CBSE Class 12 Biology Molecular Basis Of Inheritance Worksheet Set A

Question. Which one of the following steps in transcription is catalysed by RNA polymerase?

(a) Initiation
(b) Elongation
(c) Termination
(d) All of these
Answer : B

Question. In aDNAstrand the nucleotides are linked together by
(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds
Answer : B

Question. A nucleoside differs from a nucleotide. It lacks the
(a) base
(b) sugar
(c) phosphate group
(d) hydroxyl group
Answer : C

Question. Which one of the following is true with respect to AUG?
(a) It codes for methionine only
(b) It is also an initiation codon
(c) It codes for methionine in both prokaryotes and eukaryotes
(d) All of the above
Answer : D

Question. With regard to mature mRNA in eukaryotes
(a) exons and introns do not appear in the mature RNA
(b) exons appear, but introns do not appear in the mature RNA
(c) introns appear, but exons do not appear in the mature RNA
(d) Both exons and introns appear in the mature RNA
Answer : B

Question. To initiate translation, the mRNA first binds to
(a) the smaller ribosomal subunit
(b) the larger ribosomal subunit
(c) the whole ribosome
(d) No such specificity exists
Answer : A

Question. In E. coli, the lac operon gets switched on when
(a) lactose is present and it binds to the repressor
(b) repressor binds to operator
(c) RNA polymerase binds to the operator
(d) lactose is present and it binds to RNA polymerase
Answer : A

Question. Match the following
(A) f x 174 (i) 48502 bp
(B) Lambda phage (ii) 5386 Nucleotides
(C) E.Coli (iii) 6.6 x 109 bp
(D) Human somatic cell (iv) 4.6 x 106 bp
(a) A(i), B(ii), C(iv), D(iii)
(b) A(ii), B(i), C(iv), D(iii)
(c) A(i), B(ii), C(iii), D(iv)
(d) A(iv), B(iii), C(ii), D(i)

Answer : C

Question. Match the following
A. SNPs i 3164.7 million
B. Genes of ii 1.4 Million chromosome No. 1
C. Total No. of Human iii 30000 genes
D. Total nucleotides of iv 2968 human genome
A B C D
(a) ii iii iv i
(b) ii iv iii i
(c) ii iv i iii
(d) iv ii iii i

Answer : B

Question. Which of the following pyrimidine base is common in both DNA and RNA
(a) Adenine
(b) Guanine
(c) Cytosine
(d) Thymine

 Answer : C

Question. The backbone in a polynucleotide chain is formed due to
(a) Sugars and nitrogenous bases
(b) Phosphates and nitrogenous base
(c) Nitrogenous bases and histones
(d) Sugar and phosphates

Answer : D

Question. In RNA, every nucleotide residue has an additional - OH group at which of the following position
(a) 2' position of deoxyribose
(b) 1' possition of ribose sugar
(c) 3' position of ribose sugar
(d) 2' position of ribose sugar

Answer : D

Question. DNA as an acidic substance present in nucleus was first identified by
(a) Wilkins and Franklin
(b) Watson and Crick
(c) Friedich meischer
(d) Altmann

 Answer : C

Question. In addition to hydrogen bonding which of the following feature confers stability to helical structure
(a) Phosphodiester bond
(b) Pairing between one purine and one pyrimidine
(c) Glycosidic linkage between sugar and nitrogenous base
(d) The plane of one base pair stacks over the other

Answer : D

Question. Which of the following is responsible for constant distance between two polynuclestide chains in DNA
(a) Antiparallel polarity of two polynucleotide strands
(b) Hydrogen bonding
(c) Pairing between one purine and one pyrimidine
(d) All the above

Answer : C

Question. In prokaryotes predominant site for control of gene expression is the
(a) Control of rate of processing of primary transcript
(b) Control of rate of transcription initiation
(c) Control of transport of m-RNA from nucleus to cytoplasm
(d) Control of Translation

Answer : B

Question. HGP was closely associated with the rapid development of a new area in biology called as
(a) Biofortification
(b) Bioinformatics
(c) Biomining
(d) Biotransformation

Answer : B

Question. Phosphoric acid remain associated with which of the following carbon of sugar in a nucleotide :-
(a) Ist
(b) 3rd
(c) 4th
(d) 5th

Answer : D

Question. In there are 3.3 x 109 bp present in genome, then what would be the length of the DNA of any somatic cell
(a) 1.1 meter
(b) 2.2 meter
(c) 3.3 meter
(d) 6.6 meter

