datavorous/entrance-exam-dataset huggingface dataset:
'dB' is a standard abbreviation used for the quantitative expression of

Options:
A The density of bacteria in medium
B A particular pollutant
C The dominant Bacillus in a culture
D A certain pesticide

The correct answer is:
A particular pollutant

Solution:
dB is a standard abbreviation for the quantitative expression of noise. Unwanted sound is called noise. Generally, sound above 80 dB is considered as noise.

Topic: Biology, Environmental Issues, NEET
Define
\(f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & \text { if }-1 \leq x < 0 \\
\frac{2 x+1}{x-2}, & \text { if } 0 \leq x \leq 1
\end{array}\right.\)
If \(\lim _{x \rightarrow 0} f(x)\) exists, then \(p=\)

Options:
A -1
B \(-\frac{1}{2}\)
C \(\frac{1}{2}\)
D 1

The correct answer is:
\(-\frac{1}{2}\)

Solution:
Given, \(f(x)=\left\{\begin{array}{cl}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & \text { if }-1 \leq x < 0 \\ \frac{2 x+1}{x-2}, & \text { if } 0 \leq x \leq 1\end{array}\right.\)
Now, RHL \(=\) LHL (at \(x=0\))
\(\Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-2}=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}\)
Apply L'Hospital rule in LHL
\(\Rightarrow \quad-\frac{1}{2}=\lim _{x \rightarrow 0^{-}} \frac{\frac{p}{2 \sqrt{1+p x}}+\frac{p}{2 \sqrt{1-p x}}}{1}\)
\(\Rightarrow \quad-\frac{1}{2}=\frac{p}{2}+\frac{p}{2} \Rightarrow-\frac{1}{2}=p \Rightarrow p=-\frac{1}{2}\)

Topic: Mathematics, Limits, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
Define uniport, symport and antiport. Do they require energy?

Options:

None

Solution:
For movement of substances the biological membranes have many mechanism. Some are active and some are passive. Specific membrane proteins are also involved for special types of transport mechanisms. These mechanisms include: Uniport is a membranetransport system by an integral membrane protein that is involved in facilitated diffusion. These channels open in response to a stimulus for free flow of specific molecules in a specific direction. These channels transport molecule with a solute gradient without energy expenditure.
Symport involves the movement of two or more different molecules or ions, across the membrane in the same direction, with no expenditure of energy.
Antiport is called exchanger. This integral membrane protein is involved in secondary active transport of two or more different molecules or ions across the membrane in opposite directions, without affecting the transport of other molecules.

Topic: Biology, Transport in Plants
Define 'zero population growth rate'. Draw an age pyramid for the same.

Options:

None

Solution:
Zero population growth rate is when the pre-reproductive age group individuals are comparatively less and both reproductive and post-reproductive stages are almost in equal stage or at same level.
An inverted bell shaped age pyramid is obtained for zero population growth rate.

Topic: Biology, Organisms and Populations
Degeneration of a genetic code is attributed to the:

Options:
A first member of a codon
B second member of codon
C entire codon
D third member of a codon

The correct answer is:
third member of a codon

Solution:
Each codon is specific for only one amino acid (or one stop signal), yet the genetic code is described as degenerate, or redundant, because a single amino acid may be coded for by more than one codon. The codons that code for one amino acid can differ in any of their places; nevertheless, the second or third codon is more frequent.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2003
Describe the components of an ecosystem.

Options:

None

Solution:
The components of an ecosystem can be divided into two categories : biotic and abiotic.
(1) Biotic components :
(i) Producers - Green plants which can synthesize their own food.
(ii) Consumers - They donot synthesize their food. They may be
- Primary consumer/herbivores - consuming plants as food.
- Secondary and tertiary consumers or carnivores - They feed on either herbivores or carnivores.
(iii) Decomposers - These organisms breakdown the dead bodies or waste products of plants and animals into simpler inorganic compounds.
(2) Abiotic components :
(i) Climatic components
- Light
- Temperature
- Wind
- Atmospheric gases
- Rain
- Atmospheric humidity
(ii) Soil factors
- Organic materials
- Minerals
- Soil, water
- Soilair
(iii) Topographic factors
- Altitude
- Direction and steepness slope

Topic: Biology, Ecosystem
Describe the importance of the following:
(i) limestone
(ii) cement
(iii) Plaster of Paris.

