datavorous/entrance-exam-dataset huggingface dataset:
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:
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 homozygous flowers of four o'clock plant with pink flowers of four o'clock plant, the following genotypes and phenotypes are obtained:

Parent	Red (RR) X	Pink (Rr)

Gametes	R	R,r

F1 generation	RR	Rr

Phenotypes: Red (50%) and Pink (50%)
Genotypes: RR (50%) and Rr (50%)

Topic: Biology, Principles of Inheritance and Variation, NEET
In fractional distillation, which component condenses first in the fractionating column?

Options:
A The component with the higher boiling point
B The component with the lower boiling point
C Both components condense simultaneously
D None of the components condense

The correct answer is:
The component with the higher boiling point

Solution:
In fractional distillation, the component with the higher boiling point condenses first in the fractionating column.

Topic: Chemistry, General Organic Chemistry, NEET
In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 â« then the distance between the first minimum on either side of the central maximum is (Î¸ is small and measured in radian)

Options:
A 10-1 m
B 10-2 m
C 2Ã10-2 m
D 2Ã10-1m

The correct answer is:
10-2 m

Solution:
a=0.2Ã10-3 m, D=2 m
Î»=5Ã10-7 m
x=Î»Da =5Ã10-7Ã20.2Ã10-3
â´ x=5Ã10-710-4=5Ã10-7+4 m
â´ x=5Ã10-3 m
Distance between 1st minima on either side
=5Ã10-3+5Ã10-3
=10Ã10-3=10-2 m

Topic: Physics, Wave Optics, NEET
In Freundlich adsorption isotherm, adsorption is proportional to pressure P as

Options:
A P0
B P
C Pn
D P1/n

The correct answer is:
P1/n

Solution:
Freundlich adsorption isotherm is mathematically expressed as, xm=Kp1/n
It is also written as logx/m=logk+1nlog p .

Topic: Chemistry, Surface Chemistry, NEET
In Freundlich adsorption isotherm, the value of $1 / n$ is

Options:
A between 0 and 1 in all cases
B between 2 and 4 in all cases
C 1 in case of physical adsorption
D 1 in case of chemisorption

The correct answer is:
between 0 and 1 in all cases

Solution:
In Freundlich adsorption isotherm,
$\frac{x}{m}=k p^{1 / n}$
the value of $n$ is always greater than 1. So, the value of $1 / n$ lies between 0 and 1 in all cases.

Topic: Chemistry, Surface Chemistry, NEET, NEET 2012 (Screening)
In front of the lens, the aperture surrounded by the iris is called the ________.

Options:
A pupil
B retina
C ciliary body
D ganglion

The correct answer is:
pupil

Solution:
In front of the lens, the aperture surrounded by the iris is called the pupil. The diameter of the pupil is regulated by the muscle fibres of iris.

Topic: Biology, Neural Control and Coordination, NEET
In Fucus, the male and female gametes are

Options:
A motile
B non-motile
C motile and non-motile respectively
D non-motile and motile respectively

The correct answer is:
motile and non-motile respectively

Solution:
Fucus is a brown alga and it belongs to the division Phaeophyta. The thallus is flattened and small cavities called cryptostomates are scattered on the surface of the thallus. These cryptostomates have sterile hairs that help in the uptake of nutrients from the seawater. Fucus plant is a dioecious plant in which male and female plants are separate. Male and female gametes are formed in the conceptacle. Female gametes are denser than male gametes so they sink to the bottom and are attracted by sperm cells that will be motile.

Topic: Biology, Plant Kingdom, NEET
In full-wave rectification, what is the output frequency if the input frequency is 50 Hz?

Options:
A 25 Hz
B 50 Hz
C 75 Hz
D 100 Hz

The correct answer is:
100 Hz

Solution:
Correct Option is : (D)
100 Hz

Topic: Chemistry, Semiconductors, NEET
In Funaria, meiosis occurs in

Options:
A protonema
B prothallus
C spore mother cells
D spore

The correct answer is:
spore mother cells

Solution:
Funaria belongs to Mosses under the division Bryophyta (amphibians of the plant kingdom). Funaria shows a sexual and asexual reproductive mode in their life-cycle. The haploid gametophyte alternates with the diploid saprophyte in the sexual lifecycle which is halodiplontic. The spore develops into mature gametophyte bearing archegonia and antheridia. Antheridia bears male gametes that swim to the female egg and fuse to form sporophyte. Foot, setae, and capsule form parts of the sporophyte. Haploid spores are formed in the capsule by the meiosis division process.

