datavorous/entrance-exam-dataset huggingface dataset:
In ornithine cycle, which of the following wastes are removed from the blood?

Options:
A $\mathrm{CO}_2$ and urea
B ammonia and urea
C $\mathrm{CO}_2$ and ammonia
D urea and urine.

The correct answer is:
$\mathrm{CO}_2$ and ammonia

Solution:
$\mathrm{CO}_2$ and ammonia are the pair of wastes removed from the blood in Ornithine cycle. Urea is formed in Ornithine cycle or urea cycle, consisting of two molecules of ammonia and one molecule of $\mathrm{CO}_2$.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2005
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?

Options:
A 13.60 g
B 10.00 g
C 18.00 g
D 15.00 g

The correct answer is:
15.00 g

Solution:
Correct Option is : (D)
15.00 g

Topic: Chemistry, Redox Reactions, NEET
In our biosphere diversity exists at the level of

Options:
A species only
B macromolecular level
C cellular level
D all of these

The correct answer is:
all of these

Solution:
In our biosphere immense diversity exists not only at the species level but at all levels of biological organization ranging from macromolecules within cells to biomes.

Topic: Biology, Biodiversity and its Conservation, NEET
In our biosphere, the diversity of heterogeneity exists at

Options:
A species level
B genetic level
C ecological level
D all of the above

The correct answer is:
all of the above

Solution:
The biodiversity exists at many levels: genetic, species and ecosystem. Genetic diversity explains the type of chromosomes and genes present in different species and the variation seen in the same species. Species diversity is different numbers of species seen at a particular place. The number of species per unit area is known as species richness. Ecosystem diversity is variations in the different ecosystem at an area and its impact on humans. Thus diversity is seen at all hierarchy levels.

Topic: Biology, Biodiversity and its Conservation, NEET
In our galaxy, which consists of 2.5 × 10^11 stars, each with a mass of one solar mass, how long will it take for a star located 50,000 light-years away from the galactic center to complete one revolution? Given that the diameter of the Milky Way is 10^5 light-years.

Options:
A $3.9~\times ~{{10}^{8}}$years
B $3.55~\times ~{{10}^{8}}$ years
C $3.9~\times ~{{10}^{9}}$years
D $3.55~\times ~{{10}^{9}}$years

The correct answer is:
$3.55~\times ~{{10}^{8}}$ years

Solution:
Correct Option is : (B)
$3.55~\times ~{{10}^{8}}$ years

Topic: Chemistry, Gravitation, NEET
In our society women are blamed for producing female children. Choose the correct answer for the sex-determination in humans.

Options:
A Due to some defect like aspermia in man
B Due to the genetic make up of the egg
C Due to the genetic make up of the particular sperm which fertilizes the egg
D Due to some defect in the women

The correct answer is:
Due to the genetic make up of the egg

Solution:
In human beings both females and males have the same number of chromosomes (23 pairs). Females are homogametic with XX chromosomes (22 pairs + $\mathrm{XX}$ ). They produce ova having $\mathrm{X}$ chromosome. All ova are similar. Males are heterogametic with $\mathrm{X}$ and Y-chromosomes (22 pairs $+\mathrm{XY})$. They produce two kinds of sperms, one type with X-chromosome and other with Y-chromosome. On fertilization, the zygotes may have either the $X X$ or $X Y$. The zygote with $\mathrm{XX}$ becomes the female and the zygote with $X Y$ becomes the male. The sex of the offspring depends on the fertilizing sperm.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2013 (Karnataka)
In oxoacids, how is phosphorus tetrahedrally surrounded by other atoms?

Options:
A With P=O bonds only
B With P–OH bonds only
C With both P=O and P–OH bonds
D With both P=O and P–OH bonds, as well as P–P and P–H bonds

The correct answer is:
With both P=O and P–OH bonds

Solution:
The paragraph states that in oxoacids, phosphorus is tetrahedrally surrounded by other atoms and that all these acids contain at least one P=O bond and one P–OH bond.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
In ozonolysis, what is the product formed when the ozonide is cleaved?

Options:
A Alkane
B Alkene
C Aldehyde or ketone
D Alcohol

The correct answer is:
Aldehyde or ketone

Solution:
The cleavage of the ozonide in ozonolysis typically produces aldehyde or ketone products.

Topic: Chemistry, Hydrocarbons, NEET
In ozonolysis, what type of bond is cleaved in the ozonide?

