datavorous/entrance-exam-dataset huggingface dataset:
In Down's syndrome, karyotyping has shown that the disorder associated with trisomy of chromosome number 21 is usually due to

Options:
A Non-disjunction during egg-cell formation
B Non-disjunction during sperm-cell formation
C Non-disjunction during formation of egg-cells or sperm-cells
D Addition of extra chromosome during mitosis of the zygote

The correct answer is:
Non-disjunction during formation of egg-cells or sperm-cells

Solution:
Down's syndrome caused by trisomy 21 arises due to an occasional non-disjunction during meiosis when a gamete possesses an extra chromosome (n+1). Fusion with a normal gamete [n + (n + 1)] produces trisomy.

Topic: Biology, Principles of Inheritance and Variation, NEET
In Drosophila, a gene for white eye mutation is also responsible for the depigmentation of body parts. Thus a gene represents

Options:
A Oncogene
B Epistatic gene
C Hypostatic gene
D Pleiotropic gene

The correct answer is:
Pleiotropic gene

Solution:
Single gene affecting the different characters which may be physiologically unrelated. Pleiotropic gene has a multiple phenotypic effect because it can control expression of a number of characters. White eye mutation in drosophila, result in depigmentation in many other parts of the body giving a pleiotropic effect.

Topic: Biology, Principles of Inheritance and Variation, NEET
In Drosophila melanogaster, what percentage of recombination is shown between the genes white and yellow, and what percentage of linkage is shown between the genes white and miniature wing, respectively?

Options:
A 98.7% and 62.8%
B 1.3% and 37.2%
C 1.3% and 62.8%
D 98.7% and 37.2%

The correct answer is:
1.3% and 62.8%

Solution:
In Drosophila melanogaster, genes for white and yellow colour are about 1.3 map unit distant from each other so they show 1.3 % recombinations while 98.7% linkage (or parental characters). Genes for white colour and miniature wing are about 37.2 map unit distant so they show 37.2% recombinations and 62.8% linkage (or parental characters).

Topic: Biology, Principles of Inheritance and Variation, NEET
In Drosophila melanogaster,

Options:
A White eye colour is dominant over red eye colour and its gene is located on autosome
B Red eye colour is dominant over white eye colour and its gene is located on autosome
C White eye colour is dominant over red eye colour and its gene is located on the X chromosome
D Red eye colour is dominant over white eye colour and its gene is located on the X chromosome

The correct answer is:
Red eye colour is dominant over white eye colour and its gene is located on the X chromosome

Solution:
In Drosophila melanogaster, the genes for eye colour as well as body colour is located on X chromosomes. Red eye colour is dominant, whereas white eye colour is recessive. Brown body colour is dominant, whereas yellow body colour is recessive. These two genes were used by Morgan to study linkage.

Topic: Biology, Principles of Inheritance and Variation, NEET
In Drosophila, the sex is determined by,

Options:
A the ratio of number of X-chromosome to the sets of autosomes
B $\mathrm{X}$ and $\mathrm{Y}$ chromosomes
C the ratio of pairs of $\mathrm{X}$-chromosomes to the pairs of autosomes
D whether the egg is fertilized or develops parthenogenetically

The correct answer is:
the ratio of number of X-chromosome to the sets of autosomes

Solution:
In Drosophila, sex is primarily determined by the $\mathrm{X}: \mathrm{A}$ ratio, or the ratio of the number of $\mathrm{X}$ chromosomes to the number of sets of autosomes not because of the presence of a Y chromosome as in human sex determination. Although the $\mathrm{Y}$ chromosome is entirely heterochromatic, it contains at least 16 genes, many of which are thought to have male-related functions.
Related Theory
In butterflies, sex is determined by chromosome differences. But unlike in humans with the familiar $X$ and $Y$, in butterflies, it is the femates that determine the sex of offspring. Males are $\mathrm{ZZ}$, while females are $\mathrm{ZW}$

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2003
In Drosophila, the sex is determined by

Options:
A Whether the egg is fertilized or develops parthenogenetically
B The ratio of the number of X-chromosomes to the sets of autosomes
C X and Y chromosomes
D The ratio of pairs of X-chromosomes to the pairs of autosomes

The correct answer is:
The ratio of the number of X-chromosomes to the sets of autosomes

Solution:
Calvin Bridges demonstrated that in Drosophila, the sex is determined by the ratio of the number of X chromosomes to the sets of autosomes.

Topic: Biology, Principles of Inheritance and Variation, NEET
In Dryopteris, meiosis occurs at the time of

Options:
A Gamete formation
B Spore formation
C Formation of prothallus
D Formation of sex organs

The correct answer is:
Spore formation

Solution:
Inside the sporangium, there are 16 spore mother cells which on meiosis (reduction division) produce 64 spores.

