molecular biology part-2 gene geneconceptgene …

57
MOLECULAR BIOLOGY PART-2 GENE GENE CONCEPT GENE CONCEPT PROTEIN SYNTHESIS LAC OPERON CONCEPT LAC OPERON CONCEPT

Upload: others

Post on 02-Dec-2021

2 views

Category:

Documents


0 download

TRANSCRIPT

MOLECULAR BIOLOGYPART-2

GENEGENE CONCEPTGENE CONCEPTPROTEIN SYNTHESIS LAC OPERON CONCEPTLAC OPERON CONCEPT

GENE AND GENE CONCEPTGENE AND GENE CONCEPTTo know about the scientists who coined various terminologies andcoined various terminologies and proposed gene hypothesisGene was coined byGene was coined by W Johansen in 1909Gene theory was proposed byGene theory was proposed by T H Morgan in 1911L P li d I (1949)L Pauling and Ingram(1949) established the role of genes in protein synthesisprotein synthesis

The fine structure of geneThe fine structure of genewas proposed by Seymour Benzere(1962)GENE CISTRONGENE CISTRON RECON MUTON

E R Garrod (1908) proposed One gene – one product g phypothesis

G W Beadle and E L Tatum(1948) proposed

One gene – one enzymeOne gene one enzyme hypothesis

Yanofsky (1965) proposedYanofsky (1965) proposed One gene ( cistron) – one polypeptide hypothesispolypeptide hypothesis

PROKARYOTIC GENE(Unspilt gene)

EUKARYOTIC GENE(spilt gene)(Unspilt gene)

Has only coding exons(spilt gene)

Has only coding exonsand non coding intronsC1 C2 C3E E E

g

Transcription

C1 C2 C3E E EI I

E E E

(Polycistronic mRNA)

E E E

I

I ITranscription

RNA S li i( Hn – RNA)

E E E

Translation E EEI

I RNA Splicing

(Monocistronic mRNA)

Many polypeptides or t i One protein

Translation

proteins One protein

(spilt gene – Philip A Sharp and Richards Roberts- 1979, Noble prize 1993)

CENTRAL DOGMA OF MOLECULAR BIOLOGY PROPOSED BY CRICK (1958)BIOLOGY PROPOSED BY CRICK (1958)

DNAREPLICATIONREPLICATION

DNADNATRANSCRIPTION REVERSE

TRANSCRIPTIONmRNA

TRANSLATIONPolypeptideyp p

PROTEIN SYNTHESIS

Sensestrand 5’ ATGGCGCACTAA 3’Sensestrand 5 .. ATGGCGCACTAA.. 3DNA

Antisensestrand 3’ TACCGCGTGATT 5’Antisensestrand 3 ..TACCGCGTGATT.. 5

mRNA 5’ AUGGCGCAC UAA 3’TRANSCRIPTION

mRNA 5 ..AUGGCGCAC UAA..3

tRNA 3’ UACCGCGUG 5’TRANSLATION

tRNA 3 ..UACCGCGUG…….. 5

POLYPEPTIDE aa1- aa2 - aa3

PROTEIN SYNTHESIS IN PROKARYOTETRANSCRIPTION

DNADNA

RNA polymeraseMg++ g

mRNA

STEPS OF TRANSLATIONSTEPS OF TRANSLATIONa) Activation of amino acid)

Amino acidt-RNA amino acyl synthetaset-RNA amino acyl synthetase

Mg++, ATPtransformylase

amino acyl t-RNA(aa tRNA)(aa-tRNA)

b) Chain initiation:mRNA

IF3 aa-tRNAIF2GTPIF3 aa tRNA

Smaller sub-unit of ribosome (30S)GTP

IF1Larger sub-unit of ribosome (50S)Larger sub unit of ribosome (50S)

c) Chain elongation:mRNA + ribosome ( 70S)mRNA + ribosome ( 70S)

EF – TUEF – TS , GTPpeptidyl transferasepeptidyl transferaseEF – G ( translocase )

elongating polypeptide chain

d) Ch i t i tid) Chain termination:Elongating polypeptide chaing g p yp p

RF1- recognizes UAA & UAGRF2 i UAA & UGARF2- recognizes UAA & UGARF3- function unknownPeptidyl synthetaseRib di i ti f tRibosome dissociation factor

Nascent polypeptide chainp yp p

e) Chain modification:Nascent polypeptide chainDeformylaseDeformylaseExopeptidaseMolecular choperonesFunctional / Structural proteinFunctional / Structural protein( having one or more polypeptides)

LAC OPERONLAC OPERON

LAC OPERON - SWITCHED ONLAC OPERON SWITCHED ON

LAC OPERON – SWITCHED OFFLAC OPERON SWITCHED OFF

Q - Which of the following iswrongly matched?wrongly matched?

