Molecular Basis of Inheritance - 1 File Download

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Chapter 4

Molecular Basis of Inheritance

SECTION - A

Objective Type Questions

(The DNA, The Search for Genetic Material, RNA World)

1. If the sequence of one strand of DNA is 5 A T G C A T C G 3, find the sequence of complementary strand

in 5 3 direction

(1) T A C G T A G C (2) C G A T G C A T (3) A T G C A T C G (4) A T C G T A C G

Sol. Answer (2)

TA C G TA G C

AT G C AT C G

3' 5'

5 3'

2. NHC structural proteins are

(1) Basic proteins rich in lysine, arginine

(2) Regulatory proteins

(3) Catalytic proteins rich in tryptophan and arginine

(4) Required for packaging of chromatin at higher levels

Sol. Answer (4)

3. Which of the following bond is not present in DNA?

(1) – 1– 9–N–glycosidic bond (2) 3– 5 phosphodiester bond

(3) – 1–1–N–glycosidic bond (4) – 1– 2–N–glycosidic bond

Sol. Answer (4)

– 1' – 9' Glycosidic bond Sugar and Purines

– 1' – 1' Glycosidic bond Sugar and Pyrimidine

3' – 5' Phosphodiester bond Sugar and Phosphate

(Replication, Transcription, Genetic Code, Translation)

4. How many types of DNA polymerases are present in bacteria?

(1) Five (2) Three (3) Two (4) One

Solutions

Level - II

88 Molecular Basis of Inheritance Solutions of Assignment (Level-II)


Sol. Answer (2)

DNA Pol I

Pol II

Pol III

5. Synthesis of leading and lagging strand require

(1) Single primer (2) Single and many primers respectively

(3) Many and single primers respectively (4) Many primers

Sol. Answer (2)

Leading Continuous strand have polarity 3' 5' end it need single primer

Lagging Discontinuous strand have polarity 5' 3' end it need multiple primer

6. For the strand separation and stabilisation during DNA replication which of the following set of enzymes andproteins are required?

(1) SSBP, gyrase, primase (2) Topoisomerase, helicase, ligase

(3) Gyrase, ligase, primase (4) Topoisomerase, helicase, SSBP

Sol. Answer (4)

Topoisomerase – Remove twisting stress

Helicase – Unwind DNA

SSBP – Single strand binding protein

7. In eukaryotes, the RNA polymerase that synthesises tRNA is RNA polymerase ______ and is also responsiblefor formation of ______ rRNA.

(1) II, 5.8 S (2) I, 5 S (3) III, 5 S (4) II, 18 S

Sol. Answer (3)

8. What is correct for bacterial transcription?

(1) mRNA requires processing to become active

(2) Translation can begin when mRNA is fully transcribed

(3) Transcription and translation takes place in the same compartment

(4) Rho factor initiates the process

Sol. Answer (3)

Transcription and translation take place in cytoplasm.

9. Which of the following is not required during post transcriptional processing in eukaryotes?

(1) Methyl guanosine triphosphate (2) Ligase

(3) ScRNA (4) SnRNA

Sol. Answer (3)

Small cytoplasmic RNA ScRNA

10. Which of the following feature is correct for bacteria?

(1) Presence of intervening sequences in DNA

(2) DNA does not show coiling

(3) Linear ss-DNA representing single chromosome

(4) DNA can be chromosomal as well as extrachromosomal

89Solutions of Assignment (Level-II) Molecular Basis of Inheritance


Sol. Answer (4)

Nuclear DNA Nucleoid

Extrachromosomal Plasmid

11. In-vitro template independent RNA synthesis is a feature of

(1) RNA polymerase (2) Reverse transcriptase

(3) Ochoa enzyme (4) DNA polymerase

Sol. Answer (3)

Ochoa enzyme Polynucleotide phosphorylase.

12. In protein synthesis, which of the following are required for the synthesis of charged tRNA?

(1) Amino acid, GTP, initiation codon, ribosome

(2) Amino acid, ATP, Mg++, enzyme, tRNA

(3) Amino acid, ATP, K+, enzyme, mRNA

(4) Aminoacyl tRNA, ribosome, initiation codon, release factor

Sol. Answer (2)

13. Termination of polypeptide synthesis in bacteria differs from eukaryotes in

(1) Having different termination codons (2) Being GTP dependent

(3) Involving more than one type of release factors (4) All of these

Sol. Answer (3)

14. The accessibility of promoter regions of bacterial DNA in many cases regulated by the interaction of proteinswith sequences termed

(1) Regulators (2) Structural genes (3) Inhibitor genes (4) Operators

Sol. Answer (4)

If repressor is attached at operator their is no synthesis of mRNA.

15. Select the correct one (w.r.t. Wobble hypothesis)

(1) Third base of a codon lacks vibrating capacity

(2) Third base can establish H-bonds even with the non complementary anticodon

(3) Specificity of a anticodon is particularly determined by first two codon

(4) Major cause of degeneracy is the first two N-bases of codon

Sol. Answer (2)

GUG 3rd position can establish H - bond even with the non-complementary anticodon.

16. If there are 81 million bases in RNA of human cell, then calculate the total number of introns present in cDNA

(1) 27 millions (2) Zero

(3) Equal to ribonucleotides (4) Half the number of ribonucleotides

Sol. Answer (2)

Human RNA is processed RNA. So no intron in cDNA.

17. Splicing is necessary for preparing a mature transcript and its movement to cytoplasm. It requires

(1) scRNA and proteins (2) snRNA and proteins

(3) scRNA and snRNA (4) scRNA only

Sol. Answer (2)



18. Majority of unusual bases are found in tRNA, TC loop is

(1) First loop from 5-end of tRNA (2) AA - tRNA synthetase binding loop

(3) Ribosomal binding loop (4) Nodoc site

Sol. Answer (3)

TC Ribosomal binding loop

19. How many amino acids will be coded by the mRNA sequence – 5CCCUCAUAGUCAUAC3 if a adenosineresidue is inserted after 12th nucleotide?

(1) Five amino acids (2) Six amino acids (3) Two amino acids (4) Three amino acids

Sol. Answer (3)

5' CCC UCA UAG UCA UAC

AA Stop codon

AA

So only two A.A will be formed.

20. Identification and binding of RNA polymerase to the promoter sequence is a function of

(1) Rho factor (2) Sigma factor (3) Beta factor (4) Omega factor

Sol. Answer (2)

21. The DNA strand showing replication using Okazaki fragments also shows

(1) Continuous growth in 5 3 direction

(2) Discontinuous growth on 5 3 parental strand

(3) Discontinuous growth on 3 5 parental strand

(4) Involvement of one primer only

Sol. Answer (2)

22. Prokaryotic transcription mechanism requires involvement of only one polymerase type and

(a) It occurs in cytoplasm only

(b) It is often coupled with translation

(c) It does not require splicing but capping is essential

(1) All are correct (2) Both (b) & (c) are incorrect

(3) Both (a) & (c) are correct (4) Only (c) is incorrect

Sol. Answer (4)

Their is no capping and tailing in prokaryotes.

