datavorous/entrance-exam-dataset huggingface dataset:
Reaction $\mathrm{A} \rightarrow \mathrm{B}$ follows first order kinetics. The time taken for 0.8 mole of $A$ to produce 0.6 mole of $B$ is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of $\mathrm{B}$ ?

Options:
A 1 hour
B 0.5 hour
C 0.25 hour
D 2 hour

The correct answer is:
1 hour

Solution:
For first order reaction,
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)\)
Where \(k\) is rate constant
when 0.8 mol of \(A\) produces 0.6 mol of \(B\) then, \(0.8-0.6=0.2\) moles of A is left .
final moles \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.2\)
initial moles \(\mathrm{a}=0.8\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\
& \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\
& \mathrm{k}=2.303 \times 0.6020 \\
& \mathrm{k}=1.3865 \ldots \ldots . .(3)
\end{aligned}\)
In second case, 0.9 moles of A produces 0.675 moles of \(B\) so remaining moles of \(A\) are \(0.9-0.675=0.225\)
final moles of \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.225\)
initial mole \(=0.9\)
By using first order equation,
\(\begin{aligned}
\mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\
\mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\
\mathrm{t} & =1.661 \times 0.6020 \\
\mathrm{t} & =0.999=1 \mathrm{hr}
\end{aligned}\)

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2003
Reaction between C2H52Cd and CH3COCl leads to the formation of

Options:
A Diethyl ketone
B Ethyl methyl ketone
C Dimethyl ketone
D Acetaldehyde

The correct answer is:
Ethyl methyl ketone

Solution:
2C H 3 COCl+ ( C 2 H 5 ) 2 Cd→2 C 2 H 5 COC H 3 +CdC l 2

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Reaction of alkyl halides with aromatic compounds in presence of anhy \( \mathrm{AlCl}_{3} \) is known as

Options:
A Friedel-Craft's reaction
B Hofmann degradation
C Kolbe's synthesis
D Beckmann rearrangement

The correct answer is:
Friedel-Craft's reaction

Solution:
Friedel-Craft reaction : In this reaction alkyl halides react with aromatic compounds in presence of AlCl3 or FeCl3 to form alkyl substituted aromatic compounds.

Topic: Chemistry, Hydrocarbons, JEE Main
Reaction of alkyl halides with aromatic compounds in presence of anhydrous $\mathrm{AlCl}_3$ is known as

Options:
A Friedal-Craft reaction
B Hofmann degradation
C Kolbe’s synthesis
D Beckmann rearrangement

The correct answer is:
Friedal-Craft reaction

Solution:
Acylation or alkylation of aromatic compound in presence of $\mathrm{AlCl}_3$ is known as Friedal-craft reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Reaction of ammonia with excess Cl2 gives :

Options:
A NH4Cl and N2
B NH4Cl and HCl
C NCl3 and NH4Cl
D NCl3 and HCl

The correct answer is:
NCl3 and HCl

Solution:
NH3+3Cl2⟶NCl3+3HCl

limiting excess

8NH3+3Cl2⟶NH4Cl+N2

excess limiting

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (05 Sep Shift 2)
Reaction of an inorganic sulphite X with dilute H2SO4 generated compound Y. Reaction of Y with NaOH gives X. Further, the reaction of X with Y and water affords compound Z, Y and X, respectively, are :

Options:
A SO2 and Na2SO3
B SO3 and NaHSO3
C SO2 and NaHSO3
D S and Na2SO3

The correct answer is:
SO2 and NaHSO3

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (06 Sep Shift 2)
Reaction of BeCl2 with LiAlH4 gives :

(A) AlCl3

(B) BeH2

(C) LiH

(D) LiCl

(E) BeAlH4

Choose the correct answer from options given below

Options:
A A,D & E
B A,B & D
C D & E
D B,C & D

The correct answer is:
A,B & D

Solution:
All the alkali earth metals, except beryllium combine with hydrogen upon heating to form their hydrides, MH2. BeH2, however, can be prepared by the reaction of BeCl2 with LiAlH4. .

