datavorous/entrance-exam-dataset huggingface dataset:
Regarding meiosis, which of the statements is incorrect?

Options:
A DNA replication occurs in S phase of meiosis-II
B Pairing of homologous chromosomes and recombination occurs in meiosis-I
C Four haploid cells are formed at the end of meiosis-II
D There are two stages in meiosis, meiosis-I and II

The correct answer is:
DNA replication occurs in S phase of meiosis-II

Solution:
In meiosis there are two consecutive cell division events without intervening G1, S and G2 phases between meiosis I and meiosis II.

So, there is no S phase before meiosis II and no DNA replication will happen before meiosis II.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2022 (Phase 1)
Regarding mutation select how many points are correct?

(i) Random

(ii) Discontinuous

(iii) Continuous

(iv) Directionless

Options:
A Two
B Three
C Four
D Five

The correct answer is:
Three

Solution:
The mutation is a sudden and inheritable change observed in the physical and chemical nature of genetic material or DNA. The mutation is a source of discontinuous variation and is directionless, as mutation can be forward or backward. The process of mutation is random and its effect is very specific.

Topic: Biology, Evolution, NEET
Regarding the functions of our respiratory system, mark the wrong entry.

Options:
A Humidifies the air
B Warms up the air
C Exchange of gases
D Cleans up the air

The correct answer is:
Exchange of gases

Solution:
The respiratory system has no role in cleaning the air.

Topic: Biology, Breathing and Exchange of Gases, NEET
Regarding the structure of cyanamide ion, pick out the wrong statement

Options:
A It has one carbon with a negative charge
B It has two σ bonds
C It has two π bonds
D It has two negatively charged Nitrogen atoms

The correct answer is:
It has one carbon with a negative charge

Solution:
has two σ bonds, two π bonds and two negatively charged Nitrogen atoms

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Regeneration of 6 RuBP in C₃ cycle requires

Options:
A 1 ATP
B 6 ATP
C 1 ATP and 1 NADPH+H+
D 3 ATP and 2 NADPH+H+

The correct answer is:
6 ATP

Solution:
Regeneration is the step in which Ribulose-5 Phosphate takes phosphate from ATP to form Ribulose-1,5 Biphosphate. For fixation of 6 CO2 through Calvin cycle 6 ATP is consumed in regeneration step.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Regeneration of a plant cell to give rise to new plant is called:

Options:
A Reproduction
B Budding
C Totipotency
D Pleuripotency

The correct answer is:
Totipotency

Solution:
The capicity to generate a whole plant from any cell of plant is called totipotency.

Topic: Biology, Reproduction in Organisms, NEET
___ region of the brain is chiefly responsible for the regulation process of respiration.

Options:
A medulla oblongata
B medulla
C Cerebrum
D Cerebellum

The correct answer is:
medulla

Solution:
The medulla region of the brain is chiefly responsible for the regulation process of respiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
"Relationships are visualized as evolutionary trees". This statement is related to

Options:
A Identification
B Classifcation
C Sytematics
D All the three

The correct answer is:
Sytematics

Solution:
The study of the diversity of organisms and evolutionary relationships based on comparative anatomy, ecology, physiology, biochemistry and other fields is known as systematics.

Topic: Biology, The Living World, NEET
Relative permeability of iron is 5500. Its magnetic susceptibility is

Options:
A 5499
B 5500×107
C 5500×10-7
D 5501

The correct answer is:
5499

Solution:
The relation for magnetic susceptibility is given by
μr=1+χ
Or χ=ur-1
Given, μr=5500
∴ χ=5500-1=5499

Topic: Physics, Magnetic Properties of Matter, NEET
Release of pancreatic juice is stimulated by

Options:
A Gastrin
B Secretin
C Enterogastrone
D Cholecystokinin

The correct answer is:
Cholecystokinin

Solution:
Enzyme	Source	Stimulus	Action

Cholecystokinin
(CCK)/ Pancreozymin	Small intestine
(duodenum)	Food (fatty chyme and amino acids) in the duodenum.	Gall bladder, contraction of gall bladder(bile released). Pancreas- the release of pancreatic juice.

