datavorous/entrance-exam-dataset huggingface dataset:
Which element has the lowest standard electrode potential?

Options:
A Fluorine
B Hydrogen
C Lithium
D Copper

The correct answer is:
Lithium

Solution:
Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution.

Topic: Chemistry, Electrochemistry, NEET
Which element is the third most abundant in the Earth's crust?

Options:
A Iron
B Aluminium
C Carbon
D Gold

The correct answer is:
Aluminium

Solution:
Aluminium is the third most abundant element in the Earth's crust, making up approximately 8.3% by weight.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Which endangered animal is the source of the world's finest, lightest, warmest and most expensive wool-the shahtoosh?

Options:
A nilgai
B cheetah
C kashmiri goat
D chiru

The correct answer is:
chiru

Solution:
Shahtoosh comes from the short, warm fleece of the rare Tibetan antelope (Pantholops hodgsonii - Chiru), a species found almost exclusively in the Changtang area of Tibet, on the Tibetan Plateau. It takes four animals to provide enough wool for just one shahtoosh shawl or scarf.
Related Theory
Despite legal protection of the highest order, the population of chiru is constantly on the decline and today the species is extremely endangered.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2003
Which equation gives relation between concentration of ions in solution, electrode potential (E) and standard electrode potential E°?

Options:
A Faraday's equation
B Nernst's equation
C Ohm's equation
D Kohlrausch's equation

The correct answer is:
Nernst's equation

Solution:
Nernstâs equation shows relation between E and Eo.
E=Eâ-0.0591nlogQ

Topic: Chemistry, Electrochemistry, NEET
Which from following elements is most abundant on earth?

Options:
A $\mathrm{N}$
B $\mathrm{C}$
C $\mathrm{O}$
D $\mathrm{H}$

The correct answer is:
$\mathrm{O}$

Solution:
\begin{array}{|l|l|l|} \hline Std.12 & \mathrm{Ch}-7 & Subtopic-7.2 \\ \hline \end{array}

Topic: Chemistry, Hydrogen, MHT CET, MHT CET 2023 (10 May Shift 2)
Which from following equation is correct for relation between strandard cell potential and equilibrium constant?

Options:
A $\mathrm{E}_{\text {cell }}^0=\log _{10} \mathrm{~K} \frac{\mathrm{n}}{0.0592}$
B $\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$
C $\mathrm{E}_{\text {cell }}=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$
D $\mathrm{E}_{\text {cell }}=\log _{10} \mathrm{~K} \frac{\mathrm{n}}{0.0592}$

The correct answer is:
$\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$

Solution:
We know $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{Q}$
Or at equilibrium $0=\mathrm{E}_{\text {cell }}^0-\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$
Or $\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2022 (05 Aug Shift 1)
Which from the following is the correct relationship between standard Gibbs energy change and standard cell potential?

Options:
A $-\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^0$
B $\Delta \mathrm{G}^{\circ}=\frac{\mathrm{E}_{\mathrm{cel} 1}^{\circ}}{\mathrm{nF}}$
C $\mathrm{E}_{\text {cel1 }}^0=\Delta \mathrm{G}^{\circ} \times \mathrm{nF}$
D $\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}$

The correct answer is:
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}$

Solution:
Relationship between standard gibbs energy change and standard cell potential is,
$$
\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}
$$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2021 (23 Sep Shift 1)
Which group of vertebrates comprises the high number of endangered species?

Options:
A mammals
B fishes
C reptiles
D birds

The correct answer is:
mammals

Solution:
Mammals are the most complex animal that cannot easily adapt the environmental changes and some mammals are used as a source of industrial raw materials. Thus, due to climatic change and hunting their population is getting down and become endangered. Mammalian species have the highest number of endangered species (177), accounting for $3.8 \%$ of all vertebrate species.
Related Theory
Fish (including sharks) make up the largest group of vertebrates. Nearly half of all vertebrates are fish. The insect is the largest class of animals. It has over 7.00 .000 species.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2003
Which group of vertebrates comprises the highest number of endangered species?

