datavorous/entrance-exam-dataset huggingface dataset:
In a dicot leaf, palisade tissue is generally present adjacent to the epidermis

Options:
A having stomata.
B not having stomata.
C having suberin.
D having lignin.

The correct answer is:
not having stomata.

Solution:
The palisade cells are found in the mesophyll which is situated between the upper and the lower epidermis and is made up of parenchyma. The palisade parenchyma is made up of thin-walled elongated cells which lie vertically and are parallel to each other. This tissue is located below the adaxial epidermis(the upper epidermis having stomata) or adjacent to the abaxial epidermis(lower epidermis not having stomata).

Topic: Biology, Anatomy of Flowering Plants, NEET
In a dicotyledonous stem, the sequence of tissues from outside to inside is:

Options:
A Phellem - pericycle - endodermis - phloem
B Phellem - phloem - endodermis - pericycle
C Phellem - endodermis - pericycle - phloem
D Pericycle - phellem - endodermis - phloem

The correct answer is:
Phellem - endodermis - pericycle - phloem

Solution:
The sequence of tissues from the outside to the inside in a dicotyledonous stem is phellem-phellogen-phelloderm-cortex-endodermis-pericycle-phloem xylem-pith.

Topic: Biology, Anatomy of Flowering Plants, NEET
In a dihybrid cross AABB × aabb, F2 progeny of AABB, AABb, AaBB and AaBb occurs in the ratio of

Options:
A 1 : 1 : 1 : 1
B 9 : 3 : 3 : 1
C 1 : 2 : 2 : 1
D 1 : 2 : 2 : 4

The correct answer is:
1 : 2 : 2 : 4

Solution:
The genotype is the genetic make up of an organism. For a dihybrid cross, it will be:
Genotypic ratio: AABB : AaBB : AABb : AaBb : AAbb : Aabb : aaBB : aaBb : aabb
1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Topic: Biology, Principles of Inheritance and Variation, NEET
In a discharge tube ionization of enclosed gas is produced due to collisions between:

Options:
A neutral gas atoms/molecules
B positive ions and neutral atoms/ molecules
C negative electrons and neutral atoms/ molecules
D photons and neutral atoms/molecules

The correct answer is:
negative electrons and neutral atoms/ molecules

Solution:
Negative electrons and neutral atoms collides and causes the ionization of gas enclosed in a discharge tube.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2006
In a disproportionation reaction, which element undergoes simultaneous oxidation and reduction?

Options:
A Hydrogen
B Oxygen
C Chlorine
D Phosphorus

The correct answer is:
Oxygen

Solution:
Disproportionation reactions involve an element undergoing both oxidation and reduction. In the examples provided in the passage, oxygen in various compounds is shown to undergo disproportionation.

Topic: Chemistry, Redox Reactions, NEET
In a dissimilar pair of factors, which member dominates over the other?

Options:
A Both members are equal.
B The recessive member dominates.
C The dominant member dominates.
D Both members show blending.

The correct answer is:
The dominant member dominates.

Solution:
According to Mendel's Law of Dominance, in a dissimilar pair of factors, one member dominates over the other, and it is the dominant member that is expressed.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a DNA molecule, if the number of phosphodiester bonds is 198, it has ______ bases.

Options:
A 150
B 200
C 300
D 400

The correct answer is:
200

Solution:
Nucleoside is formed by a N-glycosidic linkage between nitrogenous base and pentose sugar. Phosphate group is attached to 5'-OH of a nucleoside through phosphoester bond. Two nucleotides are linked to form dinucleotide by a phophodiester bond. Therefore if there are 198 phosphodiester bonds in the DNA molecule then the number of base pairs will be two more than the bonds as the nucleotides present on the ends of the polynucleotide chain will also be counted.

Topic: Biology, Molecular Basis of Inheritance, NEET
In a DNA molecule, nitrogenous base is linked to pentose sugar by 'A' linkage and the pentose sugar is linked to phosphate by 'B' linkage. Correctly identify A and B.

