datavorous/entrance-exam-dataset huggingface dataset:
Statement I: Flower is an aggregation of sporophylls.

Statement II: Sporophylls are modified in angiosperms.

Choose the correct answer based on the above statements.

Options:
A Both statement I and statement II are true and statement II is a correct explanation of statement I
B Both statement I and statement II are true but statement II is not a correct explanation of statement I
C Statement I is true but statement II is false
D Both statement I and statement II are false

The correct answer is:
Both statement I and statement II are true but statement II is not a correct explanation of statement I

Solution:
In angiosperm, sporophylls are organised into flowers. Both microsporophylls and megasporophylls are specialised. A microsporophyll or stamen consists of a filament and an anther. A megasporophyll or carpel is rolled and partly sterilised to produce a stigma, style and ovary containing ovules.

Topic: Biology, Plant Kingdom, NEET
Statement I : For adsorption Î G , Î H , Î S all have negative values.

Statement II : Adsorption is an exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate.

Options:
A Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I.
B Statement I is true, Statement II is also true and Statement II is not the correct explanation of Statement I.
C Statement I is true, Statement II is false.
D Statement I is false, Statement II is true.

The correct answer is:
Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I.

Solution:
II is correct explanation of I as adsorption leads to arrangement i.e., decreases randomness â´ Î S = - ve at the same time energy is released due to attraction between adsorbate and adsorbent â´ Î H = - ve .
Adsorption is a spontaneous process as it occurs on its own. For all spontaneous processes, the change in free energy is negative. Or,
â´ Î G = - ve

Topic: Chemistry, Surface Chemistry, NEET
Statement I: In Spirogyra, sometimes the gametes behave directly as zygospores without fusion.
Statement II: In Spirogyra, asexual reproduction is by flagellated zoospores produced in zoosporangia.

Select the correct option regarding both the statements.

Options:
A Both the statements are correct
B Both the statements are wrong
C Statement I is correct and II is wrong
D Statement II is correct and I is wrong

The correct answer is:
Both the statements are correct

Solution:
In Spirogyra, vegetative reproduction usually takes place by fragmentation or by the formation of different types of spores. Asexual reproduction is by flagellated zoospores produced in zoosporangia.
In Spirogyra, sexual reproduction takes place through the fusion of two gametes to form zygospores. These gametes are non-flagellated (non-motile) but similar in size, and sometimes the gametes behave directly as zygospores without fusion.

Topic: Biology, Plant Kingdom, NEET
Statement I: Pregnant women may show some presence of glucose in their postprandial urine although they have no diabetes.

Statement II: In pregnant women, the glomerular filtration rate is slightly increased. As a result, the tubular load of glucose exceeds the tubular maximum for glucose reabsorption.

Options:
A Both the statements are true
B Both the statements are false
C Statement I is true and statement II is false
D Statement I is false and statement II is true

The correct answer is:
Both the statements are true

Solution:
Pregnant women may show some presence of glucose in their postprandial urine although they have no diabetes. This is because in pregnant women, the glomerular filtration rate is slightly increased. As a result, the tubular load of glucose exceeds the tubular maximum for glucose reabsorption.

Topic: Biology, Excretory Products and their Elimination, NEET
Statement I: Test cross is used to determine an unknown genotype within one breeding generation.
Statement II: Test cross is a cross between F1 hybrid and dominant parent.

Options:
A Both statements I and II are correct.
B Statement I is correct but statement II is incorrect.
C Statement I is incorrect but statement II is correct.
D Both statements I and II are incorrect.

The correct answer is:
Statement I is correct but statement II is incorrect.

Solution:
To determine the genotype of a tall plant of the F2 generation, Mendel crossed the tall plant from F2 generation with a dwarf plant. He called this a test cross. In a typical test cross an organism (pea plants) showing a dominant phenotype whose genotype is to be determined is crossed with the recessive parent instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism. Normal test cross ratio for a monohybrid cross is 1:1 and for a dihybrid cross is 1:1:1:1.

Topic: Biology, Principles of Inheritance and Variation, NEET
Statement I: The codon 'AUG' codes for methionine and phenylalanine.
Statement II: 'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below.

