datavorous/entrance-exam-dataset huggingface dataset:
The correct representation of the pyramid is,

Options:
A

Pyramid of energy in the grassland ecosystem

B

Pyramid of biomass in the grassland ecosystem

C

Pyramid of energy in the temperate ecosystem

D

Pyramid of energy in the tropical ecosystem

The correct answer is:

Pyramid of energy in the grassland ecosystem

Solution:
Upright pyramid: In the upright pyramid, the base occupied by producers is broad as compared to subsequent consumers, which gradually narrows down at the apex. It occurs in all the three kinds of pyramids. Pyramid of energy is always upright, can never be inverted, because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat at each step. The given pyramid represents the grassland ecosystem.

Topic: Biology, Ecosystem, NEET
The correct route through which pulse-making impulse travels in the heart is

Options:
A AV node → bundle of His → SA node → Purkinje fibres → heart muscles
B AV node → SA node → Purkinje fibres → bundle of His → heart muscles
C SA node → Purkinje fibres → bundle of His → AV node → heart muscles
D SA node → AV node → bundle of His → Purkinje fibres → heart muscles

The correct answer is:
SA node → AV node → bundle of His → Purkinje fibres → heart muscles

Solution:
The pulse making impulse travels in heart in the order of SA node → AV node → bundle of His → Purkinje fibres → heart muscles.

Topic: Biology, Body Fluids and Circulation, NEET
The correct sequence given below containing neutral, acidic, basic and amphoteric oxide each, respectively, is

Options:
A $\mathrm{NO}, \mathrm{CO}_2, \mathrm{ZnO}, \mathrm{CaO}$
B $\mathrm{NO}, \mathrm{CO}_2, \mathrm{CaO}, \mathrm{ZnO}$
C $\mathrm{NO}, \mathrm{ZnO}, \mathrm{CO}_2, \mathrm{CaO}$
D $\mathrm{ZnO}, \mathrm{NO}, \mathrm{CaO}, \mathrm{CO}_2$

The correct answer is:
$\mathrm{NO}, \mathrm{CO}_2, \mathrm{CaO}, \mathrm{ZnO}$

Solution:
Oxides
Chemical Nature
$\begin{array}{cc}\mathrm{NO} & \text { Neutral } \\ \mathrm{CO}_2 & \text { Acidic } \\ \mathrm{CaO} & \text { Basic } \\ \mathrm{ZnO} & \text { Amphoteric }\end{array}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2023 (Manipur)
The correct sequence of bond enthalpy of ' $\mathrm{C}-\mathrm{X}$ bond is:

Options:
A CH3-F<CH3-Cl<CH3-Br<CH3-I.
B CH3-F>CH3-Cl>CH3-Br>CH3-I
C CH3-F<CH3-Cl>CH3-Br>CH3-I
D CH3-Cl>CH3-F>CH3-Br>CH3-I

The correct answer is:
CH3-F>CH3-Cl>CH3-Br>CH3-I

Solution:
Bond enthalpy is inversely proportional to the bond length. The bond length order of C-X bond in methyl halides depends on the size of the halogen atom. Bond length order: CH3-F<CH3-Cl<CH3-Br<CH3-I

Hence, bond enthalpy order is

CH3-F>CH3-Cl>CH3-Br>CH3-I

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2021
The correct sequence of cell organelles during photorespiration is

Options:
A chloroplast-Golgi bodies-mitochondria
B chloroplast-rough endoplasmic reticulum-dictyosomes
C chloroplast-mitochondria-peroxisome
D chloroplast-vacuole-peroxisome

The correct answer is:
chloroplast-mitochondria-peroxisome

Solution:
None of the options is correct. Photorespiration required three cell organelles in sequence of chloroplast, peroxisome and mitochondria.
Option (3) may be correct if be read as said sequence.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2012 (Screening)
The correct sequence of cell organelles during photorespiration is:

Options:
A Chloroplast - Peroxisome - Mitochondria
B Chloroplast - Mitochondria - Peroxisome
C Chloroplast - Golgi bodies - Mitochondria
D Chloroplast - Peroxisomes - Dictyosomes

The correct answer is:
Chloroplast - Peroxisome - Mitochondria

Solution:
Photorespiration (also known as the oxidative photosynthetic carbon cycle, or C2 photosynthesis) refers to a process in plant metabolism where the enzyme RuBisCO oxygenates RuBP, wasting some of the energy produced by photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The correct sequence of cell organelles involved in photorespiration is

Options:
A chloroplast, mitochondria, and peroxisomes.
B peroxisomes, chloroplast, and mitochondria.
C chloroplast, peroxisome and mitochondria.
D chloroplast, vacuole, and peroxisome.