Options:

None

Solution:
(i) Limestone. Specially precipitated \(\mathrm{CaCO}_3\) is extensively used in the manufacture of high quality paper. It is also used as an antacid, mild abrasive in toothpaste, a constituent of chewing gum and as a filler in cosmetics.
(ii) Cement. Uses of cement:
- Bridge construction
- Plastering
- Most important ingredient in concrete
(iii) Plaster of Paris. Uses of Plaster of Paris:
- Used to make casts and moulds
- Used to make surgical bandages

Topic: Chemistry, s Block Elements
Describe various types of epithelial tissues with the help of labelled diagrams.

Options:

None

Solution:
Epithelial tissues
Epithelial tissues provide covering to the inner and outer lining of various organs. The cells of epithelial tissues are compactly packed with little intercellular matrix.
There are two types of epithelial tissues :
(i) Simple epithelium
(ii) Compound epithelium
(i) Simple epithelium :
Composed of single layer of cells and functions as a lining for body cavities ducts and tubes. It is further divided into three types on the basis of structure modifications-
(a) Squamous epithelium - It is made of a single layer of flattened cells with irregular boundaries.
It is found as a lining for body cavities, ducts and tubes such as in the walls of blood vessels and air sacs of lungs.
Functions - It helps in forming a diffusion boundaries.
(b) Cuboidal epithelium - It is made of single layer of cube-like cells.
It is commonly found in ducts of glands and tubular parts of nephrons in kidneys. Specialized cuboidal cells are capable of producing gametes found in gonads called germinal epithellium.
Functions - It helps in secretion and absorption and also in moving particles or mucus in a specific direction over the epithellium.
(c) Columnar epithelium - It is composed of a single layer of tall and slender cells.
Nuclei are located at base.
Its free surface may have microvilli.
It is found in lining of stomach and intestine.
Functions - It helps in secretion and absorption.
(ii) Compound epithelium :
It is made of more than one layer (multi-layered) of cells and thus it has very limited role in secretion and absorption.
It is located at dry surfaces of the skin, moist surface of buccal cavity, pharynx inner lining of ducts of salivary glands and of pancreatic ducts.
Functions - To provide protection against chemical and mechanical stresses.

Topic: Biology, Structural Organisation in Animals
Determine mean and standard deviation of first $n$ terms of an A.P. whose first term is a and common difference is $\mathrm{d}$.

Options:

None

Solution:
$$
\begin{array}{|c|c|c|}
\hline x & d=x-a & d^2 \\
\hline a & 0 & 0 \\
a+d & d & d^2 \\
a+2 d & 2 d & 4 d^2 \\
\ldots \ldots \ldots & \ldots \ldots \ldots & \ldots \ldots \ldots \\
\ldots \ldots \ldots & \ldots \ldots \ldots & \ldots \ldots \ldots \\
\ldots \ldots \ldots & \ldots \ldots \ldots & \ldots \ldots \ldots \\
\hline a+(n-1) d & (n-1) d & (n-1)^2 d^2 \\
\hline
\end{array}
$$
where $\mathrm{a}=$ Assumed Mean
$$
\begin{aligned}
\Sigma \mathrm{d} &=0+\mathrm{d}+2 \mathrm{~d}+3 \mathrm{~d}+\ldots+(\mathrm{n}-1) \mathrm{d} \\
&=\mathrm{d}\{1+2+3+\ldots+(\mathrm{n}-1)\} \\
&=\frac{\mathrm{d}(\mathrm{n}-1) \mathrm{n}}{2} \\
\Rightarrow \quad \Sigma \mathrm{d}^2 &=\mathrm{d}^2\left[1^2+2^2+3^2+\ldots .+(n-1)^2\right] \\
&=\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1) \mathrm{d}^2}{6} \\
\therefore \quad \overline{\mathrm{x}} &=\mathrm{a}+\frac{\Sigma \mathrm{d}}{\mathrm{n}}=\mathrm{a}+\frac{(\mathrm{n}-1) \mathrm{d}}{2}
\end{aligned}
$$
$$
\begin{aligned}
&\text { S.D. } \sqrt{\frac{\Sigma d^2}{n}-\left(\frac{\Sigma d}{n}\right)^2}\\
&=\sqrt{\frac{(n-1)(2 n-1) d^2}{6}-\left(\frac{n-1}{2}\right)^2 d^2}\\
&=(d) \sqrt{\left(\frac{\mathrm{n}-1}{2}\right)\left[\frac{2 \mathrm{n}-1}{3}-\frac{\mathrm{n}-1}{2}\right]}\\
&d \cdot \sqrt{\frac{(n-1)(n+1)}{12}}\\
&\Rightarrow \mathrm{SD}=\mathrm{d} \cdot \sqrt{\frac{\mathrm{n}^2-1}{12}}
\end{aligned}
$$

Topic: Mathematics, Statistics
Determine the incorrect statements regarding the process of female gamete production.