23. Pribnow box is a consensus of ________ bases, forming a binding site for E. coli RNA polymerase at promotor.

(1) TATAAT (2) AGGAGG (3) CAAT (4) GC

Sol. Answer (1)

24. In tailing, adenylate residues are added at 3′ end

(1) With the help of gyanyl transferase (2) In a template independent manner

(3) With the help of methyl transferase (4) Of hn-RNA of E.coli

Sol. Answer (2)

Poly a tail formed at 3' end to stabilize RNA.



25. For every single amino acid incorporated in peptide chain _____ ATP and _____ GTP molecules are used.

(1) 1, 4 (2) 1, 6 (3) 1, 2 (4) 1, 3

Sol. Answer (3)

1ATP Activation of tRNA

2GTP One in Peptide bond formation and other in ribosome movement to each new codon in the m-RNA.

26. In t-RNA

(1) CCA – OH is present at 5-end (2) TC loop for attaching the amino acid

(3) DHU loop for binding with AA - activating enzyme (4) There are three recognition sites

Sol. Answer (3)

(Regulation of Gene Expression, Human Genome Project, DNA Fingerprinting)

27. When the genomes of two people are cut using the same restriction enzyme, the length and number offragments obtained are different, this is called

(1) PCR (2) RFLP (3) EST (4) Northern blotting

Sol. Answer (2)

RFLP Restricted fragment length polymorphisom.

28. Which of the following does not code for any proteins?

(1) Micro-satellites (2) Exons

(3) Mini-satellites (4) More than one option is correct.

Sol. Answer (4)

(1) and (3)

29. Which statement is correct for homeotic genes?

(1) Control is exerted through homeodomain proteins

(2) Mutation in these genes not results in conversion of one body part into another

(3) Such genes have been studied extensively in humans

(4) Control oncogenesis process

Sol. Answer (1)

30. In which step of DNA profiling, nitrocellulose membrane is used?

(1) Denaturation (2) Autoradiography (3) Blotting (4) DNA amplification

Sol. Answer (3)

Southern Blotting

31. Repressor of lac-operon

(1) Is a tetrameric protein

(2) Having a molecular weight of 16,000

(3) Has only one side

(4) Is made by operator gene

Sol. Answer (1)



32. A set of genes or cDNA is immobilized on a glass slide and used in transcriptome studies is called

(1) Proteome (2) Microarray (3) DNA chip (4) Genome

Sol. Answer (2)

33. Repetitive sequences are stretches of DNA with repeated bases many times in a genome, but

(a) These sequences are of no transcriptional function

(b) These are associated with euchromatin region

(c) These helps to identify a person on the basis of its DNA specificity

(1) All are correct (2) Only (b) is incorrect

(3) Both (a) & (b) are correct (4) Both (b) & (c) are incorrect

Sol. Answer (2)

34. The microsatellites have simple sequences of repeated

(1) 11-60 bp (2) 1-6 bp

(3) 10 bp (4) 50 bp

Sol. Answer (2)

Microsatellite 1 – 6 bp

35. In tryptophan operon

(1) Non-proteinaceous aporepressor is synthesised by R-gene

(2) Normally chorismic acid is not converted into tryptophan

(3) Repression is mostly connected with a catabolic pathway

(4) Enzymes produced by structural genes normally present in the cell

Sol. Answer (4)

36. Which of the following DNA form has maximum number of base pairs per turn?

(1) A-DNA (2) B-DNA

(3) C-DNA (4) Z-DNA

Sol. Answer (4)

37. Non-histone proteins

(1) Are of five types (2) Are involved in nucleosome formation

(3) Control gene expression (4) Are basic proteins

Sol. Answer (3)

38. RNA primer is removed by

(1) DNAP-I (2) DNAP-II (3) DNAP-III (4) Primase

Sol. Answer (1)

DNA polymerase I has exonuclease activity.

39. How many types of DNA polymerases are associated with eukaryotic cell?

(1) Three (2) Four

(3) Five (4) Two

Sol. Answer (3)



40. Find the incorrect match

(1) Central dogma : F. Crick

(2) Reverse central dogma : Temin and Baltimore

(3) Split genes : Kornberg

(4) mRNA : Jacob and Monad

Sol. Answer (3)

Split gene was discovered by Richard J. Roberts and Phillip Sharp.

41. Recognition sequence for transcription in prokaryotes is

(1) TATATAT (2) TATAAT (3) TATAAAT (4) CAAT

Sol. Answer (2)

42. In the mitochondrial DNA, UGA codes for

(1) Chain termination (2) Chain initiation

(3) Tryptophan (4) Tyrosine

Sol. Answer (3)

43. Which of the following inhibits binding of amino-acyl tRNA to ribosomes?

(1) Neomycin (2) Erythromycin

(3) Streptomycin (4) Tetracycline

Sol. Answer (4)

44. Tryptophan operon is

(1) Catabolic system (2) Repressible system

(3) Inducible system (4) Having three structural genes

Sol. Answer (2)

45. Choose the correct option w.r.t. the chemical nature of apo-repressor and co-repressor respectively intrp-operon?

(1) Protein, Amino acid (2) Amino acid, Protein

(3) Lipoidal, Sugary (4) Sugary, Lipoidal

Sol. Answer (1)

46. Gene battery model was proposed by

(1) Jacob and Monad (2) Gamow

(3) H.G. Khorana (4) Britten and Davidson

Sol. Answer (4)

47. An insect leg may change into antenna due to mutation in

(1) c-oncogene (2) v-oncogene

(3) Homeotic genes (4) Proto-oncogene

Sol. Answer (3)



SECTION - B

Previous Years Questions

1. All of the following are part of an operon except [NEET-2018]

(1) an operator (2) structural genes (3) a promoter (4) an enhancer

Sol. Answer (4)

• Enhancer sequences are present in eukaryotes.

• Operon concept is for prokaryotes.

2. AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequenceof the transcribed mRNA? [NEET-2018]

(1) AGGUAUCGCAU (2) UGGTUTCGCAT (3) UCCAUAGCGUA (4) ACCUAUGCGAU

Sol. Answer (1)

Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil inmRNA.

3. Select the correct match [NEET-2018]

(1) Ribozyme - Nucleic acid

(2) F2 × Recessive parent - Dihybrid cross

(3) G. Mendel - Transformation

(4) T.H. Morgan - Transduction

Sol. Answer (1)

Ribozyme is a catalytic RNA, which is nucleic acid.

4. Select the correct match [NEET-2018]

(1) Alec Jeffreys - Streptococcus pneumoniae

(2) Alfred Hershey and Martha Chase - TMV

(3) Francois Jacob and Jacques Monod - Lac operon

(4) Matthew Meselson and F. Stahl - Pisum sativum

Sol. Answer (3)

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 – Semi-conservative DNA replication in E. coli.