Topic: Biology, Plant Kingdom, NEET
In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is

Options:
A 28
B 38
C 18
D 48

The correct answer is:
28

Solution:
% N=2822400ÃVolume of N2 at STP(ml)Weight of organic compondÃ100
% N=2822400Ã1120.5mlÃ100=28 %

Topic: Chemistry, General Organic Chemistry, NEET
In Dumas’ method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K=15 mm)

Options:
A Percentage of nitrogen= 48.22%
B Percentage of nitrogen= 17.46%
C Percentage of nitrogen= 24.45%
D Percentage of nitrogen= 38.22%

The correct answer is:
Percentage of nitrogen= 48.22%

Solution:
Correct Option is : (A)
Percentage of nitrogen= 48.22%

Topic: Chemistry, General Organic Chemistry, NEET
In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:

Options:
A 17.36
B 18.20
C 16.76
D 15.76

The correct answer is:
16.76

Solution:
\(\begin{aligned}
& n_{N_2}=\frac{P V}{R T} \\
& =\frac{(725-25) 40 \times 10^{-3}}{760 \times 0.082 \times 300} \\
& n_{N_2}=\frac{28}{760 \times 24.6} \\
& w_{N_2}=\frac{28 \times 28}{760 \times 24.6} \mathrm{~g} \text { of } N_2 \\
& \text {Weight of } \mathrm{N}_2=\frac{28 \times 28}{760 \times 24.6} \mathrm{gm} \\
& \% N_2=\frac{28 \times 28}{\frac{1}{660 \times 24.6}} \times \frac{1}{0.25} \times 100 \\
& =16.77 \%
\end{aligned}\)

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2015 (Phase 1)
In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is

Options:
A 18.20
B 16.76
C 14.76
D 20.36

The correct answer is:
16.76

Solution:
Volume of nitrogen collected at 300 K and 725 mm pressure is 40 mL actual pressure
=725-25=700 mm
Using P1V1GivenconditionsT1=P2V2AtSTPT2
V2=P1V1×T2T1×P2
Volume of nitrogen at STP (V2)=273×700×40300×760=33.52mL
As 22,400 mL of N2 at STP weight =28g
So, 33.5 mL of nitrogen weight =28×33.5222400g
N% =28×V2×10022400×W
Percentage of nitrogen (N%) =28×33.52×10022400×0.25=16.76%

Topic: Chemistry, General Organic Chemistry, NEET
In Duma's method of estimation of nitrogen $0.35 \mathrm{~g}$ of an organic compound gave $55 \mathrm{~mL}$ of nitrogen collected at $300 \mathrm{~K}$ temperature and $715 \mathrm{~mm}$ pressure. The percentage composition of nitrogen in the compound would be
(Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm}$ )

Options:
A 16.45
B 17.45
C 14.45
D 15.45

The correct answer is:
16.45

Solution:
$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
Where $p_2=$ pressure of $\mathrm{N}_2$ at $S T P=760 \mathrm{~mm}$
$$
\begin{aligned}
& T_2=\text { Temperature of } \mathrm{N}_2 \text { at } \mathrm{STP}=273 \mathrm{~K} \\
& V_2=\text { ? }
\end{aligned}
$$
Volume of $\mathrm{N}_2$ at STP (By gas equation)
$$
\left(\frac{\rho-\rho_1}{t+273}\right) V_1 \times \frac{273}{760}=V_2
$$
where $p_1=\rho-\rho_1$
$\rho=7 / 5 \mathrm{~mm}$ (pressure at which $\mathrm{N}_2$ collected).
$\rho_1=$ aqueous tension of water $=15 \mathrm{~mm}$
$T_1=t+273=300 \mathrm{~K}$
$V_1=55 \mathrm{~mL}=$ volume of moist nitrogen in nitrometer
$$
\therefore V_2=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 \mathrm{~mL}
$$
$$
\begin{aligned}
& \text { \% of nitrogen in given compound } \\
& =\frac{28}{22400} \times \frac{V_2}{W} \times 100=\frac{28}{22400} \times \frac{46.098}{0.35} \times 100 \\
& =16.45 \%
\end{aligned}
$$

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2011 (Screening)
In E. coli, DNA replication starts at:

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

The correct answer is:
Origin of replication

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

Topic: Biology, Molecular Basis of Inheritance, NEET
In each heartbeat, a heart pumps 80 ml of blood at an average pressure of 100 mm of Hg. Assuming 60 heartbeats per second, the power output of the heart is
ÏHg=13.6Ã103 kg m-3 g=9.8 m s-2

