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NEET | Botany

Name: _______________ NEET Test Series Date: 08/11/2024

Roll No: __________ Time: 01 Hour(s) Marks: 180

(1). The final proof for DNA as the genetic material came from the experiments of : (4)
A. Griffith B. Hershey and Chase

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C. Avery, Mcleod and McCarty D. Hargobind Khorana
Solution (Explanation).
As we learnt in

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Conclusion Made by Hershey and Chase -
DNA in the genetic material that is passed from virus to bacteria.
-

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(2). During DNA replication, Okazaki fragments are used to elongate: (4)
A. The leading strand towards replication fork. B. The lagging strand towards replication fork.
C. The leading strand away from replication fork. D. The lagging strand away from the replication fork.
Solution (Explanation).

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As discussed in
Replication Fork -
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For long DNA molecules since the two strands, od DNA cannot be separated, the reparation occur within a small opening
of DNA helix, referred to as replication fork.
- wherein
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TU
T
EN

Two DNA polymerase molecules work simultaneous at the DNA fork, one on the leading strand and the other on the
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lagging strand.
Each Okazaki fragment is synthesized by DNA polymerase at lagging strand in direction.
New Okazaki fragments appear as the replication fork opens further.
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As the first Okazaki fragment appears away from the replication fork, the direction of elongation would be away from
replication fork.
(3). The association of histone H1 with a nucleosome indicates: (4)
A. Transcription is occurring. B. DNA replication is occurring.
EX

C. The DNA is condensed into a Chromatin Fibre. D. The DNA double helix is exposed.
Solution (Explanation).
As we learnt in
Histones -
This is a set of positively charged basic proteins.
- wherein
Histones are rich in the basic amino acid residues lysines and arginines. They are organised to form a unit of eight
molecules called histone octamer.
The DNA is condensed into a Chromatin Fibre.
(4). Which of the following RNAs should be most abundant in animal cell? (4)
A. r-RNA B. t-RNA
 C. m-RNA D. mi-RNA
Solution (Explanation).
As we learnt in
Functions of types of RNA -
The mRNA provides the templete tRNA brings aminoacids and reads genetic code and rRNAs plays structural and
catalytic during translation.
-
rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell.
(5). The experimental proof for semiconservative replication of DNA was first shown in a (4)
A. Plant B. Bacterium

.
C. Fungus D. Virus

S..
Solution (Explanation).
As we learnt in
Observations of experiments done by meselsons and stahl -

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The DNA that was extracted from the culture of E.coli bacteria one generation after the transfer from 15 N to 14 N
medium had a hybrid or intermediate density or DNA extracted another generation (40 minutes) was composed of an
equal amount of this hybrid DNA

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-
Semi-conservative DNA replication was first shown in Bacterium Escherichia coli by Matthew Meselson and Franklin
Stahl.
(6). AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of (4)

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the transcribed mRNA ?
A. ACCUAUGCGAU B. UGGTUTCGCAT

C. AGGUAUCGCAU
IO D. UCCAUAGCGUA
Solution (Explanation).
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As we learnt in
DNA and RNA components -
Cytosine is common for both DNA and RNA and Thymine is present in DNA.
TU

- wherein
Uracil is present in RNA at the place of thymine.
Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA.
(7). All of the following are part of an operon except (4)
A. an enhancer B. structural genes
T

C. an operator D. a promoter
EN

Solution (Explanation).
As we learnt in
Haemophilia -
This is sex-linked recessive disease, which shows transmission from unaffected carrier female to some of the male
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progeny.
- wherein
In this, a simple cut will result in non-stop bleeding and no blood clotting will take place.
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Operons are a part of prokaryotes and since Enhancer sequences are present in Eukaryote, operons are not a part of
enhancer.
(8). Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant (4)
DNA technology ?
EX

A. Genetic code is not ambiguous B. Genetic code is redundant

C. Genetic code is nearly universal D. Genetic code is specific

Solution (Explanation).
Salient features of genetics code -
The code is nearly universe eg; from bactria it human, UUU would code for phenylalanine. Some exception are found in
mitochondrial codons, and in some protozoans.
-
Genetic code is nearly universal. It means that the codons code for the same amino acid in all the organisms except for
some minor deviations.This property of genetic code allows bacteria to produce human insulin through recombinant
technology.
(9). Purines found both in DNA and RNA are: (4)
 A. Adenine and thymine B. Adenine and guanine

C. Guanine and cytosine D. Cytosine and thymine

Solution (Explanation).
Nucleotide -
When a phosphaste group to - oH of a nucleoside through phosphoester linkage.
- wherein
A corresponding nucleotide (or deoxynucleotide depending upon of sugar present) is formed.
Nucleic acids are composed of 3 components:
Nitrogenous bases,
5 carbon sugars, and
one or more phospahte groups.

