datavorous/entrance-exam-dataset huggingface dataset:
During sodium-dependent glucose transport (symport or co-transport), the

Options:
A Na+ is transported passively from the lumen into the gut, glucose is transported against concentration gradient
B Na+ and glucose both are transported actively into the gut from lumen
C Na+ and glucose both are transported passively
D Na+ is transported actively and glucose passively

The correct answer is:
Na+ is transported passively from the lumen into the gut, glucose is transported against concentration gradient

Solution:
Small amounts of monosaccharides like glucose, amino acids and some of the electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon the concentration gradients. However, some of the substances such as fructose and some amino acids are absorbed with the help of the carrier ions like Na+. This mechanism is called facilitated transport.

Topic: Biology, Digestion and Absorption, NEET
During somatic hybridization in plants:

Options:
A Somaclones are produced in large numbers
B Apical meristems are cultured to get virus-free plants
C Cell wall and middle lamella are digested before fusing the cells
D Crop plants with higher levels of vitamins, proteins and minerals are hybridised

The correct answer is:
Cell wall and middle lamella are digested before fusing the cells

Solution:
Plant cells without cell wall are called protoplast. Protoplasts of two different plant can be fused to obtain a hybrid protoplast and grown into a new plant. Such process is called as somatic hybridization since somatic cells are used for fusion and not gametic cells. The hybrids formed from such technique are called somatic hybrid.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
During succession, in the successive seral stages, there is a/an:

Options:
A Change in the diversity of species of organisms
B Increase in the number of species and organisms
C Increase in the total biomass
D All of the above

The correct answer is:
All of the above

Solution:
During Ecologiccal Succession, the entire sequence of communities that successively change in a given area are called sere(s) and the individual transitional communities are termed seral stages or seral communities. In theses successive seral stages several features are changes continuously like change in the diversity of species of organisms, increase in the number of species and organisms, increase in the total biomass etc. Presently visible communities in the world have come to be because of succession that has occurred over millions of years since life started on earth.

Topic: Biology, Ecosystem, NEET
During symport two different molecules move across the membrane

Options:
A in same direction with the help of same carrier
B in opposite direction with the help of same carrier
C in same direction with the help of different carriers located at a common site
D in same direction with the help of different carriers located at different sites in the same cell

The correct answer is:
in same direction with the help of same carrier

Solution:
During symport, two different molecules move across the membrane in same direction with the help of same carrier.

Topic: Biology, Transport in Plants, NEET, NEET 2023 (Manipur)
During the adiabatic expansion of 2 mol of an ideal gas, the increase in internal energy was found to be equal to (-200 J). The work done by the gas during the process will be equal to

Options:
A 0
B 400 J
C â200Â J
D 200Â J

The correct answer is:
200 J

Solution:
According to first law of thermodynamics,
âQ=âU+âW ...(i)
Given, âU=-200 J
âQ=0 (due to adiabatic process)
Substituting in Equation (i), we get
0=-200+âW
â¹âW=200 J

Topic: Physics, Thermodynamics, NEET
During the analysis of the DNA of an organism having 5386 nucleotides, it was discovered that the percentages of the bases were A=29%, G=17%,C=32% and T=17% . Considering the Chargaff's rule it can be concluded that:

Options:
A It is a double-stranded linear DNA.
B It is a double-stranded circular DNA.
C It is a single-stranded DNA.
D Both A and B

The correct answer is:
It is a single-stranded DNA.

Solution:
According to Chargaff's rule in any double stranded DNA AT=1âandGC =1 . That is in any double stranded DNA Amount of Adenine and Thymine should be same, as well as amount of Guanine and Cytosine should be same. Although amount A+TG+C is mostly not equal to 1, it is called Dissymmetry ratio.

Topic: Biology, Molecular Basis of Inheritance, NEET
During the complete oxidation of glucose, the proportion of ATP formed by substrate-level phosphorylation is ______.

