datavorous/entrance-exam-dataset huggingface dataset:
The main difference between DNA and RNA is ____

Options:
A in the nature of sugar alone.
B in the nature of purines alone.
C in the nature of sugar and pyrimidines.
D none of these.

The correct answer is:
in the nature of sugar and pyrimidines.

Solution:
The heterocyclic compounds in nucleic acids are the nitrogenous bases named adenine, guanine, uracil, cytosine, and thymine. Adenine and guanine are substituted purines while the rest are substituted pyrimidines. The skeletal heterocyclic ring is composed of purine and pyrimidine. The sugar found in polynucleotides is either ribose (a monosaccharide pentose) or 2’ deoxyribose. A nucleic acid containing deoxyribose is called deoxyribonucleic acid (DNA), while that which contains ribose is called ribonucleic acid (RNA). Another difference between DNA and RNA is that DNA has thymine (a pyrimidine) while the RNA has uracil (a pyrimidine).

Topic: Chemistry, Biomolecules, NEET
The main difference in Gram (+ve) and Gram (-ve) bacteria resides in their:

Options:
A Cell wall
B Cell membrane
C Cytoplasm
D Flagella

The correct answer is:
Cell wall

Solution:
Gram (+)ve and Gram (-)ve bacteria separate on the basis of their cell wall composition. Christian Gram on the basis of staining behaviour of the cell wall with Gramâs stain, grouped bacteria into Gram (+)ve and Gram (â)ve type. The bacteria which retain blue or purple colour after staining are Gram (+)ve e.g. Bacillus subtilis and which loses blue colour is Gram ( â )ve and retain counter stain saffranine appear red eg. E.coli. In Gram (â)ve, the stain is washed due to high lipid content in the cell wall having a thick wall, Gram ( + )ve has a single-layered cell wall.

Gram stain â Crystal Violet

Topic: Biology, Biological Classification, NEET
The main difference in Gram +ve and Gram -ve bacteria resides in their

Options:
A Differential staining of cell wall
B Special staining of cell wall
C Simple staining of cell wall
D Positive staining of cell wall

The correct answer is:
Differential staining of cell wall

Solution:
Bacteria can be classified into two groups on the basis of the differences in the cell envelopes and the manner in which they respond to the staining procedure developed by Christian Gram. Gram staining is a differential staining method that separates bacteria into two categories based on the cell wall composition. Those that take up the Gram stain are called Gram-positive and the other that does not takes up Gram stain are called Gram-negative bacteria.

Topic: Biology, Biological Classification, NEET
The main edible part of coconut fruit is

Options:
A endosperm.
B mesocarp.
C endocarp.
D epicarp.

The correct answer is:
endosperm.

Solution:
Coconut is a type of drupe fruit. In drupe the fibrous part is mesocarp and the stony part is the endocarp. The edible part of the coconut is the seed or the endosperm which is cellular and white in a mature coconut. It is nuclear in green coconut water.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The main excretory organ of prawn is

Options:
A green gland
B flame cell
C Malpighian tubule
D nephridia

The correct answer is:
green gland

Solution:
The main excretory organs of prawn are paired green glands. They are located within the second antennae, which are the second pair of appendages, attached to the anterior part of the body. Each green gland consists of an endsac, labyrinth and a bladder.

Topic: Biology, Excretory Products and their Elimination, JIPMER, JIPMER 2005
The main factor in increasing the population is

Options:
A More reproductive capacity
B Early marriage
C Higher natality and lower mortality
D Suitable and favourable environment

The correct answer is:
Higher natality and lower mortality

Solution:
A rapidly growing population has a high birth rate and low death rate, with more number of young or pre-reproductive individuals. This type of population represents the expanding growth pattern.

Topic: Biology, Reproductive Health, NEET
The main function of lysosome is

Options:
A Sexual reproduction
B Extracellular digestion
C Intracellular digestion
D Both (Extracellular digestion) and (Intracellular digestion)

The correct answer is:
Both (Extracellular digestion) and (Intracellular digestion)

Solution:
Lysosome was discovered by C de Duve.
The main functions of lysosomes are:
(i) Digestion of large extracellular particles
(ii) Digestion of intracellular substance
(iii) Autolysis
(iv) Extracellular and intracellular digestion

Topic: Biology, Cell - The Unit of Life, NEET
The main function of mammalian corpus luteum is to produce:

Options:
A estrogen only
B progesterone
C human chorionic gonadotropin
D relaxin only

The correct answer is:
progesterone

Solution:
During the process of gametogenesis, after successful ovulation, the ruptured Graafian follicle converts into corpus luteum. It chiefly secretes progesterone which enhances the growth of the endometrial lining and helps in pregnancy.