Options:
A Both Statement I and Statement II are true
B Both Statement I and Statement II are false
C Statement I is correct but Statement II is false
D Statement I is incorrect but Statement II is true

The correct answer is:
Statement I is incorrect but Statement II is true

Solution:
AUG is the start codon that codes for amino acid methionine. Phenylalanine is coded by codons UUU and UUC. Lysine is coded by codons AAA and AAG.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2021
Statement I: The megaspore mother cell divides mitotically to produce four spores.
Statement II: Megaspore mother cells are diploid, and megaspore is haploid.

Options:
A

Both statements are true and statement II is a correct explanation of statement I.

B

Both statements are true but statement II is not a correct explanation of statement I.

C

Statement I is true but statement II is wrong.

D

Statement I is wrong but statement II is true.

The correct answer is:

Statement I is wrong but statement II is true.

Solution:
Megaspore mother cell is a prominent cell in the nucellus. It divides by meiosis and forms a row of four haploid megaspores.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Statement-1: Biolistics method of gene transfer is an example of direct gene transfer.

Statement-2: In biolistics method, pBR322 is used.

Options:
A Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
B Statement-1 is false, Statement-2 is true.
C Statement-1 is true, Statement-2 is false.
D Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

The correct answer is:
Statement-1 is true, Statement-2 is false.

Solution:
Biolistics is a commonly used method for genetic transformation of plants and other organisms. Millions of DNA-coated metal particles are shot at target cells or tissues using a biolistic device or gene gun.
pBR322 is used as a vector for E. coli transformation.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Statement-1: The innermost layer of microsporangium is tapetum.

Statement-2: Glandular tapetum provides nourishment to developing microspores.

Options:
A Only statement-1 is correct.
B Only statement-2 is correct.
C Both statement 1 & 2 is correct.
D Both statement 1 & 2 is incorrect.

The correct answer is:
Both statement 1 & 2 is correct.

Solution:
The outer three wall layers perform the function of protection and help in dehiscence of anther to release the pollen. The innermost wall layer is the tapetum. It nourishes the developing pollen grains. Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Statement-I: 2A + X0 genotype represents turner syndrome in women.
Statement-II: 2A + X0 genotype is normal male in grasshopper.

Options:
A Only Statement-I is correct.
B Only Statement-II is correct.
C Both Statement-I and Statement-II are correct.
D Both Statement-I and Statement-II are incorrect.

The correct answer is:
Both Statement-I and Statement-II are correct.

Solution:
XO type (XX-XO type): It is observed in large number of insects, grasshopper. Numbers of chromosomes are different in male and female individuals. Here, female shows homogameity (XX) while male shows hetrogameity (XO). It is clear that, all eggs (ova) bear an additional X-chromosome besides the autosomes while only 50% of the sperms bear X-chromosomes. In grasshopper, eggs fertilized by (A+X) type sperm become females while those fertilized by (A+O) type sperm become males. Therefore, sperm determines the sex of the offsprings.
TURNERâs SYNDROME:
Introduction: It was first described in 1938 by Turner. The disorder develops due to monosomy of sex chromosome. The disease is seen only in females. It is an allosomal chromosomal abnormality.
Cause: The disorder is due to monosomy. It appears due to fusion of abnormal egg (22+0) and a normal sperm (22+X) or a normal egg (22+X) and abnormal sperm (22+0).
Karyotype: [44 + XO]

Topic: Biology, Principles of Inheritance and Variation, NEET
Statement-I: Glycolysis is the first step of respiration in which glucose completely breaks into CO2 and H2O

Statement-II: In this process, there is a net gain of twenty-four molecules of ATP.

Options:
A Only statement-I is correct.
B Only statement-II is correct.
C Both statement-I and statement-II are correct.
D Both statement-I and statement-II are incorrect.

The correct answer is:
Both statement-I and statement-II are incorrect.

Solution:
Glycolysis / EMP pathway:

Discovery: It is given by Embden, Meyerhoff, and Parnas in 1930. It is the first stage of the breakdown of glucose in the cell.

Definition: Glycolysis (Gr. glykys= sweet, sugar; lysis= breaking) is a stepped process by which one molecule of glucose (6C) breaks into two molecules of pyruvic acid (3C).