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Nitrogenous bases are of two kinds i.e, purines and pyrimidines.
Adenine and guanine are purines while cytosine, thymine and uracil are pyrimidrines. In both the DNA and RNA purine
are same while among pyrimidines, thymine is present in DNA and uracil in RNA.

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(10). The nucleotide arrangement of DNA can be seen by (4)
A. Electron microscope B. Light microscope
C. Ultracentrifuge D. X-Ray Crystallography

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Solution (Explanation).
As we have already studied in Structure of DNA - X-ray diffraction data produced by Maurice Wilkins and Rosalind
Franklin provided for the nucleotide arrangement of DNA. Hence, the correct answer is option d.
(11). A nucleotide is composed of: 1) nitrogenous base 2) pentose sugar 3) phosphate group (4)

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A. 1 and 2 B. 2 and 3
C. 1 and 3
Solution (Explanation).
IO D. 1, 2 and 3

As we have already discussed in Structure of DNA -

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A nucleotide has three components – a nitrogenous base, a pentose sugar (ribose in case of RNA, and deoxyribose for
DNA), and a phosphate group.
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Hence, the correct option is (d).

T
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(12). In a nucleoside, the nitrogenous base is linked to pentose sugar through : (4)
A. Diester bond B. N-glycosidic linkage
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C. phosphate group D. Hydrogen bond

Solution (Explanation).
As discussed in the Structure of DNA -
EX

A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form a nucleoside, such as adenosine
or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.
Hence, the correct option is (b).

(13). Total number of base pairs in one complete turn of the helix of the DNA is : (4)
 A. 5 base pairs B. 10 base pairs
C. 15 base pairs D. 20 base pairs
Solution (Explanation).

As we have already discussed in Salient Features of Double-Helix Structure of DNA -

The pitch of the helix is 3.4 nm (a nanometre is one-billionth of a metre, that is 10-9 m) and there is roughly 10 bp in
each turn.
Hence, the correct option is (b).
(14). The two antiparallel strands in a DNA double helix are held together by : (4)
A. Hydrogen bond B. electrostatic force of attraction

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C. gravitational force D. van der Waals force
Solution (Explanation).

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As learnt in Salient Features of Double-Helix Structure of DNA -
The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp). Adenine forms two
hydrogen bonds with Thymine from the opposite strand and vice-versa. Similarly, Guanine is bonded with Cytosine with

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three H-bonds.
Hence, the correct option is (a).
(15). The backbone of the polynucleotide chain is made up of: (4)
A. sugar-phosphate, and the bases project inside. B. carbon-phosphate and the bases project inside

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C. sugar-phosphate, and the bases project outside. D. carbon-phosphate and the bases project outside.
Solution (Explanation). IO
As already discussed in Salient Features of Double-Helix Structure of DNA -
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DNA is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and the bases project
inside.
Hence, the correct option is (a).
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(16). The structure of which of the following types of DNA was proposed by Watson and Crick? (4)
A. B-DNA B. A-DNA

C. Z-DNA D. All of these

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Solution (Explanation).
EN

As we have discussed in Types of DNA -

Watson and Crick proposed the model of DNA that exists at 90% relative humidity and is called B-DNA.
Hence, the correct option is (a).
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(17). Nucleosome core is made of (4)

A. H1, H2A, H2B & H3 B. H1, H2A, H2B & H4
C. H1, H2A, H2B, H3 & H4 D. H2A, H2B, H3 & H4
CE

Solution (Explanation).
As we have discussed in Packaging of DNA Helix in Eukaryotes -
Core histones are H2A, H2B, H3, and H4, where two H3/H4 dimers (H3 and H4 bound together) and two H2A/H2B
EX

dimers (these two bound together) form the histone octamer. DNA wrapped around the octamer forms the nucleosome.
Hence, the correct option is (d).