Topic: Biology, Human Reproduction, NEET, NEET 2014
The main function of phloem is translocation of

Options:
A Water
B Mineral
C Air
D Food

The correct answer is:
Food

Solution:
Phloem is the chief food conducting tissue of vascular plants responsible for translocation of organic solutes.

Topic: Biology, Transport in Plants, NEET
The main function of the fimbriae of the fallopian tube in females is to

Options:
A release ovum from the graffian follicle
B make necessary changes in the endometrium for implantation
C help in the development of corpus luteum
D help in the collection of the ovum after ovulation

The correct answer is:
help in the collection of the ovum after ovulation

Solution:
The part of the fallopian tube (oviduct) closer to the ovary is the funnel-shaped infundibulum. The edges of the infundibulum possess finger-like projections called fimbriate, which help in collection of the ovum after ovulation.

Topic: Biology, Human Reproduction, NEET
The main function of tRNA with regards to protein synthesis is

Options:
A proofreading
B identifies amino acids and transports them to ribosomes
C inhibits protein synthesis
D all of these

The correct answer is:
identifies amino acids and transports them to ribosomes

Solution:
The tRNA molecules are the smallest, made up of 75 to 80 nucleotides. The transfer RNA recognizes the codons of mRNA and exhibit high affinity for 21 activated amino acids, combine with them and carry them to the site of protein synthesis. The 3I end of the polynucleotide chain ends in the CCA base sequence, the site for the attachment of activated amino acid.

Topic: Biology, Molecular Basis of Inheritance, NEET
The main functions of the root is

Options:
A absorption of water and minerals.
B to provide proper anchorage of plant.
C to store reserve food material and synthesis of plant growth regulators.
D All of the above

The correct answer is:
All of the above

Solution:
The main functions of the root system are absorption of water and minerals from soil, providing a proper anchorage to plant parts, storing reserve food material and synthesis of plant growth regulators.

Topic: Biology, Morphology of Flowering Plants, NEET
The main objective of production of herbicide-resistant GM crops is to

Options:
A eliminate weeds from the field without the use of manual labour.
B eliminate weeds from the field without the use of herbicides.
C encourage eco-friendly herbicides.
D reduce herbicide accumulation in food articles for health safety.

The correct answer is:
reduce herbicide accumulation in food articles for health safety.

Solution:
The main objective of production of herbicide-resistant GM crops is to reduce herbicide accumulation in food articles for health safety. GM plants has been useful in many ways. Genetic modifications has made crops more tolerant to abiotic stresses, reduced reliance on chemical pesticides, enhanced nutritional value of food.

Topic: Biology, Biotechnology and Its Applications, NEET
The main objective of the use of herbicide-resistant GM crops is to

Options:
A Eliminate weeds from the field without the use of manual labour
B Eliminate weeds from the field without the use of herbicides
C Encourage eco-friendly herbicides
D Reduce herbicide accumulation in food for health safety

The correct answer is:
Reduce herbicide accumulation in food for health safety

Solution:
The main objective of the production/use of herbicide-resistant genetically modified crops is to reduce herbicide accumulation in food articles for health safety.

Topic: Biology, Biotechnology and Its Applications, NEET
The main organelle involved in modification and routing of newly synthesized proteins to their destinations is:

Options:
A chloroplast
B mitochondria
C lysosome
D endoplasmic reticulum.

The correct answer is:
endoplasmic reticulum.

Solution:
Endoplasmic reticulum, along with the Golgi apparatus, is involved in the transport of proteins and enzymes to their destination, i.e., within the cell and also outside the cell. Hence this is the right answer. While the lysosome is also involved in the transport of the hydrolytic enzymes, but it is not the major transport mechanism in the cell.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2005
The main organelle involved in modification and routing of newly synthesized proteins to their destinations is

Options:
A mitochondria
B endoplasmic reticulum
C lysosome
D chloroplast

The correct answer is:
endoplasmic reticulum

Solution:
The main organelle involved in modification and routing of newly synthesized proteins to their destinations is endoplasmic reticulum. The endoplasmic reticulum serves many general functions, including the folding of protein molecules in sacs called cisternae and the transport of synthesized proteins in vesicles to the Golgi apparatus. Correct folding of newly made proteins is made possible by several endoplasmic reticulum chaperone proteins.