Options:
A 1.0 W
B 1.06 W
C 1.12Â W
D 2.16Â W

The correct answer is:
1.06 W

Solution:
Work done by the heart =pdV
Here, dV=60Ã80Ã10-6 m3
p=100âmmâofâHgÃ1.01Ã105âPa760â mm âofâHg
=1.33Ã104âpa
As, power =WorkâdoneTime
=1.33Ã104Ã60Ã80Ã10-660
=1.06â W

Topic: Physics, Work Power Energy, NEET
In each heartbeat, a heart pumps 80 mL of blood at an average pressure of 100 mm of Hg. What will the power output of the heart? (Assume 60 heartbeat per minute)

Options:
A 1 W
B 2.75 W
C 1.06 W
D 0.5 W

The correct answer is:
1.06 W

Solution:
Work done by the heart = PdV
dV = (60) x (80 x 10-6 m3)
P = (100 mm of Hg) ×1.01×105 Pa760 mm of Hg
=1.33×104 Pa
Power = Work done Time
=1.33×104 Pa×60×80×10-6 m360 s
Power = 1.06 W

Topic: Physics, Thermodynamics, NEET
In each microsporangium, pollen grains formed, are released with the _______________ of anther. These pollen grains of some plants, like _____________ cause allergies and bronchial afflictions.

Choose the option with the pair of words which correctly fills the above blanks.

Options:
A dissociation, Hibiscus
B dehydration, Viola
C dehiscence, Parthenium
D dehiscence, Rose

The correct answer is:
dehiscence, Parthenium

Solution:
As the anthers mature and dehydrate, the microspores dissociate from each other and develop into pollen grains. Inside each microsporangium, several thousands of microspores or pollen grains are formed that are released with the dehiscence of anther.
Pollen grains of many species cause severe allergies and bronchial afflictions in some people often leading to chronic respiratory disorders: asthma, bronchitis, etc. It may be mentioned that Parthenium or carrot grass that came into India as a contaminant with imported wheat, has become ubiquitous in occurrence and causes pollen allergy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In each of the following atoms or ions, electronic transition from n=4 to n=1 take place. The frequency of the radiation emitted out will be minimum for

Options:
A Hydrogen atom
B Deuterium atom
C He+ ion
D Li2+ ion

The correct answer is:
Hydrogen atom

Solution:
Î½ =cÎ»=cRZ21n12-1n22
∴ν∝ Z 2
For minimum frequency of radiation ν, atomic number Z should be minimum and it is 1 for both H and deuterium-atoms.
But from reduced mass effect, R is smaller for H-atom and hence v is minimum for H.

Topic: Physics, Atomic Physics, NEET
In each of the following options, there is one property and two structures. Among the two structures, whichever follows the property is marked 'Y' while the one doesn't is marked 'N'. In which of these options, it is incorrect?

Property	Structure

i.	Higher amount sarcoplasmic reticulum:	Red muscles	N

White muscles	Y

ii.	Smaller size in contracted myofibril:	I band	Y

A band	N

iii.	An involuntary muscle with stripes:	Skeletal muscle	N

Cardiac muscle	Y

iv.	ATP is required:	Muscle Contraction	N

Muscle Relaxation	Y

Options:
A i.
B ii.
C iii.
D iv.

The correct answer is:
iv.

Solution:
Myoglobin content is high in some of the muscles which give a reddish appearance. Such muscles are called red fibres. These muscles also contain plenty of mitochondria which can utilize a large amount of oxygen stored in them for ATP production. Some of the muscles possess a very less quantity of myoglobin and therefore, appear pale or whitish. These are the White fibers. The number of mitochondria are also few in them, but the amount of sarcoplasmic reticulum is high.
During muscle contraction, actin filaments slide over myosin in such a way that the 'I' bands get reduced, whereas the 'A' bands retain their length.
Cardiac muscles are involuntary and show faint striations. Skeletal muscles are voluntary and show well-developed striations.
ATP is required by the myosin to bind its head to actin during muscle contraction. ATP is also required during muscle relaxation to pump calcium ions back to the sarcoplasmic reticulum. Thus both options should have been underlined

Topic: Biology, Locomotion and Movement, NEET
In ECG the repolarization of ventricles is indicated by

Options:
A P wave
B QRS complex
C S wave
D T wave

The correct answer is:
T wave

Solution:
The T-wave represents the return of the ventricles from excited to a normal state (repolarisation). The end of the T-wave marks the end of the systole.

