datavorous/entrance-exam-dataset huggingface dataset:
In electrophilic substitution reactions, why does the carboxyl group act as a meta-directing group?

Options:
A Due to its activating effect on the ring
B Due to its ortho-directing effect
C Due to its resonance properties
D Due to its steric hindrance

The correct answer is:
Due to its resonance properties

Solution:
The carboxyl group's meta-directing effect in electrophilic substitution reactions is due to its resonance properties, which stabilize the meta position.

Topic: Chemistry, Aldehydes, Ketones and Carboxylic Acids, NEET
In electrophoresis, the DNA sample is loaded in the gel

Options:
A away from the wells.
B towards the positive electrode.
C towards the negative electrode.
D Both (A) and (C)

The correct answer is:
towards the negative electrode.

Solution:
Electrophoresis is an electrokinetic process that is used to separate charged particles in a fluid by using a field of electrical charge. It is often used for separation of molecules or DNA basis their size, density etc. The molecules are separated through an electric current and a gel. Electrophoresis is usually conducted in a box-like structure which has a positive charge on one end and negative on the other. The electric field around the gel causes the negatively charged molecules to move towards positive electrode and vice versa.
Since DNA fragments are negatively charged molecules they can be separated by forcing them to move towards the anode under an electric field through a medium/matrix.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In Eli Lily's method of insulin production, which of the following is correct?

Options:
A A peptide and B peptide were produced from the same host cell
B A peptide and B peptide were produced from different host cells.
C A peptide, B peptide and C peptide were produced from the same host cell.
D A peptide, B peptide and C peptide were produced from different host cells.

The correct answer is:
A peptide and B peptide were produced from different host cells.

Solution:
Insulin is a protein hormone produced by the Islets of Langerhans which help in regulating blood glucose level in our body. Mature insulin has two peptide chains - chain A and chain B joined by disulfide bonds. The mature insulin is primarily produced as inactive pro-insulin having a chain C between chain A and B. Chain C is removed for activation of the hormone. Genetic engineering process faced the challenge in developing mature insulin and it was overcome by isolating the gene for chain A and chain B and then producing it in the host bacterial cell. Eli Lily, a pharmaceutical company, prepared two separate DNA sequences each making one protein - chain A and chain B which was then produced individually in different host cell. The extracted chains were later combined by creating disulfide bonds giving the mature human insulin.

Topic: Biology, Biotechnology and Its Applications, NEET
In England, in 1920 (after industrialization) when a collection of moths was made, it was observed that:

Options:
A White winged moths were dominant
B Dark winged moths were dominant
C Dark winged moths became extinct
D Dark and white winged moths were equal in number.

The correct answer is:
Dark winged moths were dominant

Solution:
Industrial melanism
- Industrial melanism is an observation made in England that supports 'evolution by natural selection'.

Time of Moth Collection	Observation	Reason behind
1850s (Before Industrialization sets in)	On trees, White- winged moths > Dark-winged (or melanised) moths	Before industrialisation, white-coloured lichen grow and almost covered the trees because of which white winged moth survived due to white background the but the dark -coloured moth were more spotted and picked up by the predators.
1920s (After Industrialization)	On trees, Dark -winged moths > White – winged moths	Lichens do not grow in polluted areas, so the dark moths were able to camouflage (hide in the background) themselves against the tree trunks became dark due to industrial smoke and soots. Thus they survived in more numbers. White- winged moth were more spotted and picked up by the predators in comparison to the dark-winged.

Topic: Biology, Evolution, NEET
In epigynous flowers, the ovary is situated below the _________.

Options:
A Ovary, corolla and androecium.
B Calyx, corolla and ovary
C Calyx, corolla and androecium.
D None of the above

The correct answer is:
Calyx, corolla and androecium.

Solution:
In epigynous flowers, the ovary is situated below the Calyx, corolla and androecium.

Topic: Biology, Morphology of Flowering Plants, NEET
In erythroblastosis foetalis,

Options:
A Rh antibodies pass from the Rh(+ve) mother into the Rh(-ve) baby through the placenta.
B Rh antibodies pass from the Rh(-ve) mother into the Rh(+ve) baby through the placenta.
C Rh antigens pass from the Rh(-ve) mother into the Rh(+ve) baby through the placenta.
D Rh antigens pass from the Rh(+ve) mother into the Rh(-ve) baby through the placenta.

