datavorous/entrance-exam-dataset huggingface dataset:
In the 1950s, kinetin was first discovered by

Options:
A Skoog and Miller.
B Letham.
C Calvin.
D Thimman and Went.

The correct answer is:
Skoog and Miller.

Solution:
F. Skoog and his co-workers observed that from the internodal segments of tobacco stems, the callus (a mass of undifferentiated cells) proliferated only if, in addition to auxins the nutrients medium was supplemented with one of the following: extracts of vascular tissues, yeast extract, coconut milk or DNA. Miller et al. (1955), later identified and crystallised the cytokinesis promoting active substance that they termed kinetin.

Topic: Biology, Plant - Growth and Development, NEET
In the ABO system of blood groups, if both antigens are present but no antibody, the blood group of the individual would be:

Options:
A $\mathrm{B}$
B $\mathrm{O}$
C $\mathrm{AB}$
D $\mathrm{A}$.

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

Solution:
ABO Blood group system
\(\begin{array}{|c|c|c|c|}
\hline \begin{array}{c}
\text { Blood } \\
\text { group }
\end{array} & \begin{array}{c}
\text { Antigen(s) } \\
\text { present on } \\
\text { the red blood } \\
\text { cells }
\end{array} & \begin{array}{c}
\text { Antibodies } \\
\text { present } \\
\text { in the } \\
\text { serum }
\end{array} & \begin{array}{c}
\text { Geno- } \\
\text { type(s) }
\end{array} \\
\hline \text { A } & \text { A antigen } & \text { Anti-B } & \begin{array}{c}
\text { AA or } \\
\text { AO }
\end{array} \\
\hline \text { B } & \text { B antigen } & \text { Anti-A } & \begin{array}{c}
\mathrm{BB} \text { or } \\
\text { BO }
\end{array} \\
\hline \mathrm{AB} & \begin{array}{c}
\text { A antigen and } \\
\text { B antigen }
\end{array} & \text { None } & \mathrm{AB} \\
\hline \mathrm{O} & \begin{array}{c}
\text { B antigen }
\end{array} & \begin{array}{c}
\text { Anti-A and } \\
\text { Anti-B }
\end{array} & \mathrm{OO} \\
\hline
\end{array}\)
Related Theory
The ABO blood types were discovered by Karl Landsteiner in 1901.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2004
In the above circuit diagram emf of two batteries are equal, then the potential difference VA-VB between terminals A and B will be

Options:
A â36 V
B + 36 V
C + 24 V
D â 24 V

The correct answer is:
â36 V

Solution:
According to Kirchhoff's Voltage and current law
VA- E+8+12+16+E=VB
VA- VB=â 36 V

Topic: Physics, Current Electricity, NEET
In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

Options:
A $B A$ and $C D$
B $A B$ and $C D$
C $B A$ and $D C$
D $A B$ and $D C$

The correct answer is:
$A B$ and $D C$

Solution:
North of magnet is moving away from solenoid 1 so end $B$ of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end $C$ of solenoid 2 is South.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2024
In the above reaction, P and Q respectively are

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
In the above reaction product ' $\mathrm{P}$ ' is:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2002
In the absence of an external source of water, the kangaroo rat in North American deserts is capable of meeting all its water requirements through

Options:
A internal carbohydrate oxidation.
B internal fat oxidation.
C internal protein oxidation.
D Both (B) and (C)

The correct answer is:
internal fat oxidation.

Solution:
The kangaroo rats are well adapted and capable of surviving in dry conditions by oxidizing the internal fat that helps in meeting the water requirements. Fat oxidation produces water as by product. Also, it has the ability to concentrate urine.

Topic: Biology, Organisms and Populations, NEET
In the acid-catalysed dehydration of alcohol to alkene, reactive intermediate is

Options:
A free radical
B carbocation
C carbanion
D hydronium ion

The correct answer is:
carbocation

Solution:
In the acid-catalysed dehydration of alcohol to alkene, reactive intermediate is carbocation.
(i) CH3-CH2-O. .. .-H+H+âCH3-CH2-O+H2
(ii) CH 3 − CH 2 − O + H 2 → CH 3 − C + H 2 Carbocation +  H 2 O
(iii) CH3+C+H2âCH2=CH2+H+

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the acid-catalysed hydration of alkenes, what is the role of the acid catalyst?