Topic: Biology, Body Fluids and Circulation, NEET
In EcoRI, R stands for

Options:
A strain RY 13
B strain Rd
C species RY 13
D species Rd

The correct answer is:
strain RY 13

Solution:
Restriction endonucleases are named based on the bacterium from which they are isolated. The name of Restriction endonuclease is made of 4 parts. The first part is made of a single capital letter (in italics) which comes from the genus name of the bacterium The second part is made of 2 small letters which comes from the species name of the bacterium. The third part is made of a single letter (not in italics) which comes from the strain of the bacterium. The fourth part is made of a Roman numeral (not in italics) indicating the order of discovery. EcoR I is a restriction endonuclease isolated from bacterium Esherichia (E) coli (co) strain RY 13 (R) and it is the first endonuclease (I) to be isolated from this strain. Hind III is a restriction endonuclease isolated from bacterium Haemophilus (H) influenzae (in) strain Rd (d) and it is the third endonuclease (III) to be isolated from this strain.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In EcoRI, the letter ‘R’ is derived from

Options:
A Genus
B Species
C Strain
D Class

The correct answer is:
Strain

Solution:
The convention for naming these enzymes is the first letter of the name comes from the genus and the second two letters come from the species of the prokaryotic cell from which they were isolated, e.g., EcoRI comes from Escherichia coli RY 13. In EcoRI, the letter âRâ is derived from the name of the strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Topic: Biology, Biotechnology - Principles and Processes, NEET
In E.coli, an operator gene combines with

Options:
A Inducer gene to switch on structural gene transcription
B Represser protein to switch off structural gene transcription
C Regulator gene to switch on structural gene transcription
D Represser protein to switch on structural gene transcription

The correct answer is:
Represser protein to switch off structural gene transcription

Solution:
The operator gene is a part of DNA between a promoter and a structural gene, where repressor binds to stop the activity of RNA polymerase.
An operon contains operator gene, regulator gene, represser, structural genes and promoter gene. In the operon concept, the operator gene interacts with represser protein produced by the regulator gene to switch off structural gene transcription. The regulator gene functions as a repressor and controls chemical synthesis by inhibiting transcription of mRNA. In E. coli, two types of operons are recognized, the lac operon and tryptophan operon.
1. The regulation of lac operon by a repressor is called negative regulation.
2. Lac operon is switched in the presence of lactose (substrate), an inducible operon.

Topic: Biology, Molecular Basis of Inheritance, NEET
In E.coli, during lactose metabolism repressor binds to:

Options:
A regulator gene
B operator gene
C structural gene
D promoter gene

The correct answer is:
operator gene

Solution:
The repressor can bind to a short sequence of the lac gene, called operator. When repressor binds to that sequence it physically blocks access by RNA polymerase to the DNA. As a result, as long as it is bound no lac $m$ RNA can be initiated.
Related Theory
The regulatory gene codes for a regulatory protein. The lac repressor, encoded by the lac gene, is the regulatory protein of the lac operon.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2002
In E.coli, the lac operon gets switched on when:

Options:
A lactose is present and it binds to the repressor
B repressor binds to operator
C RNA polymerase binds to the operator
D lactose is present and it binds to RNA polymerase

The correct answer is:
lactose is present and it binds to the repressor

Solution:
Presence of lactose switches on lac operon as allolactose removes the repressor for RNA polymerase to bind.

Topic: Biology, Molecular Basis of Inheritance, NEET
In E.coli, the lac operon gets switched on when:

Options:
A Lactose is present and it binds to the repressor
B Repressor binds to operator
C RNA polymerase binds to the operator
D Lactose is present and it binds to RNA polymerase

The correct answer is:
Lactose is present and it binds to the repressor

Solution:
Lac operon consists of :
1. Promotor site - It is the site for the attachment of RNA polymerase enzyme.
2. One regulatory gene - i (i indicating inhibitor, not inducer). It synthesizes the active repressor protein all the time.
3. Operator (lac operator) - Active repressor protein synthesized by the regulatory gene is attached to the operator and the transcription by RNA polymerase enzyme.
4. Three structural genes (z gene, y gene and a gene).

Topic: Biology, Molecular Basis of Inheritance, NEET
In ecological succession

Options:
A species response and community response will be observed.
B species response and community response will not be observed.
C only species response is observed without community response.
D nether species response not community response is observed.

The correct answer is:
species response and community response will be observed.

Solution:
The primary succession means the development of communities only on a newly exposed habitat with no record of earlier vegetation. An important characteristic of all communities is that composition and structure constantly change in response to the changing environmental conditions. During ecological succession is a series of progressive changes in the species that make up a community is observed. This change is orderly and sequential, parallel with the changes in the physical environment. These changes lead finally to a community that is in near equilibrium with the environment.