– Alfred Hershey and Martha Chase – Proved DNA as genetic material not protein

5. The experimental proof for semiconservative replication of DNA was first shown in a [NEET-2018]

(1) Fungus (2) Bacterium (3) Virus (4) Plant

Sol. Answer (2)

Semi-conservative DNA replication was first shown in Bacterium Escherichia coli by Matthew Meselson andFranklin Stahl.

6. If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?[NEET-2017]

(1) 1 (2) 11 (3) 33 (4) 333

Sol. Answer (3)

If deletion occurs at 901st position the remaining 98 bases specifying for 33 codons of amino acids will bealtered.



7. The final proof for DNA as the genetic material came from the experiments of [NEET-2017]

(1) Griffith (2) Hershey and Chase

(3) Avery, Mcleod and McCarty (4) Hargobind Khorana

Sol. Answer (2)

Hershey and Chase gave unequivocal proof which ended the debate between protein and DNA as geneticmaterial.

8. The association of histone H1 with a nucleosome indicates [NEET-2017]

(1) Transcription is occurring (2) DNA replication is occurring

(3) The DNA is condensed into a Chromatin Fibre (4) The DNA double helix is exposed

Sol. Answer (3)

The association of H1 protein indicates the complete formation of nucleosome.

Therefore the DNA is in condensed form.

9. Spliceosomes are not found in cells of [NEET-2017]

(1) Plants (2) Fungi (3) Animals (4) Bacteria

Sol. Answer (4)

Spliceosomes are used in removal of introns during post-transcriptional processing of hnRNA in eukaryotes onlyas split genes are absent as prokaryotes.

10. DNA replication in bacteria occurs [NEET-2017]

(1) During S-phase (2) Within nucleolus

(3) Prior to fission (4) Just before transcription

Sol. Answer (3)

DNA replication in bacteria occurs prior to fission. Prokaryotes do not show well marked S-phase due to theirprimitive nature.

11. Which of the following RNAs should be most abundant in animal cell? [NEET-2017]

(1) r-RNA (2) t-RNA (3) m-RNA (4) mi-RNA

Sol. Answer (1)

rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell.

12. During DNA replication, Okazaki fragments are used to enlongate [NEET-2017]

(1) The leading strand towards replication fork (2) The lagging strand towards replication fork

(3) The leading strand away from replication fork (4) The lagging strand away from the replication fork

Sol. Answer (4)

Two DNA polymerase molecules work simultaneous at the DNA fork, one on the leading strand and the otheron the lagging strand.

Each Okazaki fragment is synthesized by DNA polymerase at lagging strand in 5 3direction. New Okazakifragments appear as the replication fork opens further.

As the first Okazaki fragment appears away from the replication fork, the direction of elongation would be awayfrom replication fork.

13. Taylor conducted the experiments to prove semiconservative mode of chromosome replication on

[NEET (Phase-2)-2016]

(1) Vinca rosea (2) Vicia faba

(3) Drosophila melanogaster (4) E. coli



Sol. Answer (2)

Semiconservative mode of chromosome replication was proved by Taylor in Vicia faba.

14. The equivalent of a structural gene is [NEET (Phase-2)-2016]

(1) Muton (2) Cistron (3) Operon (4) Recon

Sol. Answer (2)

Cistron is a segment of DNA coding for a polypeptide.

Eukaryotic structural gene is monocistronic whereas prokaryotic structural gene is polycistronic.

15. Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?


(1) 5 S rRNA (2) 18 S rRNA (3) 23 S rRNA (4) 5.8 S rRNA

Sol. Answer (3)

23S rRNA is a component of larger subunit of ribosome and it act as peptidyl transferase (ribozyme).

16. DNA-dependent RNA polymerase catalyzes transcription on one strand of the DNA which is called the


(1) Template strand (2) Coding strand (3) Alpha strand (4) Antistrand

Sol. Answer (1)

The DNA-dependent RNA polymerase catalyze the polymerisation in only one direction that is 5 3, thestrand with polarity 5 3 act as template and is called as template strand.

17. Which of the following is required as inducer(s) for the expression of Lac operon? [NEET-2016]

(1) Lactose and Galactose (2) Glucose

(3) Galactose (4) Lactose

Sol. Answer (4)

Lac operon is an inducible operon. Lactose is the substrate for the enzyme beta-galactosidase and it alsoregulates switching on and off of the operon. Hence, it is termed as inducer.

18. Which of the following is not required for any of the techniques of DNA fingerprinting available at present?

[NEET-2016]

(1) DNA-DNA hybridization (2) Polymerase chain reaction

(3) Zinc finger analysis (4) Restriction enzymes

Sol. Answer (3)

A zinc finger is a small protein structural motif that characterised by the co-ordination of one or more Zn ions inorder to stabilise the folds.

19. Which one of the following is the start codon? [NEET-2016]

(1) UAG (2) AUG (3) UGA (4) UAA

Sol. Answer (2)

AUG is the start codon.

UAA, UAG and UGA are stop codons.

20. Identify the correct order of organisation of genetic material from largest to smallest [Re-AIPMT-2015]

(1) Chromosome, genome, nucleotide, gene

(2) Chromosome, gene, genome, nucleotide

(3) Genome, chromosome, nucleotide, gene

(4) Genome, chromosome, gene, nucleotide



Sol. Answer (4)

Order of organisation of genetic material

Largest Genome (haploid set of chromosome)

Chromosome (Condensed chromatin)

Gene (Segment of DNA; unit of inheritance)

Nucleotide (Made up of pentose sugar, nitrogen base and phosphate)

Smallest

21. Satellite DNA is important because it [Re-AIPMT-2015]

(1) Code for enzymes needed for DNA replication

(2) Codes for proteins needed in cell cycle

(3) Shows high degree of polymorphism in population and also the same degree of polymorphism in an

individual, which is heritable from parents to children

(4) Does not code for proteins and is same in all members of the population

Sol. Answer (3)

Satellite DNA are the repetitive DNA which do not code for any protein. They show high degree of polymorphism

and form basis of DNA fingerprinting. Since DNA from every tissue from an individual show the same degree

of polymorphism, they become very useful identification tool in forensic applications.

22. Gene regulation governing lactose operon of E.coli that involves the lac I gene product is [AIPMT-2015]

(1) Feedback inhibition because excess of -galactosidase can switch off transcription

(2) Positive and inducible because it can be induced by lactose

(3) Negative and inducible because repressor protein prevents transcription

(4) Negative and repressible because repressor protein prevents transcription

Sol. Answer (3)

Lac operon under control of repressor is a negative regulation. The nature of operon is inducible.

23. Which one of the followings is wrongly matched? [AIPMT-2014]

(1) Transcription- Writing information from DNA to t-RNA

(2) Translation-Using information in m-RNA to make protein

(3) Repressor protein-Binds to operator to stop enzyme synthesis

(4) Operon-Structural genes, operator and promoter

Sol. Answer (4)

Operon Structural, Operator, Promoter and Regulator genes.