Options:
A Birds
B Mammals
C Fishes
D Reptiles

The correct answer is:
Mammals

Solution:
The grim scenario of extinction is the fact that more than 15,500 species worldwide are facing the threat of extinction. Presently, 12% of all bird species, 23% of all mammal species, 32% of all amphibian species and 31%t of all gymnosperm species in the world face the threat of extinction.

Topic: Biology, Biodiversity and its Conservation, NEET
Which harmonic mode of a closed-end pipe, 20 cm long, is resonantly excited by a 430 Hz source?

Options:
A First harmonic (fundamental frequency)
B Second harmonic
C Third harmonic
D Fourth harmonic

The correct answer is:
First harmonic (fundamental frequency)

Solution:
Correct Option is : (A)
First harmonic (fundamental frequency)

Topic: Chemistry, Waves and Sound, NEET
Which is a stronger reducing agent between Cr2+and Fe2+and why?

Options:
A Cr2+is a stronger reducing agent because it has a higher atomic number.
B Fe2+is a stronger reducing agent because it has a smaller ionic radius.
C Cr2+is a stronger reducing agent because it has a more positive standard reduction potential (E⊖) value.
D because d4to d3transitionoccurs in case of Cr2+to Cr3+But d6to d5transition occurs in case of Fe2+to Fe3+

The correct answer is:
because d4to d3transitionoccurs in case of Cr2+to Cr3+But d6to d5transition occurs in case of Fe2+to Fe3+

Solution:
Correct Option is : (D)
because d4to d3transitionoccurs in case of Cr2+to Cr3+But d6to d5transition occurs in case of Fe2+to Fe3+

Topic: Chemistry, d and f Block Elements, NEET
Which is a wrong pair about C4 plants?

Options:
A Mesophyll - Homogenous
B Chloroplasts - Dimorphism
C PEP carboxylase - Present in bundle sheath chloroplast
D RUBISCO - Present in abundance in chloroplasts of bundle sheath cell

The correct answer is:
PEP carboxylase - Present in bundle sheath chloroplast

Solution:
C4 plants are called so because the first product of carbon-di-oxide fixation is oxaloacetic acid, which has 4 carbon compounds. C4 plants have a special composition as their leaves show response to high light intensities and they can tolerate high temperatures. The primary acceptor of CO2 , that is phosphoenolpyruvate (PEP) is found in the mesophyll cells which is fixed with the help of an enzyme PEP carboxylase.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Which is CORRECT order of seral stages in hydrach?

Options:
A Phytoplankton-Rooted submerged stage-Rooted free floating plant-reed swamp stage-marsh meadow stage-scrub stage-forest
B Scrub stages-forest-reed swamp stage-marsh meadow stage-submerged free floating plant stage-submerged plant-phytoplankton
C Submerged plant-marsh meadow stage-scrub stages-forest-phytoplankton-submerged free floating plant stage-reed swamp stage
D Phytoplankton-red swamp stage-scrub stages-forest-submerged plant-submerged free floating plant stage-marsh meadow stage

The correct answer is:
Phytoplankton-Rooted submerged stage-Rooted free floating plant-reed swamp stage-marsh meadow stage-scrub stage-forest

Solution:
Ecological succession is defined as the process of gradual and predictable change in the species composition of an area. Primary succession occurs in those areas where no organism ever existed whereas secondary succession takes place in those areas where organisms once existed but now have lost them due to some reason. In Primary succession of wet land into forest the first to appear are the phytoplanktons. They are replaced by submerged plants and later submerged and floating plants both appear. Gradually it passes through the reed swamp and marsh meadow stages and lastly scrubs appear before forming it into a forest land.

Topic: Biology, Ecosystem, NEET
Which is CORRECTLY matched in a normal menstrual cycle?