Topic: Biology, Plant Kingdom, NEET
In Funaria, the number of peristomial teeth is

Options:
A 8
B 16
C 32
D 64

The correct answer is:
32

Solution:
The sporogonium of Funaria is composed of the apophysis, theca and operculum. The peristome lies immediately underneath the operculum. It consists of two sets of long conical teeth. There are sixteen teeth in each set (the total being 32). The teeth of both sets are on the same radii and thus opposite to each other. The peristome teeth close the opening of the spore sac. The teeth of the outer set respond to changes in the humidity and thus are said to be hygroscopic. The inner peristome teeth do not show hygroscopic movement.

Topic: Biology, Plant Kingdom, NEET
In Funaria, which of the following represents the juvenile state

Options:
A Prothallus
B Protonema
C Capsule
D All

The correct answer is:
Protonema

Solution:
The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages, the first stage is protonema stage, which develops directly from a spore. It is creeping, green, branched and frequently filamentous stage. The second stage is the leafy stage, which develops from the second protonema as a lateral bud.

Topic: Biology, Plant Kingdom, NEET
In garden pea, starch synthesis is regulated by B and b alleles. Bb genotype produces:

Options:
A Large sized round seeds
B Small sized wrinkled seeds.
C Intermediate sized round seeds
D Large sized wrinkled seeds

The correct answer is:
Intermediate sized round seeds

Solution:
A single gene product may produce more than one effect.
For example, starch synthesis in pea seeds is controlled by one gene. It has two alleles (B and b). Starch is synthesised effectively by BB homozygotes and therefore, large starch grains are produced. In contrast, bb homozygotes have lesser efficiency in starch synthesis and produce smaller starch grains. After maturation of the seeds, BB seeds are round and the bb seeds are wrinkled. Heterozygotes produce round seeds, and so B seems to be the dominant allele. But, the starch grains produced are of intermediate size in Bb seeds. So if starch grain size is considered as the phenotype, then from this angle, the alleles show incomplete dominance.

Topic: Biology, Principles of Inheritance and Variation, NEET
In gel electrophoresis, separated DNA fragments can be visualized with the help of

Options:
A Ethidium bromide in UV radiation
B Acetocarmine in UV radiation
C Ethidium bromide in infrared radiation
D Acetocarmine in bright blue light

The correct answer is:
Ethidium bromide in UV radiation

Solution:
The DNA fragments separated by gel electrophoresis can be visualized only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation. Pure DNA fragments in the visible light and without staining cannot be observed. DNA when stained with ethidium bromide appears bright orange bands under UV radiations.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2020 (Phase 1)
In gel electrophoresis, the differential mobility of DNA depends on

Options:
A helical nature of DNA
B double-stranded nature of DNA
C size of DNA
D hydrogen bonding between bases

The correct answer is:
size of DNA

Solution:
DNA fragements are negatively charged molecules they can be separated by forcing them to move towards the anode under an electric field through a medium or a matrix. The DNA fragments separates according to their size through the sieving effect provided by the agarose gel.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In gene gun method used to introduce alien DNA into host cells, microparticles of ____ metals are used.

Options:
A Silver
B Copper
C Zinc
D Tungsten or gold

The correct answer is:
Tungsten or gold

Solution:
Micro-injection is a method where the recombinant DNA is directly injected into the nucleus of an animal cell. In another method, suitable for plants, cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA in a method known as biolistics or gene gun.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2023 (All India)
In gene therapy of adenosine deaminase (ADA) deficiency, the patient requires a periodic infusion of genetically engineered lymphocytes because:

Options:
A Gene isolated from marrow cells producing ADA is introduced into cells at embryonic stages
B Lymphocytes from patient's blood are grown in culture, outside the body.
C Genetically engineered lymphocytes are not immortal cells.
D Retroviral vector is introduced into these lymphocytes.

The correct answer is:
Genetically engineered lymphocytes are not immortal cells.

Solution:
In gene therapy of the ADA, we transform the lymphocytes with the functional copy of the ADA. This process is done inside the lab in the cell culture set up.

These cells are then injected back in the patient. However, these cells are not immortal and will die after their life span is over.

