Molecular Basis of Inheritance Exam Questions Class 12 Biology

Please see Chapter 6 Molecular Basis of Inheritance Exam Questions Class 12 Biology below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 12 Biology Questions and answers for all chapters in your NCERT Book for Class 12 Biology. These solved problems for Molecular Basis of Inheritance in Class 12 Biology will help you to score more marks in upcoming examinations.

Exam Questions Chapter 6 Molecular Basis of Inheritance Class 12 Biology

Objective Questions

Question. Enzymes that joins fragments of the DNA strand synthesized by discontinuous synthesis
(a) Helicase
(b) DNA ligase
(c) DNA polymerase
(d) Telomerase

Question. Long DNA replication takes place only within a small opening called replication fork because
(a) Opening of whole DNA throughout the length needs a lot of time
(b) Opening of whole DNA throughout the length needs a lot of energy
(c) At one time in a cell only a small DNA part is replicated
(d) None of these

Question. Polyploidy takes place in a cell when
(a) If transcription takes place but cell division doesn’t take place
(b) If replication takes place but transcription doesn’t take place
(c) If replication takes place but cell division doesn’t take place
(d) If replication and cell division doesn’t take place

Question. During transcription base pairing between DNA template strand and new RNA strand is
(a) A–T,G–C
(b) A–C,G–T
(c) A–G, T–C
(c) None of these

Question. RNA synthesis takes place on a single DNA strand because
(a) The two strands of DNA have different sequences so if both are transcribed simultaneously then two different proteins will be formed
(b) As the two DNA strands are complementary to each other the RNA strands formed by transcription will also be complementary and form double stranded RNA
(c) Both of these
(d) None of these

Question. A sequence of DNA to be transcribed is 5’TTACCGAT–3’coding strand 3’AATGGCTA–5’ template strand. What will be the sequence of the transcribed m RNA?
(a) UUACCGAU
(b) AAUGGCUA
(c) TTUGGCUT
(d) All of these

Question. Main enzyme for transcription is
(a) RNA dependent – DNA Polymerase
(b) DNA dependent – DNA polymerase
(c) DNA dependent – RNA polymerase
(d) DNA ligase

Question. Transcription takes place on
(a) Both the DNA strand simultaneously
(b) Coding strand
(c) Template strand
(d) None of the above

Question. If newly transcribed RNA have sequence 5’AUUCGUA3’ the coding sequence will be :
(a) 5’CGGATT3’
(b) 3’GCCTAA5’
(c) 3’AATCCG5’
(d) 5’ATTCGTA3’

Question. Transcription unit is located on
(a) m RNA molecule
(b) DNA molecule
(c) r RNA molecule
(d) t RNA molecule

Question. Why N₁₅ isotope of nitrogen was taken in Stahl experiment?
(a) It is radioactive
(b) It is heavy than ¹⁴N– isotope and can be separated on the base of density
(c) It can be separated based on radioactivity
(d) Both (a) and (c)

Question. The E. Coli used by Meselson and Stahl for proving semiconservative nature of replication were initially grown in medium containing
(a) ³⁵S
(b) ¹⁵NH₄CI
(c) ³²P
(d) ₁₅C

Question. Taylor and colleagues in 1958 experimented on which plant for proving that replication is semiconservative
(a) Pisum sativum
(b) Vicia fabia
(c) Hibiscus rosasinesis
(d) None of the above

Question. The average replication rate is around
(a) 2000 bp/sec
(b) 2 kbp/sec
(c) 2 × 10⁶ bp/sec
(d) None of these

Question. dNTPs importance in replication is
(a) Serves as substrate
(b) Activates polymerase
(c) Gives energy for strand separation
(d) Both (a) and (c)

Very Short Answer Questions

Question. Mention the carbon positions to which the nitrogenous base and the phosphate molecule are respectively linked in the nucleotide given below:

Answer. Nitrogenous base is linked to first carbon.
Phosphate is linked to fifth carbon.

Question. Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer. Gel electrophoresis is used to separate DNA fragments.

Question. Mention one difference to distinguish an exon from an intron. 
Answer. Exon is the coded or expressed sequence of nucleotides in mRNA.
Intron is the intervening sequence of nucleotides not appearing in processed mRNA.

Question. In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates, still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why? 
Answer. DNA polymerase is highly specific to recognise only deoxyribonucleoside triphosphates.
Therefore it cannot hold RNA nucleotides.