Options:
A Carbon-carbon double bond
B Carbon-hydrogen bond
C Oxygen-oxygen bond
D Zinc-zinc bond

The correct answer is:
Carbon-carbon double bond

Solution:
In ozonolysis, the carbon-carbon double bond in the ozonide is cleaved.

Topic: Chemistry, Hydrocarbons, NEET
In P4O10 molecule, bridging P-O bond length is

Options:
A Larger than that of in P4O6
B Lesser than that of in P4O6
C Equal to that of in P4O6
D Cannot be compared

The correct answer is:
Lesser than that of in P4O6

Solution:
In P4O10 molecule, bridging P-O bond length is lesser than that of in P4O6

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what will be the ratio of yellow and green seeded plants would you expect in the F1 generation?

Options:
A 9 : 1
B 1 : 3
C 3 : 1
D 50 : 50

The correct answer is:
50 : 50

Solution:
Let YY ⇒ homozygous yellow seed plant.
Yy ⇒ heterozygous green
yy ⇒ homozygous green
According to the question :

Topic: Biology, Principles of Inheritance and Variation, NEET
In pea plants, yellow seeds are dominant to green. If heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in $\mathrm{F}_1$ generation?

Options:
A $9: 1$
B $1: 3$
C $3: 1$
D $50: 50$.

The correct answer is:
$50: 50$.

Solution:

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2007
In Periplaneta americana, on an average number of oothecae produced by females are

Options:
A 14-16
B 11-13
C 12-14
D 9-10

The correct answer is:
9-10

Solution:
On an average, females produce 9-10 oothecae, each containing 14-16 eggs.

Topic: Biology, Structural Organisation in Animals, NEET
In Periplaneta americana, the number of Malpighian tubules are

Options:
A Â 10-20
B 20-50
C 100-150
D 200-300

The correct answer is:
100-150

Solution:
Excretory system of Cockroach
-Cockroach is uricotelic.
-Following are the components of excretory system in the cockroach:
1. Malpighian tubules (most important)
-These extract water and nitrogenous waste from the haemolymph, convert into uric acid and excrete
Intothe hindgut.
2. Fat bodies
3. Nephrocytes
4. Uricose glands.

Topic: Biology, Structural Organisation in Animals, NEET
In Periplaneta americana, the ventral sclerite is called

Options:
A Pleurite
B Clitellum
C Sternite
D Tergite

The correct answer is:
Sternite

Solution:
The entire body is covered by a hard chitinous exoskeleton (brown in colour). In each segment, the exoskeleton has hardened plates called sclerites (tergites dorsally and sternites ventrally) that are joined to each other by a thin and flexible articular membrane (arthrodial membrane).

Topic: Biology, Structural Organisation in Animals, NEET
In petrochemical industry, alcohols are directly converted to gasoline by passing over heated

Options:
A Platinum
B Nickel
C Palladium
D ZSM-5

The correct answer is:
ZSM-5

Solution:
In petrochemical industry, alcohols are directly converted to gasoline by passing over heated ZSM-5(shape selective Zeolite catalysts) through dehydration.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In Phaeophyceae, the gametes are

Options:
A pyriform and bear 2 flagella (one longitudinal and another transverse)
B pear-shaped and bear 2 flagella that are laterally attached
C pyriform and bear 2-8, equal and apical flagella
D pear-shaped and bear 2-8, equal and apical flagella

The correct answer is:
pear-shaped and bear 2 flagella that are laterally attached

Solution:
Phaeophyceae or brown algae are the forms of algae usually found in the marine habitats with a wide range of varieties. The members of this algae, reproduce either vegetatively by fragmentation sexually with the help of gametes, wherein the reproduction may be either oogamous, isogamous, or anisogamous and asexual reproduction by the biflagellate zoospores which is pear-shaped and have two flagella unequal in size and laterally attached.

Topic: Biology, Plant Kingdom, NEET
In Phaeophyceae, the spores (zoospores) are

Options:
A pyriform and bear 2 flagella (one longitudinal and another transverse)
B pear-shaped and bear 2 flagella that are laterally attached
C pyriform and bear 2-8, equal and apical flagella
D pear-shaped and bear 2-8, equal and apical flagella

The correct answer is:
pear-shaped and bear 2 flagella that are laterally attached

Solution:
Phaeophyceae or the brown algae inhabit the marine environment. These organisms show sexual as well as asexual means of reproduction wherein the sexual reproduction may be monogamous, isogamous, or anisogamous whereas the asexual reproduction takes place with the help of bi-flagellate zoospores that are pear-shaped and have two unequal flagella laterally attached to it.