Options:
A To act as a reducing agent
B To act as an oxidising agent
C To facilitate the formation of a carbocation
D To increase the reaction rate

The correct answer is:
To facilitate the formation of a carbocation

Solution:
The acid catalyst in the reaction aids in the formation of a carbocation during the mechanism of hydration.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the acylation of amines, which atom in the amine is replaced by the acyl group?

Options:
A Nitrogen
B Hydrogen
C Oxygen
D Carbon

The correct answer is:
Hydrogen

Solution:
In acylation, the hydrogen atom of the -NH2 or >N-H group in the amine is replaced by the acyl group.

Topic: Chemistry, Amines, NEET
In the addition of hydrogen halides to alkenes, what type of reaction occurs?

Options:
A Nucleophilic substitution
B Electrophilic addition
C Radical substitution
D Acid-base reaction

The correct answer is:
Electrophilic addition

Solution:
Addition of hydrogen halides to alkenes is an example of electrophilic addition, where the hydrogen halide adds to the double bond, and the hydrogen attaches to one carbon, while the halide attaches to the other.

Topic: Chemistry, Hydrocarbons, NEET
In the addition of hydrogen halides to alkenes, which halide atom becomes attached to one carbon of the alkene?

Options:
A Chlorine
B Bromine
C Iodine
D Fluorine

The correct answer is:
Iodine

Solution:
In the addition of hydrogen halides to alkenes, the halide atom that becomes attached to one carbon of the alkene is iodine (I), in the case of HI addition.

Topic: Chemistry, Hydrocarbons, NEET
In the addition of hydrogen halides to alkenes, which hydrogen becomes attached to one carbon of the alkene?

Options:
A The one closer to the halogen
B The one closer to the double bond
C Both hydrogens attach equally
D None of the hydrogens attach

The correct answer is:
The one closer to the double bond

Solution:
In the addition of hydrogen halides to alkenes, the hydrogen closer to the double bond becomes attached to one carbon of the alkene.

Topic: Chemistry, Hydrocarbons, NEET
In the addition of water to alkenes in the presence of concentrated sulfuric acid, what is the role of sulfuric acid?

Options:
A It serves as a catalyst.
B It provides oxygen for the reaction.
C It helps in dehydrating the product.
D It promotes the formation of aldehydes.

The correct answer is:
It helps in dehydrating the product.

Solution:
Concentrated sulfuric acid serves as a dehydrating agent in the addition of water to alkenes, removing water molecules to promote product formation.

Topic: Chemistry, Hydrocarbons, NEET
In the addition of water to alkenes with concentrated sulfuric acid, what is the role of the concentrated sulfuric acid beyond serving as a dehydrating agent?

Options:
A It provides oxygen for the reaction.
B It acts as a reducing agent.
C It serves as a catalyst.
D It initiates radical reactions.

The correct answer is:
It serves as a catalyst.

Solution:
Concentrated sulfuric acid serves as a catalyst in addition to its role as a dehydrating agent in this reaction.

Topic: Chemistry, Hydrocarbons, NEET
In the additive nomenclature for coordination compounds, what is the first thing to be named?

Options:
A The central atom
B The ligands
C The oxidation state
D The anionic ligands

The correct answer is:
The central atom

Solution:
In additive nomenclature, the cation or central atom is named first in both positively and negatively charged coordination entities.

Topic: Chemistry, Coordination Compounds, NEET
In the adjacent circuit diagram, each resister is of 16 Î© . The equivalent resistance between A and B is

Options:
A 1 Ω
B 2 Ω
C 3 Î©
D 4 Ω

The correct answer is:
3 Î©

Solution:
RAB =3r4.r4r=316r=316 Ã 16=3 Î©

Topic: Physics, Current Electricity, NEET
In the adjacent circuit, the instantaneous current equation is