Question. When and at what end does the ‘tailing’ of hnRNA take place?
Answer. ‘Tailing’ of hnRNA takes place during conversion of hnRNA into functional mRNA after transcription. It takes place at the 3′-end.

Question. What are ‘a’ and ‘b’ in the nucleotide with purine represented below?

Answer. ‘a’ is phosphate group and ‘b’ is purine (adenine/guanine).

Question. Name the negatively charged and positively charged components of a nucleosome.
Answer. In a nucleosome, the negatively charged component is DNA and positively charged component is histone octamer.

Short Answer Questions

Question. State the dual role of deoxyribonucleoside triphosphates during DNA replication.
OR
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.
Answer. (i) Deoxyribonucleoside triphosphates act as substrates for polymerisation.
(ii) These provide energy from its two terminal phosphates for polymerisation reaction.

Question. (a) Name the molecule ‘M’ that binds with the operator.
(b) Mention the consequences of such binding.

Answer. (a) M is the repressor.
(b) When repressor binds with the operator, transcription stops.

Question. State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes.
Answer. 

S.No.	Prokaryotes	Eukaryotes
(i)	Polycistronic	Monocistronic
(ii)	No split genes present. The coding sequence is not interrupted.	Split genes present. The coding sequence is interrupted to form exon and intron.

Question. Discuss the role of enzyme DNA ligase plays during DNA replication.
Answer. DNA ligase joins or seals the discontinuous DNA fragments.

Question. A template strand is given below. Write down the corresponding coding strand and the mRNA strand that can be formed, along with their polarity.
3′ ATGCATGCATGCATGCATGCATGC 5′
Answer. Coding strand: 5′ TACGTACGTACGTACGTACGTACG 3′mRNA strand: 5′ UACGUACGUACGUACGUACGUACG 3′

Question. Draw the structure of a tRNA charged with methionine. 
Answer. 

Question. Retrovirus do not follow central dogma. Comment.
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is reverse transcribed to its complementary DNA first, then transcriped to RNA and proteins. Hence, retrovirus are not known to follow central dogma.

Question. If the sequence of the coding strand in a transcription unit is written as follows:
5′—ATGCATGCATGCATGCATGCATGCATG—3′
Write down the sequence of mRNA.
Answer. 5′—AUGCAUGCAUGCAUGCAUGCAUGCAUG—3′

Question. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer. The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest semi-conservative mechanism of DNA replication in which one strand of parent is conserved while the other complementary strand formed is new.

Question. Differentiate between the following:
(a) Repetitive DNA and satellite DNA
(b) mRNA and tRNA
(c) Template strand and coding strand

Answer. (a) Differences between repetitive DNA and satellite DNA

S.No.	Repetitive DNA	Satellite DNA
(i)	DNA in which certain base sequences are repeated many times are called repetitive DNA.	DNA in which large portion of the gene is tandemly repeated is called satellite DNA.
(ii)	Repetitive DNA sequences are transcribed.	Satellite DNA sequences are not transcribed.

(b) Differences between mRNA and tRNA

S.No.	mRNA	tRNA
(i)	It is linear.	It is clover-leaf shaped.
(ii)	It carries coded information.	It carries information for association with an amino acid and an anticodon for its incorporation in a polypeptide.
(iii)	mRNA undergoes additional processing, i.e., capping and tailing, splicing.	It does not require any processing
(iv)	Nitrogen bases are unmodified.	Nitrogen bases may be modified.

(c) Differences between template strand and coding strand

S.No.	Template strand	Coding strand
(i)	It is the strand of DNA which takes part in transcription.	It is the strand that does not take part in transcription.
(ii)	The polarity is 3′→5′.	The polarity is 5′→3′.
(iii)	Nucleotide sequence is complementary to the one present in mRNA.	The nucleotide sequence is same as the one present in mRNA except for presence of thymine instead of uracil.

Question. Draw a labelled schematic diagram of a transcription unit.
Answer. 

Question. Differentiate between the genetic codes given below:
(a) Unambiguous and Universal.
(b) Degenerate and Initiator
Answer. 

(a) Unambiguous: One codon codes for only one amino acid.	Universal: Codons are (nearly) same for all organisms (from bacteria to humans)
(b) Degenerate: More than one codon can code for the same amino acid.	Initiator: Start codon i.e., AUG is the initiation codon.