Topic: Biology, Ecosystem, NEET
In electrocardiography, P-Q interval represent:

Options:
A Atrial depolarization
B Ventricular depolarization
C Conduction time through atria, A.V. node and rest of conducting tissues
D Ventricular repolarization

The correct answer is:
Conduction time through atria, A.V. node and rest of conducting tissues

Solution:
ECG is Electrocardiogram. This represents electrical activity of the heart in graph format and is recorded by a machine known as Electro- cardiograph. This graph is obtained by connecting the patient to the machine by means of three limb leads and six chest leads. The ECG shows many peaks and each peak is denoted by an alphabet 'P' to 'T' referred as Waves. P wave marks the beginning of ECG and it denotes the conduction by Sino- Atrial node .P wave is followed by P-Q interval and this represents conduction through S.A node to AV node and rest of the conducting tissues.

Topic: Biology, Body Fluids and Circulation, NEET
In electrolysis of $\mathrm{NaCl}$ when $\mathrm{P} t$ electrode is taken then $\mathrm{H}_2$ is liberated at cathode while with $\mathrm{Hg}$ cathode it forms sodium amalgam because:

Options:
A $\mathrm{Hg}$ is more inert than $\mathrm{P} t$
B more voltage is required to reduce $\mathrm{H}^{+}$at $\mathrm{Hg}$ than at $\mathrm{P} t$
C $\mathrm{Na}$ is dissolved in $\mathrm{Hg}$ while it does not dissolve in $\mathrm{P} t$
D concentration of $\mathrm{H}^{+}$ions is larger when $\mathrm{P} t$ electrode is taken

The correct answer is:
more voltage is required to reduce $\mathrm{H}^{+}$at $\mathrm{Hg}$ than at $\mathrm{P} t$

Solution:
Sodium chloride in water dissociates as:
$\begin{aligned}
\mathrm{NaCl} & \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \\
\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{H}^{+} \mathrm{OH}^{-}
\end{aligned}$
When electric current is passed through this solution using platinum electrodes. $\mathrm{Na}^{+}$and $\mathrm{H}^{+}$move towards cathode whereas $\mathrm{Cl}^{-}$and $\mathrm{OH}^{-}$ions move towards anode. At cathode
$\begin{aligned}
\mathrm{H}^{+}+e^{-} & \rightarrow \mathrm{H} \\
\mathrm{H}+\mathrm{H} & \rightarrow \mathrm{H}_2
\end{aligned}$
An anode
$\begin{aligned}
\mathrm{Cl}^{-} & \rightarrow \mathrm{Cl}+e^{-} \\
\mathrm{Cl}+\mathrm{Cl} & \rightarrow \mathrm{Cl}_2
\end{aligned}$
If mercury is used as cathode $\mathrm{H}^{+}$ ions are not discharged at mercury cathode because mercury has a high hydrogen over voltage. $\mathrm{Na}^{+}$ ions are dischaged at cathode in preference of $\mathrm{H}^{+}$ions yielding sodium, which dissolves in mercury to form amalgam

Topic: Chemistry, Electrochemistry, NEET, NEET 2002
In electrophilic aromatic substitution, what is the primary reason that ortho and para positions are more reactive than the meta position for many substituents?

Options:
A The inductive effect of the substituent
B The resonance effect of the substituent
C The steric hindrance at the meta position
D The -I effect of the substituent

The correct answer is:
The resonance effect of the substituent

Solution:
The resonance effect of a substituent increases electron density at the ortho and para positions, making them more reactive in electrophilic aromatic substitution reactions compared to the meta position.

Topic: Chemistry, Hydrocarbons, NEET
In electrophilic aromatic substitution, which of the following factors makes the meta position less reactive compared to the ortho and para positions for many substituents?

Options:
A Steric hindrance at the meta position
B The -I effect of the substituent
C The resonance effect of the substituent
D The inductive effect of the substituent

The correct answer is:
Steric hindrance at the meta position

Solution:
The meta position is often less reactive due to steric hindrance caused by the presence of the substituent at that position, making it more challenging for electrophilic attack compared to the ortho and para positions.

Topic: Chemistry, Hydrocarbons, NEET
In electrophilic substitution reactions, to which position on the aromatic ring does the carbonyl group typically direct the incoming electrophile?

Options:
A Ortho position
B Meta position
C Para position
D None of the above

The correct answer is:
Meta position

Solution:
The carbonyl group in electrophilic substitution reactions typically directs the incoming electrophile to the meta position on the aromatic ring.

Topic: Chemistry, Aldehydes, Ketones and Carboxylic Acids, NEET