Thus, the patient will need periodic infusion of the genetically engineered lymphocytes.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2022 (Phase 1)
In general the egg apparatus of embryo sac in angiosperm consists of :

Options:
A One egg cell, two synergids, three antipodal cells, two Polar nuclei
B One egg cell, two synergids, two antipodal cells, three Polar nuclei
C One egg cell, three synergids, two antipodal cells, two Polar nuclei
D One egg cell, two synergids, two antipodal cells, two Polar nuclei

The correct answer is:
One egg cell, two synergids, three antipodal cells, two Polar nuclei

Solution:
A typical embryo sac is 8 nucleate, 7 celled structure formed in most angiosperms.
It has a 3 celled egg apparatus in which one egg cell and two synergids are present.
Note: The answer has been provided with respect to the embryo sac, not egg apparatus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2022 (Phase 2)
In general, viruses that infect plants have .............

Options:
A dsDNA
B dsRNA
C ssDNA
D ssRNA

The correct answer is:
ssRNA

Solution:
In general, viruses that infect plants have single-stranded RNA, and viruses that infect animals have either single or double-stranded RNA or double-stranded DNA.

Topic: Biology, Biological Classification, NEET
In general, what type of H-A bonds are associated with stronger acids?

Options:
A Short and polar bonds
B Long and nonpolar bonds
C Short and nonpolar bonds
D Long and polar bonds

The correct answer is:
Short and polar bonds

Solution:
In general, stronger acids are associated with short and polar H-A bonds.

Topic: Chemistry, Ionic Equilibrium, NEET
In general zoophage or the virus that infect animals does not have

Options:
A ssRNA
B ssDNA
C dsRNA
D dsDNA

The correct answer is:
ssDNA

Solution:
Viruses that infect animal cells can have double-stranded DNA, double-stranded RNA, or single-stranded RNA as genetic material. They do not have single-stranded DNA as genetic material. ssDNA is found in some bacteriophages like M-13 virus and Phi x 174 viruses.

Topic: Biology, Biological Classification, NEET
In genetic engineering, after the creation of recombinant DNA, it has to be transferred into the host using a vector. Which of the following agents can be used as a vector?

A. pBR322
B. E.coli
C. Transposons
D. Mosquito
E. A disarmed Ti plasmid

Options:
A A, C, and E
B A, B, and E
C A, B, and C
D A, B, C, D, and E

The correct answer is:
A, C, and E

Solution:
Plasmids and bacteriophages are commonly used vectors. The vector should have a few (or) at least one unique recognition site to link the foreign/alien DNA. Transposons are also frequently used as cloning vectors (the bacteriophage lambda is important in molecular biology because it is used in constructing vectors for gene cloning. Plasmids do this as well, but bacteriophages have an advantage over plasmids. Plasmids are circular, independently replicating, double-stranded DNA, most often found in bacteria. They replicate quickly and are easily manipulated in the laboratory. Plasmids are typically 2-10 thousand base pairs in size (Corbley, 1999). While this small size allows plasmids the two aforementioned attributes, it also means that plasmids are limited in the DNA fragments they can clone. They are typically limited to fragments around 5 thousand base pairs (King, 2002). While plasmids are great vector vehicles for many molecular endeavours, if the fragment to be cloned is larger than the available insertion space of the plasmid, the plasmid vector will probably not transfer the fragment of DNA successfully). Disarmed retroviruses are used as vectors in animals.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In geometrical isomerism, what is the specific feature that leads to isomerism?

Options:
A Different molecular formulas
B Different sequences of covalent bonds
C Different spatial arrangements around a double bond
D Different functional groups

The correct answer is:
Different spatial arrangements around a double bond

Solution:
Geometrical isomerism arises due to different spatial arrangements around a double bond in a molecule.