Site of occurrence: Glycolysis takes place in the cytosol and does not utilize oxygen. Thus, it is an anaerobic pathway. In fact, it occurs in both aerobic and anaerobic respiration.

Steps of glycolysis: Glycolysis consists of 10 steps divided into 2 phases preparatory phase and pay off phase. Each step is catalyzed by a specific enzyme. Most of the reactions are reversible.

Preparatory phase: In this phase, 2 molecules of ATP are utilized.

Pay-off phase: In this phase, energy is released in the form of ATP and NADH2 is formed.

Energetics of Glycolysis

Reaction Type	Reaction	Products of reaction	In terms of ATP

ATP Forming Reactions

Substrate-level phosphorylation	2 x 1,3 Biphosphoglycerate + 2ADP âBiphosphoglycerokinase 2 x 3-Phosphoglycerate + 2ATP	2 ATP	2 ATP

Substrate-level phosphorylation	2 x PEP + 2ADP âK+, Mg2+Pyruvate Kinase 2 x Pyruvate + 2 ATP	2 ATP	2 ATP

Dehydrogenation	2 x 3 Phosphoglyceraldehyde + 2 Pi + 2 NAD+ âDehydrogenase 2 x 1,3 Biphosphoglycerate + 2 NADH2	2 NADH2	6 ATP

ATP Consuming Reactions

Phosphorylation-I	Glucose + ATP âMg2+Hexokinase Glucose 6-Phosphate + ADP	-1 ATP	-1 ATP

Phosphorylation-II	Fructose 6-Phosphate + ATP âMg2+Phosphofructokinase Fructose 1,6-Biphosphate	-1 ATP	-1 ATP

Net Gain	8 ATP

Topic: Biology, Respiration in Plants, NEET
Statement-I: Indian Government has set up organisation such as GEAC (Genetic Engineering Approval Committee), which will make decisions regarding the validity of GM research and safety of introducing GM organisms for public services.

Statement-II: Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem.

Options:
A Both statements I and II are correct.
B Statement I is correct but and statement II is incorrect.
C Statement I is incorrect but and statement II is correct.
D Both statement I and II are incorrect.

The correct answer is:
Both statements I and II are correct.

Solution:
Indian Government has set up organisation such as GEAC (Genetic Engineering Approval Committee), which will make decisions regarding the validity of GM research and safety of introducing GM organisms for public services as genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem.

Topic: Biology, Biotechnology and Its Applications, NEET
Statement-I: Organic farming initiates sustainable development.
Statement-II: There is no need to use chemical fertilizers for crops which decreases the problem like biomagnification.

Options:
A Only statement-I is correct.
B Only statement-II is correct.
C Both statement-I and statement-II are correct.
D Both statement-I and statement-II are incorrect.

The correct answer is:
Both statement-I and statement-II are correct.

Solution:
Due to the green revolution, the use of inorganic fertilizers and pesticides has increased manifold for enhancing crop production. Pesticides, herbicides, fungicides, etc., are being increasingly used. These incidentally, are also toxic to non-target organisms, which are important components of the soil ecosystem. Organic farming doesn't facilitate the use of such chemical compounds and thus is helpful in both sustainable development as well as decreasing problems like biomagnification that arises due to the accumulation of such chemicals in the living organism of the aquatic system.

Topic: Biology, Environmental Issues, NEET
Statement-I: Pollen mother cells (PMCs) are the first male gametophytic cells.

Statement-II: Each PMC gives rise to two pollens.

Options:
A Both statements are right
B Both statements are wrong.
C Statement I is right and statement II is wrong
D Statement I is wrong and statement II is right

The correct answer is:
Both statements are wrong.

Solution:
Primary sporogenous cell gives rise to microspore mother cells or pollen mother cells (PMCs). They are sporophytic in nature i.e., diploid. These cells undergo meiosis (reduction division) which gives rise to 4 microspores or pollens and this formation of microspores or pollens is called microspore-genesis. Microspores represent the beginning of the gametophytic phase and they are haploid in nature.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Statement-I: Recombinant DNA technology has been less effective in therapeutic drug production.

Statement-II: Recombinant therapeutics induce unwanted immunological responses.