Question. Explain the process of transcription in a bacterium.
Answer. • In prokaryotes, the structural gene is polycistronic and continuous.
• In bacteria, the transcription of all the three types of RNA (mRNA, tRNA and rRNA) is catalysed by single DNA-dependent enzyme, called the RNA polymerase.
• All three RNA’s are needed to synthesize a protein in cell. mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNA plays structural and catalytic role durinng translation.
• The transcription is completed in three steps: initiation, elongation and termination.
• Initiation: s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion following the rule of complementarity.

Question. (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion. 
Answer. (i) George Gamow.
(ii) He proposed that there are four bases and 20 amino acids So, there should be atleast 20 different genetic codes for these 20 amino acids.
The only possible combinations that would meet the requirement is combinations of 3 bases that will give 64 codons.

Question. Name the category of codons UGA belongs to. Mention another codon of the same category.Explain their role in protein synthesis.
Answer. UGA is a stop or termination codon.
UAA, UAG are the other stop codons of the category.
They prevent the elongation of the polypeptide chain by terminating translation.

Question. Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer. Tandemness in repeats provides many copies of the sequence for fingerprinting and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Long Answer Questions

Question. How is hnRNA processed to form mRNA? 
Answer. The hnRNA undergoes the following processes to form mRNA:
(i) Capping: Addition of methyl guanosine triphosphate at 5’-end.
(ii) Tailing: Addition of 200-300 adenylate residues at 3’-end.
(iii) Splicing: Removal of introns and rejoining of exons.

Question. What is hnRNA? Explain the changes hnRNA undergoes during its processing to form mRNA.
Answer. hnRNA is the precursor of mRNA that is transcribed by RNA ploymerase II and is called heterogenous nuclear RNA.
Changes:
• The hnRNA undergoes two additional processes called capping and tailing.
• In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5’-end of hnRNA.
• In tailing, adenylate residues (about 200–300) are added at 3’-end in a template independent manner.
• Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.
• Fragments on one of the template strands.

Question. (a) Identify the polarity from a to a′ in the given diagram and mention how many more amino acids are expected to be added to this polypeptide chain.
(b) Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
(c) Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?

Answer. (a) Polarity from a to a’ is 5’→ 3′.
No more amino acid will be added to this polypeptide chain.
(b) TCA; anticodon is UCA.
(c) The untranslated sequence of bases are required for efficient translation process. They are present before the start codon at the 5′-end and after the stop codon at 3′-end.

Question. One of the codons on mRNA is AUG. Draw the structure of tRNA adapter molecule for this codon. Explain the uniqueness of this tRNA.
Answer. This tRNA is specific for amino acid Methionine and it also acts as initiator codon (initiator tRNA) 

Question. Explain the process of translation in a bacterium.
Answer. • Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
•  A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide,a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question. Unambiguous, universal and degenerate are some of the terms used for the genetic code.
Explain the salient features of each one of them. 
Answer. Unambiguous code means that one codon codes for only one amino acid, e.g., AUG codes for only methionine.
Universal code means that codon and its corresponding amino acid are the same in all organisms, e.g., from bacteria to human, UUU codes for phenylalanine.
Degenerate code means that same amino acids are coded by more than one codon, e.g., UUU and UUC code for phenylalanine.

Question. State the conditions when ‘genetic code’ is said to be
(i) degenerate,
(ii) unambiguous and specific,
(iii) universal. 
Answer. (i) Degenerate—When some amino acids are coded by more than one amino acids.
(ii) Unambiguous and speci—ficWhen one codon codes for only one specific amino acid.
(iii) A particular codon codes for same amino acid in all organisms except in mitochondria and few protozoa.

Question. 

Study the mRNA segment given above which is complete to be translated into a polypeptide chain.
(i) Write the codons ‘a’ and ‘b’.
(ii) What do they code for?
(iii) How is peptide bond formed between two amino acids in the ribosome? 
Answer. (i) a is AUG and b is UAA/UAG/UGA
(ii) AUG codes for methionine (initiation codon).
UAA/UAG/UGA do not code for any amino acid, i.e., stop or terminating codons.

(iii) There are two sites (P-site and A-site) in the large subunit of ribosome, where subsequent amino acids bind to and thus are close enough to form peptide bond by peptidyl transferase enzyme. The ribosome also acts as a catalyst for the formation of peptide bond.