Options:
A

Column A	Column B
N-Glycoside linkage	Amide linkage

B

Column A	Column B

N-Glycoside linkage	Phosphodiester linkage

C

Column A	Column B
Phosphoester linkage	Phosphoester linkage

D

Column A	Column B

Amide linkage	Phosphodiester linkage

The correct answer is:

Column A	Column B

N-Glycoside linkage	Phosphodiester linkage

Solution:
Structure of polynucleotide chain
DNA = Polynucleotide.
Nucleotide = Nitrogenous base + Pentose sugar + Phosphate group.
As nitrogenous base and pentose sugar when linked, are also termed as nucleoside, so we can also say that : Nucleotide = Nucleoside + Phosphate group.
Linkages in a nucleotide molecule:
Nitrogenous base --(N-glycosidic linkage)-- Pentose sugar --(Phosphodiester linkage)-Phosphate group. Phosphate group is linked to 5â â OH of a nucleoside through the phosphoester linkage.

Topic: Biology, Molecular Basis of Inheritance, NEET
In a DNA percentage of thymine is 20% then what will be the percentage of guanine?

Options:
A 20%
B 40%
C 30%
D 60%

The correct answer is:
30%

Solution:
According to Chargaff's rule, the ratio of purine bases to pyrimidine bases is equal
A=T, C=G.
If T is 20% then A is 20% and C and G are 30% each.

Topic: Biology, Molecular Basis of Inheritance, NEET
In a DNA strand, the nucleotides are linked together by:

Options:
A glycosidic bonds
B phosphodiester bonds
C peptide bonds
D hydrogen bonds

The correct answer is:
phosphodiester bonds

Solution:
When a phosphate group is linked to OH of 5' C of a nucleoside through phosphoester linkage, a corresponding nucleotide (or deoxynucleotide depending upon the type of sugar present) is formed.

Topic: Biology, Molecular Basis of Inheritance, NEET
In a dorsiventral leaf, protoxylem is situated towards the (I) surface and protophloem towards the (II) surface in the vascular bundle.

Options:
A I: adaxial and II: adaxial
B I: abaxial and II: abaxial
C I: abaxial and II: adaxial
D I: adaxial and II: abaxial

The correct answer is:
I: adaxial and II: abaxial

Solution:
In dorsiventral leaves, protoxylem is situated towards the upper surface (adaxial surface) and protophloem towards the lower surface (abaxial surface) in the vascular bundle. The adaxial surface is the upper surface of the leaf, whereas the abaxial surface is the lower surface.

Topic: Biology, Anatomy of Flowering Plants, NEET
In a dorsiventral leaf, the location of palisade tissue and phloem is respectively

Options:
A Abaxial and abaxial
B Adaxial and abaxial
C Adaxial and adaxial
D Abaxial and adaxial

The correct answer is:
Adaxial and abaxial

Solution:
In dorsiventral leaf there is distinct upper and lower surface, upper surface is known as adaxial surface and the lower surface is known as abaxial surface. Mesophyll tissue of dorsiventral leaf is differentiated into upper palisade parenchyma and lower spongy parenchyma. In vascular bundle xylem is present towards adaxial surface and phloem is present towards abaxial surface.

Topic: Biology, Anatomy of Flowering Plants, NEET
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? μwater=43

Options:
A 0.266°
B 0.15°
C 0.05°
D 0.1°

The correct answer is:
0.15°

Solution:
In air angular fringe width is given by, θ=λd.
Angular fringe width in water, θ'=θμ=0.243=0.15°

Topic: Physics, Wave Optics, NEET, NEET 2019 (All India)
In a double slit interference experiment, the maximum intensity of light would be _____ the maximum intensity of a single slit experiment.

