AFMC 1984]

A. Allelomorphs
B. Homozygous
C. Heterozygous
D. Polymorphs
Answer» B. Homozygous
9664.

Which statement is correct for biomolecules                                        [RPMT 2001]

A. DNA is a polymer of ribonucleotides
B. All carbohydrates break down into glucose
C. RNA is single stranded and contain different purine base than DNA
D. Sequence of amino acids determine primary structure of proteins
Answer» E.
9665.

The condition of stamens in Cruciferae family is correctly expressed as                                   [CPMT 1997]

A. A6    
B. A2+4
C. A4+2  
D. All of these
Answer» C. A4+2  
9666.

A person meets with an accident and great loss of blood has occurred. There is no time to analyse his blood group. It is safe to transfer blood of group [AIIMS 1984; Orissa JEE 2005]

A. \[AB,R{{h}^{+}}\]
B. \[AB,R{{h}^{-}}\]
C. \[O,R{{h}^{-}}\]
D. \[O,R{{h}^{+}}\]
Answer» D. \[O,R{{h}^{+}}\]
9667.

Organic compounds are present in all living cells. They all share the following characteristic    

A. Are hydrophobic
B. Are biological catalysts
C. Are used in protein synthesis
D. Are composed of carbon atom backbone surrounded by hydrogen atoms
Answer» E.
9668.

Which of the following pair is a combination of highest and lowest energy molecules

A. Glucose and pyruvic acid
B. Palmitic acid and acetyl CoA
C. Glucose and maleic acid
D. Maleic acid and acetyl CoA
Answer» C. Glucose and maleic acid
9669.

The maximum evolution of oxygen  is by greatest producers of organic matter [DPMT 1990; CBSE PMT 1989]

A. Great land area
B. Crops
C. Phytoplankton of sea
D. Forests
Answer» D. Forests
9670.

Rhizobium lives symbiotically with root nodules and fixes atmospheric nitrogen. It gives nitrogenous compound to host and inturn takes food from host. This food taken by bacterium is

A. Fats
B. Proteins
C. Carbohydrates
D. Any of these
Answer» D. Any of these
9671.

Woodward (1669) observed that plant grew better in muddy water than in rain water because

A. Muddy water had most of essential elements dissolved in it
B. Muddy water had micro nutrients dissolved in it
C. Muddy water had macro nutrients dissolved in it
D. None of these
Answer» D. None of these
9672.

Which of the following is a correct sequence for the appearance of compounds in a cell

A. Amino acids ® protein ® enzyme
B. Protein ® enzyme ® amino acids
C. Disaccharides ® monosaccharides ® polysaccharides
D. Monosaccharides ® starch ® sucrose
Answer» B. Protein ® enzyme ® amino acids
9673.

Two type of cells hyaline and green or with various shades are characteristic of bryophytes in  [BHU 1986]

A.            Funaria hygrometrica
B.            Polytrichum commune
C.          Sphagnum pappiolossum
D.            Porella pelatyphylla
Answer» D.            Porella pelatyphylla
9674.

Fungi can be stained by                       [AFMC 1993; BVP 2000]

A.                    Safranine                          
B.                    Iodine
C.                    Lactophenol                     
D.               Cotton blue
Answer» E.
9675.

Amoeba reacts                                                              [CPMT 1987]

A.          Negatively to strong light and positively to weak light
B.            Positively to strong light and negatively to weak light
C.            Unaffected by light intensity
D.            Positive to both strong and weak light
Answer» B.            Positively to strong light and negatively to weak light
9676.

Kinocilia are

A. Motile
B. Non-motile
C. Both [a] and [b] according to function
D. None of the above
Answer» B. Non-motile
9677.

E. coli bacterium possesses a single large circular DNA as its genetic material. This strands codes for

A.            1000 to 2000 different proteins
B.            2000 to 3000 different proteins
C.            3000 to 4000 different proteins
D.            More than 4000 different proteins
Answer» C.            3000 to 4000 different proteins
9678.

From a cross AABb × aaBb, the genotypes AaBB : AaBb : Aabb : aabb will be obtained in the following ratio [BHU 1994]

A. 1 : 1 : 1 : 1
B. 1 : 2 : 1 : 0
C. 0 : 3 : 1 : 0
D. 1 : 1 : 1 : 0
Answer» C. 0 : 3 : 1 : 0
9679.

The process of evolution                                             [DPMT 1985]

A. Is a continuous process
B. Is a discontinuous process
C. Was continuous in beginning but discontinuous now
D. Was discontinuous in beginning but continuous now
Answer» B. Is a discontinuous process
9680.

Parachute mechanism of fruit dispersal as found in compositae is due to structure named

A. Bract
B. Pappus
C. Coma
D. Barbs
Answer» C. Coma
9681.

Roundworms differ from flatworms in having a

A. Circulatory system
B. Pseudocoel
C. Dorsal nerve cord
D. Circular muscle layer
Answer» C. Dorsal nerve cord
9682.

Mixture of oil in water forms an emulsion. If a protein is shaken up with such an emulsion      

A. The emulsion gets stabilized
B. Protein molecules settle down
C. Coalescence of oil particles takes place
D. A thin film of protein molecules is formed on the surface
Answer» B. Protein molecules settle down
9683.