The correct answer is:
chloroplast, peroxisome and mitochondria.

Solution:
Photorespiration is the light-dependent process of oxygenation of ribulose biphosphate (RuBP) and release of carbon dioxide by the photosynthetic organs of a plant. It leads to oxidation of considerable amount of photosynthetic products to CO2 and H2O without the production of useful energy. Photorespiration occurs only in C3 plants because at high temperature and high oxygen concentration. RuBP carboxylase changes to RuBP oxygenase.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The correct sequence of cytochromes in ETC is

Options:
A Cyt a, b, c, a3
B Cyt b, c1, c , a , a3
C Cyt b,âa,â a 3 ,âc
D Cyt b,âc,â a 3 ,âa

The correct answer is:
Cyt b, c1, c , a , a3

Solution:
Cytochromes are proteins with characteristic strong absorption of visible light due to their iron-containing heme prosthetic groups. Mitochondria contain three classes of cytochromes designated a, b and c distinguished by differences in their light absorption spectra. The cytochrome of type a and b and some of type c are integral proteins of inner mitochondrial membrane. The sequence of electron carriers in ETS is:
NADH â Q â Cyt b â Cyt c1 âCyt c â Cyt a â Cyt a3 â O2

Topic: Biology, Respiration in Plants, NEET
The correct sequence of decreasing number of Ï -bonds in the structure of H2SO3, H2SO4 and H2S2O7 is:

Options:
A H2S2O7>H2SO3>H2SO4
B H2S2O7>H2SO4>H2SO3
C H2SO4>H2S2O7>H2SO3
D H2SO3>H2SO4>H2S2O7

The correct answer is:
H2S2O7>H2SO4>H2SO3

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The correct sequence of flow of electrons in the light reaction is

Options:
A PS II â Plastoquinone â Cytochromes â PS I â Ferrodoxin
B PS I â Plasotquinone â Cytochromes â PS II â Ferrodoxin
C PS I â Ferreodoxin â PS II â Cytochromes â Plastoquinone
D PS I â Plastoquinone â Cytochromes â PS II â Ferrodoxin.

The correct answer is:
PS II â Plastoquinone â Cytochromes â PS I â Ferrodoxin

Solution:
The light reaction occurs in the grana of chloroplast in which light energy causes excitation of electrons which then passes through different carriers that result in the generation of ATP. This process is known as photophosphorylation. In non-cyclic photophosphorylation electron excited from PS II is passed to plastoquinone via pheophytin, then through cytochromes to PS I. PS I than passes the electrons to Ferrodoxin and than NADP is finally reduced.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The correct sequence of four carbon acids formed during Krebs cycle is

Options:
A oxalo-acetic acid - fumaric acid - malic acid - succinic acid
B oxalo-acetic acid - malic acid - fumaric acid - succinic acid
C succinic acid - malic acid - fumaric acid - oxalo-acetic acid
D succinic acid - fumaric acid - malic acid - oxalo-acetic acid

The correct answer is:
succinic acid - fumaric acid - malic acid - oxalo-acetic acid

Solution:
During Krebs cycle, the first acid (from the given acids) to be formed is succinic acid as follows:
Then fumaric acid is formed from succinic acid as follows:
Then malic acid is formed from fumaric acid as follows:
Then oxalo-acetic acid is formed from malic acid as follows:

Topic: Biology, Respiration in Plants, NEET
The correct sequence of increasing covalent character is represented by:

Options:
A $\mathrm{LiCl} < \mathrm{NaCl} < \mathrm{BeCl}_2$
B $\mathrm{BeCl}_2 < \mathrm{LiCl} < \mathrm{NaCl}$
C $\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2$
D $\mathrm{BeCl}_2 < \mathrm{NaCl} < \mathrm{LiCl}$

The correct answer is:
$\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2$

Solution:
According to Fajan's Rule,
1. Smaller the size of cation, the larger the size of the anion, greater is the covalent character of the ionic bond.
2. Greater the charge of cation, greater is the covalent character of the ionic bond.
3. For cations with same charge and size, the one, with $(n-1) d^n n s^{\circ}$ which is found in transition elements have greater covalent character than the cation with $n s^2 n p^6$ electronic configuration, which is commonly found in alkali or alkaline earth metals.
Thus, order of covalent character is:
$$
\mathrm{NaCl} < \mathrm{LiCl} < \mathrm{BeCl}_2
$$
I Caution
Larger the anion, less is the effective nuclear charge that holds the valence electron of the ion in place. Since the last electron is loosely bound in large anions, it can easily be polarised by a cation, thereby making the compound more covalent.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2005
The correct sequence of increasing covalent character is represented by

Options:
A BeCl2<NaCl<LiCl
B BeCl2<LiCl<NaCl
C NaCl<LiCl<BeCl2
D LiCl<NaCl<BeCl2

The correct answer is:
NaCl<LiCl<BeCl2

Solution:
The covalent character of ionic compound depends on polarising power of cation. Order of polarising power of cation is Be++>Li+>Na+ , Hence order of covalent character BeCl2>LiCl>NaCl .

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The correct sequence of meninges from the inner to the outer side is:

Options:
A Duramater → arachnoid membrane → piamater
B Duramater → piamater → arachnoid membrane
C Piamater â arachnoid membrane â duramater
D Arachnoid membrane → duramater → piamater

The correct answer is:
Piamater â arachnoid membrane â duramater

Solution:
Meninges are the connective tissue membrane which protects the central nervous system and projections of its structure. These membranes are piamater (thinnest), arachnoid (thin, webbly and slightly vascular), and duramater (the outermost double layered, thick and nonvascular).

Topic: Biology, Neural Control and Coordination, NEET
The correct sequence of phases of cell cycle is:

Options:
A MâG1âG2âS
B G1âG2âSâM
C SâG1âG2âM
D G1→S→G2→M

The correct answer is:
G1→S→G2→M

Solution:
Interphase accounts for almost 95% of the total time required for the cell cycle. Interphase takes place in three stages; G1, S and G2 proceeded by mitotic phase (M phase) and cytokinesis. The nuclear content of DNA is duplicated during the synthesis phase. So, the correct sequence is -

G1→S→G2→M

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2019 (All India)
The correct sequence of plants in a hydrosere is:

Options:
A Volvox â Hydrilla â Pistia â Scirpus â Lantana â Oak
B Pistia â Volvox â Scirpus â Hydrilla â Oak â Lantana
C Oak â Lantana â Volvox â Hydrilla â Pistia â Scirpus
D Oak â Lantana â Scirpus â Pistia â Hydrilla â Volvox

The correct answer is:
Volvox â Hydrilla â Pistia â Scirpus â Lantana â Oak

Solution:
The correct sequence of plants in a hydrosere is Volvox, Hydrilla, Pistia, Scirpus, Lantana and Oak. A hydrosere is a plant succession which occurs in a freshwater lake. In time, an area of open freshwater such as a lake will naturally dry out, ultimately becoming woodland. During this change, a range of different land types such as swamp and marsh will succeed each other. The succession from open water to climax woodland is likely to take at least two hundred years.