Enterogastrone	Small intestine
(duodenum)	Monosaccharides and fats (fatty chyme) in the duodenum.	Gastric mucosa and muscles; inhibition of gastric secretion and motility. (slowing food passage).

Gastrin	Pyloric stomach	Vagus nerve (Cranial nerve X) activity peptides and proteins in the stomach	Secretory cells and muscles of the stomach: secretion of HCl and stimulation of gastric mobility.

Secretin	Intestinal wall
(Duodenum)	Food and strong acid in the stomach and intestine	Pancreas: Secretion of water and bicarbonate (NaHCO3); inhibition of gastric motility.

Topic: Biology, Digestion and Absorption, NEET
Removal of introns and joining of exons in a defined order during transcription is called

Options:
A looping
B inducing
C slicing
D splicing

The correct answer is:
splicing

Solution:
The primary transcript from a typical eukaryotic gene contains introns as well as exons. During RNA splicing, introns are removed and exons are joined in a defined order, to produce functional RNA.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2012 (Screening)
Removal of introns and joining the exons in a defined order in a transcription unit is called

Options:
A splicing
B tailing
C transformation
D capping

The correct answer is:
splicing

Solution:
In some eukaryotes, genes consist of coding nucleotids sequences, which are separated from each other by blocks of non-coding sequences. The coding sequences here are called exons and non-coding sequences are called introns. The primary transcript from a typical eukaryotic gene contains introns as well as exons. The introns are removed from this primary transcript by a process called RNA splicing.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2009 (Screening)
Removal of RNA polymerase III from nucleoplasm will affect the synthesis of

Options:
A tRNA
B hnRNA
C mRNA
D rRNA

The correct answer is:
tRNA

Solution:
RNA polymerase III transcribes tRNA, RNA poly II synthesis mRNA while RNA poly I synthesis rRNA in eukaryotes.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2012 (Screening)
Removal of shoot tips is very useful technique to boost the production of tea leaves. This is because

Options:
A gibberellins prevent bolting and are inactivated
B auxins prevent leaf drop at early stages
C effect of auxins is removed and growth of lateral buds is enhanced
D gibberellins delay senescence of leaves

The correct answer is:
effect of auxins is removed and growth of lateral buds is enhanced

Solution:
Removal of shoot tips is a very useful technique to boost the production of tea leaves. This is because effect of auxin is removed and growth of lateral bud is enhanced. This phenomenon in most higher plants in which growing apical buds (shoot tips) inhibit growth of lateral buds due to effects of auxin is called apical dominance.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2019 (Odisha)
REN is a/an

Options:
A enzyme.
B sugar.
C base.
D RNA.

The correct answer is:
enzyme.

Solution:
Molecular scissors are enzymes that cut certain restriction sites present on the DNA. These "molecular scissors" are especially known as restriction enzymes. These enzymes cut DNA at specific sites. Once it finds that sequence, it breaks the hydrogen bonds between the specific base within the site, and the DNA has been cut and exposed to the possibility of attaching with other genes to the existing DNA strand.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Rennin acts on

Options:
A Digestion of milk protein in basic pH
B Proteins in stomach
C Fat in intestine
D Digestion of milk protein in acidic pH

The correct answer is:
Digestion of milk protein in acidic pH

Solution:
Rennin: It's an enzyme that acts on milk protein casein & changes it into paracasein spontaneously precipitated into insoluble calcium paracaseinate, forming solid curd or coagulated milk. Rennin is present in gastric juice and hence works at highly acidic pH (due to HCl) in the stomach (about 1-3 pH) and absent in adult human beings.
Renin: It is a hormone that is produced by the kidney to control blood sodium levels. Renin is a hormone produced by JG cells of the kidney. the primary function is, therefore, to eventually cause an increase in blood pressure, leading to the restoration of perfusion pressure in the kidney.