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2007
The main organelle involved in modification and routing of newly synthesized proteins to their destinations is

Options:
A Chloroplast
B Glyoxysomes
C Endoplasmic reticulum
D Golgi apparatus

The correct answer is:
Golgi apparatus

Solution:
The Golgi apparatus principally performs the function of packaging materials, to be delivered either to the intracellular targets or secreted outside the cell. A number of proteins synthesized by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus before they are released from its trans face. Golgi apparatus is the important site of the formation of glycoproteins and glycolipids.

Topic: Biology, Cell - The Unit of Life, NEET
The main oxides formed on combustion of Li, Na and K in air are respectively

Options:
A Li2O,Na2O2 and KO2
B Li2O,Na2O and KO2
C LiO2,Na2O2 and K2O
D Li2O2,Na2O2 and KO2

The correct answer is:
Li2O,Na2O2 and KO2

Solution:
The stability of the oxide of alkali metals depends upon the comparability of size of cation and anion.
Therefore the main oxide of alkali metals formed on excess of air are as follows

Topic: Chemistry, s Block Elements, JEE Main
The main oxides formed on combustion of Li, Na and K in excess of air are respectively:

Options:
A Li2O2,Â Na2O2Â andÂ KO2
B Li2O, Na2O2 and KO2
C Li2O,Â Na2OÂ andÂ KO2
D LiO2,Â Na2O2Â andÂ K2O

The correct answer is:
Li2O, Na2O2 and KO2

Solution:
In 1A group
Li on reaction with excess air
4Li+O2â2Li2O
Na on reaction with excess air
2Na+O2âNa2O2
K on reaction with excess air
K+O2âKO2

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2016 (03 Apr)
The main oxides formed on combustion of Li, Na and Rb in excess of air are respectively

Options:
A Li2O2,Na2O2 and RbO2
B Li2O,Na2O and RbO2
C LiO2,Na2O2 and Rb2O
D Li2O,Na2O2 and RbO2

The correct answer is:
Li2O,Na2O2 and RbO2

Solution:
The stability of the oxide of alkali metals depends upon the comparability of size of cation and anion hence the main oxide of alkali metals formed with excess of air are as follows
Li â Li2O (oxide)
Na â Na2O2 (peroxide)
K â KO2 (superoxide)
Rb â RbO2 (superoxide)
Cs â CsO2 (superoxide)

Topic: Chemistry, s Block Elements, NEET
The main product formed in the following reaction is
(1 equivalent) $\longrightarrow$ Product.

Options:
A

B

C

D

The correct answer is:

Solution:
As, $-\mathrm{NH}_2$ group is more reactive than -OH group, $\mathrm{CH}_3 \mathrm{COCl}$ will attack on $-\mathrm{NH}_2$ group and give as product.

Topic: Chemistry, Amines, JIPMER, JIPMER 2017
The main product obtained from phenol with $\mathrm{PCl}_5$ is

Options:
A BHC
B hexachlorobenzene
C chlorobenzene
D triphenyl phosphate

The correct answer is:
triphenyl phosphate

Solution:
The reaction of phenol with $\mathrm{PCl}_5$ is as follows
$\begin{aligned}
& 3 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{POCl}_3 \longrightarrow \underset{\text { Triphenyl phosphate }}{\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{PO}_4+3 \mathrm{HCl}} \\
&
\end{aligned}$
The yield of chlorobenzene in the above reaction is very poor. Thus, the main product of the reaction is triphenyl phosphate $\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{PO}_4$.

Topic: Chemistry, Alcohols Phenols and Ethers, JIPMER, JIPMER 2019
The main product obtained when a solution of sodium carbonate reacts with mercuric chloride is

Options:
A $\mathrm{HgCO}_{3}$
B $\mathrm{HgCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2}$
C $\mathrm{Hg}(\mathrm{OH})_{2}$
D $\mathrm{HgCO}_{3}, \mathrm{HgO}$

The correct answer is:
$\mathrm{HgCO}_{3}, \mathrm{HgO}$

Solution:
The main product obtained when sodium carbonate reacts with mercuric chloride is mercuric oxide.
$\mathrm{HgCl}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{HgO}+\mathrm{CO}_{2}+2 \mathrm{NaCl} or \mathrm{HgCO}_{3}+2 \mathrm{NaCl}$

Topic: Chemistry, s Block Elements, COMEDK, COMEDK 2016
The main product of the following reaction is
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_3\right)_2 \stackrel{\text { conc. } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}$ ?