The correct answer is:
Rh antibodies pass from the Rh(-ve) mother into the Rh(+ve) baby through the placenta.

Solution:
The mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh-ve) can leak into the blood of the foetus (Rh+ve) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis.

Topic: Biology, Principles of Inheritance and Variation, NEET
In ester formation, what type of reaction is the reaction between an alcohol and a carboxylic acid?

Options:
A Substitution reaction
B Addition reaction
C Elimination reaction
D Condensation reaction

The correct answer is:
Condensation reaction

Solution:
The reaction between an alcohol and a carboxylic acid to form an ester is a condensation reaction, as water is eliminated.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In ETS, the last electron acceptor is

Options:
A cytochrome-α
B ATP
C oxygen
D water

The correct answer is:
oxygen

Solution:
NADH and FADH2 bring electrons to the ETS. As the electrons move down the system, energy is released and use to form ATP. For every pair of electrons that enters by way of NADH, three ATP result. For every pair of electrons that enters by way of FADH2 two ATP result. Oxygen, the final acceptor of the electrons, becomes a part of water.

Topic: Biology, Respiration in Plants, NEET
In eubacteria, a cellular component that resembles eukaryotic cells is

Options:
A nucleus
B ribosomes
C cell wall
D plasma membrane

The correct answer is:
plasma membrane

Solution:
The plasma membrane of eubacteria resembles to that of eukaryotic cell. It is made of phospholipid, protein and some amount of polysaccharides. However, it lacks sterol, the characteristic of eukaryotic cell membrane. Instead, there is sterol like hopanoid.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2011 (Screening)
In eukaryotes, after transcription of mRNA, some of its nucleotides are removed before it is translated into a polypeptide, The nucleotides which are removed from mRNA are called:

Options:
A Exons
B Upstream sequences
C Unusual bases
D Introns

The correct answer is:
Introns

Solution:
Eukaryote RNAs contain non-coding intervening segments called introns. The process of removal of introns through cutting and joining the essential coding sequences (exons) is called splicing.

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotes, regulation of gene expression can be done at all the following levels, except

Options:
A Regulation of splicing
B Formation of primary transcript
C Transport of hnRNA from the nucleus to the cytoplasm
D Translational level

The correct answer is:
Transport of hnRNA from the nucleus to the cytoplasm

Solution:
​Gene expression results in the formation of a polypeptide. It can be regulated at several levels. In eukaryotes, the regulation could be exerted at
(i) transcriptional level (formation of primary transcript),
(ii) processing level (regulation of splicing),
(iii) transport of mRNA from the nucleus to the cytoplasm,
(iv) translational level.

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotes, RNA polymerase II transcribes

Options:
A the precursor of tRNA
B the precursor of mRNA
C the precursor of snRNA
D the precursor of rRNA

The correct answer is:
the precursor of mRNA

Solution:
There are at least three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles). There is a clear cut division of labour. The RNA polymerase I transcribe rRNAs (28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible for transcription of tRNA, 5srRNA, and snRNAs (small nuclear RNAs). The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotes, the anticodon of tRNA that pairs with the start codon of mRNA during translation is ________.

Options:
A UAA
B UCA
C UAC
D UUU

The correct answer is:
UAC

Solution:
In eukaryotes, initial codon on mRNA is AUG. During translation, t-RNA, anticodons are exactly complementary to mRNA. Hence anticodon will be UAC. A = U , U = A , C ≡ G

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotes, the process of DNA replication is:

Options:
A Semi-conservative and semi-discontinuous
B Continuous and bidirectional
C Semi-conservative and continuous
D Both A and B

The correct answer is:
Semi-conservative and semi-discontinuous

Solution:
DNA replication is the process of making new DNA strand from the old one and this process is semi-conservative. Synthesis of new DNA occurs simultaneously and independently on both strands but in one direction as the DNA polymerase used in the process of elongation of the strand of DNA during replication process proceeds in the 5â3â direction only. Of the two strands, the one that is continuous and occurs without stopping from one end to the other is called leading strand, while the one that is discontinuous in that it occurs in several small chunks one after another is called lagging strand and each short sequence is called Okazaki fragment. The replication process is bidirectional in that it goes in both ways from the point of origin of replication in the DNA. Meselson and Stahl proved that DNA replication is semi conservative in bacteria by experimenting with E. coli.