Options:
A same as
B twice
C four times
D half

The correct answer is:
four times

Solution:
∵ in the double-slit experiment
Inet=I1+I2+2I1I2cos⁡ϕ
When I1=I2 and cos⁡ϕ=+1
we get Imax=4I1
∴ maximum intensity in the double-slit interference
=4 (maximum intensity of single slit experiment)

Topic: Physics, Wave Optics, NEET
In a double-slit experiment, the separation between the slits is d=0.25ââcm and the distance of the screen D=100ââcm from the slits. If the wavelength of light used is Î»=6000 Aâ and I0 is the intensity of the central bright fringe, the intensity at a distance x=4Ã 10 â5 ââm from central maximum is

Options:
A I0
B I02
C 3I04
D I03

The correct answer is:
3I04

Solution:
Path difference = xd D â Path difference = 4 Ã 1 0 - 5 Ã 0.25 Ã 1 0 - 2 1
Path difference = 1 Ã 1 0 - 7
Phase difference = path diff. Î» Ã 2 Ï â Phase difference = 1 Ã 1 0 - 7 6 Ã 1 0 - 7 Ã 2 Ï
Phase difference = 2 Ï 6 â Phase difference = Ï 3 â Ï = 6 0 â
I R = I 1 + I 2 + 2 I 1 I 2 cos 6 0 â â I R = I + I + 2 I Ã 1 2
I R = 3 I â I R = 3 I 0 4

Topic: Physics, Wave Optics, NEET
In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of a single-slit pattern?

Options:
A 0.2 mm

B 0.1 mm

C 0.5 mm

D 0.02 mm

The correct answer is:
0.2 mm

Solution:
d =1 mm=1×10-3 m
D=1 m., λ=500×10-9 m
Width of the central maximum = width of 10 maxima in YDSE
2λDa=10λDd
a=0.2d=0.2 mm

Topic: Physics, Wave Optics, NEET, NEET 2015 (Phase 1)
In a double-stranded DNA, one of the nitrogen base is adenine, whose amount is found to be always equal to the amount of

Options:
A 5-methyl uracil which is a purine
B 5-methyl uracil which is a pyrimidine
C Guanine which is a purine
D Guanine which is a pyrimidine

The correct answer is:
5-methyl uracil which is a pyrimidine

Solution:
The nitrogen bases in both strands of DNA are attached via hydrogen bonding forming base pairs. Adenine makes two hydrogen bonds with Thymine (5-methyl uracil) from the opposite strand and vice-versa. Similarly, guanine is bound to cytosine with three H-bonds. Hence, as a rule, pyrimidine will always bind to a purine. Erwin Chargaff gave the observation that for a double-stranded DNA, the ratios between Adenine and thymine and guanine and cytosine are constant and equals. Thus, the amount of adenine and thymine is the same and that of guanine and cytosine is the same.

Topic: Biology, Molecular Basis of Inheritance, NEET
In a dynamic acid (or base) dissociation equilibrium, which direction involves the transfer of a proton in the forward and reverse directions?

Options:
A The forward direction only
B The reverse direction only
C Both the forward and reverse directions
D Neither the forward nor the reverse direction

The correct answer is:
Both the forward and reverse directions

Solution:
In a dynamic acid (or base) dissociation equilibrium, the transfer of a proton occurs in both the forward and reverse directions.

Topic: Chemistry, Ionic Equilibrium, NEET
In a eukaryotic cell during transcription, enzymes RNA Polymerase I, II and III are used respectively for the formation of:

Options:
A m-RNA, t-RNA and 5S r-RNA
B t-RNA, 28S r-RNA and m-RNA
C 5.8S r-RNA, hn-RNA and t-RNA
D 18S r-RNA, t-RNA and m-RNA

The correct answer is:
5.8S r-RNA, hn-RNA and t-RNA

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 transcribes 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 a eukaryotic cell, during which phase of the cell cycle does the synthesis of histone proteins occur?

Options:
A G2 stage of prophase
B S phase
C The entire prophase
D Telophase

The correct answer is:
S phase

Solution:
During S phase or synthetic phase, the replication of DNA takes place. DNA combines with histone proteins to form chromosome during cell division thus the synthesis of histone proteins also occurs during s phase of the cell cycle.