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2010
The main product of the following reactions is

Options:
A
B
C
D

The correct answer is:

Solution:
Dehydration can produce two different alkenes, major product is formed according to Saytzeff rule i.e. more substituted alkene (alkene having lesser number of hydrogen atoms on the two doubly bonded carbon atoms) is the major product. E1 elimination takes place which is anti elimination.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The main products formed by the reaction of N2O5 and H2O2 are

Options:
A only HNO3
B HNO3+O3
C HNO3+HNO4
D HNO3+HNO2

The correct answer is:
HNO3+HNO4

Solution:
The main products formed by the reaction of N2O5 and H2O2 are HNO3 and HNO4 .
N2O5+H2O2âHNO3+HNO4Peroxynitric acid

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The main products formed when copper metal is reacted with concentrated HNO3 are

Options:
A CuNO32; NO
B CuNO32; H2
C CuNO32; NO2
D CuNO3; NO

The correct answer is:
CuNO32; NO2

Solution:
Copper is a transition element. HNO3 is nitric acid. When the concentrated nitric acid reacts with the copper metal then it formed the copper nitrate and nitrogen dioxide gas. So, the nitric acid will decompose into the nitrogen dioxide gas and copper nitrate.

Reaction :-

Cu + Conc. HNO3→CuNO32 + NO2

Hence, the main products formed when copper metal is reacted with concentrated HNO3 are CuNO32; NO2.

Topic: Chemistry, d and f Block Elements, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
The main products P and Q of the following unbalanced disproportionation reaction are

Se2Cl2→P+Q

Options:
A PQSeCl2SeCl3
B PQSeCl4SeCl2
C PQSeCl4Se
D PQSeCl4Se2

The correct answer is:
PQSeCl4Se

Solution:
The monohalides of 16th group elements are dimeric in nature. These dimeric halides undergo disproportionation as given below:

2Se2Cl2→∆SeCl4+3Se

*NOTE: For academic correctness the question was changed from dihalide to mono halide.

Topic: Chemistry, Redox Reactions, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)
The main reactants involved in Etard reaction are

Options:
A Toluene $+\mathrm{CrO}_2 \mathrm{Cl}_2$
B Toluene $+\mathrm{CrO}_3+\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}$
C Toluene $+\mathrm{Cl}_2 / \mathrm{hv}$
D Benzene $+\mathrm{CO}+\mathrm{HCl}$ Anhy. $\mathrm{AlCl}_3$

The correct answer is:
Toluene $+\mathrm{CrO}_2 \mathrm{Cl}_2$

Solution:
In Etard reaction, chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.

Topic: Chemistry, Hydrocarbons, TS EAMCET, TS EAMCET 2023 (14 May Shift 2)
The main reason for larger number of oxidation states exhibited by the actinides than the corresponding lanthanides, is

Options:
A lesser energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals
B larger atomic size of actinides than the lanthanide
C more energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals
D greater reactive nature of the actinides than the lanthanides

The correct answer is:
lesser energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals

Solution:
The main reason for larger number of oxidation states exhibited by actinides than corresponding lanthanides is lesser energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and 5d-orbitals.

Topic: Chemistry, d and f Block Elements, JIPMER, JIPMER 2007
The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids is:

Options:
A more energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals
B lesser energy difference between $5 \mathrm{f}$ and $6 \mathrm{~d}$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals
C larger atomic size of actinoids than the lanthanoids
D greater reactive nature of the actinoids than the lanthanoids

The correct answer is:
lesser energy difference between $5 \mathrm{f}$ and $6 \mathrm{~d}$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals

Solution:
In actinides, both $4 f$ and $5 f$ orbitals are present. These orbitals poorly shields the nuclear charge due to which the energy difference between the $5 f$ and the $6 d$ orbitals will be very low due to which the actinides will show a large number of oxidation states.
Caution
The elements in which the extra electron enters 5 - orbitals of $(n-2)^{\text {th }}$ shell (5f-block elements, actinides) shows the maximum oxidation state i.e. +7.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2005
The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids is:

Options:
A more energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals
B lesser energy difference between $5 \mathrm{f}$ and $6 \mathrm{~d}$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals
C larger atomic size of actinoids than the lanthanoids
D greater reactive nature of the actinoids than the lanthanoids

The correct answer is:
lesser energy difference between $5 \mathrm{f}$ and $6 \mathrm{~d}$ orbitals than between $4 f$ and $5 \mathrm{~d}$ orbitals

Solution:
In actinides, both $4 f$ and $5 f$ orbitals are present. These orbitals poorly shields the nuclear charge due to which the energy difference between the $5 f$ and the $6 d$ orbitals will be very low due to which the actinides will show a large number of oxidation states.
Caution
The elements in which the extra electron enters 5 - orbitals of $(n-2)^{\text {th }}$ shell (5f-block elements, actinides) shows the maximum oxidation state i.e. +7.