I. The oogonia keep on adding in the ovary till menopause.
II. The primary oocytes are halted at prophase-I of meiotic division.
III. The fluid-filled cavity present in the secondary follicle is called the atrium.
IV. The secondary oocyte forms a membrane called zona glomerulosa around it.

Options:
A II, III and IV
B I, III and IV
C I and II
D III and IV

The correct answer is:
I, III and IV

Solution:
I. No more oogonia are added to the ovary after birth.III. The fluid filled cavity present in the tertiary follicle is called antrum.IV. The secondary oocyte forms a membrane called zona pellucida around it.

Topic: Biology, Human Reproduction, NEET
Determine the wrong statement/s with respect to the nucleated blood cells.

I. The most abundant leucocyte is a granulocyte while the least abundant leucocyte is an agranulocyte.
II. The largest leucocyte is a granulocyte while the smallest leucocyte is an agranulocyte.
III. Phagocytic leucocytes may be agranulocyte or granulocyte.
IV. Polymorphonuclear leucocytes are the largest granulocyte while the largest agranulocyte has a kidney-shaped nucleus.

Options:
A I, II and III
B II and IV
C III and IV
D I and II

The correct answer is:
I and II

Solution:
The most abundant leucocyte (neutrophil) is a granulocyte while the least abundant leucocyte (basophil) is also a granulocyte. The largest leucocyte (monocyte) is an agranulocyte while the smallest leucocyte (lymphocyte) is also an agranulocyte. Phagocytic leucocytes are neutrophils (granulocyte) and monocyte (agranulocyte). Polymorphonuclear leucocytes (neutrophils) are the largest granulocyte while the largest agranulocyte (monocyte) has a kidney shaped nucleus.

Topic: Biology, Body Fluids and Circulation, NEET
Dicot stem has open vascular bundles. They are called 'open' because they

Options:
A Are surrounded by pericycle but no endodermis
B Are capable of producing secondary xylem and phloem
C Possess conjunctive tissue between xylem and phloem
D Are not surrounded by pericycle

The correct answer is:
Are capable of producing secondary xylem and phloem

Solution:
In the vascular bundles of dicotyledonous stems, cambium is present between phloem and xylem. This cambium has the ability to form secondary xylem and phloem tissues. This the vascular bundles can grow continuously. Hence they are called open vascular bundles.

Topic: Biology, Anatomy of Flowering Plants, NEET
Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.

Options:
A Dumas method and Kjeldahls method are not applicable for estimation of nitrogen in organic compounds
B In the Dumas method, the organic compound is heated in a combustion tube with copper oxide (CuO) to convert all nitrogen to nitrogen gas (N2), which is collected and measured. In Kjeldahl's method, the organic compound is digested with concentrated sulphuric acid (H2SO4), converting the nitrogen to ammonium sulphate (NH4)2SO4.The liberated ammonia (NH3) is then titrated with a standardized acid.
C Both the Dumas and Kjeldahl's methods involve heating the organic compound to convert nitrogen into nitrogen oxides and then titrating the resulting solution with a standardized acid. However, the specific reagents and procedures used differ.
D The Dumas method relies on the decomposition of the organic compound in the presence of a catalyst to release nitrogen gas, which is measured. In Kjeldahl's method, the organic compound is digested with sulphuric acid to convert the nitrogen to ammonia, which is subsequently distilled and titrated.

The correct answer is:
In the Dumas method, the organic compound is heated in a combustion tube with copper oxide (CuO) to convert all nitrogen to nitrogen gas (N2), which is collected and measured. In Kjeldahl's method, the organic compound is digested with concentrated sulphuric acid (H2SO4), converting the nitrogen to ammonium sulphate (NH4)2SO4.The liberated ammonia (NH3) is then titrated with a standardized acid.

Solution:
Correct Option is : (B)
In the Dumas method, the organic compound is heated in a combustion tube with copper oxide (CuO) to convert all nitrogen to nitrogen gas (N2), which is collected and measured. In Kjeldahl's method, the organic compound is digested with concentrated sulphuric acid (H2SO4), converting the nitrogen to ammonium sulphate (NH4)2SO4.The liberated ammonia (NH3) is then titrated with a standardized acid.

Topic: Chemistry, General Organic Chemistry, NEET
Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Duma's method
(ii) Kjeldahl's method.