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotes, which of these rRNA is not synthesized by RNA Polymerase I?

Options:
A 28S rRNA
B 5S rRNA
C 5.8S rRNA
D 18S rRNA

The correct answer is:
5S rRNA

Solution:
In eukaryotes, there are 3 types of DNA dependent RNA polymerases
RNA polymerase I: Transcribes rRNA (28S, 18S, 5.8S)
RNA polymerase II: Transcribes hnRNA (heterogeneous nuclear RNA) a precursor of mRNA
RNA polymerase III: Transcribes tRNA, 5S rRNA & snRNAs (small nuclear RNAs).

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotic cells, the proteins that will function as ribonucleoproteins are made by

Options:
A Ribosomes in nucleolus
B Ribosomes in mitochondria
C Ribosomes in cytoplasm
D Ribosome in chloroplast

The correct answer is:
Ribosomes in nucleolus

Solution:
Ribosomes are made up of ribonucleoproteins, rRNA and proteins. Nucleolus is called ribosome factory of cell. Thus synthesis of ribonucleoprotein which makes ribosomes occurs in nucleolus.

Topic: Biology, Cell - The Unit of Life, NEET
In eukaryotic rRNA, mRNA and tRNA synthesis is catalysed by:

Options:
A RNA polymerase I and DNA polymerase I
B RNA polymerase II and DNA polymerase II
C DNA polymerase I and II
D RNA polymerase I and II

The correct answer is:
RNA polymerase I and II

Solution:
In Eukaryotes, RNA polymerase I catalyses the transcription of rRNA, RNA polymerase II catalyses the transcription of hnRNA, the precursor of mRNA and RNA polymerase III catalyses transcription of tRNA, snRNA and 5S RNA.

Topic: Biology, Molecular Basis of Inheritance, NEET
In eukaryotic transcription, heteronuclear RNA (hnRNA) is transcribed by

Options:
A RNA polymerase I
B RNA polymerase II
C RNA polymerase III
D All of these

The correct answer is:
RNA polymerase II

Solution:
RNA polymerase II (RNAP II or Pol II) is a multiprotein complex that transcribes DNA into precursors of messenger RNA (mRNA) i.e. hnRNA and most small nuclear RNA (snRNA) and microRNA. It is one of the three RNAP enzymes found in the nucleus of eukaryotic cells.

Topic: Biology, Molecular Basis of Inheritance, NEET
In every experiment with potentiometer in the null point state, the potential difference between the ends of the galvanometer is

Options:
A zero
B infinite
C equal to the potential difference of the cell
D unknown

The correct answer is:
zero

Solution:
As in the null point state, no current will flow through a galvanometer.
Therefore, the potential difference between the two ends of the galvanometer is VG=IG
As IG=0
hence VG=0

Topic: Physics, Current Electricity, NEET
In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be ______m.

Options:
A 2Vemh
B h2Vem
C 2Vemh
D h2Vem

The correct answer is:
h2Vem

Solution:
Kinetic energy 12mv2=eV
v=2eVm
Wavelength λ=hmv
λ=hm2eVm=h2eVm

Topic: Physics, Dual Nature of Matter, NEET
In face centred cubic arrangement of A and B atoms, A atoms are at the corners of the unit cell and B atoms at the face centres. One of the A atoms is missing from one corner in the unit cell. The simplest formula of the compound is

Options:
A A7B3
B AB3
C A7B24
D A7/8B5

The correct answer is:
A7B24

Solution:
If one atom (A) is missing from one corner,
no. of atoms A in unit cell = 7 Ã 1 8 = 7 8
no. of atoms B in unit cell = 6 Ã 1 2 = 3
Aâ¡ : Bâ¡ = 7 8 : 3 so simplest formula is A7B24â

Topic: Chemistry, Solid State, NEET
In facilitated diffusion

Options:
A the hydrolysis of ATP is required to facilitate diffusion.
B specific substances are transported against their concentration gradient.
C membrane phospholipids create a channel that permits diffusion of a specific substance.
D a membrane protein binds to a specific solute and then undergoes a conformational change.

The correct answer is:
a membrane protein binds to a specific solute and then undergoes a conformational change.