(18). Select the correct statement: (4)

A. Loosely packed chromatin is transcriptionally active B. Loosely packed chromatin is transcriptionally active
and is called heterochromatin. and is called euchromatin.
 C. Tightly packed chromatin is transcriptionally active D. Tightly packed chromatin is transcriptionally
and is called heterochromatin. inactive and is called euchromatin.
Solution (Explanation).

Aas learnt in Packaging of DNA Helix in Eukaryotes -

Loosely packed is said to be euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is
inactive.
Hence, the correct option is (b).
(19). A segment of DNA has 120 adenine and 120 cytosine bases. The total number of nucleotides present in the (4)
segment is
A. 60 B. 120

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C. 240 D. 480
Solution (Explanation).

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As we have already studied in Chargaff’s Rule -
A = 120, so T will also be 120; C = 120, so G will also be 120
Hence, the total number of nucleotides will be 120 X 4 = 480

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Hence, the correct option is (d).
(20). Read the following statements and select the correct one: (4)
A. S-type strains of Streptococcus pneumonia is B. R-type strains of Streptococcus pneumonia is
virulent due to the absence of mucus. virulent due to the absence of mucus.

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C. S-type strains of Streptococcus pneumonia is D. R-type strains of Streptococcus pneumonia is
virulent due to the presence of mucous. virulent due to the presence of mucous.
Solution (Explanation).
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As we have already studied in The Search For Genetic Material: Transforming Principle -
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S-type strain of Streptococcus pneumonia is virulent due to the presence of mucus over the cells.
Hence, the correct option is (c).
(21). Which of the following is correct w.r.t to transforming experiment? (4)
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1. It was based on transforming principle that transformed non-virulent strain to virulent strain.
2. The transformed strain resulted in a healthy and live mice.
3. The transformed strain caused the death of mice.
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4. Griffith successfully inferred the nature of transforming principle.

EN

A. 1, 2 and 4 B. 1 and 3
C. 3 and 4 D. 2, 3 and 4
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Solution (Explanation).

As discussed in The Search For Genetic Material: Transforming Principle -

The transforming experiment was based on transforming principle that transformed non-virulent strain to virulent strain
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and caused mice to die even when the virulent strain was heat-killed and non-virulent was live.
Hence, the correct option is (b).
(22). Read the following statements : 1. The transforming principle was characterized as DNA by Griffith. 2. Oswald (4)
EX

Avery, Colin MacLeod and Maclyn McCarty decoded the nature of transforming principle in Griffith’s experiment.
3. Protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) resulted in transformation. 4. DNA-
digesting enzyme (DNase) caused inhibition of transformation, which suggests that the DNA caused the
transformation. Select the correct option :
A. 1, 2 and 4 B. 2, 3 and 4

C. 2 and 4 D. 1 and 4
Solution (Explanation).
As we have already discussed in Biochemical Characterisation of Transforming Principle - Oswald Avery, Colin MacLeod
and Maclyn McCarty (1933-44) worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s
experiment in an in vitro system. The DNA-digesting enzyme (DNase) caused inhibition of transformation, which
suggests that the DNA caused the transformation. Hence, the correct option is (c).
(23). Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation nd Termination in the (4)
process of transcription in prokaryotes ?
A. DNA dependnet DNA polymerase B. DNA dpendent RNA polymerase
C. DNA Ligase D. DNase
Solution (Explanation).
Correct option is (b)
DNA dpendent RNA polymerase
(24). Identify the correct statement : (4)
A. In capping, methyl guanosine triphosphate is added B. RNA polymerase binds with Rho factor to
to the 3' end of hnRNA terminate the process of transcription in bacteria
C. The coding strand in a transcription unit is copied to D. Split gene arrangement is characteristic of

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an mRNA prokaryotes
Solution (Explanation).
In the cell, Rho binds to untranslated naked RNAs and terminates the synthesis of mRNA at the end of a significant

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number of operons.
(25). Which one of the following is wrongly matched? (4)
A. Transcription - Writing information From DNA to B. Translation - Using information in m-RNA to make

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t-RNA. protein.
C. Repressor protein - Binds to operator to stop enzyme D. Operon - Structural genes, operator and promoter.
synthesis.
Solution (Explanation).