Topic: Biology, Ecosystem, NEET
The correct sequence of plants in a hydrosere is

Options:
A Oak $\rightarrow$ Lantana $\rightarrow$ Scirpus $\rightarrow$ Pistia $\rightarrow$ Hydrilla $\rightarrow$ Volvox
B Volvax $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus $\rightarrow$ Lantana $\rightarrow$ Oak
C Pistia $\rightarrow$ Volvox $\rightarrow$ Scirpus $\rightarrow$ Hydrilla $\rightarrow$ Oak $\rightarrow$ Lantana
D Oak $\rightarrow$ Lantana $\rightarrow$ Volvox $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus

The correct answer is:
Volvax $\rightarrow$ Hydrilla $\rightarrow$ Pistia $\rightarrow$ Scirpus $\rightarrow$ Lantana $\rightarrow$ Oak

Solution:
The various stages in a hydrosere are well studied in ponds, pools or lakes. The various stages of hydrosere are:
(i) Phytoplankton stage, eg, Some blue-green algae, green algae (Volvox), diatoms and bacteria, etc.
(ii) Rooted submerged stage, eg Hydrilla, Vallisneria, etc.
(iii) Floating stage, eg, Nelumbo, Nymphaea, etc. Some free floating species are Pistia, Azolla, Lemna, etc.
(iv) Red-swamp stage, eg, Species of Scirpus, Typha, etc.
(v) Sedge-meadow stage, eg, Species of Cyperaceae and Gramineae.
(vi) Woodland stage, eg, Lantana, Salix, Populus, etc.
(vii) Forest stage, eg, Tropical rain forests, mixed forests of Almus, Acer, Quercus (oak), tropical deciduous forests.

Topic: Biology, Ecosystem, NEET, NEET 2009 (Screening)
The correct sequence of plants in a hydrosere is

Options:
A Oak â Lantana â Scirpus â Pistia â Hydrilla â Volvox
B Volvox â Hydrilla â Pistia â Scirpus â Lantana â Oak
C Pistia â Volvox â Scirpus â Hydrilla â Oak â Lantana
D Oak â Lantana â Volvox â Hydrilla â Pistia â Scirpus

The correct answer is:
Volvox â Hydrilla â Pistia â Scirpus â Lantana â Oak

Solution:
The various stages in a hydrosere are well studied in ponds, pools or lakes. The various stages of hydrosere are :
(i) Phytoplankton stage, e.g., Some blue-green algae, green algae (Volvox), diatoms and bacteria, etc.
(ii)Rooted submerged stage, e.g., Hydrilla, Vallisneria, etc.
(iii) Floating stage, e.g., Nelumbo, Nymphaea, etc. Some free floating species are Pistia, Azolla, Lemna, etc.
(iv) Red-swamp stage, e.g., Species of Scirpus, Typha, etc.
(v) Sedge-meadow stage, e.g., Species of Cyperaeae and Gramineae.
(vi) Woodland stage, e.g., Lantana, Salix, Populus, etc.
(vii) Forest stage, e.g., Tropical rain forests, mixed forests of Almus, Acer, Quercus (oak), tropical deciduous forests.

Topic: Biology, Ecosystem, NEET
The correct sequence of reactions in Krebs cycle is:

Options:
A Isocitric acid → α -ketoglutaric acid → oxalosuccinic acid
B Isocitric acid → oxalosuccinic acid → α -ketoglutatric acid
C Oxalosuccinic acid → citric acid → α -ketoglutatric acid
D α -ketoglutaric acid → isocitric acid → oxalosuccinic acid

The correct answer is:
Isocitric acid → oxalosuccinic acid → α -ketoglutatric acid

Solution:
Acetyl-CoA donates its acetyl group to the four-carbon compound oxaloacetate to form the six-carbon citrate. Citrate is then transformed into isocitrate, also a six-carbon molecule, which is dehydrogenated with loss of CO2 to yield the five-carbon compound Î±â ketoglutarate. The latter undergoes loss of CO2 and ultimately yields the four-carbon compound succinate and a second molecule of CO2. Succinate is then enzymatically converted in three steps into the four-carbon oxaloacetate, with which the cycle began; thus, oxaloacetate is ready to react with another molecule of acetyl-CoA to start a second turn. In each turn of the cycle, one acetyl group (two carbons) enters as acetyl-CoA and two molecules of CO2 leave. In each turn, one molecule of oxaloacetate is used to form citrate but-after a series of reactions the oxaloacetate is regenerated. Therefore no net removal of oxaloacetate occurs; one molecule of oxaloacetate can theoretically suffice to bring about oxidation of an infinite number of acetyl groups. Four of the eight steps in this process are oxidations, in which the energy of oxidation is conserved, with high efficiency, in the formation of reduced cofactors (NADH and FADH2).