Answer : B

Question. Positive charge and basic nature of histone is due to abundance of
(a) Lysines and tryptophanes
(b) Arginines & threonines
(c) Lysines and arginines
(d) Tryptophanes and threonines

Answer : C

Question. Negative charge of DNA is due to which of the following constituent
(a) Sugar
(b) Nitrogenous base
(c) Phosphoric acid
(d) Hydroxyl group (-OH) present on sugar

Answer : C

Question. A typical nucleosome contains how much amount of DNA
(a) 100 bp
(b) 146 bp
(c) 200 bp
(d) 346 bp

Answer : C

Question. In a mammalian somatic cell how many nucleosomes are present
(a) 6.6 x 109
(b) 3.3 x 109
(c) 3.3 x 107
(d) 3.3 x 105

Answer : C

Question. Which of the following is actual sequence of packaging of DNA in eukaryotic cells
(a) DNA → Chromatin → Nucleosome → Chromosome
(b) DNA → Nucleosome → Chromosome → Chromatin
(c) DNA → Nucleosome → Chromatin → Chromosome
(d) DNA → Chromosome → Chromatin → Nucleosome

Answer : C

Question. The packaging of chromatin at higher level requires additional set of proteins that is known as
(a) Histone proteins
(b) NHC proteins
(c) Homeotic proteins
(d) Domain proteins

Answer : B

Question. The unequivocal proof that DNA is the genetic material came from the experiments of
(a) Griffith
(b) Avery, Macleod & Mccarty
(c) Hershey and Chase
(d) Watson and Crick

Answer : C

Question. Radioactive (35S) was detected in
(a) Supernatant
(b) Sediment
(c) Both
(d) Either 1 or 2

 Answer : A

Question. Regarding to RNA which of the following feature is wrong
(a) Catalytic property
(b) Labile and easily degradable
(c) Absence of thymine
(d) Presence of methylated uracil

Answer : D

Question. Which fo the folloiwng reason is suitable to expalin that RNA is best for expression of characters
(a) It shows catalytic properties
(b) Presence of 2'-OH group on ribose sugar
(c) It can directly code for the synthesis of protein
(d) Presence of uracil

Answer : C

Question. Which of the following is not the feature of human genome ?
(a) Less than 2 percent of the genome code for protein
(b) Chromosome 1 has fewest gene (231)
(c) Repetitive sequences make up very large portion of human genome
(d) The functions are unknown for over 50% of the discovered genes

Answer : B

Question. The sequence of which chromosome number was completed in May 2006 ?
(a) Chromosome number 1
(b) Chromosome number 2
(c) Chromosome number 5
(d) Chromosome number 10

Answer : A

Question. The repressor of the operon is synthesized :
(a) All the time
(b) Certain time
(c) Non constitutively
(d) None of these

Answer : A

Question. Which of the following is responsible for short life span and fast rate of mutation and evolution
(a) Presence of DNA
(b) Presence of highly reactive RNA
(c) Double stranded genetic material
(d) Single stranded genetic material

Answer : B

Question. Double helix model of DNA proposed by watson and crick was based on
(a) X-ray diffraction data of Meischer
(b) X-ray crystallography data of Wilkins and Franklin
(c) X-ray diffraction data of Watson and Crick
(d) X-ray diffraction data of Chargaff

Answer : B

Question. Approach of HGP focused on identifying all the genes that expressed as RNA is known as
(a) Expressed sequence tags
(b) Sequence annotation
(c) Polymerase chain reaction
(d) Dermatoglyphics

Answer : A

Question. Automated DNA sequencers worked on the principle of a method developed by
(a) Watson
(b) Chargaff
(c) Frederick sanger
(d) Singer and Nicolson

Answer : C

Question. Regarding to features of double helix struture of DNA which of the following is wrong
(a) Two polynucleotide chains have antiparallel polarity
(b) The bases in two strands are paired through phosphodiester bonds
(c) Adenine form two hydrogen bonds with thymine
(d) The pitch of the helix is 3.4 nm

Answer : B

Important Questions for NCERT Class 12 Biology Molecular Basis of Inheritance

Very Short Answers Type Questions

Question What is the function of histones in DNA packaging?
Answer : Functions of histones in DNA packaging are
(i) Histones as units of octamer participate in primary packaging of DNA.
(ii) Basic histone proteins neutralise the acidic DNA molecule.