Options:
A 100%
B 89.5%
C 10.5%
D 50%

The correct answer is:
10.5%

Solution:
Respiration is the process or mechanism to break down food molecules to release energy in the form of ATP. During the process of aerobic respiration, phosphorylation of ADP occurs forming the energy currency molecule - Adenosine triphosphate. The energy required for the production of proton gradient which is needed for phosphorylation is obtained from the oxidation-reduction of the molecules. One of the steps of the respiration process is glycolysis where 6C-glucose is broken down to 3C-pyruvate.
In the first step of aerobic respiration, glycolysis takes place where the glucose molecule is broken down into two molecules of 3C-pyruvic acids. Each of these pyruvates enters the next phase, that is the Kreb's cycle. In each round of the TCA cycle, three molecules of NADH2 , one GTP, and one molecule of FADH2 are generated using 1 Acetyl CoA. Thus, for each glucose molecule in a total of 6 molecules of NADH2, 2GTP's and 2 molecules of FADH2 are formed in 2 turns of Kreb's cycle.
Aerobic respiration results in formation of 38 ATPs of which 4 ATPs are produced in glycolysis which is approximately 10.5%.

Topic: Biology, Respiration in Plants, NEET
During the dark reaction, 3-PGA is derived from

Options:
A RuBP only
B CO2 only
C RuBP + CO2
D RuBP + CO2 + PEP

The correct answer is:
RuBP + CO2

Solution:
For fixation of carbon, 3-phosphoglyceric acid is formed from CO2 and RuBP. CO2 first combines with RuBP to form unstable 6 carbon compound which immediately breaks into two molecules of 3 C - PGA.

Topic: Biology, Photosynthesis in Higher Plants, NEET
During the development of helobial endosperm,

Options:
A the first nuclear division of primary endosperm nucleus is followed by equal cytoplasmic division
B the first nuclear division of primary endosperm nucleus is followed by another nuclear division
C the first nuclear division of primary endosperm nucleus is followed by unequal cytoplasmic division with the larger cell towards the chalaza
D the first nuclear division of primary endosperm nucleus is followed by unequal cytoplasmic division with the larger cell towards the micropyle

The correct answer is:
the first nuclear division of primary endosperm nucleus is followed by unequal cytoplasmic division with the larger cell towards the micropyle

Solution:
Helobial endosperm formation occurs in monocots. During the development of helobial endosperm, the first nuclear division of primary endosperm nucleus is followed by unequal cytoplasmic division with the larger cell towards the micropyle. After this free nuclear divisions occur in each cell which become multi nucleated. Eventually cytoplasmic division occurs and it is finally changed into cellular endosperm

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is

Options:
A 55 minutes
B 110 minutes
C 220 minutes
D 330 minutes

The correct answer is:
110 minutes

Solution:
For 1 mole of Cl2, 2F electricity is required.
∴ For 0.1 mole of Cl2, 0.2 F electricity will be required.
∴Q=I×t
or, t=QI=0.2×965003×60≈110 minutes.

Topic: Chemistry, Electrochemistry, NEET, NEET 2016 (Phase 2)
During the electron transport in mitochondria which of the following changes are observed.

(1) The pH of the matrix increases.
(2) ATP synthase synthesizes one molecule of ATP when a pair of the proton is transferred back to the matrix.
(3) NAD+ is oxidised.

Options:
A only 1
B 1 and 2
C 2 and 3
D 1, 2 and 3

The correct answer is:
1, 2 and 3

Solution:
pH of matrix increases as protons are transferred to perimitochondrial space. NAD+ is oxidized as it gives its hydrogen to complex I of ETS. One molecule of ATP is synthesised by ATPase when a pair of one pair proton moves from its higher concentration in PMS to matrix.

Topic: Biology, Respiration in Plants, NEET
During the formation of bread it becomes porous due to the release of $\mathrm{CO}_2$ by the action of:

Options:
A yeast
B bacteria.
C virus
D protozoans

The correct answer is:
yeast

Solution:
Yeast Saccharomyces cerevisiae is also known baker's yeast. Upon fermentation, yeast release ethanol and carbon dioxide as byproducts. $\mathrm{CO}_2$ forms bubbles which give the dough the perfect consistency and structure for it to rise and porous.
Related Theory
During the baking process, most of the alcohol in the dough evaporates into the atmosphere. This is basically the same thing that happens to much of the water in the dough as well. And it has long been known that bread contains residual alcohol, up to $1.9 \%$ of it. Thus, bread isn't alcoholic.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2002
During the formation of gametophytes and embryo in angiosperms, the cells undergo two types of cell divisions. The diagram given below is a representation of them. The red lines indicate their plane of division. What is the correct type of X and Y?

Options:
A X is periclinal division, while Y is an anticlinal division
B X is anticlinal division while Y is an periclinal division
C Both X and Y are periclinal divisions
D Both X and Y are anticlinal division

The correct answer is:
X is periclinal division, while Y is an anticlinal division

Solution:
Periclinal division is a type of division in which the plane of division is parallel to the plane of the tissue. Anticlinal division is a type of division in which the plane of division is perpendicular to the plane of the tissues. Due to the periclinal division, we get many rows of cells.