24. Select the correct option [AIPMT-2014]

(1)

(2)

(3)

(4)

5 – 3

5 – 3

3 – 5

3 – 5

Direction ofRNA synthesis

Direction of reading of the template DNA strand

3 – 5

5 – 3

3 – 5

5 – 3

Sol. Answer (1)

25. Transformation was discovered by [AIPMT-2014]

(1) Meselson and Stahl (2) Hershey and Chase

(3) Griffith (4) Watson and Crick

Sol. Answer (3)

On Diplococcus pneumoniae

26. The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.

DNAA

mRNAB

ProteinProposed by

C [NEET-2013]

(1) A-translation, B-transcription, C-Erevin Chargaff

(2) A-transcription, B-translation, C-Francis Crick

(3) A-translation, B-extension, C-Rosalind Franklin

(4) A-transcription, B-replication, C-James Watson

Sol. Answer (2)

DNA mRNA Protein Proposed by

CReplicationA

Transcription

B

TranslationFrancis Crick

Central Diagram

27. Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac Y gene?

[NEET-2013]

(1) Lactose permease (2) Transacetylase

(3) Lactose permease and transacetylase (4) -galactosidase

Sol. Answer (4)

Z Y A Transacetylase

Permease

Z –Galactosidase

Y is mutated so Z gene will give its effects.

28. Removal of RNA polymerase III from nucleoplasm will affect the synthesis of [AIPMT (Prelims)-2012]

(1) mRNA (2) rRNA (3) tRNA (4) hnRNA

Sol. Answer (3)

Polymerase III Synthesize tRNA

29. Removal of introns and joining of exons in a defined order during transcription is called [AIPMT (Prelims)-2012]

(1) Slicing (2) Splicing (3) Looping (4) Inducing

Sol. Answer (2)



30. Which one of the following is not a part of a transcription unit in DNA? [AIPMT (Prelims)-2012]

(1) A promoter (2) The structural gene (3) The inducer (4) A terminator

Sol. Answer (3)

Inducer is a chemical which will trigger the DNA to switch on gene in its presence.

31. If one strand of DNA has the nitrogenous base sequence as ATCTG, what would be the complementary RNAstrand sequence? [AIPMT (Prelims)-2012]

(1) AACTG (2) ATCGU (3) TTAGU (4) UAGAC

Sol. Answer (4)

A T C T G

U A G A C

32. Read the following four statements (A-D)

(A) In transcription, adenosine pairs with uracil

(B) Regulation of lac operon by repressor is referred to as positive regulation

(C) The human genome has approximately 50,000 genes

(D) Haemophilia is a sex-linked recessive disease

How many of the above statements are right? [AIPMT (Mains)-2012]

(1) Two (2) Three

(3) Four (4) One

Sol. Answer (1)

(A) and (D)

33. PCR and Restriction Fragment Length Polymorphism are the methods for [AIPMT (Prelims)-2012]

(1) DNA sequencing (2) Genetic fingerprinting

(3) Study of enzymes (4) Genetic transformation

Sol. Answer (2)

PCR and RFLP

34. What is it that forms the basis of DNA fingerprinting? [AIPMT (Mains)-2012]

(1) Satellite DNA occurring as highly repeated short DNA segments

(2) The relative proportions of purines and pyrimidines in DNA

(3) The relative difference in the DNA occurrence in blood, skin and saliva

(4) The relative amount of DNA in the ridges and grooves of the fingerprints

Sol. Answer (1)

Satellite DNA occur in highly repeated member.

35. Which one of the following also acts as a catalyst in a bacterial cell? [AIPMT (Prelims)-2011]

(1) 23 S rRNA (2) 5 S rRNA

(3) sn RNA (4) hn RNA

Sol. Answer (1)

36. What are those structures that appear as 'beads – on – string' in the chromosomes when viewed under electronmicroscope? [AIPMT (Prelims)-2011]

(1) Base pairs (2) Genes (3) Nucleotides (4) Nucleosomes

Sol. Answer (4)



37. In history of biology, human genome project led to the development of [AIPMT (Mains)-2011]

(1) Bioinformatics (2) Biosystematics (3) Biotechnology (4) Biomonitoring

Sol. Answer (1)

38. The unequivocal proof of DNA as the genetic material came from the studies on a [AIPMT (Mains)-2011]

(1) Viroid (2) Bacterial virus (3) Bacterium (4) Fungus

Sol. Answer (2)

Hershy and Chase experiment on T2 Bacteriophage

39. Which one of the following does not follow the central dogma of molecular biology? [AIPMT (Prelims)-2010]

(1) HIV (2) Pea (3) Mucor (4) Chlamydomonas

Sol. Answer (1)

Reverse transcriptase enzyme is found in HIV

40. Select the two correct statements out of the four (a–d) given below about lac operon.

(a) Glucose or galactose may bind with the repressor and inactivate it.

(b) In the absence of lactose the repressor binds with the operator region.

(c) The z-gene codes for permease.

(d) This was elucidated by Francois Jacob and Jacque Monod.

The correct statements are [AIPMT (Prelims)-2010]

(1) (a) and (b) (2) (b) and (c) (3) (a) and (c) (4) (b) and (d)

Sol. Answer (4)

Z - gene - galactosidase

41. The one aspect which is not a salient feature of genetic code, is its being [AIPMT (Prelims)-2010]

(1) Specific (2) Degenerate (3) Ambiguous (4) Universal

Sol. Answer (3)

Non-ambiguous

42. In eukaryotic cell transcription, RNA splicing and RNA capping take place inside the [AIPMT (Mains)-2010]

(1) Ribosomes (2) Nucleus (3) Dictyosomes (4) ER

Sol. Answer (2)

Post transcriptional changes take place in nucleus.

43. The lac operon consists of [AIPMT (Mains)-2010]

(1) Four regulatory genes only (2) One regulatory gene and three structural genes

(3) Two regulatory genes and two structural genes (4) Three regulatory genes and three structural genes

Sol. Answer (2)

i P O Z Y A

Regulator 3 Structural genes

44. The 3 — 5 phosphodiester linkages inside a polynucleotide chain serve to join [AIPMT (Mains)-2010]

(1) One DNA strand with the other DNA strand (2) One nucleoside with another nucleoside

(3) One nucleotide with another nucleotide (4) One nitrogenous base with pentose sugar

Sol. Answer (3)



45. Satellite DNA is useful tool in [AIPMT (Prelims)-2010]

(1) Genetic engineering (2) Organ transplantation

(3) Sex determination (4) Forensic science

Sol. Answer (4)

DNA fingerprinting

46. Removal of introns and joining the exons in a defined order in a transcription unit is called:

[AIPMT (Prelims)-2009]

(1) Tailing (2) Transformation (3) Capping (4) Splicing

Sol. Answer (4)

Take place while post transcriptional change in RNA.

47. Semiconservative replication of DNA was first demonstrated in [AIPMT (Prelims)-2009]

(1) Escherichia coli (2) Streptococcus pneumoniae

(3) Salmonella typhimurium (4) Drosophila melanogaster

Sol. Answer (1)

48. What is not true for genetic code? [AIPMT (Prelims)-2009]

(1) It is nearly universal

(2) It is degenerate

(3) It is unambiguous

(4) A codon in mRNA is read in a noncontiguous fashion

Sol. Answer (4)

49. Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a "triplet"?