Question.  Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer : Densely packed and dark stained chromatin regions are called hetorochromation, while, loosely packed light stained regions are called euchromatin.
Euchromation is transcriptionally active and is transcribed into mRNA. Due to very tight coiling heterochromatin can not be transcribed and is inert/inactive form.

Question. Suggest a techni-e to a researcher who needs to separate fragments of DNA.
Answer : Gel electrophoresis is used to separate DNA fragments.

Question. Name the transcriptionally active region of chromatin in a nucleus. 
Answer : Euchromatin or exon.

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. How does a degenerate code differ from an unambigu-s one? 
Answer : Degenerate code means that one amino acid can be coded by more than one codon. Unambigu-s code means that one codon codes for only one amino acid.

Question. Mention two functions of the codon AUG. 
Answer : Two functions of the codon AUG are:
(i) It acts as a start codon during protein synthesis.
(ii) It codes for the amino acid methionine.

Question. Give an example of a codon having dual function. 
Answer : AUG acts as an initiation codon and also codes for methionine.

Question. Why are proteins either positively or negatively charged? 
Answer : If the proteins are rich in basic amino acids, they are positively charged, and if the proteins are rich in acidic amino acids, they are negatively charged.

Question. Mention how does DNA polymorphism arise in a population. 
Answer : DNA polymorphism in a population arise due to presence of inheritable mutations at high fre-ency.

Question. How many base pairs w-ld a DNA segment of length 1.36 nm have? 
Answer : Distance between two base pairs = 0.34 nm or 0.34×10^–6 nm
Number of base pairs in 1.36 nm DNA segment
= 1/(0.34×10^-6) × 1.36
= 4 ×10  bp

Question. In an experiment, DNA is treated with a comp-nd which tends to place itself amongst the stacks of nitrogen-s base pairs. As a result of which the distance between two consecutive base increases, from 0.34 nm to 0.44 nm. Calculate the length of DNA d-ble helix (which has 2 × 109 bp) in the presence of saturating am-nt of this comp-nd. 
Answer : 2 × 10⁹ × 0.44 nm.

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

Question. What do ‘X’ and ‘Y’ represent in the transcription unit of the DNA molecule shown?

Answer : X—Template Strand Y—Terminator

Question. Why hnRNA is re-ired to undergo splicing?
Answer : hnRNA undergoes splicing in order to remove introns which are intervening or non-coding se-ences and exons are joined to form functional mRNA.

Question. The enzyme DNA polymerase in E.coli is a DNA dependent polymerase and also has the ability to proofread the DNA strand being synthesised Explain. Discuss the dual polymerase.
Answer : In bacteria, three types of DNA polymerases are there. All of them can add nucleotides in 5¢® 3¢ direction. They process exonuclease activity as well. DNA polymerase III can proofread the newly synthesised strand and senses the wrong base insertions. It deletes wrong bases and helps correct the mistake by putting in the right one, DNA polymerase. The only mistake it cannot corrects substitution of uracil in place of thymin. It can repair any damages done to DNA by UV exposure, etc., or the left over proofreading mistakes. It detects mutation caused by UV, removes mismatched pairs and puts back the right ones.

Question. What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?
Answer : Synthesis of DNA always takes place in 5¢® 3¢ direction. In a double stranded DNA both strands are anti parallel and complementary. During DNA synthesis as both strands act as templates, only one strand, i.e., 3¢® 5¢ can synthesis complementary strand in 5¢® 3¢ direction.
The other strand, i.e., 5¢® 3¢ has to be synthesised in small stretches in opposite direction as replication fork moves to right. That is why DNA synthesis is discontinuous on one of the
parental strands of DNA. These small stretches called Okazaki fragments are joined together by DNA ligase enzyme that closes the nicks.

Question. Given below is the sequence of coding strand of DNA in a transcription unit 3¢ AATGCAGCTAT TAGG-5¢ write the sequence of
(a) its complementary strand
(b) the mRNA
Answer : According to base complementary rules,
(a) 5'TTACGTCGATAATCC-3' (b) 5'CGAUUAUCGACGUAA-3'
RNA uses the base uracil (U) rather than thymine (T). So, in RNA the base pairs are Adenine (A) pairs with uracil (U) Guanine (G) pairs with cytosine (C).