Topic: Biology, Anatomy of Flowering Plants, NEET
During the formation of leaves and elongation of the stem, some cells âleft behindâ from shoot apical meristem; they are known as

Options:
A shoot apical meristems
B accessory buds
C axillary bud
D root apical meristems

The correct answer is:
axillary bud

Solution:
During the formation of leaves and elongation of the stem, some cells âleft behindâ from shoot apical meristem, constituting the axillary bud. Such buds are present in the axils of leaves and are capable of forming a branch or a flower.

Topic: Biology, Anatomy of Flowering Plants, NEET
During the fusion of organic compound with sodium metal, nitrogen present in the organic compound is converted into

Options:
A NaNO2
B NaNH2
C NaCN
D NaNC

The correct answer is:
NaCN

Solution:
The carbon and nitrogen present in an organic compound during fusion with sodium metal give sodium cyanide (NaCN).

Topic: Chemistry, s Block Elements, NEET
During the initial days of lactation, the milk produced is termed as

Options:
A Colostrum
B Baby's milk
C Rostrum
D Cholesterum

The correct answer is:
Colostrum

Solution:
Lactation is the process of releasing milk from the breast which helps the mother to feed the baby. The initial milk secretion is called colostrum and it contains several antibodies important for resistance development in the baby.

Topic: Biology, Human Reproduction, NEET
During the kinetic study of the reaction, $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$, following results were obtained
\(\begin{array}{|c|c|c|c|}
\hline \text { Run } & [A] / \mathrm{mol~} \mathrm{L}^{-1} & [B] / \mathrm{mol~} \mathrm{L}^{-1} & \begin{array}{cc} \text{Initial rate of} \\ \text{formation of} \\ D / \mathrm{mol~} \mathrm{L}^{-1} \mathrm{~min}^{-1} \end{array} \\
\hline \text { I } & 0.1 & 0.1 & 6.0 \times 10^{-3} \\
\hline \text { II } & 0.3 & 0.2 & 7.2 \times 10^{-2} \\
\hline \text { III } & 0.3 & 0.4 & 2.88 \times 10^{-1} \\
\hline \text { IV } & 0.4 & 0.1 & 2.40 \times 10^{-2} \\
\hline
\end{array}\)
Based on the above data which one of the following is correct?

Options:
A rate $=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}]$
B rate $=k[A][B]$
C rate $=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}]^2$
D rate $=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2$

The correct answer is:
rate $=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2$

Solution:
Let the order of reaction with respect to $\mathrm{A}$ is $x$ and with respect to $B$ is $y$. Thus,
$$
\text { rate }=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}
$$
For the given cases,
(I) rate $=\mathrm{k}(0.1)^{\mathrm{x}}(0.1)^{\mathrm{y}}=6.0 \times 10^{-3}$
(II) rate $=\mathrm{k}(0.3)^{\mathrm{x}}(0.2)^{\mathrm{y}}=7.2 \times 10^{-2}$
(III) rate
$$
=\mathrm{k}(0.3)^{\mathrm{x}}(0.40)^{\mathrm{y}}=2.88 \times 10^{-1}
$$
(IV) rate $=\mathrm{k}(0.4)^{\mathrm{x}}(0.1)^{\mathrm{y}}=2.40 \times 10^{-2}$
On dividing Eq. (I) by (IV), we get
$$
\left(\frac{0.1}{0.4}\right)^x\left(\frac{0.1}{0.1}\right)^y=\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}
$$
or
$$
\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^1
$$
$$
\therefore \quad \mathrm{x}=1
$$
On dividing Eq. (II) by (III), we get
$$
\begin{aligned}
& & \left(\frac{0.3}{0.3}\right)^x\left(\frac{0.2}{0.4}\right)^y & =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}} \\
\text { or } & & \left(\frac{1}{2}\right)^y & =\frac{1}{4} \\
\text { or } & & \left(\frac{1}{2}\right)^y & =\left(\frac{1}{2}\right)^2 \\
\therefore & & y & =2
\end{aligned}
$$
$\therefore$ Thus, rate law is,
$$
\text { rate }=\mathrm{k}[\mathrm{A}]^1[\mathrm{~B}]^2 \text { or }=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2010 (Screening)
During the kinetic study of the reaction, 2A+B→C+D , following results were obtained

Run	[A]/molL-1	[B]/molL-1	Initial rate of formation of D/molL-1min-1

I	0.1	0.1	6.0×10-3

II	0.3	0.2	7.2×10-2

III	0.3	0.4	2.88×10-1

IV	0.4	0.1	2.40×10-2

Based on the above data which one of the following is correct?