Options:
A 2sin⁡100t-π4
B 2sin⁡100t-π4
C 2sin⁡200t-π4
D 2100t+π4

The correct answer is:
2sin⁡100t-π4

Solution:
Let Ï is the phase difference between V and i
tanâ¡Ï=XLR=ÏLR=100Ã1100=1
â Ï=Ï4
Since this is an LR circuit, V will lead i
â´ The instantaneous current equation will be
i=i0 sinâ¡100 t-Ï where Ï=Ï4 and i0=V0Z
i0=V0R2+ÏL2=2001002+100Ã12=2
â´i=2sinâ¡100 t-Ï4

Topic: Physics, Alternating Current, NEET
In the adjoining circuit diagram, E=5 V, r=1 Î©, R2=4 Î©, R1=R3=1 Î© and C=3 mF. Then the numerical value of the charge on each plate of the capacitor is

Options:
A 24 μC
B 12 μC
C 6 μC
D 3 μC

The correct answer is:
6 μC

Solution:
Current through resistors R1 and R3 in steady state is zero.
Therefore current through R2 is
i2=ER2+r=54+1=1 A
Therefore charge on capacitor is
q=Ci2R22=6 μC

Topic: Physics, Current Electricity, NEET
In the adjoining figure, the electric field lines for charges q1 and q2 are shown. Identify the sign of the charges.

Options:
A both negative
B upper charge is negative and lower is positive
C both positive
D upper charge is positive and lower is negative

The correct answer is:
both negative

Solution:
Field lines terminate into negative charge. Therefore both q1 and q2 are negatively charged.

Topic: Physics, Electrostatics, NEET
In the adsorption of oxalic acid on activated charcoal, the activated charcoal is called

Options:
A Adsorbent
B Adsorbate
C Adsorber
D Absorber

The correct answer is:
Adsorbent

Solution:
In the adsorption of oxalic acid on activated charcoal, the activated charcoal is called adsorbent as adsorbent is a substance on which adsorption takes place.

Topic: Chemistry, Surface Chemistry, NEET
In the alkane H3C–CH2–C(CH3)2–CH2–CH(CH3)2,identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.

Options:
A 15 H attached to 1° carbons,8 H attached to 2° carbons and 1 H attached to 3° carbons
B 14 H attached to 1° carbons,6 H attached to 2° carbons and 1 H attached to 3° carbons
C 16 H attached to 1° carbons,4 H attached to 2° carbons and 2 H attached to 3° carbons
D 15 H attached to 1° carbons,4 H attached to 2° carbons and 1 H attached to 3° carbons

The correct answer is:
15 H attached to 1° carbons,4 H attached to 2° carbons and 1 H attached to 3° carbons

Solution:
Correct Option is : (D)
15 H attached to 1° carbons,4 H attached to 2° carbons and 1 H attached to 3° carbons

Topic: Chemistry, Hydrocarbons, NEET
In the aluminothermite process, aluminium acts as

Options:
A an oxidizing agent
B a flux
C a reducing agent
D a solder

The correct answer is:
a reducing agent

Solution:
In the aluminothermite process, aluminium acts as a reducing agent.
2Al+Fe2O3→Al2O3+2Fe+Heat
Al reduces Fe2O3 to Fe.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
In the aromatization or reforming reaction, what is the primary product obtained from the conversion of n-alkanes?

Options:
A Alkenes
B Alkynes
C Aromatic hydrocarbons
D Alkanes

The correct answer is:
Aromatic hydrocarbons

Solution:
The primary product obtained from the conversion of n-alkanes in the aromatization or reforming reaction is aromatic hydrocarbons.

Topic: Chemistry, Hydrocarbons, NEET
In the arrangement of 3 bulbs of 100 W each as shown in the figure, total power consumption will be

Options:
A 300Â W
B 50Â W
C 150 W
D 25Â W

The correct answer is:
150 W

Solution:
For the two bulbs in series, equivalent power is
Pâ²=P1P2P1+P2=PPP+P
Pâ²=100100100+100
Pâ²=50 W
This Pâ² is in parallel with other bulb. Therefore overall equivalent power is
Peq=P'+100 W
Peq=150 W

Topic: Physics, Current Electricity, NEET
In the arrangement shown in the figure, slits S3 and S4 are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of S1S2 and S3S4. (Assume D>>d)