Options:
A Endometrium regenerates 5th to 10th day
B Release of egg- 5th day
C Endometrium secretes nutrients for implanation - 11th to 18th day
D Rise in progesterone level 1st to 15th day

The correct answer is:
Endometrium regenerates 5th to 10th day

Solution:
Menstrual cycle consists of changes in the Ovary and the Uterus simultaneously. It begins with Menstrual or bleeding phase. This represents shedding of endometrium of uterus and unfertilized secondary oocyte. This phase is followed by follicular phase as follicles start developing in the ovary. At the same time the shedded endometrium starts its preparation and this is termed as proliferative phase. At the same time the follicles develop in the ovary from day 5th to 10th day. The mid of 28 day cycle shows ovulatory phase that is graffian follicle from ovary releases secondary oocyte in the oviduct. This is followed by formation of corpus luteum and is known as Luteal phase and this shows high level of estrogen and progesterone.

Topic: Biology, Human Reproduction, NEET
Which is false for vexillary or papilionaceous aestivation?

Options:
A Two smallest anterior petals are known as keel.
B Two lateral petals are known as wing.
C The largest standard petal is posterior.
D The largest standard petal is anterior and overlaps wing.

The correct answer is:
The largest standard petal is anterior and overlaps wing.

Solution:
In vexillary aestivation the largest posterior petal is called as standard or vexillum, two lateral petals are called as wings and the two anterior smallest petals which are fused are known as keel. Vexillary aestivation thus shows 1+2+B arrangement of petals.

Topic: Biology, Morphology of Flowering Plants, NEET
Which is not a secondary messenger in hormone mediated action?

Options:
A IP3
B Ca2+
C cAMP
D NaHCO3

The correct answer is:
NaHCO3

Solution:
Hormonal action depends on whether the hormone is Lipid or water soluble. Lipid soluble hormones can cross the target cell membrane easily, get attached to the mobile receptors within the cytoplasm and reach the nucleus to perform their functions. However water soluble hormones are proteinaceous and these include insulin, oxytocin etc. These cannot cross the cell membrane easily. These get attached to the static receptors on the cell membrane. This attachment triggers the formation of secondary messengers that help these hormones to act on cells. These also cause amplification of hormonal action so that many targets are affected by single hormone. This concept was explained by Sutherland. Secondary messengers commonly seen in cells are Cyclic AMP. Other examples are Inositol tri phosphate (IP3), Calcium and Cyclic GMP.

Topic: Biology, Chemical Coordination and Integration, NEET
Which is not true about borax ?

Options:
A It is a useful primary standard for titrating against acids
B One mole of borax can be used as a buffer
C Aqueous solution of borax can not be used as buffer
D It is made up of two triangular BO3 units and two tetrahedral BO4 units

The correct answer is:
Aqueous solution of borax can not be used as buffer

Solution:
In borax Na 2 B 4 O 7 · 1 0 H 2 O , among 10 water molecules 2 molecules are part of structure, i.e., exists Na 2 B 4 O 5 OH 4 · 8 H 2 O

Na 2 B 4 O 5 OH 4 · 8 H 2 O + 2 HCl → 2 NaCl + 4 H 3 BO 3 + 5 H 2 O
At end point pH is about 4 and, therefore, Methyl orange (works between pH = 3.1 - 4.4) is used to detect end point. Aq. solution of borax acts as buffer, as borax is salt of strong base NaOH and weak acid H3BO3.
B 4 O 5 ( OH ) 4 2− ( s )+5 H 2 O( l )⇌ 2B ( OH ) 3 ( aq ) ( weak  acid ) + 2 [ B ( OH ) 4 ] − ( aq ) ( salt )

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Which is the correct relationship between equilibrium constant and standard reduction potential of cell?

Options:
A Ecell0=0.059nlogKc
B Ecell0=0.059nlogKc
C Ecell0=n0.059logKc
D Ecell0=logKcn

The correct answer is:
Ecell0=0.059nlogKc

Solution:
The equilibrium constant and standard reduction potential of cell is given by Ecello=2.303RTnFlogK=0.0591nlogKcat298 K

Topic: Chemistry, Electrochemistry, NEET
Which is true according to Werner's coordination theory?