Options:
A rate=k[A][B]2
B rate=k[A]2[B]
C rate=k[A][B]
D rate=k[A]2[B]2

The correct answer is:
rate=k[A][B]2

Solution:
2A+B→C+D
Let rate=K[A]m[B]n
To get m- value divide (IV) by I and on solving m=1
Similarly, to get n devide (III) by (II) and on solving n=2
Hence Rate=K[A]1[B]2

Topic: Chemistry, Chemical Kinetics, NEET
During the life-cycle, Fasciola hepatica (liver fluke) infects its intermediate host and primary host at the following larval stage respectively :

Options:
A redia and miracidium
B cercaria and redia
C metacercaria and cercaria
D miracidium and metacercaria

The correct answer is:
miracidium and metacercaria

Solution:
Fasciola hepatica is called as digenetic because its life cycle is completed in two hosts (sheep and invertebrate host). It is endoparasite because it lives inside the bile passage of sheep. (Ref. to Fig. 29.1)
Related Theory
The infective stage of liverfluke is metacercariae (larval stage).

Topic: Biology, Human Health and Diseases, NEET, NEET 2003
During the mechanism of vision, the cells which generate the first potential on stimulation be light is/are:

Options:
A Rods and Cones
B Bipolar neurons
C Optic nerve
D Ganglion cells

The correct answer is:
Rods and Cones

Solution:
Mechanism of Vision
Light rays (of visible spectrum) are focussed on the retina via passing through the cornea and lens → Light induced dissociation of photosensitive compounds [(photopigments composed of opsin (a protein) and retinal (an aldehyde of vitamin A)] into retinal and opsin → Structural changes in opsin → Changes in membrane permeability → Potentials are generated in photoreceptor cells (rods and cones) → Production of a signal crosses the bipolar cells and reaches the ganglion cells → Generation of action potentials in the ganglion cells → Transmission of these action potentials (impulses) by the optic nerves upto the visual cortex area of the brain for the analysis.

Topic: Biology, Neural Control and Coordination, NEET
During the metagenesis of Obelia,

Options:
A polyps produce medusae asexually and medusae form the polyps sexually.
B polyps produce medusae sexually and medusae form the polyps asexually.
C both polyps and medusae are produced sexually by each other.
D both polyps and medusae are produced asexually by each other.

The correct answer is:
polyps produce medusae asexually and medusae form the polyps sexually.

Solution:
Cnidarians exhibit two basic body forms called polyp and medusa. The former is a sessile and cylindrical form like Hydra, Adamsia, etc. whereas, the latter is umbrella-shaped and free-swimming like Aurelia or jellyfish. Those cnidarians which exist in both forms exhibit alternation of generation (Metagenesis), i.e., polyps produce medusae asexually and medusae form the polyps sexually (e.g., Obelia).

Topic: Biology, Animal Kingdom, NEET
During the metaphase stage of mitosis, spindle fibres attach to chromosomes at:

Options:
A kinetochore
B both centromere and kinetochore
C centromere, kinetochore and are as adjoining centromere
D centromere

The correct answer is:
kinetochore

Solution:
During metaphase, the kinetochore microtubules pull the sister chromatids back and forth until they align along the equator of the cell, called the equatorial plane.
Related Theory
The centromere is a DNA region where they join sister chromatids together, whereas kinetochore is an assembling protein complex in the centromere where they attach to opposite poles of the mitotic spindle.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2013 (Karnataka)
During the past century, the temperature of Earth has increased by ____ Â°C, most of it during the last ______ decades.

Options:
A 10.6, four
B 1.6, three
C 0.6, three
D 6.0, four

The correct answer is:
0.6, three

Solution:
Increase in the level of greenhouse gases has led to considerable heating of Earth leading to global warming. During the past century, the temperature of Earth has increased by 0.6Â°C, most of it during the last three decades. Scientists believe that this rise in temperature is leading to deleterious changes in the environment and resulting in odd climatic changes (e.g. El Nino effect) , thus leading to increased melting of polar ice caps as well as of other places like the Himalayan snow caps. Over many years, this will result in a rise in sea level that can submerge many coastal areas.