The minimum value of Z for which the intensity at O is zero is

Options:
A 3λD2d
B λD2d
C λD3d
D λDd

The correct answer is:
λDd

Solution:
Intensity at O is due to S3 and S4.
∴ Intensity at S3 = zero and Intensity at S3 is given by
I=K cos 2 ( ϕ 2 )
Intensity at O is zero if ϕ2=π2
ϕ=2πβ×Z2=π ⇒Z=β ;
Z=β=λDd
ϕ=phase difference between waves at S3 and S4

Topic: Physics, Wave Optics, NEET
In the arrangement shown, the pulleys, strings and the spring balance, all are ideal. If m1>m2, then the reading of the spring balance is

Options:
A m1-m2
B 12m1+m2
C 4m1m2m1+m2
D 2m1m2m1+m2

The correct answer is:
2m1m2m1+m2

Solution:
The acceleration of the system = m 1 - m 2 m 1 + m 2 g and hence the tension T in the string is 2 m 1 m 2 m 1 + m 2 g
The reading of the spring balance is T (in units of force) and T/g(in units of mass).

Topic: Physics, Laws of Motion, NEET
In the Arrhenius concept, what is the role of strong bases in an aqueous medium?

Options:
A They act as proton donors.
B They act as proton acceptors.
C They partially dissociate into ions.
D They are almost completely dissociated into ions, producing hydroxyl ions.

The correct answer is:
They are almost completely dissociated into ions, producing hydroxyl ions.

Solution:
In the Arrhenius concept, strong bases are described as almost completely dissociated into ions in an aqueous medium, giving rise to hydroxyl ions (OH-).

Topic: Chemistry, Ionic Equilibrium, NEET
In the artery,

Options:
A systolic blood pressure is lesser than diastolic blood pressure
B the duration of systolic blood pressure is greater than the duration of diastolic blood pressure
C the duration of systolic blood pressure is lesser than the duration of diastolic blood pressure
D the duration of systolic blood pressure is same as the duration of diastolic blood pressure

The correct answer is:
the duration of systolic blood pressure is lesser than the duration of diastolic blood pressure

Solution:
The maximum pressure exerted by blood on the walls of arteries when the heart contracts is called Systolic blood pressure. Its value is 120 mm Hg. The minimum pressure exerted by blood on the walls of arteries when the heart relaxes is called Diastolic blood pressure. Its value is 80 mm Hg. Systolic blood pressure in arteries is due to ventricular systole which lasts for 0.3 seconds while the diastolic blood pressure in arteries is due to ventricular diastole which lasts for 0.5 seconds.

Topic: Biology, Body Fluids and Circulation, NEET
In the Auger process, an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom (an Auger electron). Assuming the nucleus to be massive, what is the kinetic energy of an n = 4 Auger electron emitted by Chromium after absorbing the energy from an n = 2 to n = 1 transition?

Options:
A 5385.6 eV
B 6321.8 eV
C 7210.4 eV
D 4214.2 eV

The correct answer is:
5385.6 eV

Solution:
Correct Option is : (A)
5385.6 eV

Topic: Chemistry, Atomic Physics, NEET
In the Avery, MacLoed, and McCarty experiment, which of the following enzyme did not affect the process of transformation?

Options:
A Protease
B RNase
C DNase
D Both (A) and (B)

The correct answer is:
Both (A) and (B)

Solution:
Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein. They worked to determine the biochemical nature of ‘transforming principle’ in Griffith's experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S-cells to see which ones could transform live R-cells into S-cells. They discovered that DNA alone from S-cells bacteria caused R bacteria to become transformed. They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.

Topic: Biology, Molecular Basis of Inheritance, NEET
In the balanced chemical equation
IO3-+aI-+bH+→cH2O+dI2
a, b, c and d are respectively.