2BeCl2+LiAlH4→2BeH2+LiCl+AlCl3

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Reaction of benzene with $\mathrm{Me}_{3} \mathrm{CCOCl}$ in the presence of anhydrous AlCl ${ }_{3}$ gives

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Reaction of $\mathrm{Br}_2$ with $\mathrm{Na}_2 \mathrm{CO}_3$ in aqueous solution gives sodium bromide and sodium bromate with evolution of $\mathrm{CO}_2$ gas. The number of sodium bromide molecules involved in the balanced chemical equation is

Options:

The correct answer is:
5

Solution:
$\mathrm{Br}_2$ is disproportionated in basic medium as
$$
3 \mathrm{Br}_2+3 \mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow 5 \mathrm{NaBr}+\mathrm{NaBrO}_3+3 \mathrm{CO}_2
$$

Topic: Chemistry, Redox Reactions, JEE Main
Reaction of CoH2O62+ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t2g-orbitals of the product is

Options:

The correct answer is:
6

Solution:
CoH2O62++6NH3→airCoNH363++6H2O+e−

Electronic configuration of CoNH363+=t2g6eg0

Air oxides the Co(II) to Co(III)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known as

Options:
A
a Schiff’s base

B
an enamine

C
an imine

D
an amine

The correct answer is:
an enamine

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2005
Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known as

Options:
A
a Schiff’s base

B
an enamine

C
an imine

D
an amine

The correct answer is:
an enamine

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Reaction of ester with Grignard reagent give rise to

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

The correct answer is:
Tertiary alcohol

Solution:
All esters (except HCOOC2H5) on reaction with Grignard's reagent give rise to tertiary alcohol.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Reaction of methyl bromide with aqueous sodium hydroxide involves

Options:
A racemisation
B $S_{N} 1$ mechanism
C retention of configuration
D $\mathrm{S}_{\mathrm{N}} 2$ mechanism

The correct answer is:
$\mathrm{S}_{\mathrm{N}} 2$ mechanism

Solution:
Since, the reaction rate depends upon the concentration of both reactant and nucleophile, it is a $\mathrm{S}_{\mathrm{N}} 2$ reaction. It involves inversion of configuration.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Reaction of nitrous acid with aliphatic primary amine in the cold gives

Options:
A A diazonium salt
B An alcohol
C A nitrite
D A dye

The correct answer is:
An alcohol

Solution:
Aliphatic primary amine reacts with nitrous acid in cold to give alcohol.
$\mathrm{RNH}_2+\mathrm{HNO}_2 \rightarrow \mathrm{R}-\mathrm{OH}+\mathrm{N}_2+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Amines, JEE Main
Reaction of one molecule of $\mathrm{HBr}$ with one molecule of 1,3 -butadiene at $40^{\circ} \mathrm{C}$ gives predominantly

Options:
A
3-bromobutene under kinetically controlled conditions

B
1-bromo-2-butene under thermodymically controlled conditions

C
3-bromobutene under thermodynamically controlled conditions

D
1-bromo-2-butene under kinetically controlled conditions

The correct answer is:
1-bromo-2-butene under thermodymically controlled conditions

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2005
Reaction of phenol with chloroform/sodium hydroxide to give o -hydroxy benzaldehyde involves the formation of

Options:
A Dichloro carbene
B Trichloro carbene
C Chlorine atoms
D Chlorine molecules

The correct answer is:
Dichloro carbene

Solution:
This is Reimer-Tiemann reaction where the electrophile is dichlorocarbene generated from chloroform by the action of a base.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Reaction of phenol with dil, $\mathrm{HNO}_3$ gives

Options:
A $p$ and $m$-nitrophenols
B 0 - and $p$-nitrophenols
C Picric acid
D 0 - and $m$-nitrophenols

The correct answer is:
0 - and $p$-nitrophenols

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Reaction of propanamide with Br2/KOH (aq) produces:

Options:
A Ethylnitrile
B Propylamine
C Propanenitrile
D Ethylamine

The correct answer is:
Ethylamine

Solution:
It is Hoffmann bromamide degradation reaction in which propanamide is converted to an ethylamine containing one carbon atom lesser than in the amide.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Reaction of $\mathrm{R} \mathrm{OH}$ with $\mathrm{R}^{\prime} \mathrm{MgX}$ produces :

Options:
A $\mathrm{RH}$
B $\mathrm{R}^{\prime} \mathrm{H}$
C $\mathrm{R}-\mathrm{R}$
D $R^{\prime}-R^{\prime}$