Topic: Biology, Environmental Issues, NEET
During the preparation of colloidal sol by Bredig's are method, traces of alkali (NaOH or Na2CO3 ) is added to

Options:
A homogenise the sol
B stabilize the sol
C act as preptizing agentact
D act as coagulating agent

The correct answer is:
stabilize the sol

Solution:
During the preparation of colloidal sol by Bredig's are method, traces of alkali (NaOH or Na2CO3 ) is added to stabilize the sol.the OH- ions get absorbed on the surface of metal sol particles making them –vely charged hence, the charged sol particles are prevented from coming together.

Topic: Chemistry, Surface Chemistry, NEET
During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of $\mathrm{Fe}^{2+}$ ion?

Options:
A concentrated sulphuric acid
B dilute nitric acid
C dilute sulphuric acid
D dilute hydrochloric acid

The correct answer is:
dilute sulphuric acid

Solution:
During the preparation of Mohr's salt, dilute sulphuric acid is added to prevent the hydrolysis of $\mathrm{Fe}^{2+}$ ion.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2024
During the process of aerobic respiration, (i) gets oxidized and its electrons get transferred to the electron transport chain while in photosynthesis, (ii) gets oxidized to transfer molecules to the electron transport chain.

Options:
A (i)-glucose; (ii)-xanthophyll
B (i)-carbon dioxide, (ii) -xanthophyll
C (i)-carbon dioxide, (ii) -chlorophyll-a
D (i) -glucose, (ii) -chlorophyll-a

The correct answer is:
(i) -glucose, (ii) -chlorophyll-a

Solution:
The electron transport chain is a series of reactions taking place in the mitochondria in both the cells as well as in photosynthesis. In aerobic cellular respiration, the glucose molecule gets oxidized and the electrons released get transferred to the electron transport chain, whereas in chloroplasts, the light helps in the conversion of water to oxygen and the chlorophyll a donates two electrons to the electron transport chain after oxidation.

Topic: Biology, Respiration in Plants, NEET
During the process of digestion, the proteins present in food material are hydrolysed to amino acids. The two enzymes involved in the process:
are respectively:

Options:
A invertase and zymase
B amylase and maltase
C diastase and lipase
D pepsin and trypsin

The correct answer is:
pepsin and trypsin

Solution:
In the process of digestion, the proteins present in food material are hydrolysed to amino acids.
Related Theory
In the absence of enterokinase, trypsinogen is not converted into the trypsin and thus the digestion of protein will be incomplete.

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2006
During the process of digestion, the proteins present in food materials are hydrolysed to amino acids. The two enzymes involved in the process
proteins→enzyme(A)polypeptides→enzyme(B)aminoacids,
are respectively

Options:
A invertase and zymase
B amylase and maltase
C diastase and lipase
D pepsin and trypsin

The correct answer is:
pepsin and trypsin

Solution:
The two enzymes involved in the process are pepsin and trypsin.
proteins→pepsinpolypeptides→trypsinaminoacids,

Topic: Chemistry, Biomolecules, NEET
During the process of isolation of DNA, chilled ethanol is added to:

Options:
A precipitate DNA
B break open the cell to release DNA
C facilitate action of restriction enzymes
D remove proteins such as histones.

The correct answer is:
precipitate DNA

Solution:
Chilled ethanol is used to precipitate DNA out of a mixture of biomolecules. Low temperature protects DNA by slowing down the activity of enzymes that could break DNA apart while ethanol, being a dehydrating agent removes water molecules bound to DNA and making it visible. Chloroform is used in the extraction process of DNA to remove proteins bound to DNA.
Related Theory
There are five basic steps of DNA extraction that are consistent across all the possible DNA purification chemistries: 1) disruption of the cellular structure to create a lysate. 2) separation of the soluble DNA from cell debris and other insoluble material, 3) binding the DNA of interest to a purification matrix, 4) washing proteins and other contaminants away from the matrix and 5) elution of the DNA.
Caution
The option Chilled isopropanol inslead of isopropanol is also correct option.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2013 (Karnataka)
During the processing of the prohormone ''proinsulin" into mature and functional ''insulin", the

Options:
A C- peptide is added to proinsulin
B C- peptide is removed from proinsulin
C B- peptide is added to proinsulin
D B- peptide is removed from proinsulin

The correct answer is:
C- peptide is removed from proinsulin

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