Short Answer Type Questions

Question. Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl’s experiment.
Answer : They performed experiments on E. coli to prove that DNA replication is semi-conservative. They first grew the bacteria in a medium containing 15 4 NH Cl (in which 15N is the heavy istope of nitrogen) for many generations. Then they transferred the cells into a medium with normal 14 4 NH Cl (in which 14N is the lighter isotope) and took the samples at various definite time intervals as the cells multiplied. The extracted DNAs were centrifuged and measured to get their densities. The DNA extracted from the culture after one generation of transfer from then 15N mediun to 14N mediun, (i.e., after 20 minutes E.coli divides every 20 minutes) showed an intermediate hybrid density, i.e., both heavy and light nitrogen, which proved the semi-conservative nature of DNA.

Question. Define a cistron. Giving examples differentiate between monocistronic and polycistronic unit.
Answer : A cistron is stretch of base sequences that codes for one polypeptide chain including adjacent control regions. It may also code for a tRNA, rRNA molecule or may perform other specific functions including regulating functions of other cistrons. This term has replaced the definition of a gene. Monocistronic transcription unit will have all the regulatory and coding sequences for a single polypeptide, whereas polycistronic may have coding sequences for more than one polypeptide. In eukaryotic cells almost all the messenger RNAs are monocistronic. In prokaryotes, lac operon coding sequence would be an example of polycistronic DNA region.

Question. Give any six features of the human genome.
Answer : Salient features of human genome
(i) The human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 30000 the largest know human gene being dystrophin at 2.4 Million bases.
(iii) The total number of genes is estimated to be 30000 and 99.9% nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50% of the discovered genes.
(v) Less than 2% of the genome codes for proteins.
(vi) The human genome contains large repeated sequences.
(vii) The repeated sequence is thought to have no direct coding functions but they throw light on chromosome structures, dynamics and evolution.
(viii) Chromosome I 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 Polymorphisms occur in humans

Question. You are repeating the Hershey-Chase experiment and are provided with two isotopes 32P and 15N (in place of 35S in the original experiment). How does you expect your results to be different?
Answer : Use of 15N will be inappropriate because method of detection of 32P and 15N different (32P being a radioactive isotope while 15N is non-radioactive but is the heavier isotope of nitrogen). Even if 15Nwas radioactive then its presence would have been detected, both inside the cell (15N incorporated as introgenous base in DNA) as well as in the supernatant, because 15N would also get incorporated in amino group of amino acids in proteins. Hence, the use of 15N would not give any conclusive results.

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. (a) Draw a clover leaf structure of tRNA showing the following:
(i) Tyrosine attached to its amino acid site.
(ii) Anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(b) What does the actual structure of tRNA look like?
Answer : 

(b) The actual structure of tRNA looks like inverted L.

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. (a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below:

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.
Answer : 

Question. (i) Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA need to undergo changes? List the changes that hnRNA undergoes and where in the cell such changes take place.
Answer : (i) RNA polymerase II.
(ii) hnRNA has non-functional introns in between the functional exons. To remove these, it undergoes changes. The changes that hnRNA undergoes include capping, i.e., methyl guanosine triphosphate is added to 5′ end; tailing in which poly A tail is added at 3′ end; and splicing by which introns are removed and exons are joined.

Question. Draw a schematic representation of dinucleotide. Label the following:
(i) The components of a nucleotide
(ii) 5′ end
(iii) N-glycosidic linkage
(iv) Phosphodiester linkage. 
Answer : Nucleotide = Ribose sugar + Base + phosphate gr-p.

Question. Describe the structure of a RNA polynucleotide chain having f-r different types of nucleotides.
Answer : 

Question. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its anticodon.
Answer : The dual function of AUG codon:
(a) It codes for amino acid methionine.
(b) It is an initiation codon.
The sequence of bases from which it is transcribed is TAC. Its anticodon is UAC.

Question. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do y- think the statement is correct? Defend y-r Answer :
Answer : The statement is correct because of degeneracy of codons, mutations at third base of codon, usually doe not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.

Question. State the functions of the following in a prokaryote:
(i) tRNA (ii) rRNA CB
Answer : (i) tRNA reads the genetic codes, carries amino acids to the site of protein synthesis and act as an adaptor molecule.
(ii) rRNA plays structural and catalytic role during 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.