The correct answer is:
$\mathrm{R}^{\prime} \mathrm{H}$

Solution:
Hints:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

Options:
A
4-phenylcyclopentene

B
2-phenylcyclopentene

C
1-phenylcyclopentene

D
3-phenylcyclopentene

The correct answer is:
3-phenylcyclopentene

Solution:
According to ${E}_2$ mechanism.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2006
Reaction rate between two substance $A$ and $B$ is expressed as following: rate $=k[A]^n[B]^m$
If the concentration of $\mathrm{A}$ is doubled and concentration of $\mathrm{B}$ is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

Options:
A
$m+n$

B
$n-m$

C
$\frac{1}{\left.2^{(m+n}\right)}$

D
$\left.2^{(n-m}\right)$

The correct answer is:
$\left.2^{(n-m}\right)$

Solution:
$$
\text { } \begin{aligned}
\text { Rate }_1 & =k[A]^n[B]^m \\
\text { Rate }_2 & =k[2 A]^n\left[\frac{1}{2} B\right]^m \\
\therefore \frac{\text { Rate }_2}{\text { Rate }_1} & =\frac{k[2 \mathrm{~A}]^n\left[\frac{1}{2} \mathrm{~B}\right]^m}{k[\mathrm{~A}]^n[\mathrm{~B}]^m}=(2)^n\left(\frac{1}{2}\right)^m \\
& =2^n \cdot(2)^{-m}=2^{n-m}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (07 May Online)
Reactions in Zeolite catalysts depend on

Options:
A Pores
B Apertures
C Size of cavities
D All of these

The correct answer is:
All of these

Solution:
It is a shape-selective catalyst.

Topic: Chemistry, Surface Chemistry, JEE Main
Reactivity of borazole is greater than that of benzene because

Options:
A Borazole is non-polar compound
B Borazole is a polar compound
C Borazole is electron deficient compound
D Not isostructural with benzene

The correct answer is:
Borazole is a polar compound

Solution:
Borazole is compound because of electronegativity difference between B and N but benzene is non - polar (all C - C bonds). Hence borazole is more reactive than benzene.
Borazole (borazine) is highly polar molecular due to high electronegativity difference between boron and nitrogen. The pi bonds in borazine are highly polarised than pi bonds in benzene. It is, therefore, more nucleophilic and hence more reactive than C 6 H 6 .

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Reactivity order of primary(p), secondary(s) and tertiary(T) alcohols towards esterification is

Options:
A T > S > P
B S > T < P
C P > S > T
D None of these

The correct answer is:
P > S > T

Solution:
In esterification, alcohol behaves as a nucleophile. So due to bulkyness, 3o alcohols show relatively poor nucleophilicity.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Read each of the statements carefully and choose the one that is true for the cell wall.

Options:
A The function of the middle lamella is to hold different neighbouring cells together.
B The cell wall is only known to provide shape and protection to the plant cell from infections.
C The primary wall disintegrates as the cell matures and the secondary wall is capable of growth.
D The algal cell wall is made up of cellulose, hemicellulose, pectins and proteins.

The correct answer is:
The function of the middle lamella is to hold different neighbouring cells together.

Solution:
The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together. The cell wall and middle lamellae may be traversed by plasmodesmata which connect the cytoplasm of neighbouring cells.

Topic: Biology, Cell - The Unit of Life, NEET
Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity
(c) with constant speed must have zero acceleration
(d) with positive value of acceleration must be speeding up

Options:

None

Solution:
(a) True: when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.
(b) False: because velocity is the speed of body in given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.
(c) True: when a particle is moving along a straight line with speed, its velocity remains constant with time. Therefore, acceleration (= change in velocity/time) is zero.
(d) False: True only when the chosen position direction is along the direction of motion.

Topic: Physics, Motion In One Dimension
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.

Options:

None

Solution:
(a) False, because kinetic energy is a scalar but does not remain conserved in an inelastic collision.
(b) False, because potential energy in a gravitational field may have negative values.
(c) False, because mass, length, time, speed, work, etc., all have dimensions.
(d) False, because speed, energy, etc., vary from point to point in space.
(e) True, because a scalar quantity will have the same value for observers with different orientations of axes since a scalar has no direction of its own.

Topic: Physics, Motion In Two Dimensions
Read following statements and identify the incorrect one in regards to Hardy-Weinberg equilibrium.