1) cistron – functional unit of gene2) it f bi ti f2) recon – unit of recombination of a

gene3) muton – unit of mutation of a gene 4) exon – unit of split gene which ) p g

does not carry information

Q - The splicing of Hn RNAQ - The splicing of Hn RNAhelps in

1) removal of exons2) removal of coding segments2) removal of coding segments 3) removal of introns4) removal of coding exons

Q -The hypothesis proposed by G W Beadle and E L Tatum basedG.W.Beadle and E.L. Tatum based on their experiment in Neurospora crassa isNeurospora crassa is1) one gene- one enzyme2) one gene- one polypeptide 3) one gene- one product3) one gene one product 4) one gene- one function

E R Garrod(1908) proposed One gene – one product g phypothesis

G W Beadle and E L Tatum(1948) G ead e a d atu ( 9 8)proposed

One gene – one enzymeOne gene one enzyme hypothesis

Yanofsky(1965) proposedYanofsky(1965) proposed One gene ( cistron) – one polypeptide hypothesispolypeptide hypothesis

Q -The hypothesis proposed by G.W.Beadle and E.L. Tatum based on their experiment in Neurospora crassa isp1) one gene- one enzyme2) one gene one polypeptide2) one gene- one polypeptide 3) one gene- one product 4) one gene- one function

Q - In Eukaryotes, the mRNA is y ,transcribed from DNA with the help of the enzymehelp of the enzyme

1) RNA primase2) RNA poly I 3) RNA poly II3) RNA poly II 4) RNA poly III

Q During transcription mRNA isQ - During transcription mRNA is synthesized in ---- direction

d th t l t t d fand the template strand of DNA is read in ---- direction and during translation mRNA is read in ---- direction.1) 5’ →3’, 3’ → 5’ and 5’ → 3’2) 5’ →3’ 5’ → 3’ and 5’ → 3’2) 5 →3 , 5 → 3 and 5 → 33) 3’ →5’, 3’ → 5’ and3’ → 5’4) 3’ →5’, 5’ → 3’ and 5’ → 3’

PROTEIN SYNTHESISSensestrand 5’.. ATGGCGCACTAA.. 3’

DNADNAAntisensestrand 3’..TACCGCGTGATT.. 5’(Template strand)(Template strand)

mRNA 5’ AUGGCGCACUAA 3’

TRANSCRIPTION

mRNA 5 .. AUGGCGCACUAA.. 3

tRNA 3’ UACCGCGUG 5’TRANSLATION

tRNA 3 .. UACCGCGUG.. 5

POLYPEPTIDE aa1—aa2—aa3POLYPEPTIDE aa1 aa2 aa3

Q D i t i ti RNA iQ - During transcription mRNA is synthesized in ---- direction and the template strand of DNA is read in ---- direction and during translation mRNA is read in ---- direction.1) 5’ →3’, 3’ → 5’ and 5’ → 3’1) 5 3 , 3 5 and 5 32) 5’ →3’, 5’ → 3’ and 5’ → 3’3) 3’ 5’ 3’ 5’ d3’ 5’3) 3’ →5’, 3’ → 5’ and3’ → 5’4) 3’ →5’, 5’ → 3’ and 5’ → 3’) ,

Q - What sequence in the template strand of DNA correspondsstrand of DNA corresponds to the first amino acid inserted i t t i ?into protein?

1) TAC1) TAC2) AUG3) UAG4) AAA4) AAA

PROTEIN SYNTHESISPROTEIN SYNTHESIS

Sensestrand 5’.. ATGGCGCACTAA.. 3’

Antisensestrand 3’..TACCGCGTGATT.. 5’

mRNA 5’ AUGGCGCACUAA 3’

TRANSCRIPTION

mRNA 5 .. AUGGCGCACUAA.. 3

tRNA 3’ UACCGCGUG 5’TRANSLATION

tRNA 3 .. UACCGCGUG.. 5

POLYPEPTIDE aa1—aa2—aa3POLYPEPTIDE aa1 aa2 aa3

Q - What sequence in the templateQ What sequence in the template strand of DNA corresponds to the first amino acid inserted intofirst amino acid inserted into protein?