(1) Hershey and Chase (2) Morgan and Sturtevant

(3) Beadle and Tatum (4) Nirenberg and Mathaei

Sol. Answer (4)

50. Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with the categorymentioned against it? [AIPMT (Prelims)-2008]

(1) Adenine, Thymine - Purines (2) Thymine, Uracil - Pyrimidines

(3) Uracil, Cytosine - Pyrimidines (4) Guanine, Adenine - Purines

Sol. Answer (1)

51. In the DNA molecules [AIPMT (Prelims)-2008]

(1) There are two strands which run antiparallel one in 53' direction and other in 35

(2) The total amount of purine nucleotides and pyrimidine nucleotides is not always equal

(3) There are two strands which run parallel in the 53' direction

(4) The proportion of adenine in relation to thymine varies with the organism

Sol. Answer (1)

52. Which one of the following pairs of codons is correctly matched with their function or the signal for theparticular amino acid? [AIPMT (Prelims)-2008]

(1) UUA, UCA - Leucine (2) GUU, GCU - Alanine

(3) UAG, UGA - Stop (4) AUG, ACG - Start / Methionine



Sol. Answer (3)

(i) UUA Leucine UCA Serine

(ii) GUU Valine GCU Alanine

(iii) AUG Methionine ACG Threonine

53. One gene-one enzyme relationship was established for the first time in [AIPMT (Prelims)-2007]

(1) Diploccus pneumoniae

(2) Neurospora crassa

(3) Salmonella typhimurium

(4) Escherichia coli

Sol. Answer (2)

54. A sequential expression of a set of human genes occurs when a steroid molecule binds to the:


(1) Ribosome (2) Transfer RNA (3) Messenger RNA (4) DNA sequence

Sol. Answer (4)

55. The Okazaki fragments in DNA chain growth: [AIPMT (Prelims)-2007]

(1) Polymerize in the 5 to 3 direction and explain 3 to 5 DNA replication

(2) Result in transcription

(3) Polymerize in the 3 to 5 direction and forms replication fork

(4) Prove semi-conservative nature of DNA replication

Sol. Answer (1)

56. Molecular basis of organ differentiation depends on the modulation in transcription by:


(1) Anticodon (2) RNA polymerase (3) Ribosome (4) Transcription factor

Sol. Answer (4)

57. Telomere repetitive DNA sequence control the function of eukaryote chromosomes because they:


(1) Prevent chromosome loss (2) Act as replicons

(3) Are RNA transcription initiator (4) Help chromosome pairing

Sol. Answer (1)

They prevent the loss of DNA from the terminal part of DNA.

58. During transcription, RNA polymerase holoenzyme binds to a gene promoter and assumes a saddle -likestructure. What is it's DNA-binding sequence? [AIPMT (Prelims)-2007]

(1) TATA (2) TTAA (3) AATT (4) CACC

Sol. Answer (1)

59. The two polynucleotide chains in DNA are [AIPMT (Prelims)-2007]

(1) Semiconservative (2) Parallel (3) Discontinuous (4) Antiparallel

Sol. Answer (4)

Antiparallel nature means one strand has polarity 5' 3' another strand has polarity 3' 5'



60. The length of DNA molecule greatly exceeds the dimensions of the nucleus in eukaryotic cells. How is thisDNA accommodated? [AIPMT (Prelims)-2007]

(1) Through elimination of repetitive DNA (2) Deletion of non-essential genes.

(3) Super-coiling in nucleosomes (4) DNAse digestion

Sol. Answer (3)

DNA is coiled to form a solanoid.

61. One turn of the helix in a B-form DNA is approximately: [AIPMT (Prelims)-2006]

(1) 20 nm (2) 0.34 nm (3) 3.4 nm (4) 2 nm

Sol. Answer (3)

3.4 nm or 34 Å

62. Antiparallel strands of a DNA molecule means that: [AIPMT (Prelims)-2006]

(1) One strand turns anti-clockwise

(2) The phosphate groups of two DNA strands, at their ends, share the same position

(3) The phosphate groups at the start of two DNA strands are in opposite position (pole)

(4) One strand turns clockwise

Sol. Answer (3)

One strand has polarity 5' 3'

Another strand has polarity 3' 5'

63. Amino acid sequence, in protein synthesis is decided by the sequence of [AIPMT (Prelims)-2006]

(1) t-RNA (2) m-RNA (3) c-DNA (4) r-RNA

Sol. Answer (2)

mRNA is a cistron which carry information for synthesis of protein.

64. One gene – one enzyme hypothesis was postulated by [AIPMT (Prelims)-2006]

(1) R. Franklin (2) Hershey and Chase (3) A. Garrod (4) Beadle and Tatum

Sol. Answer (4)

Beadle and Tatum experimented on Neurospora crassa.

65. During transcription holoenzyme RNA polymerase binds to a DNA sequence and the DNA assumes a saddlelike structure at that point. What is that sequence called? [AIPMT (Prelims)-2005]

(1) CAAT box (2) GGTT box (3) AAAT box (4) TATA box

Sol. Answer (4)

66. E. coli cells with a mutated Z gene of the lac operon cannot grow in medium containing only lactose as thesource of energy because [AIPMT (Prelims)-2005]

(1) In the presence of glucose, E. coli cells do not utilize lactose

(2) They cannot transport lactose from the medium into the cell

(3) The lac operon is constitutively active in these cells

(4) They cannot synthesize functional â-galactosidase

Sol. Answer (4)

67. Nucleotide are building blocks of nucleic acids, nucleotide is a composite molecule formed by :


(1) (Base-sugar-phosphate)n

(2) Base-sugar-OH

(3) Base-sugar-phosphate (4) Sugar-phosphate

Sol. Answer (3)



68. Which one of the following makes use of RNA as a template to synthesize DNA? [AIPMT (Prelims)-2005]

(1) Reverse transcriptase (2) DNA dependant RNA polymerase

(3) DNA polymerase (4) RNA polymerase

Sol. Answer (1)

It is RNA dependent DNA polymerase

69. Uridine, present only in RNA is a

(1) Pyrimidine (2) Nucleoside (3) Nucleotide (4) Purine

Sol. Answer (2)

70. Which of the following is not a property of the genetic code?

(1) Universal (2) Non-overlapping (3) Ambiguous (4) Degeneracy

Sol. Answer (3)

Genetic code is non-ambiguous.

71. One of the most frequently used techniques in DNA fingerprinting is

(1) AFLP (2) VNTR (3) SSCP (4) SCAR

Sol. Answer (2)

72. In an inducible operon, the genes are

(1) Always expressed

(2) Usually not expressed unless a signal turns them "on"

(3) Usually expressed unless a signal turns them "off"

(4) Never expressed

Sol. Answer (2)

It needs a inducer to trigger and switch on gene.

73. A single strand of nucleic acid tagged with a radioactive molecule is called

(1) Plasmid (2) Probe (3) Vector (4) Selectable marker

Sol. Answer (2)

Probe Radioactive DNA help to make out sequence of unknown DNA.