Question. “The codon is a triplet and is read in a contiguous manner without punctuations.” Provide the genetic basis for the statement.
Answer. Since there are only four bases which code for twenty amino acids, the code should be made up of three bases, i.e., (4 × 4 × 4) = 64 codons; a number more than the required.
If the codon consists of four letters, only (4 × 4), only sixteen codons are possible, which is less than the required. Hence the codon is a triplet.
As the ribosome moves on mRNA, continuously without break, the codons are read in a contiguous manner.

Question. Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated from it.
AUG UUU UCU UUU UUU UCU UAG
Met – Phe – Ser – Phe – Phe – Ser
Answer. 

S.No.	Salient features of genetic code	Reason
(i)	The codon is a triplet.	e.g., AUG, UUU, etc, are triplets
(ii)	One codon codes for only one amino acid,hence it is unambiguous and specific	e.g., UUU codes for serine, AUG for methionine, etc.
(iii)	AUG has dual function as it codes for methionine and it also acts as initiator codon.	AUG is seen at the beginning of the polypeptide chain.
(iv)	UAG does not code for any amino acid hence is called stop codon and leads to end of translation.	No amino acid is coded by UAG in the polypeptide chain given.

Question. (a) Explain the observations of Meselson and Stahl when
(i) they cultured E. coli in a medium containing 15NH4Cl for a few generations and centrifuged the content.
(ii) they transferred one such bacterium to the normal medium of NH4Cl and cultured for 2 generations?
(b) What does the above experiment prove?
(c) Which is the first genetic material identified?
Answer. (a) (i) Meselson and Stahl observed that in the E. coli bacterium the DNA becomes completely labelled with 15N medium by centrifugation for few generations.
(ii) After two generations, density changed and showed equal amount of light DNA (14N) and dark hybrid DNA (15N–14N).
(b) They concluded that DNA replicates semi-conservatively.
(c) Ribonucleic acid (RNA) was the first genetic material.

Question. (a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it?Support your answer with a reason and an example.
(b) Explain how the biochemical characterisation (nature) of ‘Transforming Principle’ was determined, which was not defined from Griffith’s experiments.
Answer. (a) Francis Crick proposed the central dogma of molecular biology which states that genetic information flows from DNA to mRNA (transcription) and then from mRNA to protein (translation) always unidirectionally (except bidirectionally in some viruses and the process is called reverse transcription).

Yes, there are some exceptions to it. In some viruses flow of information is in reverse direction (reverse transcription).

Question. (a) Name the molecule ‘X’ synthesised by ‘i’ gene. How does this molecule get inactivated?
(b) Which one of the structural genes codes for b-galactosidase?
(c) When will the transcription of this gene stop?

Answer. (a) The molecule ‘X’ is repressor. It gets inactivated when lactose (inducer) binds with the repressor molecule.
(b) z gene codes for b-galactosidase.
(c) Transcription of the gene stops when lactose is absent and thus repressor is free to bind with the operator.

Question. Draw a schematic diagram of lac operon in its ‘switched off’ position. Label the following:
(i) The structural genes (ii) Repressor bound to its correct position
(iii) Promoter gene (iv) Regulatory gene. 
Answer.
(i) z, y and a are structural genes.
(iii) p is the promoter sequence.
(iv) i is the regulatory gene.

Question. (a) Explain the process of aminoacylation of tRNA. Mention its role in translation.
(b) How do ribosomes in the cells act as factories for protein synthesis?
(c) Describe ‘initiation’ and ‘termination’ phases of protein synthesis.
Answer. (a) Aminoacylation is the process by which amino acids become activated by binding with its aminoacyl tRNA synthetase in the presence of ATP.
If two charged tRNAs come close during translation process, the formation of peptide bond between them in energetically favourable.
(b) The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists
as two subunits: a large subunit and a small subunit. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in
the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst (23S rRNA in
bacteria is the enzyme ribozyme) for the formation of peptide bond.
(c) • Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
• A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide, a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question. (a) Name the scientist who postulated the presence of an adapter molecule that can assist in protein synthesis.
(b) Describe its structure with the help of a diagram. Mention its role in protein synthesis.
Answer. (a) Francis Crick
(b) Structure 

• The secondary structure of tRNA is cloverleaf like but the three-dimensional tertiary structure depicts it as a compact inverted L-shaped molecule.
• tRNA has five arms or loops:
(i) Anticodon loop, which has bases complementary to the code.
(ii) Amino acid acceptor end to which amino acids bind.
(iii) T loop, which helps in binding to ribosome.
(iv) D loop, which helps in binding aminoacyl synthetase.
(v) Variable arm.
The adapter molecule reads code with the help of anticodon loop and on the other end binds specific amino acids by peptide bond.