1) TAC2) AUG2) AUG3) UAG)4) AAA

Q - In protein synthesisQ In protein synthesis, translation involves

1) decoding of triplet codons of tRNA by anticodons of mRNA y

2) decoding of triplet codons of mRNA by anticodons of tRNAmRNA by anticodons of tRNA

3) decoding of triplet codons of rRNA by anticodons of tRNA

4) synthesis of mRNA from DNA4) synthesis of mRNA from DNA

Q - Which of the following statement is wrong with respect to chain initiation step of protein synthesis in E.coli?

1) 5’ end of mRNA attaches to 30S )ribosome with the help of IF3

2) Formyl methionine tRNA decodes ) yAUG with the help of IF2

3) 50S ribosome attaches to the initiation )complex with the help of IF1

4) This process derives the energy from ) p gyATP

Chain initiationmRNA

IF3 aa-tRNAIF2GTPIF3 aa tRNA

Smaller sub-unit of ribosome (30S)GTP

IF1 (INITIATION COMPLEX)

Larger sub-unit of ribosome (50S)

Q - Which of the following statement is wrong with respect to chain initiation g pstep of protein synthesis in E.coli

1) 5’ end of mRNA attaches to 30S ribosome )with the help of IF3

2) Formyl methionine tRNA decodes AUG with the help of IF2

3) 50S ribosome attaches to the initiation l ith th h l f IF1complex with the help of IF1

4) This process derives the energy from ATPATP

Q - Which of the following factors are labelled as translocase that help in the movement of ribosome on mRNA?ribosome on mRNA?

1) EF – TU and EF- TS2) EF- G and eEF23) eEF1 and eEF1B3) eEF1 and eEF1B4) 23S and 28S rRNA

Q - Peptide bond formation occurs during protein synthesis with theduring protein synthesis with the help of the enzyme

1)Amino acyl tRNA synthetase2)Peptidyl transferase) p y3)Peptidyl synthetase4)Transformylase4)Transformylase

Q - If a segment of an mRNA has a sequence 5’ GUACCGAUC 3’ , whichqof the following could have been the template DNA molecule?template DNA molecule?

1) 3’CAUGGCUAG 5’ 2) 5’CATGGCTAG3’ 3) 5’CAUGGCUAG3’ 3) 5 C UGGCU G34) 5’GATCGGTAC3’

Sensestrand 5’..… GTACCGATC….. 3’(CODOGEN)(CODOGEN)

Antisensestrand 3’..… CATGGCTAG….. 5’(CRYPTOGRAM)(CRYPTOGRAM)

mRNA 5’ GUACCGAUC 3’mRNA 5 ..… GUACCGAUC….. 3(CODON)

tRNA 3’..… CAUGGCUAG….. 5’ (ANTICODON)(ANTICODON)

Q - If a segment of an mRNA has a sequence 5’ GUACCGAUC 3’ , which of the following could have been the template DNA molecule?template DNA molecule?

1) 3’CAUGGCUAG 5’ 2) 5’CATGGCTAG3’2) 5’CATGGCTAG3’ 3) 5’CAUGGCUAG3’ )4) 5’GATCGGTAC3’

Q - A fragment of DNA nucleotide arrangement is

5’ ATGGCAGCTTAT 3’5 ….ATGGCAGCTTAT….3What will be the anticodons in tRNAs which will translate the codons produced by this part of p y pDNA?