Topic: Biology, Cell Cycle and Cell Division, NEET
In a face-centred cubic lattice, a unit cell is shared equally by how many unit cells?

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
In a face centred cubic lattice a unit cell is shared equally by six unit cells.

Topic: Chemistry, Solid State, NEET, NEET 2005
In a family, the father had a trait but the mother did not. All their sons and daughters had this trait. The same trait was found in some grand daughters, though daughters were married to normal persons.

In this pedigree the genotypes of the father, mother, and husband of daughters are

Options:
A father is AA, the mother is aa, husbands are aa
B father is AA, the mother is aa, and husbands are AA
C father is aa, the mother is Aa, husbands AA
D father is AA, mother AA, one husband is Aa, and second husband aa.

The correct answer is:
father is AA, the mother is aa, husbands are aa

Solution:
The pedigree analysis provides information regarding inheritance of particular traits or diseases and using it we can predict the possibility of appearance of any traits in the new generation. An analysis chart is a visual tool for observing biological relationships within and between families and to understand the mode of inheritance of genetic diseases.
Father has the trait (homozygous) and the mother is not having the trait (homozygous). Their children are having the trait, which means the trait is dominant as it can be expressed in heterozygous condition.
Also, it passed to some grand-children proving it to be dominant.
Thus, the genotype of the father is AA, the mother is aa and the husbands married to daughters have the genotype aa.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a fcc unit cell, atoms are numbered as shown below.

The atoms not touching each other are (atom numbered 3 is face centre of front face)

Options:
A 3 and 4
B 1 and 3
C 1 and 2
D 2 and 4

The correct answer is:
1 and 2

Solution:
In fcc structure, conrer atoms do not touch each other (atoms 1 and 2), but every face centre atom touches corner. Moreover, every face centre atom touches every other face centre atom provided it is not the opposite face centre atom in an fcc unit cell.
I. Atoms 3 and 4 are touching each other where centre-to-centre distance =a2
II. Atoms 1 and 2 are not touching each other
III. Atoms 2 and 4 are touching each other where centre-to-centre distance =a2 .

Topic: Chemistry, Solid State, NEET
In a first order reaction $\mathrm{A} \rightarrow \mathrm{B}$ if $k$ is rate constant and initial concentration of the reactant $\mathrm{A}$ is $0.5 \mathrm{M}$, then the half-life is:

Options:
A $\frac{\log 2}{k}$
B $\frac{\log 2}{k \sqrt{0.5}}$
C $\frac{ln 2}{k}$
D $\frac{0.693}{0.5 k}$

The correct answer is:
$\frac{ln 2}{k}$

Solution:
For a first order reaction.
$k=\frac{2.303}{t} \log _{10} \frac{a}{a-x}$
when $t=t_{1 / 2}$
$k=\frac{2.303}{t_{1 / 2}} \log _{10} \frac{a}{a-\frac{a}{2}}$
or $t_{1 / 2}=\frac{2.303}{k} \log _{10} 2=\frac{\ln 2}{k}$
A Caution
The half-life offirst order reaction is constant because the initial concentration of the reactant has no influence on the half-life of the reaction, i.e. the half-life is constant regardless of the concentration of the reactant.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2007
In a first order reaction the concentration of reactant decreases from 800  mol/dm 3 50  mol/dm 3 in 200 sec . The rate constant of reaction in s −1 is

Options:
A 2Ã 10 &minus;4 ââ s â1
B 1.386Ã 10 â2 ââ s â1
C 3.45× 10 5    s −1
D 2× 10 4   s −1

The correct answer is:
1.386× 10 −2    s −1

Solution:
C t  =  C 0 2 n
2 n = 800 50 =16
2 n = 2 4
So, n = 4
4T 1 2  = 200 sec
T 1 2  = 50 sec
K =  0.693 T 1 2  =  0.693 50  = 1.386 ×  10 −2   sec −1