Topic: Chemistry, d and f Block Elements, JEE Main
The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids is

Options:
A Lesser energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals
B Larger atomic size of actinoids than the lanthanoids
C More energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals
D Greater reactive nature of the actinoids than the lanthanoids

The correct answer is:
Lesser energy difference between $5 f$ and $6 d$ orbitals than between $4 f$ and $5 d$ orbitals

Solution:
Due to lesser energy difference between the \(5 \mathrm{f}\) and \(6 \mathrm{~d}\) orbital of actinoids, more electrons can participate in bond formation, so more oxidation states are exhibited by actinoids.

Topic: Chemistry, d and f Block Elements, JEE Main
The main reason that SiCl4 is easily hydrolysed as compared to CCl4 is that

Options:
A Si-Cl bond is weaker than C-Cl bond
B SiCl4 can form hydrogen bonds
C SiCl4 is covalent
D Si can extend its coordination number beyond four

The correct answer is:
Si can extend its coordination number beyond four

Solution:
Silicon has got vacant 'd' oribtals but carbon does not have that moreover Silicon, also has larger size than carbon so silicon can have extended valency rather extended coordination number 6 also.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The main reason that $\mathrm{SiCl}_{4}$ is easily hydrolysed as compared to $\mathrm{CCl}_{4}$ is that

Options:
A $\mathrm{Si}-\mathrm{Cl}$ bond is weaker than $\mathrm{C}-\mathrm{Cl}$ bond
B $\mathrm{SiCl}_{4}$ can form hydrogen bonds
C $\mathrm{SiCl}_{4}$ is covalent
D Si can extend its coordination number beyond four

The correct answer is:
Si can extend its coordination number beyond four

Solution:
It is less reactive than other oxides of chlorine and does not react with $P, S,$ coal or paper at room temperature.

The main reason that $\mathrm{SiCl}_{4}$ is easily hydrolysed as compared to $\mathrm{CCl}_{4}$, is that Si can extend its coordination number beyond four because it has vacant $d$ -orbital. On the other hand, carbon has no $d$ -orbital thus, it cannot extend its coordination number beyond four, so its halides are not attacked (hydrolysed) by water. The vacant $d$ -orbitals of Si can coordinate with water molecules and hence their halides are hydrolysed by water.
The hydrolysis of $\mathrm{SiCl}_{4}$ occurs due to coordination of $-\mathrm{OH}$ with empty $3 d$ -orbitals of Si-atom of $\mathrm{SiCl}_{4}$ molecule.

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2018
The main reason why antibiotics could not solve all the problems of bacteria mediated diseases is

Options:
A insensitivity of the individual following prolonged exposure to antibiotics
B inactivation of antibiotics by bacterial enzymes
C decreased efficiency of the immune system
D the development of mutant strains resistant to antibiotics.

The correct answer is:
the development of mutant strains resistant to antibiotics.

Solution:
A major problem in the treatment of bacterial mediated diseases is that many bacteria have been found to show resistance to antibiotics. The emergence of antibiotic-resistant bacteria is closely linked to the extent that antibiotics are used in humans and items of human diet. Resistant strains may appear rapidly or slowly, according to the amount or type of antibiotic used. Bacteria occur in such large numbers that there is a high chance of a mutant individual eventually appearing in the population. As soon as it does, use of the antibiotic to which it is resistant will give it a selective advantage over non-resistant types and it will multiply and eventually become the dominant type.
Antibiotics may also be destroyed by enzymes inside the cells being targetted. A well-known and important example is the group of enzymes known as penicillinases which hydrolyses and destroys penicillins and cephalosporins.

Topic: Biology, Biotechnology and Its Applications, AIIMS, AIIMS 2008
The main reduction product of the following compound with NaBH4 in methanol is:

Options:
A
B
C
D

The correct answer is:

Solution:
NaBH4 reduce the carbonyl into alcohol

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2018 (15 Apr)