Options:
A Primary valency can be ionized
B Secondary valency can be ionized
C Primary and secondary valencies both cannot be ionized
D Only primary valency cannot be ionized

The correct answer is:
Primary valency can be ionized

Solution:
According to Werner's coordination theory prmary valencies are ionisable while secondary valency which satisfy by ligands are non ionisable.

Topic: Chemistry, Coordination Compounds, NEET
Which isotope of hydrogen is the most abundant?

Options:
A Protium
B Deuterium
C Tritium
D Hydrogen-2

The correct answer is:
Protium

Solution:
Correct Option is : 1
Protium

Topic: Chemistry, Hydrogen, NEET
Which level of protein structure involves the spatial arrangement of subunits in proteins composed of multiple polypeptide chains?

Options:
A Primary structure
B Secondary structure
C Tertiary structure
D Quaternary structure

The correct answer is:
Quaternary structure

Solution:
Quaternary structure involves the spatial arrangement of subunits in proteins composed of multiple polypeptide chains.

Topic: Chemistry, Biomolecules, NEET
Which level of protein structure represents the overall folding of polypeptide chains?

Options:
A Primary structure
B Secondary structure
C Tertiary structure
D Quaternary structure

The correct answer is:
Tertiary structure

Solution:
The tertiary structure of proteins represents the overall folding of polypeptide chains, further shaping the secondary structure.

Topic: Chemistry, Biomolecules, NEET
Which ligand among the following has highest splitting power of d-orbitals of central metal ion?

Options:
A $\mathrm{S}^{2-}$
B $\mathrm{OH}^{-}$
C $\mathrm{NCS}$
D $\mathrm{CO}$

The correct answer is:
$\mathrm{CO}$

Solution:
The order of boiling point of amines is as follows: Primary $>$ Secondary $>$ Tertiary

Topic: Chemistry, Coordination Compounds, MHT CET, MHT CET 2020 (15 Oct Shift 1)
Which ligand can bind through two nitrogen and four oxygen atoms to a central metal ion?

Options:
A Unidentate
B Didendate
C Polydentate
D Hexadentate

The correct answer is:
Hexadentate

Solution:
Hexadentate ligands, like EDTA4–, can bind through multiple donor atoms, such as two nitrogen and four oxygen atoms.

Topic: Chemistry, Coordination Compounds, NEET
Which ligand type can bind through two donor atoms?

Options:
A Polydentate
B Unidentate
C Hexadentate
D Didendate

The correct answer is:
Didendate

Solution:
Ligands that can bind through two donor atoms are referred to as didentate ligands.

Topic: Chemistry, Coordination Compounds, NEET
Which ligand type is characterized by its ability to bind through multiple donor atoms?

Options:
A Didendate
B Hexadentate
C Polydentate
D Unidentate

The correct answer is:
Polydentate

Solution:
Polydentate ligands can bind through multiple donor atoms.

Topic: Chemistry, Coordination Compounds, NEET
Which ligand type is known for having only one donor atom per molecule?

Options:
A Didendate
B Hexadentate
C Polydentate
D Unidentate

The correct answer is:
Unidentate

Solution:
Unidentate ligands have one donor atom per molecule and can bind to a metal ion through that atom.

Topic: Chemistry, Coordination Compounds, NEET
Which method can be used to find out the strength of reducant\oxidant in a solution? Explain with an example.

Options:

None

Solution:
Strength of a reductant (reducing agent) or oxidant (oxidising agent) can be found out by measuring the relative electrode potential when it's connected in a solution using a cell.
For example, $\mathrm{Fe}^{3+} / \mathrm{Fe}$ is the element we want to test with the Standard Hydrogen electrode (SHE). The half-cell reaction for $\mathrm{Fe}$ and $\mathrm{H}$ are given below:
$\begin{array}{ll}\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{H}_2 ; & \mathrm{E}^{\circ}=0.0 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} ; & \mathrm{E}^{\circ}=0.77 \mathrm{~V}\end{array}$
When any element needs to be evaluated it is placed as an electrode with SHE. The amount of emf it generates in the cell can be considered as the potential of the element.
$\mathrm{E}^{\circ}$ cell $=0-0.77$
$\mathrm{E}^{\circ}$ cell $=0.77$
The above assumed configuration of $\mathrm{Fe}$ being anode can be reversed and hence strength $\mathrm{Fe}$ as a reductant can be established. Hence the strength of an oxidant can be determined.