Topic: Chemistry, General Organic Chemistry, NEET
In germinating seeds fatty acids are degrad exclusively in the

Options:
A proplastids
B glyoxisomes
C peroxisomes
D mitochondria

The correct answer is:
glyoxisomes

Solution:
Glyoxysomes are found to occur in the cells of yeast, Neurospora and oil rich seeds of many higher plants. During germination of oily seeds, the stored lipid molecules of sphaerosomes are hydrolysed by the enzyme lipase to glycerol and fatty acids. The long chain fatty acids are then broken down by successive removal of two carbon fragments in the process of $\beta$-oxidation.
Peroxisomes are present in all photosynthetic cells of higher plants in etiolated leaf tissue. It is the site of hydrogen peroxide metabolism and glycolate cycle.
Mitochondria is the site of aerobic respiration in eukaryotic cell. It is called power house of the cell.
During the development of chloroplast the first structure to appear is so called proplastid, which has a double membrane.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2008 (Screening)
In ginger, vegetative propagation occurs through:

Options:
A Rhizome
B Offsets
C Bulbils
D Runners

The correct answer is:
Rhizome

Solution:
Rhizome: It is a thick stem growing horizontally, beneath the soil surface. It has distinct nodes and internodes. The nodes bear small scale leaves with buds in their axils. These buds develop into aerial shoot during favourable conditions. The tip of the rhizome ends in a terminal bud. The lower surfaces of the nodes give out small slender adventitious roots. E.g. ginger, turmeric, etc.

Topic: Biology, Reproduction in Organisms, NEET, NEET 2015 (Phase 1)
In given reaction [X] and [Y] respectively are

Options:
A
B
C
D

The correct answer is:

Solution:
In Bouvealt blanc reduction ester reduces to alcohols
R'COOR→Na/EtOHR'CH2OH+ROH

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
In global biodiversity of plants, the

Options:
A angiosperms represent the maximum number of species.
B algae represent the minimum number of species.
C fungi represent the minimum number of species.
D mosses represent the maximum number of species.

The correct answer is:
angiosperms represent the maximum number of species.

Solution:
Angiosperm, any of about 300,000 species of flowering plants, the largest and most diverse group within the kingdom Plantae. Angiosperms represent approximately 80 percent of all the known green plants now living.

Topic: Biology, Biodiversity and its Conservation, NEET
In glucose pentaacetate, how are the five –OH groups attached to carbon atoms?

Options:
A They are attached to the same carbon atom.
B They are attached to two different carbon atoms.
C They are attached to three different carbon atoms.
D They are attached to five different carbon atoms.

The correct answer is:
They are attached to five different carbon atoms.

Solution:
Since glucose pentaacetate is a stable compound, the five –OH groups should be attached to different carbon atoms.

Topic: Chemistry, Biomolecules, NEET
In Glycine max, the product of biological nitrogen fixation is transported from the root nodules to other parts as:

Options:
A Ammonia
B Glutamate
C Nitrates
D Ureides

The correct answer is:
Ureides

Solution:
Along with the transpiration stream the nodules of some plants like glycine max or soyabean export the fixed nitrogen in the form of ureides. These compounds have a high nitrogen to carbon ratio.

Topic: Biology, Mineral Nutrition, NEET, NEET 2020 (Phase 2)
In glycogen, the right end of the polysaccharide chain is called

Options:
A Non-reducing end
B C-terminal
C Reducing end
D N-terminal

The correct answer is:
Reducing end

Solution:
In glycogen, the right end is called the reducing end, and the left end is called the non-reducing end. It has branches as shown in the form of a cartoon

Topic: Chemistry, Biomolecules, NEET
In glycolysis, ATP is used in which of the following reactions?

Options:
A Conversion of fructose 6-phosphate to fructose 1, 6-bisphosphate
B Conversion of 3-phosphoglyceraldehyde is converted to 1, 3-bisphosphoglycerate
C Conversion of glucose into glucose 6-phosphate
D Both (A) and (C)

The correct answer is:
Both (A) and (C)

Solution:
ATP is utilised at two steps: first in the conversion of glucose into glucose 6-phosphate and second in the conversion of fructose 6-phosphate to fructose 1, 6-bisphosphate.
One step where NADH + H+ is formed from NAD+; this is when 3-phosphoglyceraldehyde (PGAL) is converted to 1, 3-bisphosphoglycerate (BPGA).

Topic: Biology, Respiration in Plants, NEET
In glycolysis, during oxidation electrons are removed by:

Options:
A ATP
B glyceraldehyde-3-phosphate
C $\mathrm{NAD}^{+}$
D molecular oxygen.