Options:
A Only statement-I is correct
B Only statement-II is correct
C Both statements are correct
D Both statements are incorrect

The correct answer is:
Both statements are incorrect

Solution:
The recombinant DNA technology has been very effective in therapeutic drug production, removing most of the unwanted immunological responses.

Topic: Biology, Biotechnology and Its Applications, NEET
Statement-I: Sponges do not show any animal nature.
Statement-II: Sponges are sessile with no apparent way of capturing food or eliminating water.

Options:
A

Both statements are correct.

B

Both statements are incorrect.

C

Only statement-I is correct

D Only statement-II is correct

The correct answer is:
Only statement-II is correct

Solution:
The animal nature of sponges was well established on the following grounds: (i) Sponges feed on inwafted solid particles. Their mode of nutrition is truly holozoic.
(ii) Sponge cells are devoid of cellulose cell walls.
(iii) The life cycle of sponges includes swimming ciliated larval stages resembling those of other marine animals. Sponges are sessile and digestion is very simple without any apparent way of capturing food or eliminating wastes.

Topic: Biology, Animal Kingdom, NEET
Statement-I: The honey bee queen copulates only ones in her lifetime.
Statement-II: The honey bee queen can lay fertilized as well as unfertilized eggs.

Options:
A Only statement-I is correct.
B Only statement-II is correct.
C Both statements are correct.
D Both statements are incorrect.

The correct answer is:
Both statements are correct.

Solution:
Honey bee queen copulates only once in her life span (nuptial flight) and stores all the sperms in her seminal receptacles. It depends on her whether she releases sperms while laying eggs or not, so there are both fertilized and unfertilized eggs.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Statement-I: The largely tropical Amazonian rain forest in North America has the greatest biodiversity on Earth.
Statement-II: Amazon forest is home to more vertebrates than invertebrates.

Options:
A Both the statements are true
B Both the statements are false
C Statement I is true and statement II is false
D Statement II is true and statement I is false

The correct answer is:
Both the statements are false

Solution:
The largely tropical Amazonian rain forest in South America has the greatest biodiversity on earth- it is home to more than 40,000 species of plants, 3,000 of fishes, 1,300 of birds, 427 of mammals, 427 of amphibians, 378 of reptiles and of more than 1,25,000 invertebrates. Scientists estimate that in these rain forests there might be at least two million insect species waiting to be discovered and named

Topic: Biology, Biodiversity and its Conservation, NEET
Statements related to human Insulin are given below: Which statement(s) is/are correct about genetically engineered Insulin?

(a) Pro-hormone insulin contain extra stretch of C-peptide.

(b) A-peptide and B-peptide chains of insulin were produced separately in E. coli, extracted and combined by creating disulphide bond between them.

(c) Insulin used for treating Diabetes was extracted from cattle and pigs.

(d) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone.

(e) Some patients develop allergic reactions to the foreign insulin.

Choose the most appropriate answer from the options given below.

Options:
A (b) only
B (c) and (d) only
C (c), (d) and (e) only
D (a), (b) and (d) only

The correct answer is:
(b) only

Solution:
In the year 1983, the American company Eli Lilly prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2022 (Phase 1)
Statins are a competitive inhibitor of:

Options:
A Streptokinase
B Cholesterol synthesis
C Pectinase
D Cellulase

The correct answer is:
Cholesterol synthesis

Solution:
The statins-that lower cholesterol levels in blood and reduce the frequency of heart attacks.statin a cholesterol-lowering agentis HMG-CoA reductase inhibitor. Mevalonate is converted into cholesterol, the enzyme (β -Hydroxy β -methylglutaryl-CoA (HMG-CoA) working on it is inhibited by statin.

Topic: Biology, Microbes in Human Welfare, NEET
Statins lower the blood cholesterol level by _________.

Options:
A allosteric inhibition of enzyme is responsible for the synthesis of cholesterol.
B competitive inhibition of enzyme is responsible for the synthesis of cholesterol.
C irreversible inhibition of enzyme is responsible for the synthesis of cholesterol.
D non-competitive inhibition of enzyme is responsible for the synthesis of cholesterol.

The correct answer is:
competitive inhibition of enzyme is responsible for the synthesis of cholesterol.