Solution:
Facilitated diffusion is a kind of diffusion wherein there is passive transport(cannot transport via self) of molecules spontaneously across a biological membrane with the help of the protein molecules of the membrane which act as the carriers for the polar molecules in this diffusion process. In this process, the membrane carrier proteins get bounded to the specific solute to be transported and undergo a series of conformational changes (macromolecular shape change) to transfer the solute across the membrane.

Topic: Biology, Transport in Plants, NEET
In Fe(CO)5 , the Fe-C bond possesses

Options:
A Ï- character only
B Ionic character
C Ï- character only
D Both Ï and Ï characters

The correct answer is:
Both Ï and Ï characters

Solution:
In a metal carbonyl, the metal-carbon bond possesses both Ï and Ï characters. A Ï -bond between metal and carbon atom is formed when a vacant hybrid orbital of the metal atom overlaps with an orbital of C atom of carbon monoxide containing a lone pair of electrons .
Formation of Ï-bond is caused when a filled orbital of the metal atom overlap with a vacant antibonding Ï * orbital of C atom of CO. This overlap is aslo called back donation of electrons by metal atom to carbon.
(a) The formation of the metal â carbon Ï - bond using an unshared pair of the C atom . (b) The formation of the metal â carbon Ï - bond.
The Ï- Ï -bond. overlap is perpendicular to the nodal plane of

Topic: Chemistry, Coordination Compounds, NEET
In Fe₃O₄, the oxidation number of Iron is:

Options:
A +2
B +3
C +4
D +5

The correct answer is:
+4

Solution:
Correct Option is : (C)
+4

Topic: Chemistry, Redox Reactions, NEET
In fern, sporangia are borne on the

Options:
A Margin of leaf
B Abaxial side of leaf
C Adaxial side of leaf
D Only on the tip of leaf

The correct answer is:
Margin of leaf

Solution:
Sporangia develop in groups called sori. Sori develop on the under surface of leaves and leaflets, laterally or marginally.

Topic: Biology, Plant Kingdom, NEET
In fern, spores are formed in

Options:
A Sporangium
B Oogonium
C Archegonium
D Stomium

The correct answer is:
Sporangium

Solution:
The plants of pteridophytes are sporophytes. They reproduce asexually by forming spores in sporangia.

Topic: Biology, Plant Kingdom, NEET
In five kingdom classification, ascomycetes are included in

Options:
A Monera
B Protista
C Animalia
D Fungi

The correct answer is:
Fungi

Solution:
Ascomycetes are commonly called sac fungi and are included under kingdom fungi (Mycota).

Topic: Biology, Biological Classification, NEET
In five kingdom classification, the kingdom that includes the blue-green algae, nitrogen-fixing bacteria and methanogenic archaebacteria, is the kingdom:

Options:
A Plantae
B Fungi
C Protista
D Monera

The correct answer is:
Monera

Solution:
In the five-kingdom classification system, the kingdom Monera includes all the prokaryotic celled organisms. Prokaryotes are organisms whose cells lack a well-defined nucleus. All the bacterial groups are included in Monera as well since bacteria have a prokaryotic cell structure. Cyanobacteria are a group of free-living photosynthetic bacteria that perform oxygenic photosynthetic process

Topic: Biology, Biological Classification, NEET
In flame tests for alkali metals, why is it important to use an oxidizing flame?

Options:
A To create a larger flame
B To minimize the heat generated
C To prevent the emission of characteristic colors
D To provide the conditions for excitation of electrons

The correct answer is:
To provide the conditions for excitation of electrons

Solution:
An oxidizing flame is used in flame tests for alkali metals because it provides the necessary conditions, including oxygen, to excite the outermost orbital electrons of alkali metal atoms, leading to the emission of characteristic colors.

Topic: Chemistry, s Block Elements, NEET
In flowering plants, double fertilization and triple fusion is a unique phenomenon. For a given plant, during the study of the endosperm, it was found that it has 24 chromosomes. What will the number of chromosomes in microspore mother cells?