Options:

None

Solution:
(i) Duma's method: The nitrogen containing organic compounds, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water. \% nitrogen is measured as follows :
\(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{N}_2+\left(\frac{2 \mathrm{x}+\mathrm{y}}{2}\right) \mathrm{CuO} \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}(\mathrm{~N}_2+\left(\frac{2 \mathrm{x}+\mathrm{y}}{2}\right) \mathrm{Cu}\)
\(\%\) of Nitrogen \(=\frac{28 \times \mathrm{z} \times 100}{2} \)\
\(2400 \times \mathrm{m}\)
(ii) Kjeldahl method: Nitrogen containing compound is treated with conc. \(\mathrm{H}_2 \mathrm{SO}_4\) to get ammonium sulphate which liberates ammonia on treating with \(\mathrm{NaOH}\); ammonia is absorbed in known volume of standard acid with the help of this \(\%\) nitrogen is calculated.
- Organic compound \(+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \stackrel{2 \mathrm{NaOH}}{\longrightarrow} \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{NH}_3+2 \mathrm{H}_2 \mathrm{O}\)
\(\begin{aligned}
&2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \\
&\% \text { of nitrogen }=\frac{1.4 \times \mathrm{M} \times 2\left(\mathrm{~V}-\mathrm{V}_1 / 2\right)}{\mathrm{m}}
\end{aligned}\)
\(-\mathrm{V}\) and \(\mathrm{V}_1\) are volume \((\mathrm{mL})\) of \(\mathrm{H}_2 \mathrm{SO}_4\) and volume \((\mathrm{mL})\) of \(\mathrm{NaOH}\) respectively.

Topic: Chemistry, General Organic Chemistry
Differentiation of organs and tissues in a developing organism, is associated with:

Options:
A differential expression of genes
B lethal mutations
C deletion of genes
D developmental mutations.

The correct answer is:
differential expression of genes

Solution:
Differential expression of different genes at different stages of development leads to organ and tissues differentiation, for example, size and pattern of puffs of polytene chromosomes, which are site of transcription, varies at different developmental stages, indicating differential gene activity. Transcription factor is the molecular basis of organ differentiation.
Related Theory
The fate of a cell describes what it will become in the course of normal development. The fate of a particular cell can be discovered by labelling that cell and observing what structures it becomes a part of. The developmental potential, or potency, of a cell describes the range of different cell types it can become. The zygote and its very early descendants are totipotent - these cells have the potential to develop into a complete organism. The determination of different cell types (cell fates) involves progressive restrictions in their developmental potentials. Differentiation follows determination, as the cell elaborates a cell-specific developmental program. Differentiation results in the presence of cell types that have clear-cut identities, such as muscle cells, nerve cells, and skin cells.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2007
Dikaryon formation is characteristics of

Options:
A Ascomycetes and Basidiomycetes
B Phycomycetes and Ascomycetes
C Basidiomycetes and Zygomycetes
D Phycomycetes and Deuteromycetes.

The correct answer is:
Ascomycetes and Basidiomycetes

Solution:
The formation of secondary mycelium is the feature of Ascomycetes and Basidiomycetes. A secondary mycelium forms upon conjugation of two sexually compatible hyphae. The secondary mycelium is dikaryotic, in that it has two haploid nuclei, one from each parent. As the dikaryotic mycelium grows, the cells divide and more septa are formed between the new cells.

Topic: Biology, Biological Classification, NEET
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid wastes

Options:

None

Solution:
(a) Radioactive waste : Radioactive waste include materials that are radioactive \& for which there is no further practical use. These are generated by nuclear reactor, nuclear fallout, man made (refining and mining of platinum and thorium), natural radioactive waste and release of radiation in radiation therapy.
Increased risk of cancer, birth defects \& infertility are few harmful effects caused by nuclear waste. So, nuclear waste is an extremely potent pollutant.
(b) Defunct ships \& e-wastes - The dismantling of defunct ship is a technically complex process, which is potentially harmful to the environment \& human health. Defunct ships contain toxicants like asbestos, mercury, etc. The workers breaking the ships are not suitably protected and are exposed to toxic chemicals. The coastal areas in the vicinity of the ship-breaking yard also becomes polluted. At the international level, it is accepted that there is uncertainty about the relevant controls for the dismantling of such vessels $\&$ there is an urgent need to establish a specific enforceable control framework.
Electronic waste comprised of irrepairable computer and other electronic goods, generated by developed countries.
It is valuable source of secondary raw materials, if treated properly, however if not treated properly it is the major source of toxins. Eventually recycling is the only solution for the treatment of e-wastes provided it is carried out in an environment friendly manner.
(c) Municipal solid wastes : These are commonly known as trash or garbage. It consists of everyday items such as product packaging, furniture, clothing, bottles, food scraps, newspapers.
Appliances, paints, batteries etc. Source reduction, recycling and compositing are several municipal social waste management practices. Source reduction involves altering the design, manufacture or use of products \& materials to reduce the amount and toxicity of what gets thrown away. Recycling diverts items such as paper, glass, plastic \& metals into a new products. Composting decomposes organic waste such as food scraps \& yard trimmings with micro-organisms, producing a humus-like substance.