Topic: Biology, Digestion and Absorption, NEET
Replica plate experiment by Lederberg showing antibiotic resistance proves

Options:
A genetic drift
B natural selection
C chemical mutation
D founder's effect

The correct answer is:
natural selection

Solution:
Replica plate experiment was conducted by Joshua Lederberg and Esther Lederberg in 1952. This demonstrated the genetic basis of drug-resistant mutation in Esherichia coli bacteria. This experiment proved that he antibiotic-resistant strains of bacteria were able to survive antibiotics because these bacteria has the resistance before their exposure to it. This is an example of natural selection.

Topic: Biology, Evolution, NEET
Replication in E. coli DNA starts in a definite region known as

Options:
A Promoter region
B Origin of replication
C Replication fork
D Template strand

The correct answer is:
Origin of replication

Solution:
DNA replication does not initiate randomly at any place in DNA. Replication starts at a specific region termed as the origin of replication. In Eukaryotes, there is multiple origins of replication but in prokaryotes, there is an only a single origin of replication.

Topic: Biology, Molecular Basis of Inheritance, NEET
Replum is present in the ovary of flower of

Options:
A lemon
B mustard
C sunflower
D pea

The correct answer is:
mustard

Solution:
The gynoecium in family-Cruciferae (Brassicaceae) is bicarpellary, syncarpous, ovary superior, unilocular with parietal placentation. At maturity ovary becomes bilocular due to formation of false septum (replum) eg, mustard.
In family-Asteraceae (Compositae) the gynoecium is bicarpellary, syncarpous, ovary inferior, unilocular, one ovule in each locule, basal placentation and stigma branched, eg, sunflower.
In family-Fabaceae (Leguminosae) the gynoecium is monocarpellary, unilocular with superior ovary and marginal placentation, eg, pea.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Screening)
Representation of age pyramids for the human population is given below. Identify A, B and C by selecting the correct option.

Options:
A

A	B	C
Declining	Stable	Expanding

B

A	B	C
Stable	Expanding	Declining

C

A	B	C
Expanding	Stable	Declining

D

A	B	C
Stable	Declining	Expanding

The correct answer is:

A	B	C
Expanding	Stable	Declining

Solution:
Age pyramid having higher population of pre-reproductive individuals, moderate number of reproductive individuals and low post-reproductive individual count forms the proper upright triangular shape. It represents growing population or expanding one.
If the count of reproductive population is higher than the pre-reproductive population, the age pyramid forms an urn shape. It represents declining or diminishing population with low natality.
For a bell-shaped age pyramid, the population of pre-reproductive and reproductive groups is almost equal. Post reproductive population is low. Such age pyramid shows stable or stationary population, neither growing nor diminishing.

Topic: Biology, Organisms and Populations, NEET
Represented below is the inheritance pattern of a certain type of traits in humans.

Which one of the following conditions could be an example of this pattern?

Options:
A Thalassemia
B Haemophilia
C Phenylketonuria
D Sickle cell anaemia

The correct answer is:
Haemophilia

Solution:
The inheritance pattern in the given figure shows criss-cross inheritance (X-linked disorder), an example of such a pattern is haemophilia. Haemophilia is a group of inherited blood disorders in which the blood does not clot properly.

Topic: Biology, Principles of Inheritance and Variation, NEET
Reproduction is characteristic of living organisms. Which of these organisms do not reproduce by fragmentation?

Options:
A unicellular algae
B filamentous algae
C protonema of mosses
D flatworms

The correct answer is:
unicellular algae

Solution:
In multicellular organisms, reproduction refers to the production of progeny possessing features more or less similar to those of parents. Invariably and implicitly we refer to sexual reproduction. Organisms reproduce by asexual means also. In Planaria (flatworms), we observe true regeneration, i.e., a fragmented organism regenerates the lost part of its body and becomes a new organism. The fungi, the filamentous algae, the protonema of mosses, all easily multiply by fragmentation. When it comes to unicellular organisms like bacteria, unicellular algae or Amoeba, reproduction is synonymous with growth, i.e., increase in number of cells.