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As we learnt in
Transcription -
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The process of copying genetic information from one strand of The DNA into RNA is turned as transcription.
- wherein
The principle of complementarity guesses the process of transcription.
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Transcription is writing information from DNA to m-RNA not DNA to t-RNA.
(26). Select the correct option (4)
Direction of RNA Direction of reading of the template
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Synthesis DNA strand

A. 5' - 3' 3' - 5' B. 3' - 5' 5' - 3'
C. 5' - 3' 5' - 3' D. 3' - 5' 3' - 5'
T

Solution (Explanation).
As we learnt in
EN

Process of Transcription -
There is s single DNA dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria.
- wherein
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CE

Hence, the correct option is (a).

(27). Which one -of the following is not a part of a transcription unit in DNA? (4)
A. The inducer B. A terminator
EX

C. A promoter D. The structural gene

Solution (Explanation).
As we learnt in
Transcription Unit -
The transcription unit in DNA is defined primarily by three regions in DNA.
(1) A promoter
(2) The structural gene
(3) A Terminator
- wherein
(28). Removal of introns and joining of exons in a defined order during transcription is called : (4)
A. Looping B. Inducing
C. Slicing D. Splicing
Solution (Explanation).
As we learnt in Splicing - The primary transcripts are subjected to splicing where the interons are removed and exons are

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joined in a defined order.
(29). The first phase of translation is (4)
A. Binding of mRNA to the ribosome B. Recognition of DNA molecule

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C. Aminoacylation of tRNA D. Recognition of an anti-codon
Solution (Explanation).

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The first phase of translation is an initiation in which there are the two ribosomal subunits (small and large), The mRNA
to be translated, the first (formyl) aminoacyl tRNA (the tRNA charged with the first amino acid),
Hence, the correct option is (a).
(30). The process of translation of mRNA to proteins begins as soon as : (4)
A. The larger subunit of ribosome encounters mRNA B. Both the subunits join together to bind with mRNA

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C. The tRNA is activated and the larger subunit of D. The small subunit of ribosome encounters mRNA
ribosome encounters mRNA
Solution (Explanation).
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When the smaller submit of ribosome encounter m-RNA, the process of translation of m-RNA to protein begins. This
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process is followed by the binding of a larger subunit. The activated t-RNA molecule carrying the amine acid methionine
binds to the start codon of the mRNA sequence.
Hence, the correct option is (4).
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(31). Given below are the steps of protein synthesis . Arrange them in correct sequence andd select the correct option (4)
. (i) Codon - anticodon reaction between mRNA and aminoacyl tRNA complex. (ii) Attachment of mRNA and
smaller sub-unit of ribosome . (iii) Charging of aminoacylation of tRNA . (iv) Attachment of larger sub-unit of
ribosome to the mRNA -tRNAMet complex. (v) Linking of adjacent amino acids. (vi) Formation of polypeptide
chain.
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A. (ii) (i) (iii) (v) (iv) (vi) B. (v) (ii) (i) (iii) (iv) (vi)
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C. (iii) (ii) (iv) (i) (v) (vi) D. (iii) (ii) (i) (iv) (v) (vi)
Solution (Explanation).
As we learnt ,
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Initiator t-RNA -
For initiation, there is another specific t-RNA that is referred to as initiator t-RNA.
-
(32). Which RNA carries the amino acids fro the amino acid pool to mRNA during protein synthesis ? (4)
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A. rRNA B. mRNA
C. tRNA D. hnRNA
EX

Solution (Explanation).
As we learnt ,
Translation -
Translation refers to the process of polymerisation of amino acids to form a polypeptide.
-
(33). Select the correct match : (4)
A. Matthew Meselson and F. Stahl - Pisum sativum B. Alfred Hershey and Martha Chase - TMV
C. Alec Jeffreys - Streptococcus pneumoniae D. Francois Jacob and Jacques Monod - Lac operon
Solution (Explanation).
As we learnt in
The lac operon -
In lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory genes.
- wherein
Such arrangements are very common in bacteria and is referred to as operon.
Francois Jacob and Jacque Monod proposed model of gene regulation known as operon model/lac operon.
Alec Jeffreys – DNA fingerprinting technique.
Matthew Meselson and F. Stahl – Semiconservative DNA replication in E. coli.
Alfred Hershey and Martha Chase – Proved DNA as genetic material not protein
(34). Match column I with column II and select the correct option from the given codes. Column I (4)
Column II

1. Operator site i)Binding site for RNA polymerase

2. Promoter site ii) Binding site for repressor molecule
3. Regulator gene iii) Codes for protein / enzyme

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4. Structural gene iv) Codes for repressor molecule