Options:
A The allelic frequency in a population are stable and constant from generation to generation
B Gene migration or gene flow is essential for the maintenance of Hardy - Weinberg principle
C Evolution occurs when Hardy - Weinberg principle in disturbed
D When frequency measured, differs from expected values, the difference indicates the extent of evolutionary charge

The correct answer is:
Gene migration or gene flow is essential for the maintenance of Hardy - Weinberg principle

Solution:
Gene migration or gene flow is one of the factors affecting Hardy-Weinberg principles. When a population is following the principle and staying in equilibrium, it is not evolving. Thus, the allele frequencies will not change throughout different generations. Gene migration is movement of species out of a population or going into it causing change in the gene pool, thus breaking the equibrium. Gene flow does not help in maintaining the equilibrium.

Topic: Biology, Evolution, NEET
Read the below statements and select the appropriate option.

a. Experimental removal of Pisaster resulted in the extinction of more than ten species of invertebrates.
b. Nearly 20 percent of all insects are known to be phytophagous.
c. Predators help to maintain species diversity.

Options:
A Both a and c are correct
B b is correct but c is incorrect
C a is correct but b is incorrect
D only c is correct

The correct answer is:
Both a and c are correct

Solution:
The biological interaction where one organism (predator) kills and eats another organism (prey) is called predation. Predation can be considered as a natural way of energy transfer between organisms in the trophic level. Predators are found to support species diversity by reducing competition among preys that are competing for resources. Predators have an important role in the ecosystem and their absence can have a detrimental effect. Example, in the American Pacific coast, Pisaster, a starfish is an important predator. On its removal from the ecosystem caused extinction of more than 10 species of invertebrates.
Nearly 25 percent of all insects are known to be phytophagous (feeding on plant sap and other parts of plants).

Topic: Biology, Organisms and Populations, NEET
Read the following analogies and state the one which is wrong.

Options:
A Starch: bluish black colour with iodine:: Glycogen : Red colour with iodine
B DNA: 4 types of monomers:: Protein: 20 types of monomers
C Oleic acid: MUFA :: Palmitic acid: PUFA
D Phospholipid: Lecithin:: Glycolipid: Ganglioside

The correct answer is:
Oleic acid: MUFA :: Palmitic acid: PUFA

Solution:
Starch, glycogen and cellulose are homopolysaccharides as these consist of only one type of monomer. Starch and glycogen consists of branched and unbranched chains while cellulose has only unbranched chains. Hence when Iodine is added to starch and glycogen, it forms complexes with the branched part which appears blue or red respectively. These complexes are not seen in cellulose. DNA is made up of many nucleotides (monomers) which consist of phosphate, a pentose sugar and four types of nitrogen bases. Thus, DNA consists of four types of monomers while macromolecule protein consists of 20 different amino acids out of which ten are essential and remaining ten are non-essential. Fatty acids can be saturated or unsaturated depending on the absence or presence of double bonds. Unsaturated fatty acids can be monounsaturated fatty acids (MUFA) or polyunsaturated fatty acids (PUFA) as it can have only one double bond or multiple double bonds respectively. Examples of MUFA are oleic acid, palmitic acid etc., while omega-3 fatty acid is PUFA.

Topic: Chemistry, Biomolecules, NEET
Read the following five statements (A - E) and answer as asked next to them.
(A) In Equisetum, the female gametophyte is retained on the parent sporophyte.
(B) In Ginkgo, male gametophyte is not independent.
(C) The sporophyte in Riccia is more developed than that in Polytrichum.
(D) Sexual reproduction in Volvox is isogamous.
(E) The spores of slime moulds lack cell wall.

Options:
A Two
B Three
C Four
D One

The correct answer is:
One

Solution:
Equisetum is a pteridophyte and in pteridophytes, the main plant body is sporophyte which is differentiated into true root, stem and leaves. Gametophytes are small or inconspicuous and free living, mostly photosynthetic thalloid called prothallus.
Riccia is a liverwort and Polytrichum is a moss. The sporophyte in mosses is more elaborate than that in liverworts.
Volvox shows anisogamous fusion i.e., fusion between one large, non motile (static) female gamete and a smaller, motile male gamete.
During unfavourable conditions, the slime mould diferrentiates and forms fruiting bodies bearing spores at their tips. The spores possess true walls. They are extremely resistant and survive for many years, even under adverse conditions.