Options:
A 5, 6, 3, 3
B 5, 3, 6, 3
C 3, 5, 3, 6
D 5, 6, 5, 6

The correct answer is:
5, 6, 3, 3

Solution:
The balanced equation is given below
IO3-+5I-+6H+→3H2O+3I2
So a = 5, b = 6, c = 3 and d = 3

Topic: Chemistry, Redox Reactions, NEET
In the balanced chemical reaction,

IO3-+aI-+bH+â¶cH2O+d I2

a, b, c and d respectively correspond to -

Options:
A 5, 6, 3, 3
B 5, 3, 6, 3
C 3, 5, 3, 6
D 5, 6, 5, 5

The correct answer is:
5, 6, 3, 3

Solution:
IO3-+aI-+ 6H+⟶cH2O+dI2
step-1, I-⟶ I2Oxidn , IO3-⟶I2Redn
step-II 2IO3-+ 12H+⟶ I2+6H2O
step-III, 2IO3–+12H++10e–⟶ I2+ 6H2O
2I–⟶I2+2e–
step-IV, 2IO3-+ 12H++10e–→ I2+ 6H2O
2I→I2+2e– × 5
step-V, 2IO3–+10 I–+12H+→6I2+6H2O
IO3–+5I–+ 6H+→3I2+3H2O
On comparing a=5, b=6, c=3, d=3

Topic: Chemistry, Redox Reactions, NEET
In the below circuit, AB is a wire of length 100 cm with 5 Î© resistance. If there is no deflection in the galvanometer, the current flowing in the wire AB is

Options:
A 0.1 A
B 0.5 A
C 1.0 A
D 1.4 A

The correct answer is:
0.1 A

Solution:
Since no current flows through galvanometer,
â´ Equivalent circuit will be
â´ iAB1.510+5 Î©0.1 A

Topic: Physics, Current Electricity, NEET
In the below diagram, identify the a, b, c and d.

Options:
A a - Strobilus, b - Node, c - Internode, d - Rhizoids
B a - Cone, b - Stem, c - Leaves, d - Rhizods
C a - Strobilus, b - Leaves, c - Stem, d - Rhizome
D a - Strobilus, b - Internode, c - Node, d - Rhizome

The correct answer is:
a - Strobilus, b - Internode, c - Node, d - Rhizome

Solution:
Above diagram is of Equisetum which is a pteridophyte and here, a- Strobilus, b - Internode, c - Node, d - Rhizome

Topic: Biology, Morphology of Flowering Plants, NEET
In the below diagrams, A and B are initial areas of leaves. A' and B' are areas of leaves after growth. Which of the following is correct?

Options:
A Both relative and absolute growth of leaves are same
B Absolute growth is same, but relative growth is different
C Absolute growth is different, but relative growth is the same
D Both absolute and relative growth of leaves are different

The correct answer is:
Absolute growth is different, but relative growth is the same

Solution:
Quantitative comparisons between the growth of a living system can also be made in two ways: (i) measurement and the comparison of total growth per unit time is called the absolute growth rate. (ii) The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter, is called the relative growth rate.
Here, absolute growth of leaf A to Aâ is 2.5 cm2, whereas that of leaf B to Bâ is 10cm2. Hence, absolute growth is different. Since 2.5 is 25% of 10, there is a total 25% increase in growth in leaf A, and in leaf B, 10 is 25% of 40, so there is also 25% increase in growth. Hence, the relative growth of two leaves is the same, that is 25%.

Topic: Biology, Plant - Growth and Development, NEET
In the borax bead test, which compound is formed?

Options:
A Orthoborate
B Metaborate
C Double oxide
D Tetraborate

The correct answer is:
Metaborate

Solution:
In borax bead test the coloured Metaborates are formed by transition metal salts.
$\begin{aligned}
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} & 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3 \\
\text { Salt of } \mathrm{Co} \stackrel{\Delta}{\longrightarrow} & \mathrm{CoO}+\text { gas } \\
\mathrm{CoO}+\mathrm{B}_2 \mathrm{O}_3 \stackrel{\Delta}{\longrightarrow} & \mathrm{Co}\left(\mathrm{BO}_2\right)_2 \\
& \begin{array}{c}
\text { cobalt metaborate } \\
\text { (blue colour) }
\end{array}
\end{aligned}$

Topic: Chemistry, Practical Chemistry, NEET, NEET 2002
In the Born-Haber cycle for the formation of solid common salt NaCl , the largest contribution comes from