Topic: Biology, Respiration in Plants, NEET
The correct sequence of substances appeared during the course of origin of life on earth was

Options:
A glucose, ammonia, amino acids, nucleic acids, proteins.
B ammonia, amino acids, proteins, nucleic acid.
C nucleotides, ammonia, nucleic acids, enzymes.
D amino acids, ammonia, phosphates, nucleic acids.

The correct answer is:
ammonia, amino acids, proteins, nucleic acid.

Solution:
The initial atmosphere consisted of H, N, C and O. Hydrogen was most reactive and thus reacted with nitrogen to form ammonia, with carbon to form methane and with oxygen to form water. Thus these three are most important substances formed in the beginning. Due to high heat from sun water was in vapour form. These three substances constantly reacted with each other in the presence of sunlight and thus resulted in formation of various micromolecules. The first substance formed was amino acids. These together formed proteins. Due to UV rays there were many nucleotides formed which resulted in the formation of nucleic acids (RNA) and these acted as protobionts for formation of first cell. The first genetic material was RNA, from which DNA developed.

Topic: Biology, Evolution, NEET
The correct sequence of the arrangement of following parts in the mammary glands is

Options:
A Mammary alveolus, mammary duct, ampulla and mammary lobe
B Ampulla, mammary lobes, mammary alveolus and mammary ducts
C Mammary lobes, mammary ducts, mammary alveolus and ampulla
D Ampulla, mammary duct, mammary alveolus and mammary lobe

The correct answer is:
Ampulla, mammary duct, mammary alveolus and mammary lobe

Solution:
A functional mammary gland is characteristic of all female mammals. The mammary glands are paired structures (breasts) that contain glandular tissue and variable amount of fat. The glandular tissue of each breast is divided into 15-20 mammary lobes containing clusters of cells called alveoli. The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous duct through which milk is sucked out.

Topic: Biology, Human Reproduction, NEET
The correct sequence of the first six amino acids in HbA beta globulin peptide is

Options:
A Valine - Histidine - Leucine - Threonine - Proline - Glutamic acid
B Valine - Histidine - Leucine - Proline - Glutamic acid - Threonine
C Leucine - Valine - Histidine - Threonine - Proline - Glutamic acid
D Valine - Histidine - Leucine - Threonine - Proline - Valine

The correct answer is:
Valine - Histidine - Leucine - Threonine - Proline - Glutamic acid

Solution:
HbA is adult hemoglobin. The first six peptides on its beta-globin chain are Valine - Histidine - Leucine - Threonine - Proline - Glutamic acid. In sickle cell anemia, HbS is formed. It has valine instead of glutamic acid in the 6th position.

Topic: Biology, Principles of Inheritance and Variation, NEET
The correct set of four quantum numbers for the valence electron of rubidium atom $(Z=37)$ is

Options:
A $5,1,1,+\frac{1}{2}$
B $6,0,0,+\frac{1}{2}$
C $5,0,0,+\frac{1}{2}$
D $5,1,0,+\frac{1}{2}$

The correct answer is:
$5,0,0,+\frac{1}{2}$

Solution:
${ }_{37} \mathrm{Rb}={ }_{36}[\mathrm{Kr}] 5 s^1$
Its valence electron is $5 s^1$.
So,
$\begin{aligned}
& n =5 \\
& l =0 \text { (For } s \text { orbital) } \\
& m =0 (\text { As } m=-l \text { to }+l) \\
& s =+\frac{1}{2}
\end{aligned}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2012 (Screening)
The correct solubility order is/are
I. CaCO3>SrCO3>BaCO3
II. Li2CO3<Na2CO3<K2CO3
III. K2CO3<Rb2CO3<Cs2CO3
IV. Na2CO3>K2CO3>Rb2CO3