Topic: Biology, The Living World, NEET
Reptiles of different shape and size dominated Earth. Which reptile went back into water to evolve into fish like reptiles 200 mya?

Options:
A Ichthyosaurs
B Tyrannosaurus
C Hemidactylus
D Pelycosaurs

The correct answer is:
Ichthyosaurs

Solution:
The first cellular form appeared on earth about 2000 million years ago (mya) which then formed into multicellular form following which there was an evolution of fish. Fish evolved into amphibians which in turn evolved into reptiles. Ichthyosaurus is a fish-like reptile that evolved from land reptiles which went back into the water. Tyrannosaurus is the biggest dinosaurs, which are land reptiles, and are about 20 feet in height.

Topic: Biology, Evolution, NEET
Required amount of crystalline oxalic acid (eq. wt.=63 ) to prepare N10, 250 ml oxalic acid solution is

Options:
A 0.158 g
B 1.575 g
C 15.75Â g
D 6.3Â g

The correct answer is:
1.575 g

Solution:
N = WB Ã 1000E Ã Vâ N10 = x Ã 100063 Ã 250
Where, WB is the weight of sample, E is the equivalent weight of substance and V is volume of solution.
â´ x=1.575 grams

Topic: Chemistry, Redox Reactions, NEET
Reserpine, the active chemical produced by Rauwolfia vomitoria is used in the treatment of

Options:
A diabetes mellitus
B urinary tract infections
C high blood pressure
D arthritis

The correct answer is:
high blood pressure

Solution:
Reserpine is used in the treatment of hypertension. It reduces the stores of nor-epinephrine reducing the heart rate and blood pressure.

Topic: Biology, Biodiversity and its Conservation, NEET
Residual air mostly occurs in

Options:
A alveoli
B bronchus
C nostrils
D trachea

The correct answer is:
alveoli

Solution:
The amount of air left in the alveoli after maximal exhalation is called reserve air. There is always some air left in the lungs after a maximal exhalation. In the lungs, never the alveoli, trachea, or bronchi can be left completely empty without air. If this residual is not there the lung tissues would stick together.

Topic: Biology, Breathing and Exchange of Gases, NEET
Residual volume is

Options:
A Lesser than tidal volume
B Greater than inspiratory volume
C Greater than vital capacity
D Greater than tidal volume

The correct answer is:
Greater than tidal volume

Solution:
Residual volume (1200mL) is greater than tidal volume (500mL).

Topic: Biology, Breathing and Exchange of Gases, NEET
Resistance genes present in pBR322 are

Options:
A Penicillin and ampicillin
B Ampicillin and tetracycline
C Tetracycline and erythromycin
D Erythromycin and ampicillin.

The correct answer is:
Ampicillin and tetracycline

Solution:
Plasmids are able to replicate in the host cell independent of the control of chromosomal DNA. Plasmids like pBR322 is used as a cloning vector in E. coli organisms. The pBR322 plasmid has resistance site for two antibiotics - tetracycline and ampillicin. If we ligate a foreign DNA at the site of tetracycline resistance the recombinant plasmids will not be tetracycline resistant. We can obtain the recombinants by growing them on ampicillin resistant medium.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Resistance $n$, each of $r$ ohm, when connected in parallel give an equivalent resistance $R \mathrm{~ohm}$. If these resistances were connected in series, the combination would have a resistance in ohms, equal to :

Options:
A $n^2 R$
B $\frac{R}{n^2}$
C $\frac{R}{n}$
D $n R$

The correct answer is:
$n^2 R$

Solution:
According the question:
$\begin{aligned}
R & =\frac{r}{n} \\
\Rightarrow \quad r & =n R \\
\therefore \quad R \text { series } & =n r=n^2 R
\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2004
Resistance of a carbon resistor determined from colour codes is (22000±5%) Ω. The colour of the third band must be :

Options:
A Yellow
B Red
C Green
D Orange

The correct answer is:
Orange

Solution:
Resistance using the colour codes is written as
R=[(numbers)×(multiplier) ±(tolerance(%))].