The correct answer is:
$\mathrm{NAD}^{+}$

Solution:
When 3-phosphoglyceraldehyde is converted into 1,3 diphosphoglyceric acid, two electrons and two protons are released which are utilised to convert $\mathrm{NAD}^{+}$to $\mathrm{NADH}$.
$$
\mathrm{NAD}^{+}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{NADH}+\mathrm{H}^{+}
$$
Related Theory
Glycolysis is a partial breakdown of a sixcarbon glucose molecule into two, threecarbon molecules of pyruvate, $2 \mathrm{NADH}, 2 \mathrm{H}^{+}$, and 2 net ATP as a result of substrate-level phosphorylation. Glycolysis occurs in the cytoplasm of the cell.

Topic: Biology, Respiration in Plants, NEET, NEET 2004
In grasses, bulliform cells are

Options:
A large bubble-shaped epidermal cells, present on the lower surface of leaves
B large bubble-shaped epidermal cells, present on the upper surface of leaves
C small pyramidal epidermal cells, present on the upper surface of leaves
D small pyramidal epidermal cells, present on the lower surface of leaves

The correct answer is:
large bubble-shaped epidermal cells, present on the upper surface of leaves

Solution:
Bulliform cells spread around the leaf on the upper surface play a role for water storage. These cells assist during transpiration process to reduce water loss. During stress, these cells lose water causing the leaves to curl and prevent damage. They are commonly found in the grasses.

Topic: Biology, Anatomy of Flowering Plants, NEET
In grasses, certain adaxial epidermal cells along the veins modify themselves into large empty colourless cells. These cells are called

Options:
A Subsidiary cells
B Bulliform cells
C Cork cells
D Guard cells

The correct answer is:
Bulliform cells

Solution:
In grasses and some other monocots cetain adaxial cells becomes large and colourless known as bulliform cells. When these cells absorbs water and are turgid, the leaf surface is exposed and when they are flaccid due to water stress, they make the leaves curl inward to minimize water loss.

Topic: Biology, Anatomy of Flowering Plants, NEET
In gravity-free environments, how do drops behave?

Options:
A They become perfectly spherical
B They stay flattened
C They evaporate
D They become dense

The correct answer is:
They become perfectly spherical

Solution:
In gravity-free environments, drops become perfectly spherical.

Topic: Chemistry, States of Matter, NEET
In gravity-free space, a man of mass M standing at a height h above the floor throws a ball of mass m straight down with a speed u. When the ball reaches the floor, the distance of the man above the floor will be

Options:
A h1+mM
B h 2–mM
C 2h
D a function of m, M, h and u

The correct answer is:
h1+mM

Solution:
Due to conservation of momentum in the vertical direction, the man will begin to move up with speed muM. The ball will reach the floor in time hu. In this time the man will move up by muM hu. Total distance of man from floor =h + mhM.

Topic: Physics, Center of Mass Momentum and Collision, NEET
In green plants, during photosynthesis water act as:

Options:
A Hydrogen acceptor
B Oxygen acceptor
C Hydrogen donor
D Electron acceptor

The correct answer is:
Hydrogen donor

Solution:
In green plants, H2O is the hydrogen donor and is oxidised to O2. Some organisms do not release O2 during photosynthesis. When H2S, instead of H2O is the hydrogen donor for purple and green sulphur bacteria, the ‘oxidation’ product is sulphur or sulphate depending on the organism and not O2. Hence, he inferred that the O2 evolved by the green plant comes from H2O, not from carbon dioxide. While the hydrogen donated results in formation of NADPH + H+.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In Griffith's transforming principle, if heat-killed S - strain is injected into mice, then:

Options:
A Transformation of S - strain into R - strain will occur.
B Transformation of R - strain into S - strain will occur.
C Mice will die.
D Mice will live.

The correct answer is:
Mice will live.

Solution:
Frederick Griffith has done experiments on bacteria Streptococcus pneumonia. This bacteria causes pneumonia.

Step 1	S –strain injected into the mice	Mice die
Step 2	R –strain injected into the mice	Mice live
Step 3	Heat killed S –strain injected into the mice	Mice live
Step 4	Heat killed S –strain + R –strain injected into the mice	Mice die

Topic: Biology, Molecular Basis of Inheritance, NEET
In ground to ground projectile motion under gravity, which of the following doesn't affect the time of flight?

Options:
A rotation of earth
B air resistance
C curvature of earth
D all of these

The correct answer is:
rotation of earth

Solution:
Since the time of flight only depends upon the vertical component of projection velocity, rotation of the earth doesn't affect the time of flight

Topic: Physics, Motion In Two Dimensions, NEET