The fruit of Apricot undergoes dispersal by

A. Compensated zoochory
B. By man
C. Forced zoochory
D. By wind
Answer» B. By man
9684.

The larva of Ascaris undergoes the migration in which of the following course [BHU 1999]

A. Alimentary canal, liver, intestine
B. Alimentary canal, heart, lungs, trachea, intestine
C. Alimentary canal, heart, liver, lungs, trachea, mouth, intestine
D. Alimentary canal, liver, heart, lungs, trachea, pharynx, intestine
Answer» D. Alimentary canal, liver, heart, lungs, trachea, pharynx, intestine
9685.

How many nucleosomes are found in helical coil of 30 nm chromatin fibre [RPMT PMT 2000]

A. 10
B. 12
C. 6
D. 9
Answer» D. 9
9686.

Most of the unique properties of water result from the fact that water molecules         

A. Are very small                                     
B. Tend to stick together
C. Easily separate from one another
D. Are held together by covalent bonds
Answer» E.
9687.

In hydrochory small fruits and seeds are carried over the surface of water through

A. Surface tension
B.                   Light weight of fruits
C. Buoyancy mechanism
D. None of these
Answer» B.                   Light weight of fruits
9688.

If the number of chromosomes in most body cells of a mammal is 40, the cells in the seminiferous tubule will have [AFMC 1994]

A. 40 chromosomes
B. 20 chromosomes
C. 10 chromosomes
D. While some other will have 20
Answer» E.
9689.

If a pair of shared electrons are more attracted to one nucleus than to the other, they form a/an......bond.               

A. Polar covalent                                     
B. Ionic
C. Nonpolar covalent                             
D. Hydrogen
Answer» B. Ionic
9690.

Given below are four matchings of an animal and its kind of respiratory organ [CBSE PMT  2003] 1.  Silver Fish -- trachea,  2.  Scorpion - book lung, 3.  Sea squirt - pharyngeal gills, 4. Dolphin -- skin

A. 3 and 4
B. 1 and 4
C. 1, 2 and 3
D. 2 and 4
Answer» D. 2 and 4
9691.

Dust seeds, flattened fruits, balloon fruits are ideal for the

A. Autochory
B. Anemochory
C. Hydrochory
D. Zoochory
Answer» C. Hydrochory
9692.

A substance that...........is considered to be a chemical compound   

A. Has covalent bonds
B. Has a chemical bond
C. Has a stable valence shell                  
D. Contains at least two different elements
Answer» E.
9693.

Fruit of custard apple is                                [AFMC 2004]

A. Etaerio of berries
B. Etaerio of follicles
C. Etaerios of achenes
D. Ethaerio of drups
Answer» B. Etaerio of follicles
9694.

The formation of a male child depends on the sperms because [CPMT 1992]

A. Sperms may be X and Y
B. Sperms are all Y
C. The eggs from the other ovary may be Y
D. Sperms are more active
Answer» B. Sperms are all Y
9695.

Molecules are always moving. Some molecules move faste than others,.........is a measure of their average velocity of movement      

A. Density     
B. Polarity
C. Electronegativity                                 
D. Temperature
Answer» E.
9696.

During its life cycle, Fasciola hepatica (Liver Fluke) infects its intermediate host and primary host at the following larval stages respectively [CBSE PMT 2003]

A. Redia and miracidium
B. Cercaria and redia
C. Metacercaria and cercaria
D. Miracidium and metacercaria
Answer» E.
9697.

Coleorrhiza is a cap like covering over [AMU 1997]

A. Plumule in a dicot
B. The radicle in a monocot           
C. Radicle in dicot
D. Plumule in a monocot
Answer» C. Radicle in dicot
9698.

Frequency of crossing over in a chromosome is an index of

A. Relative distance between its genes only
B. Structural affinity between its genes only
C. Strength linkage between its genes only
D. [a] and [b] both are correct
Answer» B. Structural affinity between its genes only
9699.

Which of the following hierarchy is correct with reference to levels of biological organization   

A. Species \[\to \]Communities \[\to \] Populations \[\to \] Ecosystems \[\to \]Landscapes \[\to \]Biosphere
B. Species \[\to \] Populations \[\to \] Communities \[\to \] Ecosystems \[\to \] Landscapes \[\to \] Biosphere
C. Species \[\to \] Populations \[\to \] Ecosystems \[\to \] Landscapes \[\to \] Communities \[\to \] Biosphere 
D. Species \[\to \] Population \[\to \] Communities \[\to \] landscapes \[\to \] Biosphere \[\to \] Ecosystems
Answer» C. Species \[\to \] Populations \[\to \] Ecosystems \[\to \] Landscapes \[\to \] Communities \[\to \] Biosphere 
9700.

Which of the following is a non-endospermic monocot seed                                       [Bihar MDAT 1991]

A. Plumbago
B. Castor
C. Linseed   
D. Alisma
Answer» E.