Solution:
Statins are produced by yeast Monascus purpurea are commercially used as blood cholesterol-lowering agent. They bind to the enzyme responsible for the synthesis of cholesterol and competitively inhibits the enzyme.

Topic: Biology, Microbes in Human Welfare, NEET
Statment I: National parks are meant for the welfare of the wild life.

Statment II: National parks are controlled by State Governments.

Options:
A Both the statements are wrong
B Both the statements are right
C Statement I is right and statement II is wrong
D Statement II is right and statement I is wrong

The correct answer is:
Both the statements are right

Solution:
A national park is an area which is strictly reserved for the welfare of wild life. No human activities are allowed there. Central governments control them but are administered by state government.

Topic: Biology, Biodiversity and its Conservation, NEET
Statues and monuments in India are affected by acid rain. How?

Options:
A Acid rain corrodes the surfaces of the statues and monuments.
B Acid rain makes the statues and monuments more resistant to weathering.
C Acid rain provides a protective layer on the surfaces of statues and monuments.
D Acid rain has no effect on statues and monuments.

The correct answer is:
Acid rain corrodes the surfaces of the statues and monuments.

Solution:
Correct Option is : (A)
Acid rain corrodes the surfaces of the statues and monuments.

Topic: Chemistry, Environmental Chemistry, NEET
Steam at 100 Â°C is passed into 1.1ââkg of water contained in a calorimeter of water equivalent 0.02ââkg at 15 Â°C , till the temperature of the calorimeter and its contents rises to 80âÂ°C. The mass of steam condensed (in kg) is
(Take latent heat of steam = 540 cal g-1):

Options:
A 0.130
B 0.065
C 0.260
D 0.135

The correct answer is:
0.130

Solution:
Heat release by steam = Heat gained by calorimeter and its contents
1.12Ã80-15=mÃ540+mÃ20
m=0.130 kg

Topic: Physics, Thermal Properties of Matter, NEET
Steam at 100 oC is passed into 20 g of water at 10 oC. When water acquires a temperature of 80oC, the mass of water present will be:
[Take specific heat of water =1 cal g-1 oC-1 and latent heat of steam =540 cal g-1]

Options:
A 24Â g
B 31.5Â g
C 42.5Â g
D 22.5 g

The correct answer is:
22.5 g

Solution:
Given: Specific heat of water Sw=1 cal g-1°C-1, Latent heat of steam Ls=540 cal g-1

Let the mass of steam =m g

So heat lost in change of state from steam to water equal to sum of latent heat and heat use in raise of the temperature of water.

Q1 =mLs +msw∆T

Q1=m×540 +m×1×(100-80)

Q1 =540m +20m =560m

Heat gained by water to change its temperature

Q2=mwSw∆T

Q2=20×1×(80-10)=1400.

Now we know from the principle of calorimetry is that total heat lost by the hot body is equal to total heat gained by cold body i.e it follows law of conservation energy. Hence

Heat lost Q1= Heat gained Q2

560m =1400m =1400560 =2.5 g
Therefore, total mass of water at 80°C

m=(20+2.5)g=22.5 g

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2014
Steam is passed into 54 g of water at 30 Â°C till the temperature of the mixture becomes 90 Â°C. If the latent heat of steam is 536 cal g-1, the mass of the mixture will be nearly

Options:
A 80Â g
B 60 g
C 50Â g
D 24Â g

The correct answer is:
60 g

Solution:
Steam 100Â°CâHeat Lost water 100Â°CâHeat Lost water 90Â°C= [Heat gain by water 30Â°C to 90Â°C]

m536+m1100-90=54190-30
m546=5406
m=5405466â6 g
â´ mass of mixture =54+6=60 g

Topic: Physics, Thermal Properties of Matter, NEET
Stele includes all of these except

Options:
A pericycle
B vascular bundles
C pith
D endodermis

The correct answer is:
endodermis

Solution:
The central part present in the root or stem having the tissues derived from the procambium is known as stele in plants. The stele contains the vascular bundles, xylem and phloem for the storage and transport of water and nutrients. Pith is a ground tissue which stores and transports nutrients in the plants and is located in the centre of the stem in some plants. The pericycle is a layer of cells which surrounds the stele.

Topic: Biology, Anatomy of Flowering Plants, NEET