A. A-(ii), B-(i), C-(iii), D- (iv) B. A-(ii), B-(i), C-(iv), D- (iii)

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C. A-(iv), B-(iii), C-(i), D- (ii) D. A-(ii), B-(iii), C-(i), D- (iv)
Solution (Explanation).
Operators -

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The accessibility of promoters region of prokaryotic DNA is in many cases regulated by the interaction of protein with
sequence termed operators.
-
Correct option is 2

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(35). In Lac operon, gene ”a” codes for: (4)
A. Permease B. -galactosidase

C. Transacetylase
IO D. None of these
Solution (Explanation).
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The a gene -
The a gene encodes a transacetylase.
-
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Correct option is 3
(36). Inducible operon system usually occurs in (i)___________ pathways. Nutrient molecules serve as (4)
(ii)__________ to stimulate production of the enzymes necessary for their breakdown. Genes for inducible operon
are usually switched (iii)__________ and the repressor is synthesised in an (iv)___________ form.
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A. (i) (ii) (iii) B. (i) (ii) (iii)

(iv) Anabolic corepressor on (iv) Anabolic inducer off
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inactive active
C. (i) (ii) (iii) D. (i) (ii) (iii)
(iv) Catabolic inducer off (iv) Catabolic corepressor on
active inactive
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Solution (Explanation).
Inducer -
Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. hence it is
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termed as inducer.
-
Correct option is 3
(37). Match the following genes of the Lac operon with their respective products : (4)
EX

Select the correct option.

A. B.
C. D.
Solution (Explanation).
The lac operon -
In lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory genes.
- wherein
Such arrangements are very common in bacteria and is referred to as operon.
(a) codes for Repressor protein
(b) codes for -galactosidase
(c) codes for Transacetylase
(d) codes for Permease
(38). The inducer for switching ‘on’ the lac operon in bacteria is (4)
A. presence of lactose B. number of bacteria

C. presence of structural genes in the bacteria D. presence of sucrose

Solution (Explanation).

As discussed in Lac Operon -

Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the lac operon. Also,
it is termed as an inducer. Hence, the correct option is (a).

.
(39). Match the following w.r.t Lac Operon 1. beta-galactosidase a. joining of DNA fragments (4)

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2. permease b. peptide bond formation 3. ligase c.
hydrolysis of lactose 4. ribozyme d. increase the permeability of beta-galactosidase
A. 1.b.,2.a.,3.d.,4.c. B. 1.c.,2.d.,3.a.,4.b.

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C. 1.b.,2.d.,3.a.,4.c. D. 1.a.,2.b.,3.d.,4.c.
Solution (Explanation).

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As discussed in Lac Operon -
1.beta-galactosidase - hydrolysis of lactose
2.permease - increase the permeability of beta-galactosidase
3. ligase - joining of DNA fragments
4. ribozyme - peptide bond formation

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Hence, the correct option is (b)
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(40). Gene regulation governing lactose operon of E.coli that involves the lac I gene product is: (4)
A. negative and repressible because repressor protein B. Feedback inhibition because excess of -
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prevents transcription galactosidase can switch off transcription
C. Positive and inducible because it can be induced by D. negative and inducible because repressor protein
lactose prevents transcription.
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Solution (Explanation).
Lac I gene produces an inhibitor or repressor and negative regulation of lac operon is induced. The repressor binds to the
operator gene and stops its working. The repressor is meant to block the operator gene so that structural genes are unable
to form mRNA thus stopping the transcription of genes.
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(41). Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac gene? (4)
A. Lactose permease and transacetylase B. -galactosidase
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C. Lactose permease D. Transacetylase

Solution (Explanation).
As we learnt in
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Z gene code -
The Z gene code for Beta-galactosidase , which is primarily responsible for the hydrolysis of the disaccharide,
lactose, into its mono metric units, galactose and glucose.
-
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and
The y-gene -
The y-gene code for permease which increases permeability of the cell to ? - galactosides.
EX

-
(42). Select the two correct statements out of the four (a-d) given below about lac operon. (4)

1. Glucose or galactose may bind with the repressor and inactivate it

2. In the absence of lactose the repressor binds with the operator region
3. The z-gene codes for permease
4. This was elucidated by Francois Jacob and Jacque Monod