Topic: Biology, Plant Kingdom, NEET
Read the following five statements (A to E) and select the option with all correct statements:
(a) Mosses and lichens are the first organisms to colonize bare rock.
(b) Selaginella is a homosporous pteridophyte.
(c) Coralloid roots in Cycas have VAM.
(d) Main plant body in bryophytes is gametophytic, whereas in pteridophytes it is sporophytic.
(e) In gymnosperms, male and female gametophytes are present within sporangia located on sporophytes.

Options:
A (a), (c) and (d)
B (b), (c) and (d)
C (a), (d) and (e)
D (b), (c) and (e)

The correct answer is:
(a), (d) and (e)

Solution:
Mosses are found in moist, shady places, moist rocks, and damp walls or on tree trunks as well as lichens grow on rock surfaces & named by Theophrastus are the first organisms to colonize bare rock.

Selaginella is a heterosporous pteridophye, meaning both male and female spores are produced by the plant.

Coralloid roots of Cycas have nitrogen-fixing cyanobacteria. Mycorrhiza (VAM) is seen in the roots of Pinus.
Main plant body in bryophytes is gametophyte because gametophyte is dominant over sporophyte as gametophyte is branched, haploid, long-lived and independent whereas sporophyte is diploid short-lived and dependent upon gametophyte. Main plant body in pteridophyte is sporophyte.

Unlike bryophytes and pteridophytes, the male and the female gametophytes of gymnosperms do not have an independent free-living existence. They remain within the sporangia retained on the sporophytes.

Topic: Biology, Plant Kingdom, NEET, NEET 2015 (Phase 1)
Read the following five statements (A-E) and answer the question.

A Pollination by water is more common amongst abiotic pollinating agent.
B Pollination by water is quite rare in flowering plants.
C Water hyacinth and water lily, the flowers emerge above the level of water and are pollinated by water.
D Wind pollinated plants numerous flowers packed into an inflorescence.
E Self incompatibility is an outbreeding device which prevents inbreeding.

How many of the above statements are correct?

Options:
A Four
B One
C Two
D Three

The correct answer is:
Three

Solution:
Pollination by water is quite rare in flowering plants and it occurs only in 30 genera, mainly monocots. Air is more common amongst abiotic pollinating agent. Water hyacinth and water lily, the flowers emerges above the water surface and are pollinated by insects. In wind pollination small flowers are packed into inflorescence to make them more conspicuous. Self incompatibility is the mechanism which prevents self pollen from germinating on the stigma, thus inhibits inbreeding.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Read the following five statements and answer as asked next to them
I. In Equisetum the female gametophyte is retained on the parent sporophyte.
II. In Ginkgo male gametophyte is not independent.
III. The sporophyte in Riccia is more developed than that in Polytrichum.
IV. Sexual reproduction in Volvox is isogamous.
V. The spores of slime molds lack cell walls How many of the above statements are correct?

Options:
A Two
B Three
C Four
D One

The correct answer is:
One

Solution:
The spore of Equisetum germinate shortly after shedding and gives rise to form prothallus. Prothalli of Equisetum are basically monoecious and the gametophytes are rarely seen.
The simplest known sporophyte among bryophytes is that of Riccia while highest evolved known is Bryopsida (Polytrichum, Pogonatum).
Sexual reproduction in Volvox is oogamous type.
The spores in slime molds (e.g., Physarum) are haploid, unicellular, usually globose with smooth, spiny or reticulately thickened cell walls.

Topic: Biology, Plant Kingdom, NEET, NEET 2012 (Mains)
Read the following four statement.
I. In transcription, adenosine pairs with uracil.
II. Regulation of lac operon by repressor is referred to as positive regulation.
III. The human genome has approximately 50000 genes.
IV. Haemophilia is a sex-linked recessive disease.
How many of the above statements are right?

Options:
A Two
B Three
C Four
D One

The correct answer is:
Two

Solution:
Transcription is the process of synthesis of RNA on the DNA template. During transcription adenosine pairs with uracil.
Repressor gene determines the transcription of structural gene. It codes for repressor protein. After synthesis the repressor molecule is diffused from the ribosome and bind to the operator in absence of inducer.
The human genome has approximately 30000 genes with $\sim 3.2$ billion base pairs.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2012 (Mains)