Topic: Biology, Plant Kingdom, NEET
In phenylketonuria, the patient has a deficiency of any enzyme which

Options:
A causes hydroxylation of phenylalanine to form tyrosine
B causes hydroxylation of tyrosine to form phenylalanine
C causes carboxylation of phenylalanine to form tyrosine
D causes carboxylation of tyrosine to form phenylalanine

The correct answer is:
causes hydroxylation of phenylalanine to form tyrosine

Solution:
Phenylketonuria is an autosomal recessive disorder. In this disease, the patient cannot produce enzyme phenylalanine hydroxylase, which helps in conversion of phenylalanine to tyrosine. Thus, there is accumulation of phenylalanine and its derivatives in the body. Accumulation of these in brain results in mental retardation. These are also excreted through urine because of its poor absorption by kidney. The gene coding for phenylalanine hydroxylase is located on chromosome number 12.

Topic: Biology, Principles of Inheritance and Variation, NEET
In photoelectric effect, the photo current

Options:
A increases with increase of frequency of incident photon
B decreases with increase of frequency of incident photon
C does not depend on the frequency of photon but depends only on intensity of incident light
D depends both on intensity and frequency of incident beam

The correct answer is:
does not depend on the frequency of photon but depends only on intensity of incident light

Solution:
When intensity of incident photon increases number of electrons emitted from the surface increases. Thus current increases. It is independent of frequency of photon.

Topic: Physics, Dual Nature of Matter, NEET
In photoelectric effect the slope of straight line graph between stopping potential v0 and frequency of incident light v gives

Options:
A charge on electrons
B work function of emitter
C planck's constant
D ratio of Planck's constant to charge on electron

The correct answer is:
ratio of Planck's constant to charge on electron

Solution:
We know that
E=ϕ+kE
⇒hυ=W+eV0
⇒eV0=hυ-W
V0=heυ+-We
Compare this with equation of line y=mx+c
m= slope of graph =he

Topic: Physics, Dual Nature of Matter, NEET
In photoelectric effect, the threshold wavelength of sodium is 5000 â«. Find its work function(h=6.6Ã10-34 J s, c=3Ã108 m s-1, 1 eV=1.6Ã10-19 J)

Options:
A 7.5 eV
B 2.5 eV
C 10 eV
D 5.0 eV

The correct answer is:
2.5 eV

Solution:
The work function of sodium
W= hc Î» 0
W=6.6Ã10-34Ã3Ã1085Ã10-7
or W=3.96Ã10-19 J
or W=2.5 eV

Topic: Physics, Dual Nature of Matter, NEET
In photoelectric emission process from a metal of work function $1.8 \mathrm{eV}$, the kinetic energy of most energetic electrons is $0.5 \mathrm{eV}$. The corresponding stopping potential is

Options:
A $1.3 \mathrm{~V}$
B $0.5 \mathrm{~V}$
C $2.3 \mathrm{~V}$
D $1.8 \mathrm{~V}$

The correct answer is:
$0.5 \mathrm{~V}$

Solution:
Stoping potential $=$ Maximum KE
$$
\mathrm{eV}=\mathrm{KE}_{\max }
$$
So, option (b) is correct.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Screening)
In photorespiration, the first stable product is

Options:
A oxalo acetate.
B phosphoglycerate.
C phosphoglycerate and phosphoglycolate.
D None of these

The correct answer is:
phosphoglycerate and phosphoglycolate.

Solution:
In photorespiration, the RuBP instead of being converted to 2 molecules of PGA binds with O2 to form one molecule of phosphoglycerate and phosphoglycolate (2 Carbon). In the photorespiratory pathway, there is neither synthesis of sugars, nor of ATP. Rather it results in the release of CO2 with the utilisation of ATP. In the photorespiratory pathway there is no synthesis of ATP or NADPH.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, 12 molecules of water are utilized and 6 molecules of water are released while the formation of one molecule of glucose. The oxygen atom present in glucose came from

Options:
A the fixed carbon dioxide
B the water released
C the water utilised
D Both (A) & (C)

The correct answer is:
the fixed carbon dioxide

Solution:
Martin, Ruben, and Kamen proved that the source of oxygen evolved during photosynthesis is water. They used heavy isotope of oxygen O18 as follows:
This experiment also proves that oxygen present in glucose comes from carbon dioxide, and the water produced during photosynthesis also has oxygen which comes from carbon dioxide.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, energy from light reaction to dark reaction is transferred in the form of:

Options:
A ADP
B ATP
C RUBP
D chlorophyll

The correct answer is:
ATP

Solution:
As a result of light reaction, oxygen, NADPH and ATP are formed. Oxygen is released into the atmosphere while NADPH and ATP are utilised for reduction of $\mathrm{CO}_2$ to carbohydrate in dark reaction.
Related Theory
In the light reactions, light energy is converted into the chemical energy of ATP, while water is broken down for its hydrogen atoms. The ATP and hydrogen atoms are fed to the dark reaction which uses them to produce glucose sugar from carbon dioxide. Ultimately the light energy ends up in the glucose, but it does not get their easily. The light reaction trapped the solar energy and transferred it to the dark reaction via $A T P$ so that the dark reaction had the energy to make glucose. Essentially, the solar energy is being packaged into the chemical energy of glucose.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2002
In photosynthesis, for fixation of one molecule of glucose, the number of ATP and NADPH2 required is

Options:
A 12 and 18
B 18 and 12
C 6 and 12
D 18 and 18

The correct answer is:
18 and 12

Solution:
During photosynthesis one Calvin cycle takes in only one carbon (as CO2) at a time so, it takes six turns of the cycle to produce six-carbon compound (glucose). For the formation of one molecule of hexose sugar (glucose) 18 ATP and 12 NADPH2 are used, whereas 30 ATP and 12 NADPH2 are used during the C4 cycle.
These ATP and NADPH2 (assimilatory power) are produced during the light reaction of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, light reaction includes:

Options:
A Cyclic photophosphorylation
B Non-cyclic photophosphorylation
C Both a and b
D None of these

The correct answer is:
Both a and b

Solution:
Photophosphorylation is the synthesis of ATP from ADP and inorganic phosphate in the presence of light. When two photosystems work in a series, first PS II and then PS I, a process called noncyclic photophosphorylation occurs. When only PS I is functional, the electron is circulated within the photosystem and phosphorylation occurs due to cyclic flow of electrons, it is cyclic photophosphorylation.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, the first step is

Options:
A Conversion of PG to PGA
B Carboxylation of RUBP
C Electron release by chlorophyll a
D Photolysis of water.

The correct answer is:
Electron release by chlorophyll a

Solution:
The light reaction begins with the absorption of the sunlight and it causes an electron of reaction center to be in an excited state. Since reaction center is a specialized chlorophyll-a molecule therefore first step in the light reaction is excitation and release of an electron from chlorophyll-a.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In photosynthesis, the light-independent reactions take place in

Options:
A Photosystem I
B Photosystem II
C Stromal matrix
D Both (A) and (B)

The correct answer is:
Stromal matrix

Solution:
Light-independent reactions or dark reactions occur in the stroma/ stromal matrix. During these reactions carbon dioxide is reduced to carbohydrates.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2015 (Phase 2)
In photosynthesis, the light-independent reactions take place in

Options:
A Photosystem I
B Photosystem II
C Stromal matrix
D Both (A) and (B)

The correct answer is:
Stromal matrix

Solution:
Light-independent reactions or dark reactions occur in the stroma/ stromal matrix. During these reactions carbon dioxide is reduced to carbohydrates.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In pigs, white coat (W) is dominant to black (w). Two white pigs are bred to produce nine white and three black pigs. What are the genotypes of the parents?

Options:
A WW×WW
B WW×Ww
C Ww×Ww
D ww×ww

The correct answer is:
Ww×Ww

Solution:
In the pig, white coat is dominant over black coat, thus, the white coat will express in heterozygous condition. When the hybrid white is selfed-
Ww× Ww

W	w
W	WW	Ww
w	Ww	ww

The phenotypic ratio is 3:1
Thus for, 9 whites and 3 blacks, the ratio is 3:1 and the genotype will be Ww× Ww

Topic: Biology, Principles of Inheritance and Variation, NEET
In Pinus, gametophytic generation is represented by

Options:
A Microspores
B Megaspore/Macrospores
C Male and female cones
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
The gymnosperms are heterosporous; they produce haploid microspores and megaspores. The two kinds of spores are produced within sporangia that are borne on sporophylls which are arranged spirally along an axis to form lax or compact strobili or cones. The strobili bearing microsporophylls and microsporangia are called microsporangiate or male strobili. The microspores develop into a male gametophytic generation which is highly reduced and is confined to only a limited number of cells. This reduced gametophyte is called a pollen grain. The development of pollen grains take place within the microsporangia. The cones bearing megasporophylls with ovules or megasporangia are called macrosporangiate or female strobili. The male or female cones or strobili may be
borne on the same tree like in Pinus.