Topic: Chemistry, Chemical Kinetics, NEET
In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is

Options:
A 15 min
B 75 min
C 60 min
D 30 min

The correct answer is:
30 min

Solution:
Order 1
Concentration changes from 0.8 M to 0.4 M in (50%) 15 minutes this half life is = 15 minutes = t50
A change from 0.1 M to 0.025 M is 75% and for first order reaction
t75=2Ãt50=2Ã15=30 minutes
t50=15 minutes
k=2.303log2t50=2.303log215
Î±=0.1 M
Î±-x=0.025 M
for first order:
k=2.303logtÎ±Î±-x
2.303log215=2.303tlog0.10.025=2.303tlog4
2.303log215=2Ã2.303log2t
t = 30 minutes.

Topic: Chemistry, Chemical Kinetics, NEET
In a first-order reaction, what is the ratio between time required for completion of 99.9% of reaction and its half-life (t1/2) ?

Options:
A 20
B 10
C 30
D 15

The correct answer is:
10

Solution:
Correct Option is : (B)
10

Topic: Chemistry, Chemical Kinetics, NEET
In a fixed quarter circular track of radius R which lies in a vertical plane, a block is released from point A and it leaves the path at point B. The radius of curvature of its trajectory when it just leaves the path will be

Options:
A R
B R 4
C R 2
D None of these

The correct answer is:
R 2

Solution:
By energy conservation between A and B
â Mg 2 R 5 + 0 = MgR 5 + 1 2 Mv 2
v = 2 gR 5
Now, radius of curvature after just leaving B is
r = v â¥ 2 a r = 2 gR / 5 g cos 37 = R 2

Topic: Physics, Laws of Motion, NEET
In a flexible balloon, 2 moles of SO2 having an initial volume of 1 kL at a temperature of 27 Â°C is filled. The gas is first expanded to thrice its initial volume isobarically and then further expanded adiabatically so as to attain its initial temperature. Assuming the gas to be ideal, the work done by the gas in the whole process is [Î³SO2=97, R=253 J mol-1 K-1]

Options:
A 10 kJ
B 35 kJ
C 45 kJ
D 15 kJ

The correct answer is:
45 kJ

Solution:
Wtotal=Wisobaric+Wadiabatic

Wtotal=nR∆T+nR∆Tγ-1
Wtotal=2×253×6001+197-1=45 kJ

Topic: Physics, Thermodynamics, NEET
In a flowering plant 2n = 24, the number of chromosomes in its endosperm will be:

Options:
A 18
B 36
C 20
D 17

The correct answer is:
36

Solution:
Double fertilization is a unique characteristic observed in the Angiosperm group of plants. In this, two fertilizations are observed - 1st between the female and male gametic cells forming zygote and 2nd is between the second male gamete and polar nuclei forming the primary endosperm nucleus. The ploidy of PEN is 3n (male gamete[1n] + polar nuclei [2n]).
Plant chromosome number is 24 2n . Thus, n= 12 .
Endosperm is triploid, 3n. Therefore, its chromosome number will be 36 .

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In a flowering plant, archesporium give rise to

Options:
A Only the wall of the sporangium
B Both wall and the sporogenous cells
C Wall and the tapetum
D Only tapetum and sporogenous cells

The correct answer is:
Both wall and the sporogenous cells

Solution:
In a flowering plant, archesporium gives rise to both the wall and the sporogenous cells. Archesporium cell divides into the primary parietal cell which later forms the anther wall and primary sporogenous cells that form the microspore mother cell.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In a flowering plant, archesporium gives rise to:

Options:
A only the wall of the sporangium
B both wall and the sporogenous cells
C wall and the tapetum
D only tapetum and sporogenous cells

