MCQ Of Molecular basis of Inheritance | Chapter 6 | Class 12 | Biology |

MCQ Of Molecular basis of Inheritance, Chapter 6, Class 12, Biology

 

Question 1: The length of DNA usually depends on nucleotides

  1. Position of nucleotides
  2. Number of nucleotides
  3. Both A and B
  4. None of the above

Answer : B (Number of nucleotides)

Question 2: Haploid content of human DNA has

  1. 3.3 × 10⁷ bp
  2. 3.3 × 10⁸ bp
  3. 3.3 × 10⁹ bp
  4. 3.3 × 10¹⁰ bp

Answer : C (3.3 × 10⁹ bp)

Question 3: Nitrogenous bases are linked to sugar by

  1. Hydrogen bond
  2. phosphodiester bond
  3. N – glycosidic bond
  4. O – glycosidic bond

Answer : C (N – glycosidic bond)

Question 4: What is the difference between adenosine and deoxyadenosine?

  1. Only sugar
  2. Only purine
  3. Only phosphate
  4. All of these

Answer : A (Only sugar)

Question 5: When a phosphate group is linked to …A… group of nucleoside through …B… bond, a corresponding …C… is formed.

Choose the correct option for A, B and C.

  1. A – 5′ OH, B – phosphodiester bond, C – nucleotide
  2. A – 3′ OH, B – phosphodiester bond, C – nucleotide
  3. A – 2′ OH, B – phosphodiester bond, C – nucleotide
  4. A – 5 OH, B – phosphodiester bond, C – nucleoside

Answer : A (A – 5′ OH, B – phosphodiester bond, C – nucleotide )

 

Question 6: Backbone of DNA is formed by

  1. Sugar
  2. Phosphates
  3. Both A and B
  4. Nitrogenous bases (purine and pyrimidine)

Answer : C (Both A and B)

 Question 7: Which additional group is present at the 2′ position of the ribose sugar in RNA ?

  1. R—H
  2. CHO
  3. OH
  4. COOH

Answer : C (OH)

Question 8: In some viruses, the flow of information is in reverse direction, i.e. from RNA to DNA. Can you suggest a simple name to the process?

  1. Transcription
  2. Transception
  3. Reverse transcription
  4. Translation

Answer : C (Reverse transcription)

Question 9: Find out the number of base pairs in in E. coli DNA if its DNA is 1.36 mm long.

  1. 4 × 10⁶ bp
  2. 3 × 10⁶ bp
  3. 2 × 10⁶ bp
  4. 7 × 10⁶ bp

Answer : A (4 × 10⁶ bp)

Question 10: Positively charged basic proteins that are found in eukaryotes are called

  1. Histones
  2. Protamine
  3. Arginine
  4. Lysine

Answer : A (Histones)

You may also read MCQ of Reproduction in Organisms, MCQ of Sexual Reproduction in Flowering Plants, MCQ of Human Reproduction, MCQ of Reproductive Health, MCQ Of Principles of Inheritance and Variation for better understanding of the chapters.

 Question 11: In the given diagram, identify A, B and C.

  1. A – DNA, B – Histone, C – Histone octamer
  2. A – RNA, B – Histone, C – Histone octamer
  3. A – DNA, B – Histone, C – Histone tetramer
  4. A – RNA, B – Histone, C – Histone tetramer

Answer : A (A – DNA, B – Histone, C – Histone octamer)

Question 12: Lightly stained part of chromatin which remains loosely packed and its transcriptionally active named as

  1. Euchromatin
  2. Heterochromatin
  3. Chromatosome
  4. Chromonemata

Answer : A (Euchromatin)

Question 13: Experiment organism of Frederick and Griffith was

  1. Variola virus
  2. Tuberculosis bacteria
  3. Actinomycetes
  4. Streptococcus pneumoniae

Answer : D (Streptococcus pneumoniae)

Question 14: In Griffith’s experiment, mice infected with the …A… die from pneumonia infection but mice infected with …B… don’t develop pneumonia. Choose the correct option for A and B.