Colour 1: First significant figure
Colour 2: Second significant figure
Colour 3: Decimal multiplier
Colour 4(or no colour): Tolerance

For the given case:

Resistance can be written as =22×103±5% Ω.

Colour of the third band corresponds to decimal multiplier which is 103, hence colour is Orange.

Topic: Physics, Current Electricity, NEET, NEET 2023 (All India)
Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity.

Options:
A 24.0 S/m ;10-4 S / m
B 124.0 S/m; 124 × 10-4 S m2
C 1.24 S/m; 100× 10-4 S m2 mol-1
D 0.248 S m-1 ; 124 × 10-4 S m2 mol-1

The correct answer is:
0.248 S m-1 ; 124 × 10-4 S m2 mol-1

Solution:
The cell constant is given by the equation: Cell constant = G* = conductivity × resistance = 1.29 S/m × 100 Ω = 129 m-1 = 1.29 cm-1. Conductivity of 0.02 mol L-1 KCl solution = cell constant / resistance= G* R = 129 m-1 / 520 Ω= 0.248 S m-1 ; Molar conductivity = Λm = 𝜅/C = 248 ×10-3 S m-1 / 20 mol m-3 = 124 × 10-4 S m2 mol-1.

Topic: Chemistry, Electrochemistry, NEET
Resistance of a wire at 20âÂ°C is 20 Î© and at 500âÂ°C is 60 Î©. At what temperature its resistance is 25 Î©?

Options:
A 160°C
B 250°C
C 100°C
D 80°C

The correct answer is:
80°C

Solution:
Here, R20=20, R500=60Î©,Rt=25Î©,
âµ Rt=R01+Î±t (where R 0 = Resistance at 0 o C )
Where Î± is the temperature coefficient of resistance.
â´ R20=R01+Î±Ã20
or 20=R01+Î±Ã20 ..i
R500=R01+Î±Ã500
or 60=R01+Î±Ã500 â¦ii
Dividing Eq. (ii) by Eq. (i), we get
6020=1+Î±Ã5001+Î±Ã20
or 3+60Î±=1+500Î±
or Î±=2440 = 1 220 . o C −1
Again, R20=R01+Î±Ã20
or 20=R01+1220Ã20â¦(iii)
Rt=R01+Î±t
25=R01+1220Ãt ..iv
Dividing Eq. (iv) by Eq (iii), we get
2520=1+1220Ãt1+1220Ã20â¹4+4t220=5+100220
or t=80Â°C

Topic: Physics, Current Electricity, NEET
Resistance to jassids in cotton and cereal leaf beetle in wheat is due to:

Options:
A Solid stem
B Hairy leaves
C Nectarless nature
D Both A and B

The correct answer is:
Hairy leaves

Solution:
Insect and pest infestation are one of the major causes of large-scale destruction of crop plants and its produce. Crops plants show insect resistance which can be due to morphological, biochemical or physiological characteristics. Hairy leaves in several plants are associated with resistance to insect pests, e.g, resistance to jassids in cotton and cereal leaf beetle in wheat. In wheat, solid stems lead to non preference by the stem sawfly and smooth leaved and nectar-less cotton varieties do not attract bollworms.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Resistances 1 Î©, 2 Î© and 3 Î© are connected to form a triangle. If a 1.5 V cell of negligible internal resistance is connected across the 3 Î© resistor, the current flowing through this resistor will be

Options:
A 0.25 A
B 0.5 A
C 1.0 A
D 1.5 A

The correct answer is:
0.5 A

Solution:
Equivalent resistance between A and B = series combination of 1 Ω and 2 Ω in parallel with 3 Ω resistor.
1R=13+13=23 or R=1.5 Ω.
∴ Current in the circuits ,I = VR= 1.5 1.5= 1 A
Since the resistance in arm ACB = resistance in arm AB=3 Ω, the current divided equally in the two arms. Hence the current through the 3 Ω resistor = I2=0.5 A.