Topic: Chemistry, Redox Reactions
Which noble gas is most abundant in atmosphere?

Options:
A He
B Ne
C Ar
D Kr

The correct answer is:
Ar

Solution:
Argon constitutes 0.93 % by volume of air.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which of following alcohols gives white turbidity almost immediately with the Lucas reagent at room temperature?

(i) n-Butanol

(ii) tertiary-Butanol

(iii) Benzyl alcohol

(iv) Allylic alcohol

Options:
A (i), (ii) and (iii)
B (i), (iii) and (iv)
C (ii), (iii) and (iv)
D (i), (ii) and (iv)

The correct answer is:
(ii), (iii) and (iv)

Solution:
The reagent used in Lucas test are conc. HCl+ZnCl2. It gives the white turbidity or cloudiness with alcohols. It is used to distinguish between primary, secondary and tertiary alcohol. Primary alcohol give the white turbidity in 30 min. Secondary alcohol give the white turbidity in 5 min. Tertiary alcohol give the white turbidity immediately. But phenol and enol can not give Lucas test.

n-butanol is a primary alcohol and tertiary alcohol, benzyl alcohol,allylic alcohol give the turbidity immediately.

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
Which of the compounds HCHO (I), CH3CH2CHO (II), CH3COCH3 (III) and HCOOC2H5 (IV) will give a secondary alcohol on reaction with excess Grignard reagent, followed by hydrolysis? Select the correct answer using the codes given below

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

The correct answer is:
II and IV

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Which of the following acts as a filter of the blood by trapping blood borne micro-organisms and also has a large reservoir of erythrocytes?

Options:
A Bone marrow
B Thymus
C Spleen
D Tonsils

The correct answer is:
Spleen

Solution:
Lymphoid organs are Primary and secondary. Primary lymphoid organs are those in which immature lymphocytes become antigen sensing cells. These organs are bone marrow and thymus. Secondary lymphoid organs are those that receive cells from bone marrow and thymus. These organs are lymph nodes, Spleen, peyer's patches etc.Spleen is known as the blood bank of the body. Histologically it shows two regions: the red pulp and the white pulp. The red pulp contains damaged erythrocytes as speen is the graveyard for these cells. White pulp contains lymphocytes and phagocytes and these cells trap the microorganisms.

Topic: Biology, Human Health and Diseases, NEET
Which of the following acts is not a part of the 'Evil Quartet'?

Options:
A Destruction of rainforests for cultivation
B Shifting endangered species out of their natural habitat to special setting
C Introduction of Nile perch which led to the destruction of species of cichlid fish
D Extinction of passenger pigeon led to extinction of its parasitic bird lice

The correct answer is:
Shifting endangered species out of their natural habitat to special setting

Solution:
Evil Quartet is four major causes of biodiversity loss which include: Habitat loss and fragmentation, over-exploitation of natural resources by humans example: rainforests destroyed for cultivation. Alien species invasion-example: introduction of Nile perch led to the destruction of species of cichlid fish. Co-extinctions, for example, co-extinction of the passenger pigeon and it's parasitic bird lice Columbicola extinctus and Campanulotes defectus. Shifting endangered species out of their natural habitat to a special setting is a type of ex situ conservation.

Topic: Biology, Biodiversity and its Conservation, NEET
Which of the following alcohols is a secondary alcohol?

Options:
A 1-Propanol
B 2-Propanol
C 1-Butanol
D 2-Butanol

The correct answer is:
2-Butanol

Solution:
2-Butanol is a secondary alcohol because the hydroxyl group is attached to a carbon atom bonded to two other carbon atoms.