Options:
A The low ionisation energy of Na
B The high electron affinity of Cl
C The low ΔHvap of Nas
D The lattice energy

The correct answer is:
The lattice energy

Solution:
According to Born Haber cycle enthalpy of formation is ΔHf=ΔHsubl.+12ΔHdiss.+IE-EA+LE

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
In the bottom of a vessel with mercury of density ρ there is a round hole of radius r . At what maximum height of the mercury layer will the liquid still not flow out through this hole? (Surface tension = T )

Options:
A Trρg
B T2rρg
C 2Trρg
D 4Trρg

The correct answer is:
2Trρg

Solution:
Because mercury meniscus is convex, the pressure just inside the hole will be less than the outside pressure by 2Tr
∴ hρg=2Tr
or h=2Trρg

Topic: Physics, Mechanical Properties of Fluids, NEET
In the Brönsted-Lowry theory, what is the primary role of ammonia molecules?

Options:
A To act as strong acids
B To act as strong bases
C To act as proton donors
D To act as proton acceptors

The correct answer is:
To act as proton acceptors

Solution:
In the Brönsted-Lowry theory, ammonia molecules primarily act as proton acceptors.

Topic: Chemistry, Ionic Equilibrium, NEET
In the bromination of anisole, what role does iron (III) bromide play as a catalyst?

Options:
A It activates the benzene ring
B It stabilises the bromine atoms
C It prevents side reactions
D It increases the reaction rate

The correct answer is:
It increases the reaction rate

Solution:
Iron (III) bromide acts as a Lewis acid catalyst, accelerating the bromination reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the C4 cycle, the primary CO2 acceptor is a 3C molecule ...(i)........ and is present in the mesophyll cells. The enzyme responsible for this fixation is .....(ii).....

Options:
A (i) PEP (ii) PEP carboxylase
B (i) PEP (ii) RuBiSCO
C (i) OAA (ii) PEP carboxylase
D (i) PGA (ii) RuBisCO

The correct answer is:
(i) PEP (ii) PEP carboxylase

Solution:
C4 plants shows Kranz anatomy in which mesophyll chloroplast differs from bundle sheath chloroplast. In C4 cycle the primary CO2 acceptor is a 3C molecule PEP and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the C4 pathway

Options:
A Chloroplasts are of same type
B Kranz anatomy occurs where mesophyll have large chloroplasts whereas bundle sheath have granular chloroplasts
C Kranz anatomy occurs where mesophyll have small chloroplasts whereas bundle sheath have larger agranular chloroplasts
D Kranz anatomy where mesophyll cells are diffused

The correct answer is:
Kranz anatomy occurs where mesophyll have small chloroplasts whereas bundle sheath have larger agranular chloroplasts

Solution:
C4 plants have 'Kranz type' anatomy. Vascular bundles are surrounded by bundle sheath cells. The chloroplasts are dimorphic. The cells of bundle sheath have very large chloroplasts. They lack grana and contain starch grains. They are centripetally arranged white mesophyll cells have small and granular chloroplast.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the Calvin cycle for the fixation of 5 molecules of CO2, how many ATP and NADPH are required in the reduction step?

Options:
A 18 ATP and 12 NADPH
B 15 ATP and 10 NADPH
C 10 ATP and 10 NADPH
D 3 ATP and 2 NADPH

The correct answer is:
10 ATP and 10 NADPH

Solution:
During the Calvin cycle, for fixing one molecule of CO2, 2 ATP and 2 NADPH are required in the reduction step. Hence, for the fixation of 5 molecules of CO2, 10 ATP, and 10 NADPH are required in the reduction step.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the Cannizzaro reaction given below –
2Ph–CHO→Ph–CH2OH+PhCOO– in the presence of OH– the slowest step is –

Options:
A The attack of OH– at the carbonyl group
B The transfer of hydride to the carbonyl group
C The abstraction of proton from the carboxylic group
D The deprotonation of pH-CH2OH

The correct answer is:
The transfer of hydride to the carbonyl group

Solution:
In Cannizzaro’s reaction attack of hydride is RDS .

Topic: Chemistry, Aldehydes and Ketones, NEET