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

The correct answer is:
I, II, III

Solution:
As in case of alkali metals solubility increases down the group but decreases for alkaline earth metals.
Here the correct orders of solubility are
I. CaCO3>SrCO3>BaCO3
II. Li2CO3<Na2CO3<K2CO3
III. K2CO3<Rb2CO3<Cs2CO3

Topic: Chemistry, s Block Elements, NEET
The correct stability order for the following species is

Options:
A (II)>(IV)>(I)>(III)
B (I)>(II)>(III)>(IV)
C (II)>(I)>(IV)>(III)
D (I)>(III)>(II)>(IV)

The correct answer is:
(I)>(III)>(II)>(IV)

Solution:
Carbocations with more no of the hyperconjugation structures and charge delocalization are more stable.

Topic: Chemistry, General Organic Chemistry, NEET
The correct stability order of the following three quinones is

Options:
A I>III>II
B I=III>II
C I=II>III
D III>I>II

The correct answer is:
III>I>II

Solution:
The keto group has considerable resonance energy; hence more the number of keto groups in a compound, higher is its stability. However, here all the three compounds have same (two) number of keto groups. Compound II is least stable because two adjacent keto groups destabilize the molecule due to positive charge on the adjacent carbon atoms. Relative stability of I and II can be ascertained by the stability of their fully reduced forms (phenols), since, fully reduced form of III has two benzene rings, it will be more stable than the I whose form has only one benzene ring.
Thus

Topic: Chemistry, General Organic Chemistry, NEET
The correct statement about diseases of the cardiovascular system is

Options:
A Blood pressure of about 140/100 mm Hg causes rupturing of blood vessels of eyes.
B Hypertension significantly increases the after load on the heart, causes decrease in the heart size and finally heart failure.
C Coronary artery disease refers to a condition such as arteriosclerosis which causes formation of plaque which decreases the size of lumen of coronary arteries.
D The pain of Angina pectoris is severe, in the chest, and referred to the neck, lower jaw, left arm and left shoulder. It may disappear with rest.

The correct answer is:
The pain of Angina pectoris is severe, in the chest, and referred to the neck, lower jaw, left arm and left shoulder. It may disappear with rest.

Solution:
Blood pressure of 210/100 causes rupturing of blood vessels in brain. High blood pressure is also known as hypertension. It significantly causes increase in load of the heart. In Coronary artery disease there is accumulation of cholesterol which narrows the lumen of the heart. The pain of Angina pectoris is severe, in the chest and referred to the neck, lower jaw, left arm, and left shoulder. It may disappear with rest.

Topic: Biology, Body Fluids and Circulation, NEET
The correct statement about orthoboric acid is

Options:
A It is a strong monobasic acid
B It is not a proton donor, but a weak Lewis acid
C It is a tribasic acid
D It is harmful for eyes

The correct answer is:
It is not a proton donor, but a weak Lewis acid

Solution:
Orthoboric acid is a weak mono basic acid with Ka=1.0 x 10-9. It does not act as protonic acid (i.e. proton donor) but behaves as Lewis acid by accepting a pair of electrons from OH- ion.
B OH 3 + 2 H - O - H → B OH 4 - + H 3 O +

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The correct statement about ptyalin and pancreatic amylase is:

Options:
A Ptyalin breaks down only cooked starch in acidic medium while pancreatic amylase breaks down only uncooked starch in alkaline medium.
B Ptyalin breaks down only cooked starch in alkaline medium while pancreatic amylase breaks down only uncooked starch in acidic medium.
C Ptyalin breaks down only uncooked starch in acidic medium while pancreatic amylase breaks down only cooked starch in alkaline medium.
D Ptyalin breaks down only cooked starch in acidic medium while pancreatic amylase breaks down cooked as well as uncooked starch in alkaline medium.

The correct answer is:
Ptyalin breaks down only cooked starch in acidic medium while pancreatic amylase breaks down cooked as well as uncooked starch in alkaline medium.

Solution:
Ptyalin, i.e., salivary amylase, produced by parotid glands breaks down only cooked starch at pH 6.8, whereas amylopsin, i.e., pancreatic amylase, breaks down cooked as well as uncooked starch at pH 7.8.

Topic: Biology, Digestion and Absorption, NEET