74. The reaction, Amino acid + ATP Aminoacyl-AMP + P–P depicts

(1) Amino acid assimilation (2) Amino acid transformation

(3) Amino acid activation (4) Amino acid translocation

Sol. Answer (3)

Amino acid get activated to attach to tRNA.

75. The transcription of any gene is the indication of its

(1) Induction (2) Activity (3) Stimulation (4) Hypersensitivity

Sol. Answer (2)

76. mRNA directs the building of proteins through a sequence of

(1) Introns (2) Codons (3) Exons (4) Anticodons

Sol. Answer (2)

Codon will carry sequence in which AA will attach.

77. Beadle and Tatum showed that each kind of mutant bread mould they studied lacked a specific enzyme. Theirexperiments demonstrated that

(1) Cells need specific enzymes in order to function

(2) Genes are made of DNA

(3) Genes carry information for making proteins

(4) Enzymes are required to repair damaged DNA information



Sol. Answer (3)

Beadle & Tatum experiment was done on Nurospora Crassa. They proposed one gene one enzyme hypothesis

78. Which of the following necleotide sequences contain 4 pyrimidine bases?

(1) GATCAATGC (2) GCUAGACAA (3) UAGCGGUAA (4) Both (2) & (3)

Sol. Answer (1)

GATCAATCG 4 1 2 34

GCUAGACAA 2

UAGCGGUAA 2

Pyrimidine

79. The 1992 Nobel Prize for medicine was awarded to Edmond H. Fischer and Edwin J. Krebs for their workconcerning

(1) Reversible protein phosphorylation as a biological regulation mechanism

(2) Isolation of the gene for a human disease

(3) Human genome project

(4) Drug designing involving inhibition of DNA synthesis of the pathogen

Sol. Answer (1)

80. Initiation codon in eukaryotes is

(1) GAU (2) AGU (3) AUG (4) UAG

Sol. Answer (3)

AUG Methionine

81. ‘Lac operon’ in E. coli, is induced by

(1) “I” gene (2) Promoter gene (3) -galactosidase (4) Lactose

Sol. Answer (4)

82. There are special proteins that help to open up DNA double helix in front of the replication fork. These proteinsare

(1) DNA ligase (2) DNA topoisomerase I (3) DNA gyrase (4) DNA polymerase I

Sol. Answer (3)

DNA gyrase Show helicase as well as topoisomerase activity.

83. In protein synthesis, the polymerization of amino acids involves three steps. Which one of the following is notinvolved in the polymerization of protein?

(1) Termination (2) Initiation (3) Elongation (4) Transcription

Sol. Answer (4)

Transcription is the process.

84. Anticodon is an unpaired triplet of bases in an exposed position of

(1) t-RNA (2) m-RNA (3) r-RNA (4) Both (2) & (3)

Sol. Answer (1)

tRNA contain anticodon.

85. An environmental agent, which triggers transcription from an operon, is a

(1) Depressor (2) Controlling element (3) Regulator (4) Inducer

Sol. Answer (4)



86. In split genes, the coding sequences are called

(1) Exons (2) Cistrons (3) Introns (4) Operons

Sol. Answer (1)

Exon Coding Intron Noncoding

87. The lac operon is an example of

(1) Repressible operon (2) Overlapping genes (3) Arabinose operon (4) Inducible operon

Sol. Answer (4)

88. If the DNA codons are ATG ATG ATG and a cytosine base is inserted at the beginning, then which of thefollowing will result?

(1) CAT GAT GATG (2) A non-sense mutation (3) C ATG ATG ATG (4) CA TGA TGA TG

Sol. Answer (1)

ATG ATG ATG

CAT GAT GATG

C Addition

89. The wild type E.coli cells are growing in normal medium with glucose. They are transferred to a mediumcontaining only lactose as sugar. Which of the following changes take place?

(1) The lac operon is induced (2) E.coli cells stop dividing

(3) The lac operon is repressed (4) All operons are induced

Sol. Answer (1)

In the presence of lactose lac operone is induced.

90. If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will be

(1) GUAGCUUA (2) AUUCGAUG (3) CAUCGAAU (4) UAAGCUAC

Sol. Answer (4)

ATTCGATG

UAAGCTAC

DNA

91. Which of the following serves as an stop codon?

(1) UAG (2) AGA (3) AUG (4) GCG

Sol. Answer (1)

UAG Stop codon

92. The codons causing chain termination are

(1) AGT, TAG, UGA (2) UAG, UGA, UAA (3) TAG, TAA, TGA (4) GAT, AAT, AGT

Sol. Answer (2)

UAG Amber UGA Opal UAA Ocher

93. DNA synthesis can be specifically measured by estimating the incorporation of radio-labelled

(1) Thymidine (2) Deoxyribose sugar (3) Uracil (4) Adenine

Sol. Answer (1)

Thymidine in Vicia faba by Taylor.

94. Which of the following step of translation does not consume a high energy phosphate bond?

(1) Peptidyl transferase reaction (2) Aminoacyl t-RNA binding to A-site

(3) Translocation (4) Amino acid activation



Sol. Answer (1)

Peptide bond is independent.

95. DNA elements, which can switch their position, are called

(1) Cistrons (2) Transposons (3) Exons (4) Introns

Sol. Answer (2)

Transposon jumping genes

96. Sequence of which of the following is used to know the phylogeny?

(1) m-RNA (2) r-RNA (3) t-RNA (4) DNA

Sol. Answer (4)

97. Genes that are involved in turning on or off the transcription of a set of structural genes are called

(1) Redundant genes (2) Regulatory genes (3) Polymorphic genes (4) Operator genes

Sol. Answer (2)

Regulatory genes Repressor

98. In operon concept, regulator gene functions as

(1) Inhibitor (2) Repressor (3) Regulator (4) All of these

Sol. Answer (2)

99. In DNA, when AGCT occurs, their association as per which of the following pair?

(1) AT-GC (2) AG-CT (3) AC-GT (4) All of these

Sol. Answer (1)

1AGCT

A – T

G – C

100. Irregularity is found in Drosophila during the organ differentiation. For example, in place of wing Iong legs areformed. Which gene is responsible?

(1) Double dominant gene (2) Homeotic gene (3) Complimentary gene (4) Plastid gene

Sol. Answer (2)

101. Method of DNA replication in which two strands of DNA separate and synthesise new strands, is

(1) Dispersive (2) Conservative (3) Semi conservative (4) Non conservative

Sol. Answer (3)

Semi conservative half parental strand is retained half newly formed.

102. In negative operon

(1) Co-repressor binds with repressor (2) Co-repressor does not bind with repressor

(3) Co-repressor binds with inducer (4) cAMP have negative effect on lac operon

Sol. Answer (1)

Co-repressor bind with apo-repressor Apo-corepressor complex which completely block the site of operatorgene.