Topic: Physics, Current Electricity, NEET
Resistors of resistance 20 Ω and 30 Ω are joined in series with a battery of emf 3 V. It is desired to measure current and voltage across the 20 Ω resistor with the help of an ammeter and a voltmeter. Identify the correct arrangement of ammeter (A) and voltmeter (V) out of four possible arrangements shown in the figure given below.

Options:
A
B
C
D

The correct answer is:

Solution:
As ammeter must be connected in series of 20 Ω resistor; and the voltmeter in parallel to 20 Ω resistor, the correct arrangement is as shown in figure (3).

Topic: Physics, Current Electricity, NEET
Resistors P and Q are connected in the gaps of the meter bridge. The balancing point is obtained 13m from the zero end. If a 6 Î© resistance is connected in series with P; the balance point shifts to 23m from the same end. P and Q are

Options:
A 4, 2
B 2, 4
C Both (a) and (b)
D Neither (a) nor (b)

The correct answer is:
2, 4

Solution:
â P:Q=1:2
P=k, Q=2k
P+6Q=231-23
â P+6Q=21
k+62k=21
â k+6=4k
â k=2
â´ P=2 Î©, Q=4 Î©

Topic: Physics, Current Electricity, NEET
Resolving power of telescope increases when

Options:
A wavelength of light decreases.
B wavelength of light increases.
C focal length of eye-piece increases.
D focal length of eye-piece decreases.

The correct answer is:
wavelength of light decreases.

Solution:
Resolving power of telescope is given by
P=d1.22λ
When λ decrease, resolving power increases and vice versa

Topic: Physics, Wave Optics, NEET
Resonance structures of CH2=CH–CHO are given below. Indicate relative stability of the contributing structures.

Options:
A Stability: I > II > III
B Stability: II > I >III
C Stability: I > III > II
D Stability: III > II > I

The correct answer is:
Stability: I > II > III

Solution:
Correct Option is : (A)
Stability: I > II > III

Topic: Chemistry, General Organic Chemistry, NEET
Resonance structures of propenal are given below. Which of these resonating structures is more stable?

Options:
A II > I
B I > II
C I = II
D None of the above

The correct answer is:
I > II

Solution:
Correct Option is : (B)
I > II

Topic: Chemistry, General Organic Chemistry, NEET
Resource partitioning in an environment with limited resources and interbreeding species leads to

Options:
A convergent evolution.
B divergent evolution.
C parallel evolution.
D non-stabilizing evolution.

The correct answer is:
divergent evolution.

Solution:
The occurrence of interspecific competition in nature, they point out that the species facing competition might evolve mechanisms that promote co-existence rather than exclusion. One such mechanism is âresource partitioningâ. If two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns. One classical example is Darwin's Finches. Darwin's finches are small birds which were first observed by Darwin at Galapagos Islands during his voyage in his ship H.M.S.Beagle. Darwin noticed that all the islands had these small finches which were very similar except for their diverse beak shapes. This was surprising for him and so Darwin observed these birds and thus came to the conclusion that beak shapes were different because of their different feeding habits and the habitat. Thus the finches got adapted to the new habitat and habits. This is called adaptive radiation. This was also an example of divergent evolution as according to Darwin these finches being similar had a common ancestor but due to habitat changes got adapted separately. Hence these finches are examples of adaptive and divergent evolution.

Topic: Biology, Evolution, NEET