1) 5’UAC3’ 5’CGU3’ 5’CGA3’ 5’AUA3’ 2) 3’UAC5’ 3’CGU5’ 3’CGA5’ 3’AUA5’2) 3 UAC5 3 CGU5 3 CGA5 3 AUA5 3) 5’AUG3’ 5’GCA3’ 5’GCU3’ 5’UAU3’ 4) 3’AUG5’ 3’GCA5’ 3’GCU5’ 3’UAU5’

5’ ATGGCAGCTTAT 3’( Sensestrand )( Sensestrand )

noncoding

Anticodons of tRNA are complementary to

t d ( di )sense strand (noncoding)

3’ UACCGUCGAAUA 5’(tRNA)

1) 5’UAC3’ 5’CGU3’ 5’CGA3’ 5’AUA3’ 2) 3’UAC5’ 3’CGU5’ 3’CGA5’ 3’AUA5’ 3) 5’AUG3’ 5’GCA3’ 5’GCU3’ 5’UAU3’ )4) 3’AUG5’ 3’GCA5’ 3’GCU5’ 3’UAU5’

Q - The reading frame of mRNAQ The reading frame of mRNA having 1002 nucleotides translated into a nascenttranslated into a nascent polypeptide having amino acids numbering

1) 3331) 3332) 666 3) 999 4) 10014) 1001

Q- Consider the statements given below and find the answer amongbelow and find the answer among the options given below.

Statement A: UAA, UAG and UGA d i RNA t i t thcodons in mRNA terminate the

synthesis of poylpeptide chain Statement B: UAA, UAG and UGA codons are not recognized by tRNAcodo s a e ot ecog ed by t

1) Both statements A and B are correct1) Both statements A and B are correctand B is not the reason for A

2) Both statements A and B are correct2) Both statements A and B are correct and B is the reason for A

3) Statement A is correct and B is wrong3) Statement A is correct and B is wrong4) Statement A wrong and B is correct

Q-The replacement of any one-nucleotide base in a codonnucleotide base in a codonchanging the specificity for amino acid isamino acid is

1)Mis-sense mutation1)Mis sense mutation 2) non-sense mutation 3) silent mutation3) silent mutation 4) frame shift/gibberish mutation

Q- Which among the following is l t h d?wrongly matched?

1) Lac Z gene β galactosidase1) Lac Z gene – β galactosidase2) Lac y gene – β galactoside permease3) Lac A gene – Transacetylase4) Operator gene – Repressor protein) p g p p

Q- Operator gene of Lac operoni t d h l tis turned on when lactose molecules bind to1) promoter site2) operator site2) operator site3) regulator gene4) repressor protein

LAC OPERON - SWITCHED ON

Q- Operator gene of Lac operonis turned on when lactose molecules bind to

1) t it1) promoter site2) operator site)3) regulator gene4) repressor protein4) repressor protein

Q- A synthetic mRNA ofQ- A synthetic mRNA of repeating sequence 5’ CA CA CA CA CA CA CA C 3’5 CA CA CA CA CA CA CA C…3is used for a cell-free protein synthesizing system like the onesynthesizing system like the one used by Nirenberg. If we assume that protein synthesis can beginthat protein synthesis can begin without the need for an initiator codon what product or productscodon, what product or products would you expect to occur after protein synthesis?protein synthesis?

1) one protein with an alternating1) one protein, with an alternating sequence of two different types of amino acidsof amino acids

2) one protein, consisting of a single type of amino acidtype of amino acid

3) two proteins with an alternatingsequence of two different types ofsequence of two different types of amino acids

4) t i ith lt ti4) one protein with an alternating sequence of three different types of amino acidsamino acids

mRNA 5’ CACACACACACA 3’(codon)tRNA 3’ GUGUGUGUGUGU 5’tRNA 3’ GUGUGUGUGUGU 5’(anticodon)( )POLYPEPTIDE Val – Cys – Val – Cys( i id )(aminoacids)

1) one protein with an alternating1) one protein, with an alternating sequence of two different types of amino acidsof amino acids

2) one protein, consisting of a single type of amino acidtype of amino acid

3) two proteins with an alternating sequence of two different types ofsequence of two different types of amino acids

4) t i ith lt ti4) one protein with an alternating sequence of three different types of amino acidsamino acids

Column I Column II

Q- Match the following operon genes

Column I

A) Structural gene

Column II

i) Binding site for ) g

B) Operator gene

) grepressor gene

ii) Codes for repressor

C) Promoter geneprotein

iii) Codes for enzymet i

D) Regulator geneproteins

iv) Binding site for RNApolymerasepolymerase

A B C DA B C D1) i iii ii iv2) ii i iii iv3) iv i ii iii3) iv i ii iii4) iii i iv ii

Q-The Pribnow box is represented by

1) 5` ATATTA 3` 2) 5` AATAAT 3`2) 5 AATAAT 3 3) 5` TATAAT 3` 4) 5` TATAA 3`