Topic: Biology, Environmental Issues
Distinction between primary, secondary and tertiary alcohol is done by

Options:
A Oxidation method
B Lucas test
C Victor Meyer method
D All of these

The correct answer is:
All of these

Solution:
Distinction between primary, secondary and tertiary alcohol is done by all three methods : oxidation, Victormeyer and Lucas test.

Victor Meyer's method : The given alcohol is converted into an iodide by concentrated HI or red phosphorous and iodine.If a blood red colour is obtained, the original alcohol is primary. If a blue colour is obtained, the alcohol is secondary. If no colour is produced, the alcohol is tertiary.

Oxidation Test : Primary alcohol gets easily oxidized to an aldehyde and can further be oxidized to carboxylic acids too. Secondary alcohol gets easily oxidized to ketone but further oxidation is not possible. Tertiary alcohol doesn't get oxidized in the presence of sodium dichromate.

Lucas test : Primary alchol give white turbidity immediatly.

secondary alcohol gives white turbidity in 5-10 min.

Tertiary alcohol given white turbidity in 30 min.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Diversification in plant life appeared:

Options:
A due to long periods of evolutionary changes
B due to abrupt mutations
C suddenly on earth
D by seed dispersal.

The correct answer is:
due to long periods of evolutionary changes

Solution:
Diversification in plant life appeared due to long periods of evolutionary changes. Earlier, the plants were thalloid and there were no differentiation among root, stem and leaves. Vascular tissues were absent as in alglae and bryophytes. In pteridophytes, seed habit was observed for the first time. Gymnosperm consists of vascular tissue and naked seed habitat. Angiosperms showed secondary growth, along with fruit formation.

Topic: Biology, Evolution, NEET, NEET 2004
Do secondary metabolites have any ecological importance?

Options:
A No, they do not have any ecological importance.
B Yes, they have some ecological importance.
C The ecological importance of secondary metabolites is not yet known.
D Secondary metabolites only have ecological importance in plant and fungal cells.

The correct answer is:
Yes, they have some ecological importance.

Solution:
some secondary metabolites have ecological importance.

Topic: Biology, Biomolecules (B), NEET
Do sexually reproducing organisms exhibit events and processes that have remarkable fundamental similarity?

Options:
A Yes
B No
C It depends
D None of the above

The correct answer is:
Yes

Solution:
After attainment of maturity, all sexually reproducing organisms exhibit events and processes that have remarkable fundamental similarity.

Topic: Biology, Reproduction in Organisms, NEET
Double fertilization means

Options:
A fusion of nucleus of male gamete with the egg nucleus
B fusion of two polar nuclei
C fusion of sperm nucleus with secondary nucleus
D fusion of one sperm nucleus with nucleus and fusion of other sperm nucleus with secondary nucleus

The correct answer is:
fusion of one sperm nucleus with nucleus and fusion of other sperm nucleus with secondary nucleus

Solution:
Double fertilization is the process the one sperm fuses with the egg while the other sperm fuses with the two polar nuclei. mien pollen tube enters ovule, it strikes one of the synergids and burst open to release the two male gametes which fuses with two different structures in the same female gametophyte. The fusion two polar nuclei forms endosperm. So, the correct answer is option D.

Topic: Biology, Sexual Reproduction in Flowering Plants, JIPMER, JIPMER 2004
Double fertilization occurs in

Options:
A Riccia
B Pteridium
C Cycas
D Nostoc

The correct answer is:
Nostoc

Solution:
Double fertilization is the characteristic feature of angiosperms, e.g., Capsella. It includes fusion of one male gamete with egg and second male gamete with secondary nucleus (this forms endosperm).

Topic: Biology, Sexual Reproduction in Flowering Plants, JIPMER, JIPMER 2011
Draw a standard ECG and explain the different segments in it.

Options:

None

Solution:
The recording of electrical potential generated by the spread of cardiac impulse, is called electrocardiogram (ECG). ECG is the graphic record of electronic current produced by the excitation of cardiac muscles. A normal electrocardiogram is composed of $\mathrm{P}$ wave, $\mathrm{QRS}$ complex and $\mathrm{T}$ wave, $\mathrm{P}$ wave indicate the depolarisation of the atria. QRS complex expresses the ventricular depolarisation. T wave indicate an repolarisation of ventricles.

Topic: Biology, Body Fluids and Circulation