The correct statements are

A. (a) and (b) B. (b) and (c)
C. (a) and (c) D. (b) and (d)
Solution (Explanation).
The lac operon is the operon responsible for lactose transport and metabolism in Escherichia coli and other bacteria.
François Jacob and Jacques Monad were the first to explain the regulatory mechanism responsible for the expression of the
genes coding for the enzymes concerned. The ?-galactoside permease gene, responsible for transporting lactose into the
cell, is encoded by the lac y gene rather than the z gene. Statement 4 is correct. In the presence of lactose, allolactose, an
isomer of lactose, binds to the repressor and causes an allosteric transition. The modified repressor cannot bind to the
operator, allowing RNA polymerase to transcribe the lac gene. Glucose/galactose are the preferred sugars and their
presence inhibits the expression of the lac operon by catabolite repression. This sugar does not bind the repressor. Hence
option 4 is correct.
(43). The lac operon consists of (4)
A. Four regulatory genes only B. One regulatory gene and three structural genes
C. Two regulatory genes and two structural genes D. Three regulatory genes and three structural genes

.
S..
Solution (Explanation).
The Lac operon consists of a regulator gene, an operator gene and three structural genes. The three structural genes are
LacZ, LacY and LacA. LacZ encodes ?-galactosidase (LacZ), an intracellular enzyme that cleaves the disaccharide lactose

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into glucose and galactose. LacY codes for a ?-permease that pumps lactose into cells. Lac A codes for a beta-
transacetylase that removes toxic substances from cells.Regulatory genes produce mRNA which is translated into a
repressor. The repressor binds to the operator and blocks expression of the structural gene in the absence of the inducer.

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In the presence of an inducer, the repressor binds to the inducer, leaving the operator unblocked, allowing expression of
the structural gene. An operator gene is a region of the DNA sequence that attaches to a repressor molecule and stops
expression of the structural gene. Hence, option 2 is correct
(44). Which of the following statements about the lac operon are INCORRECT? Choose TWO options. (i) Glucose (4)
or galactose may bind to the repressor and inactivate it. (ii) In the absence of lactose, the repressor binds to the

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operator region. (iii) The z-gene codes for permease. (iv)This was elucidated by Francois Jacob and Jacque Monod.
A. (ii) and (iii) B. (i) and (iii)
C. (ii) and (iv)
IO D. (i) and (ii)
Solution (Explanation).
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(i) Glucose or galactose may bind to the repressor and inactivate it. (Incorrect): Lactose, not glucose or galactose, binds to
the repressor and inactivates it, allowing transcription of the lac operon genes.
(ii) In the absence of lactose, the repressor binds to the operator region. (Correct): This statement is true. When lactose is
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absent, the repressor remains active and binds to the operator, blocking RNA polymerase from transcribing the lac operon
genes.
(iii) The z-gene codes for permease. (Incorrect): The z-gene actually codes for β-galactosidase, an enzyme that breaks
down lactose. The y-gene codes for permease, a protein that allows lactose to enter the cell.
(iv) This was elucidated by Francois Jacob and Jacque Monod. (Correct): Francois Jacob and Jacque Monod are indeed
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credited with elucidating the lac operon model.

Therefore, statements (i) and (iii) about the lac operon are incorrect.
EN

(45). Statement I: Lactose, if provided in the growth medium one of E. coli, will be transported into the cells (4)
through the action of permease. Statement II: In the presence of lactose, the repressor is inactivated by its
interaction.
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A. Both Statement I and Statement II are true. B. Both Statement I and Statement II are false.
C. Statement I is correct but Statement II is false. D. Statement I is incorrect but Statement II is true.
Solution (Explanation).
CE

In E. coli, the lac operon is necessary for the transport and metabolism of lactose. Because allolactose binds to the Lac
repressor protein and prevents it from binding to the Lac operator, when lactose is given to the growth media, the lac
genes are expressed. One isomer of lactose is called allolactose. Lactose penetrates E. Coli and forms small amounts of
allolactose. The conformational shift in the repressor protein is brought about by allolactose binding to it. The repressor
EX

falls off as a result of its inability to bind to the operator region. The lac genes can then be transcribed by RNA
polymerase when it has bound to the promoter. As lactose is fully digested by enzymes over time, the regulator gene
produces the repressor as the amount of lactose decreases. By attaching itself to the operator gene, this repressor stops
transcription by preventing RNA polymerase from trancribing the operon. Negative regulation is the term for this kind of
regulation.