Topic: Biology, Plant Kingdom, NEET
In Pinus, male and female cones occur on

Options:
A different plants
B same branch of the same plant
C different branches of the same plant
D different branches of different plants

The correct answer is:
different branches of the same plant

Solution:
Pinus is a gymnosperm which produces cones bearing the male and the female reproductive gametes. The male cones are smaller and are found on the lower branches of the tree and the female cones larger than the male cones are on the upper branches of the same tree. Therefore the cones occur on the different branches of the same plant.

Topic: Biology, Plant Kingdom, NEET
In plant nutrition, elements are classified as major or minor depending on

Options:
A Their availability in the soil
B Their relative production in the ash obtained after burning the plants
C The relative amounts required by the plants
D Their relative importance in plant growth

The correct answer is:
The relative amounts required by the plants

Solution:
Major elements are required by plants in the quantity of more than 1 milligram/gram of dry matter, and minor elements are required in the quantity of less than 1 milligram/gram of dry matter.

Topic: Biology, Mineral Nutrition, NEET
In plants, a low concentration of N, S and Mo causes

Options:
A inhibition of cell division
B delay in flowering
C death of tissue
D both A and B

The correct answer is:
both A and B

Solution:
Low concentration of these minerals causes both, inhibition of cell division and delay in flowering.

Topic: Biology, Mineral Nutrition, NEET
In plants meiosis occurs in

Options:
A Anther
B Root tip
C Cambium
D Pollen grain

The correct answer is:
Anther

Solution:
The process of the formation and differentiation of microspores (pollen grains) from microspore mother cells (MMC) by reductional division is called microsporogenesis. Meiosis occurs in pollen mother cells of anther.

Topic: Biology, Reproduction in Organisms, NEET
In plants, most diffusible sugar from source to sink is

Options:
A Glucose
B Sucrose
C Starch
D Fructose

The correct answer is:
Sucrose

Solution:
Sucrose is the form of sugar transported through phloem during translocation. Since it is soluble and less reactive, it is an ideal form of sugar to transport.

Topic: Biology, Transport in Plants, NEET
In plants,

potential in the water column in the xylem plays a major role in water transport up a stem.

Options:
A Positive pressure
B Negative pressure
C Positive water
D Negative water

The correct answer is:
Negative pressure

Solution:
Pressure potential is usually positive, though in plants negative potential or tension in the water column in the xylem plays a major role in water transport up a stem.

Topic: Biology, Transport in Plants, NEET
In plants, the process of gamete formation involves

Options:
A mitosis
B meiosis
C both mitosis and meiosis
D no cell division

The correct answer is:
both mitosis and meiosis

Solution:
In angiosperms, pollen grains are formed in the pollen sacs by meiosis. These pollen grains undergo two mitotic divisions to form two male gametes. In the ovule, megaspore is formed by meiosis which undergoes three mitosis to form an embryo sac which has one male gamete.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In plants, which of the following disease is caused by bacteria?

Options:
A Mosaic disease of tobacco
B Citrus canker
C Late blight of potato
D Potato spindle tuber disease

The correct answer is:
Citrus canker

Solution:
In plants, several diseases can be produced by different types of pathogens like viruses, fungus, mycoplasma, bacteria, viroids, etc. Citrus canker is a bacterial disease which is caused by bacteria Xanthomonas citri in members of the citrus family (lemon family). The mosaic disease of tobacco is caused by Tobacco Mosaic Virus (TMV). Late blight of potato disease is caused by a fungus named Phytophthora infestans. Potato spindle tuber disease is caused by viroids (infectious low molecular weight RNA).

Topic: Biology, Strategies for Enhancement in Food Production, NEET
In Plasmodium, cyst around the ookinete is

Options:
A partly secreted by the intestinal wall and partly by salivary gland.
B completely secreted by the stomach wall.
C completely secreted by the ookinete itself.
D partly secreted by the stomach wall and partly by the ookinete itself.

The correct answer is:
partly secreted by the stomach wall and partly by the ookinete itself.

Solution:
The encysted zygote is called oocyst (ookinete). The cyst wall of the oocyst is partly secreted by the zygote and partly by the stomach wall of mosquito.

Topic: Biology, Human Health and Diseases, NEET