Topic: Chemistry, Hydrocarbons, NEET
Which of the following alcohols is most likely to undergo dehydrogenation to form an alkene?

Options:
A Primary alcohol
B Secondary alcohol
C Tertiary alcohol
D Aromatic alcohol

The correct answer is:
Secondary alcohol

Solution:
Secondary alcohols are most likely to undergo dehydrogenation to form alkenes.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Which of the following alkali metals has the least negative standard electrode potential (Eo)?

Options:
A Sodium (Na)
B Potassium (K)
C Lithium (Li)
D Rubidium (Rb)

The correct answer is:
Sodium (Na)

Solution:
Sodium (Na) has the least negative standard electrode potential (Eo) among alkali metals, indicating its lower reducing power compared to the other alkali metals.

Topic: Chemistry, s Block Elements, NEET
Which of the following alkyl halides is most likely to undergo dehydrohalogenation rapidly?

Options:
A A primary alkyl halide
B A secondary alkyl halide
C A tertiary alkyl halide
D A quaternary alkyl halide

The correct answer is:
A tertiary alkyl halide

Solution:
Tertiary alkyl halides are more likely to undergo dehydrohalogenation rapidly due to their increased reactivity.

Topic: Chemistry, Hydrocarbons, NEET
Which of the following amines, can give \(\mathrm{N}\)-nitrosoamine on treatment with \(\mathrm{HNO}_2\) ?

Options:
A

B

C

D

The correct answer is:

Solution:
Secondary amines form nitrosoamines which are yellow oily liquids insoluble in water
Dialkyl nitrosoamine

Topic: Chemistry, Amines, AIIMS, AIIMS 2008
Which of the following amines gives yellow oily liquid with $\mathrm{HNO}_{2}$ ?

Options:
A Ethyl methyl amine
B Aniline
C 3 -methyl benzyl amine
D Methyl amine

The correct answer is:
Ethyl methyl amine

Solution:
Secondary amines, (containing - NH-group) when treated with $\mathrm{HNO}_{2}$ give nitrosoamines which separate as yellow oily liquids. The structures of the given compounds are
(a)
(b)
(c)
(d)
$\mathrm{CH}_{3} \mathrm{NH}_{2}$
methyl amine
$\left(1^{\circ}\right)$
Among the given compounds, only ethyl methyl amine is a secondary amine, thus it gives a yellow oily liquid with $\mathrm{HNO}_{2}$.

Topic: Chemistry, Amines, MHT CET, MHT CET 2010
Which of the following amines is most basic in nature in aqueous phase?

Options:
A Ammonia
B Trimethylamine
C Methylamine
D Dimethylamine

The correct answer is:
Dimethylamine

Solution:
(B)
Secondary amine is most basic in aqueous solution.

Topic: Chemistry, Amines, MHT CET, MHT CET 2020 (19 Oct Shift 2)
Which of the following amines undergoes acylation reaction?

Options:
A Ethyldimethylamine
B N-Methylaniline
C N,N-Dimethylmethanamine
D N,N-Dimethylaniline

The correct answer is:
N-Methylaniline

Solution:
Aliphatic and aromatic primary and secondary amines undergo acylation reaction.
Among the given options, N-Methylaniline is a secondary amine and hence, it undergoes acylation reaction. All other amines given are tertiary amines.

Topic: Chemistry, Amines, MHT CET, MHT CET 2023 (09 May Shift 1)
Which of the following amino-acids was not found to be synthesized in Miller's experiment?

Options:
A alanine
B glycine
C aspartic acid
D glutamic acid

The correct answer is:
glutamic acid

Solution:
Miller and Urey through their experiment abiotically synthesised amino acids and bases. They synthesised glycine, aspartic acid and alanine in abundant quantities. However, glutamic acid could not be synthesised in their experiment.

Topic: Biology, Evolution, NEET, NEET 2006