103. Gene and cistron words are sometimes used synonymously because

(1) One cistron contains many genes (2) One gene contains many cistrons

(3) One gene contains one cistron (4) One gene contains no cistron



Sol. Answer (3)

Cistron is functional unit of DNA that produce or code for polypeptide.

104. m-RNA is synthesised on DNA template in which direction?

(1) 5 3 (2) 3 5 (3) Both (1) & (2) (4) Any of these

Sol. Answer (1)

5 3' Polymerase activity.

105. At the time of organogenesis, genes regulate the process at different levels and at different time due to

(1) Promoter (2) Regulator (3) Intron (4) Exon

Sol. Answer (4)

106. A mutant strain of T4 - bacteriophage, R-II, fails to lyse the E. coli but when two strains R-IIX and

R-IIY are mixed then they lyse the E. coli. What may be the possible reason?

(1) Bacteriophage transforms in wild (2) It is not mutated

(3) Both strains have similar cistrons (4) Both strains have different cistrons

Sol. Answer (4)

107. In E. coli, during lactose metabolism repressor binds to

(1) Regulator gene (2) Operator gene (3) Structural gene (4) Promoter gene

Sol. Answer (2)

108. In a DNA, percentage of thymine is 20% then what will be percentage of guanine?

(1) 20% (2) 40% (3) 30% (4) 60%

Sol. Answer (3)

A + G

2 0 + 3 0

= C + T

= 3 0 + 2 0

109. Out of 64 codons, 61 codons code for 20 types of amino acid. It is called

(1) Degeneracy of genetic code (2) Overlapping of genes

(3) Wobbling of codons (4) Universality of codons

Sol. Answer (1)

More than one codon code for a amino acid.

110. Jacob and Monad studied lactose metabolism in E. coli and proposed operon concept. Operon concept isapplicable for

(1) All prokaryotes (2) All prokaryotes and some eukaryotes

(3) All prokaryotes and all eukaryotes (4) All prokaryotes and some protozoans

Sol. Answer (1)

111. Exon part of hn-RNA have code for

(1) Protein (2) Lipid (3) Carbohydrate (4) Phospholipid

Sol. Answer (1)

Hn – RNA (Heterogeneous nuclear RNA).

112. Which form of RNA has a structure resembling with clover leaf?

(1) rRNA (2) hn-RNA (3) mRNA (4) tRNA

Sol. Answer (4)

Clover leaf



113. Which of the following reunites the exon segments after RNA splicing?

(1) RNA polymerase (2) RNA primase (3) RNA ligase (4) RNA proteoses

Sol. Answer (3)

RNA Ligase

114. During initiation of translation in prokaryotes, a GTP molecule is needed in

(1) Formation of formyl-met-tRNA

(2) Binding of 30 S subunit of ribosome with mRNA

(3) Association of 30 S mRNA with formyl-met-tRNA

(4) Association of 50 S subunit of ribosome with initiation complex

Sol. Answer (3)

GTP is needed

mRNA

30 S

115. In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids?

(1) 20 (2) 64 (3) 61 (4) 60

Sol. Answer (3)

61 codons specify 20 amino acids.

3 codons are stop codons i.e., UAA, UAG, UGA.

116. The telomeres of eukaryotic chromosomes consist of short sequences of

(1) Thymine rich repeats (2) Cytosine rich repeats

(3) Adenine rich repeats (4) Guanine rich repeats

Sol. Answer (4)

117. What does “lac” refer to in what we call the lac operon?

(1) Lactose (2) Lactase

(3) Lac insect (4) The number 1,00,000

Sol. Answer (1)

Lactose operon

118. Degeneration of a genetic code is attributed to the

(1) First member of a codon (2) Second member of codon

(3) Entire codon (4) Third member of a codon

Sol. Answer (4)

GUG Third position is Wooble position

119. During transcription, the DNA site at which RNA polymerase binds is called

(1) Promoter (2) Regulator

(3) Receptor (4) Enhancer

Sol. Answer (1)

Promoter site RNA bind



120. What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated toUAA ?

(1) A polypeptide of 24 amino acids will be formed

(2) Two polypeptides of 24 and 25 amino acids will be formed



Sol. Answer (1)

UAA Stopcodon

24 Codon 25 th Codon

Amino Acid

121. Which one of the following triplet codon, is correctly matched with its specificity for an amino acid in proteinsynthesis or as ‘start’ or ‘stop’ codon ?

(1) UCG - start (2) UUU - stop (3) UGU – leucine (4) UAC - tyrosine

Sol. Answer (4)

UUU Phenylalanine

UGU Cystine

UCG

UCGSerine

122. DNA fingerprinting refer to

(1) Molecular analysis of profiles of DNA samples

(2) Analysis of DNA samples using imprinting devices

(3) Techniques used for molecular analysis of different specimens of DNA

(4) Techniques used for identification of fingerprints of individuals

Sol. Answer (1)

123. During transcription, if the nucleotide sequence of the DNA strand that is being coded is ATACG then thenucleotide sequence in the mRNA would be

(1) TATGC (2) TCTGG (3) UAUGC (4) UATGC

Sol. Answer (3)

A T A C G DNA

U A U G C RNA

124. After a mutation at a genetic locus the character of an organism changes due to change in

(1) Protein structure (2) DNA replication

(3) Protein synthesis pattern (4) RNA transcription pattern

Sol. Answer (1)

125. During replication of a bacterial chromosome DNA synthesis starts from a replication origin site and

(1) RNA primers are involved (2) Is facilitated by telomerase

(3) Moves in one direction of the site (4) Moves in bi-directional way

Sol. Answer (4)



126. The following ratio is generally constant for a given species

(1) A + G / C + T (2) T + C / G + A (3) G + C / A + T (4) A + C / T + G

Sol. Answer (3)

G + C

A + T

127. What is true for E. coli with inhibited lac-z gene?

(1) They cannot synthesize permease

(2) They cannot synthesize functional beta galactosidase

(3) They cannot synthesize transacetylase

(4) They cannot transport lactose from the medium into the cell

Sol. Answer (2)

Z - gene - Galactosidase

128. Which antibiotic inhibits interaction between tRNA and mRNA during bacterial protein synthesis?

(1) Tetracycline (2) Erythromycin (3) Neomycin (4) Streptomycin

Sol. Answer (3)

Erythromycin Inhibits translocation of mRNA along ribosomes.

Streptomycin Inhibits initiation of translation and causes misreading.

Tetracycline Inhibit binding of aminoacyl-tRNA to ribosome.

129. During transcription, RNA polymerase holoenzyme binds to a gene promoter and assumes a saddle - likestructure. What is its DNA-binding sequence?