Question. In a maternity clinic, for some reasons the authorities are not able to hand over the two newborns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter. 
Answer. The technique is DNA fingerprinting. It includes the following steps:
Methodology and Technique
(i) DNA is isolated and extracted from the cell or tissue by centrifugation.
(ii) By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. 
(iii) DNA is cut into small fragments by treating with restriction endonucleases.
(iv) DNA fragments are separated by agarose gel electrophoresis.
(v) The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye.
(vi) The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting.
(vii) VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation.

Question. Describe the elongation process of transcription in bacteria. 
Answer. After initiation, RNA polymerase loses the s factor but continues the polymerisation of ribonucleotides to form RNA. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion, following the rule of complementarity. For diagram,

Question. Describe the termination process of transcription in bacteria. 
Answer. Once the RNA polymerase reaches the termination region of DNA, the RNA polymerase is separated from DNA–RNA hybrid, as a result nascent RNA separates. This process is facilitated by a termination factor r (rho). In prokaryotes, mRNA does not require any processing, so transcription and translation both occur in the cytosol.
For diagram,

Question. Write any six salient features of the human genome as drawn from the human genome project. 
Answer. (i) The human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 3000 bases; the largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated to be 30,000 and 99.9 per cent nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50 per cent of the discovered genes.
(v) Less than 2 per cent of the genome codes for proteins.
(vi) The human genome contains large repeated sequences, repeated 100 to 1000 times.
(vii) The repeated sequence is thought to have no direct coding functions but they throw light on chromosome structures, dynamics and evolution.
(viii) Chromosome 1 has most genes (2968) and the Y has the fewest genes (231).
(ix) Scientists have identified about 1.4 million locations where single base DNA sequence differences called SNPs or single nucleotide polymorphism (pronounced as ‘snips’) occur in humans.This information promises to revolutionise the processes of finding chromosomal locations for disease—associated sequences and tracing human history.

Question. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and write the procedure that would help in the identification of kinship. 
OR
A number of passengers were severely burnt beyond recognition during a train accident.
Name and describe a modern technique that can help to hand over the dead to their relatives.
Answer. DNA fingerprinting can help in identification of kinship.
For procedure, • Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases.
• Human genome has 3 × 109 bp. 99.9% of base sequences among humans are the same, which makes every individual unique in phenotype.

Question. (a) Draw a schematic representation of the structure of a transcription unit and show the following in it:
(i) Direction in which the transcription occurs
(ii) Polarity of the two strands involved
(iii) Template strand
(iv) Terminator gene
(b) Mention the function of promoter gene in transcription.
Answer. (a) (i) Transcription occurs in 5′→3′.

(b) Promotor gene has DNA sequence that provide binding site for RNA polymerase.

Question. Describe the initiation process of transcription in bacteria. 
Answer. In bacteria, the transcription of all the three types of RNA (mRNA, tRNA, rRNA) is catalysed by single DNA-dependent enzyme called the RNA polymerase. The RNA polymerase has cofactors that catalyse the process. During initiation, s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription.
For diagram,

Question. Observe the representation of genes involved in the lac operon given below: (Img 254)

(a) Identify the region where the repressor protein will attach normally.
(b) Under certain conditions repressor is unable to attach at this site. Explain.
(c) If repressor fails to attach to the said site what products will be formed by z, y and a?
(d) Analyse why this kind of regulation is called negative regulation.
Answer. (a) The repressor protein will attach to operator region, o.
(b) In presence of an inducer, lactose, repressor is unable to attach.
(c) z—b galactosidase.
y—Permease
a—Transacetylase
(d) It is called negative regulation as it involves constitutive (all the time) repressor. The operon is always in off position due to presence of repressor and is switched on only in presence of an inducer. Inducer Lactose or allolactose interacts with repressor making it inactive.