Options:
A 12
B 48
C 16
D 24

The correct answer is:
16

Solution:
The endosperm is triploid in nature in angiosperms, while microspore mother cells are diploid structures.
Since the number of chromosomes in the endosperm is 24 = 3n. Therefore n = 8.
Hence, the number of chromosomes in the microspore mother cell will be 2n = 2x8 =16.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In flowering plants, meiosis takes place at the time of

Options:
A Seed germination
B Bud formation
C Formation of pollen grains
D Formation of root tip

The correct answer is:
Formation of pollen grains

Solution:
"In flowering plants, meiosis occurs during gamete formation". All pollen mother cells are diploid and produce haploid pollen grains after meiosis. The pollen grain is the male gametophyte and it consists of two male gametes.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In flowering plants, the substances that would need to be transported are:

Options:
A Water and mineral nutrients
B Organic nutrients
C Plant growth regulators
D All of these

The correct answer is:
All of these

Solution:
In a flowering plant, the substances that would need to be transported are water, mineral nutrients, organic nutrients and plant growth regulators.

Topic: Biology, Transport in Plants, NEET
In fluid mosaic model of plasma membrane:

Options:
A upper layer is non-polar and hydrophilic.
B polar layer is hydrophobic.
C phospholipids form a bimolecular layer in the middle part.
D proteins form a middle layer.

The correct answer is:
phospholipids form a bimolecular layer in the middle part.

Solution:
S. J Singer and G. L Nicholson proposed fluid mosaic model of cell membranes. According to this model, the plasma membrane is formed of "protein icebergs in the sea of lipids". The membrane is not solid but is quasi-fluid. The lipids of the membrane form a bilayer which is interrupted by the embedded protein molecules.
Related Theory
Depending on the degree of association and the ability of separation, the membrane proteins are classified into two types, which are: internal proteins also known as intrinsic or integral proteins and the external proteins also known as extrinsic or peripheral proteins. The integral or intrinsic proteins account for $70 \%$ of the total membrane proteins and pass into the lipid bilayer to different depths. They cannot be easily separated from the membrane without disrupting it.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2002
In fluorite structure (CaF2)-

Options:
A Ca2+ ions form ccp & F- ions are present in all the tetrahedral voids
B Ca2+ ions form ccp & F- ions are present in all the octahedral voids
C Ca2+ ions form ccp & F- ions are present in half of the octahedral voids and the rest half ions in the tetrahedral voids
D None

The correct answer is:
Ca2+ ions form ccp & F- ions are present in all the tetrahedral voids

Solution:
Ca2+ ions form ccp and F- ions are present in the tetrahedral voids. So, the no. of Ca2+ ions is 4 and no. of F- ions is 8.
So, the formula of the Calcium fluoride Ca4F8 or, the simplest formula of calcium fluoride is CaF2 .

Topic: Chemistry, Solid State, NEET
In forward biasing of the $p-n$ junction

Options:
A the positive terminal of the battery is connected to $n$-side and the depletion region becomes thin
B the positive terminal of the battery is connected to $n$-side and the depletion region becomes thick
C the positive terminal of the battery is connected to $p$-side and the deplection region become thin
D the positive terminal of the battery is connected to $p$-side and the depletion region becomes thick

The correct answer is:
the positive terminal of the battery is connected to $p$-side and the deplection region become thin

Solution:
In forward biasing of $p-n$ junction the positve terminal of the battery is connected to $p$-side and the deplection region becomes thin.

Topic: Physics, Semiconductors, NEET, NEET 2011 (Screening)
In four o'clock plants, the gene for red flower colour (R) is incompletely dominant over the gene for white flower colour (r), hence the plants heterozygous for flower colour (Rr) have pink flowers. What will be the ratio of offsprings in a cross between red flowers and pink flowers?

Options:
A 75% red flowers, 25% pink flowers
B All red flowers
C 50% red flowers, 50% pink flowers
D Red : pink : white : : 1 : 2 : 1

The correct answer is:
50% red flowers, 50% pink flowers

Solution:
As Mendel, on crossing homozygous members for contrasting characters, the dominant trait is expressed in the heterozygous or hybrid condition. Incomplete dominance is a deviation of Mendel's law of dominance where the heterozygous individual instead of expressing the dominant trait, it expresses a trait which is a mix of the parent traits.
On crossing, red(RR) flower plant with white(rr) flower plant, the offspring flower is pink(Rr) flower.
When a red flower plant is crossed with pink flower plant we get:

R	R

R	RR
(Red)	RR
(Red)

r	Rr
(Pink)	Rr
(Pink)

Genotype of F1 - RR and Rr 1:1
Phenotype of F1 - Red and Pink 1:1 (50% and 50%)

Topic: Biology, Principles of Inheritance and Variation, NEET