Question. A DNA segment has a total of 1500 nucleotides, -t of which 410 are Guanine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
Answer : According to Chargaff’s rule
A/G   = T/G =1

G = C, G = 410, hence C = 410
G + C = 410 + 410
= 820
So, A + T = 1500 – 820
= 680

A = T, so T = 680/2
= 340
So, Pyrimidines = C + T
= 410 + 340
= 750

Question. What is aminoacylation? State its significance.
Answer : Aminoacylation of tRNA or charging of tRNA is the activation of amino acids in the presence of ATP and their linkage to their cognate tRNA.
If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically.

Question. During in vitro synthesis of DNA, a researcher used 2′, 3′-dideoxycytidine triphosphate as raw nucleotide in place of 2′-deoxy cytidine triphosphate, other conditions remaining as standard.
Will further polymerisation of DNA continue up to the end or not? Explain.
Answer : Further polymerisation up to the end would not occur as the 3 –OH on sugar molecule is not for forming ester bond, which is re-ired to add another nucleotide.

Question. Differentiate between codon and an anticodon.
Answer :  Differences between codon and anticodon
Codon 
The sequence of 3 nitrogen bases on mRNA that codes for a particular amino acid during translation is called codon.
Anticodon
The sequence of 3 nitrogenous bases on tRNA that are complementary to the codon on mRNA for a particular amino acid during translation is called anticodon.

Question. What would happen if histones were to be mutated and made rich in amino acids aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer : If histone proteins were rich in acidic amino acids instead of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

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. (a) Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis. 
Answer : (a) RNA polymerase III is responsible for transcription of tRNA and the initiator tRNA gets linked with the amino acid methionine.
(b) The initiator tRNA, which is charged with amino acid methionine, reaches the smaller subunit of ribosome. Its anticodon UAC recognises the codon AUG on mRNA and binds by forming complementary base pairs. The large subunit of ribosome joins the smaller subunit and initiates translation.

Question. One of the salient features of the genetic code is that it is nearly universal from bacteria to humAnswer : Mention two exceptions to this rule. Why are some codes said to be degenerate?
Answer : The genetic code is universal except in mitochondria and few protozoAnswer :
Some codes are said to be degenerate because some amino acids are coded by more than one code.

Question. Given below is a schematic representation of lac operon:

(a) Identify i and p.
(b) Name the ‘inducer’ for this operon and explain its role. 
Answer : (a) i is the regulatory gene and p is the promoter gene.
(b) Lactose is the inducer. It is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon.

Question. “Genes contain the information that is required to express a particular trait.” Explain.
Answer : The genes present in an organism show a particular trait by way of forming certain product.
This is facilitated by the process of transcription and translation (according to central dogma of Biology)

Question. A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena? 
Answer : In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducer. Hence, cannot relieve the lac operon from its repressed state. Therefore, lac operon is always expressed.

Question. Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?
Answer : Bacteriophage does not have repetitive sequence such as VNTR in its genome as its genome is very small and have all the codon sequenced. Therefore, DNA fingerprinting is not done for bacteriophages.

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. Write the full form of VNTR. How is VNTR different from a probe?
Answer : Full form of VNTR is Variable Number of Tandem Repeats
VNTR 
The segment of DNA which shows very high no. of repetitive nucleotide sequences which vary from person to person are called variable number tandem repeats.
Probe
The small fragments of DNA strands which are highly selective and specific to be complementary to VNTR sequences are called probe.

Question. Mention two applications of DNA polymorphism. 
Answer : DNA polymorphism is applicable in genetic mapping and DNA finger printing.

Question. Where is an ‘operator’ located in a prokaryote DNA? How does an operator regulate gene expression at transcriptional level in a prokaryote? Explain. 
Answer : The operator region is located adjacent to promoter elements or prior to structural gene.
The operator regulates switching on and off the operon when the repressor binds to the operator region it is switched off and prevents transcription.
In the presence of inducer the repressor is inactivated and operator allows RNA polymerase to access the promoter. The operon is switched on and transcription proceeds.

1.Why is regulation of lac operon said to be negative control?
2.Given below is a part of the template strand of a structural gene TAC CAT TAG GAT
a) Write its transcribed mRNA strand with its polarity.
Explain the mechanism involved in initiation of transcription of this strand
3.a)Explain Griffith’s experiments where he witnessed transformation in bacteria he worked with. b) Name the scientists responsible for determining the biochemical nature of “transforming principle” in Griffith’s experiments. What did they prove?
1. Describe the experiment conducted by Alfred Hershey and Martha Chase for identification of genetic material.
2. Why is it considered path breaking in the field of molecular biology.
Where are anticodons seen? What is the codon of anticodon AUG?