(1) AATT (2) CACC (3) TATA (4) TTAA

Sol. Answer (3)

TATA box

130. Differentiation of organs and tissues in a developing organism, is associated with

(1) Differential expression of genes (2) Lethal mutations

(3) Deletion of genes (4) Developmental mutations

Sol. Answer (1)

131. The nuclease enzyme, which begins its attack from free end of a polynucleotide, is

(1) Polymerase (2) Endonuclease (3) Exonuclease (4) Kinase

Sol. Answer (3)

132. Radio-tracer technique shows that DNA is in

(1) Multi-helix stage (2) Single-helix stage (3) Double-helix stage (4) None of these

Sol. Answer (3)

133. Genes are packaged into a bacterial chromosome by

(1) Acidic protein (2) Actin (3) Histones (4) Basic protein

Sol. Answer (4)



134. The hereditary material present in the bacterium E.coli is

(1) Single-stranded DNA (2) Double-stranded DNA

(3) DNA and RNA (4) RNA

Sol. Answer (2)

Doble strand circular DNA

135. The Pneumococcus experiment proves that

(1) Bacteria do not reproduce sexually

(2) RNA sometime controls the production of DNA and proteins

(3) DNA is the genetic material

(4) Bacteria undergo binary fission

Sol. Answer (3)

Experiment was done by Griffith on Pneumococcus pneumoniae.

136. E.coli about to replicate was placed in a medium containing radio active thymidine for five minutes. Then itwas made to replicate in a normal medium. Which of the following observation shall be correct?

(1) Both the strands of DNA will be radioactive (2) One strand radioactive

(3) Each strand half radioactive (4) None is radioactive

Sol. Answer (2)

DNA replication is semiconservative.

137. Types of RNA polymerase required in nucleus of eukaryotes for RNA synthesis is/are

(1) 1 (2) 2 (3) 3 (4) 4

Sol. Answer (3)

RNA Polymerase I

RNA Polymerase II

RNA Polymerase III

138. Transformation experiment was first performed on which bacteria?

(1) E. coli (2) Diplococcus pneumoniae

(3) Salmonella (4) Pasteurella pestis

Sol. Answer (2)

Diplococcus pneumoniae by Griffth

139. Telomerase is an enzyme which is a

(1) Simple protein (2) RNA (3) Ribonucleoprotein (4) Repetitive DNA

Sol. Answer (3)

It synthesise the terminal or telomeric DNA.

140. In transgenics, expression of transgene in target tissue is determined by

(1) Enhancer (2) Transgene (3) Promoter (4) Reporter

Sol. Answer (3)



141. A nutritionally wild type organism, which does not require any additional growth supplement, is known as

(1) Osmotroph (2) Mixotroph (3) Auxotroph (4) Prototroph

Sol. Answer (4)

Used in Beadel and Tatum experiment.

142. What is not true for genetic code?

(1) It is unambiguous

(2) A codon in mRNA is read in a non-contiguous fashion

(3) It is nearly universal

(4) It is degenerate

Sol. Answer (2)

143. Semiconservative replication of DNA was first demonstrated in

(1) Salmonella typhimurium (2) Drosophila melanogaster

(3) Escherichia coli (4) Streptococcus pneumoniae

Sol. Answer (3)

By Meselson and Stahl

144. Which one of the following statements about the particular entity is true?

(1) Nucleosome is formed of nucleotides

(2) DNA consists of a core of eight histones

(3) Centromere is found in animal cells, which produces aster during cell division

(4) The gene for producing insulin is present in every body cell

Sol. Answer (4)

SECTION - C

Assertion-Reason Type Questions

1. A : RNA polymerase is of three types in eukaryotes for the synthesis of all types of RNAs.

R : RNA polymerase consists of six types of polypeptides alongwith rho factor which is involved in termination ofRNA synthesis.

Sol. Answer (3)

2. A : 5S rRNA and surrounding protein complex provides binding site of tRNA.

R : tRNA is soluble RNA with unusual bases.

Sol. Answer (2)

3. A : Operator gene is functional when it is not blocked by repressor.

R : Regulator gene produces active protein only which acts on operon system in E.coli.

Sol. Answer (3)

Regulator gene produce its product. It is inducer which will bind with repressor.

4. A : Peptidyl transfer site is contributed by larger sub-unit of ribosome.

R : The enzyme peptidyl transferase is contributed by both 23S and 16S ribosomal sub-units.

Sol. Answer (3)



5. A : Teminism is unidirectional flow of information.

R : It requires DNA dependent RNA polymerase enzyme.

Sol. Answer (4)

Teminisms is revese transcription so no unidirectional flow of information.

6. A : In bacterial translation mechanism, two tRNA are required by methionine.

R : AUG codes for methionine and it shows nonambiguity also.

Sol. Answer (2)

7. A : Nutritional mutant strain of pink mould is auxotroph.

R : It is not able to prepare its own metabolites from the raw materials obtained from outside.

Sol. Answer (1)

Auxotrops are mutants which have specific nutritional demand

8. A : In DNA fingerprinting, hybridization is done with molecular probe.

R : Molecular probe is small DNA segment synthesized in laboratory with known sequence that recognisecomplementary sequence in RNA.

Sol. Answer (3)

Probe help in identification of unknown sequence of DNA.

9. A : c-DNA libraries are important to scientists in human genomics.

R : c-DNA is synthetic type of DNA generated from mRNA.

Sol. Answer (2)

10. A : SNPs-pronounced "snips" are common in human genome.

R : It is minute variations that occurs at a frequency of one in every 300 bases.

Sol. Answer (1)

Single nucleotide polymorphism.

11. A : Catalytic functions were assigned to RNA molecule during evolution.

R : The rate of mutation is quite fast in RNA.

Sol. Answer (2)

12. A : Kornberg enzyme is associated with the removal of primers and thyminedimer.

R : DNA polymersase I does exonuclease activity in 5 3 and 3 5 directions.

Sol. Answer (1)

13. A : Wobbling reduces the number of tRNAs required for polypeptide synthesis.

R : It increases the effect of code degeneracy.

Sol. Answer (3)

One tRNA can read non-complementary sequence by Wobble hypothesis.

14. A : Unknown DNA after hybridization with VNTR probe, the autoradiogram gives many bands of differing sizesin DNA profiling.

R : These bands represents DNA fingerprint of organism.

Sol. Answer (2)

15. A : Polypeptide sequences are dictated by DNA and represented by mRNA.

R : Sequence of amino acids in a polypeptide can be predicted by the exact sequence of nucleotides on themRNA and template DNA.

Sol. Answer (3)



16. A : Triplet genetic code can be confirmed by frame shift mutations.

R : Frame shifting involves the change in protein product coded by triplet codons.

Sol. Answer (2)

By additio or deletion of one Nucleotide cause frame shift mutation.

17. A : Lac operon exerts negative control only.

R : The operator is occupied by aporepressor during regulation.

Sol. Answer (4)

Lac openon is +ve and –ve control.

18. A : Single DNA dependent RNA polymerase catalyses transcription of all types of RNA in all except bacteria.

R : Structural genes in bacteria are monocistronic.

Sol. Answer (4)

19. A : Sigma factor of RNA polymesase recognizes the start signal region in prokaryotes.

R : Promotor region lies at 5 of template strand.

Sol. Answer (3)

20. A : HGP was completed in 2003 by sequencing all genes of all chromosomes.

R : All coding and noncoding genes were sequenced by ESTs.

Sol. Answer (4)

HGP completed by 2006.

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