The correct answer is:
both wall and the sporogenous cells

Solution:
Archesporium is a cell or cell structure in a sporophyte from which spores may later develop during the alternate generation. Archesporium cell divides into the primary parietal cell which later forms the anther wall and primary sporogenous cells which form the microspore mother cell. Primary parietal wall after few more periclinal divisions forms anther wall and sporogenous cells give rise to sporogenous tissue.
Related Theory
The pollen grain grows into the male gametophyte on germination which is initiated before pollination takes place. Inside the microsporangium. Pollen mother cell (PMC) undergoes meiosis and results in four microspores which eventually mature into pollen grains.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2003
In a flowering plant, archesporium gives rise to

Options:
A only the wall of the sporangium
B tetrad of haploid microspores
C wall and the tapetum
D only tapetum and sporogenous cells

The correct answer is:
tetrad of haploid microspores

Solution:
The archesporium cells during microsporogenesis give rise to sporogenous cells on the inner side of microsporangium. These sporogenous cells later form the microspore mother cell which divide meiotically to form four haploid pollen cells called as tetrad.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In a food chain, the largest population is that of

Options:
A primary consumers
B tertiary consumers
C producers
D decomposers

The correct answer is:
producers

Solution:
Plants and animals inter-related in the form of the organism being eaten as food by the other is referred to as the food chain. Producers form the largest population and they make their own food. Producers form the first trophic level.

Topic: Biology, Ecosystem, NEET
In a forward biased p-n junction diode, which of the following graphs is most appropriately represented the potential barrier in the depletion region?

Options:
A
B
C
D

The correct answer is:

Solution:
In a forward biased p-njunction, the applied potential is opposite to the junction barrier potential The consequence of this is the effective barrier potential reduces. Hence, the graph (d) is correctly shown.

Topic: Physics, Semiconductors, NEET
In a fully turgid cells, the values of DPD, OP and TP will be

Options:
A DPD = 10 atm, OP = 15 atm, TP = 5 atm
B DPD = 5 atm, OP = 12 atm, TP = 7 atm
C DPD = 2 atm, OP = 7 atm, TP = 5 atm
D DPD = 0 atm, OP = 15 atm, TP = 15 atm

The correct answer is:
DPD = 0 atm, OP = 15 atm, TP = 15 atm

Solution:
A fully turgid cell, experiences equilibrium with the water movement. The net movement of water from outside to inside and inside to outside will be negligible or nil. Thus, the water potential becomes zero, the osmotic potential is negative and pressure potential becomes positive. Hence, OP is equal to TP and thus DPD is zero.
DPD = OP - TP

Topic: Biology, Transport in Plants, NEET
In a functional breast, milk is temporarily stored in:

Options:
A Mammary duct
B Mammary alveoli
C Mammary ampulla
D Lactiferous duct

The correct answer is:
Mammary alveoli

Solution:
Mammary glands
- 15 -20 mammary lobes are found in glandular tissue of each breast.
- Each mammary lobe is divided into numerous alveoli. Alveoli is a group of cells having a lumen within.
- Alveolar cells secrete milk in the lumen of alveoli where it is temporarily stored.

Topic: Biology, Human Reproduction, NEET
In a galvanic cell, what is the direction of current flow compared to the flow of electrons?

Options:
A The same as the flow of electrons.
B Opposite to the flow of electrons.
C Unrelated to the flow of electrons.
D It depends on the specific cell.

The correct answer is:
Opposite to the flow of electrons.

Solution:
The direction of current flow in a galvanic cell is opposite to the flow of electrons.

Topic: Chemistry, Electrochemistry, NEET
In a galvanic cell, what is the term for the half-cell where oxidation takes place?

Options:
A Anode
B Cathode
C Electrolyte
D Salt bridge

The correct answer is:
Anode

Solution:
In a galvanic cell, the half-cell where oxidation takes place is called the anode.

Topic: Chemistry, Electrochemistry, NEET
In a galvanometer, 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance S connected to the galvanometer is

Options:
A 19 G

B G19
C 20 G

D G20

The correct answer is:
G19

Solution:
The shunt of an ammeter,
S=Ig×GI-Ig
=5×G100-5=G19

Topic: Physics, Current Electricity, NEET