  1. A – S strain; B – S + R strain
  2. A – S strain; B – R strain
  3. A – R strain; B – S strain
  4. A – R strain; B – S + R strain

Answer : B (A – S strain; B – R strain)

Question 15: A molecule that can act as a genetic material must fulfill the traits given below, except

  1. It should be able to express itself in the form of ‘Mendelian characters’
  2. It should be able to generate its replica
  3. It should be unstable structurally and chemically
  4. It should provide the scope for slow changes that are required for evolution

Answer : C (It should be unstable structurally and chemically)

Question 16: In Hershey and Chase experiment, Bacteriophage nucleic acids were labelled as

  1. ³²P labelled phosphate
  2. ³H labelled H2O
  3. ³⁵S labelled sulphate
  4. ¹⁴C labelled CO2

Answer : A (³²P labelled phosphate)

Question 17: Hershey and Chase concluded that viral infecting agent in their experiment was

  1. Protein
  2. DNA
  3. RNA
  4. Both B and C

Answer : B (DNA)

Question 17: RNA is the genetic material in

  1. All bacteria
  2. Tobacco Mosaic Viruses (TMV)
  3. QB bacteriophage
  4. Both B and C

Answer : D (Both B and C)

Question 18: Which group present in RNA nucleotide is very reactive and makes RNA liable and easily degradable than DNA?

  1. 3-OH’ group at every nucleotide
  2. 2-OH group on ribose sugar
  3. 3-OH’ group on ribose sugar
  4. 4 OH’ group on ribose sugar

Answer : B (2-OH group on ribose sugar)

Question 19: Stability of DNA is impacted by to

  1. Deoxyribose sugar
  2. Presence of thymine in place of uracil
  3. Both A and B
  4. None of the above

Answer : C (Both A and B)

Question 20: Which one of the following is not applicable to RNA ?

  1. Complementary base pairing
  2. 5′ phosphoryl and 3′ hydroxyl ends
  3. Heterocyclic nitrogenous bases
  4. Chargaff’s rule

Answer : D (Chargaff’s rule)

 Question 21: DNA replication is semiconservative. It was first shown in

  1. Fungi
  2. Bacteria
  3. Vicia faba
  4. Algae

Answer : B (Bacteria)

Question 22: Name the heavy isotope used by Meselson and Stahl for proving the semiconservative mode of DNA.

  1. ¹⁵NH4Cl
  2. ¹⁴NH3Cl2
  3. ¹³NH2Cl3
  4. All of these

Answer : A (¹⁵NH4Cl)

Question 23: Heavy DNA can be differentiated from normal DNA by which centrifugation technique?

  1. AgCl density gradient
  2. CaSO4 density gradient
  3. CsCl density gradient
  4. KCl density gradient

Answer : C ( CsCl density gradient)

Question 24: In Meselson and Stahl’s experiment, DNA extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid (or intermediate density). Why?

  1. Because the generation time of E. coli (culture) was about 20 minutes
  2. Because it would take 20 minutes for RNA replication
  3. Because it would take 20 minutes for replication of DNA to RNA (transcription)
  4. Because it would take 20 minutes for translation RNA to protein

Answer : A (Because the generation time of E. coli (culture) was about 20 minutes)

Question 25: Radioisotope used by Taylor in his experiment was

  1. Iron
  2. Titanium
  3. Thymidine
  4. Copper

Answer : C (Thymidine)

Question 26: DNA polymerase is

  1. DNA dependent
  2. DNA independent
  3. RNA dependent
  4. RNA independent

Answer : A (DNA dependent)

Question 27: DNA dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the

  1. Template strand
  2. Coding strand
  3. Alpha strand
  4. Anti – strand

Answer : A (Template strand )

Question 28: DNA dependent DNA polymerases catalyses polymerization in which direction ?

  1. 3′ — 5′
  2. 5′ — 2′
  3. 5′ — 3′
  4. 2′ — 5′

Answer : C (5′ — 3′)

Question 29: Which among the following statements does not stands true for DNA replication ?

  1. Replication initiate randomly at any place in DNA by DNA polymerase
  2. In eukaryotes, DNA replication takes place at S – phase of cell cycle
  3. Failure of cell division after DNA replication results in polyploidy
  4. E.coli.having 4.6 × 10⁶ base pair completes DNA replication within 38 minutes

Answer : A (Replication initiate randomly at any place in DNA by DNA polymerase)

Question 30: On which strand of DNA, replication is continuous ?

  1. 5′ — 3′ polarity strand
  2. 3′ — 5′ polarity strand
  3. 3′ — 2′ polarity strand
  4. 3′ — 4′ polarity strand

Answer : B (3′ — 5′ polarity strand)

 Question 31: Identify A, B and C strands.

  1. A – Continuous strand, B – Discontinuous strand, C – Template strand
  2. A – Leading strand, B – Lagging strand, C – Parental strand
  3. A – 5′ — 3′ strand B, 3 – 5′ strand, C – Parental strand
  4. All of the above

Answer : A (A – Continuous strand, B – Discontinuous strand, C – Template strand)

Question 32: Why both the strands of DNA are not copied during transcription?

  1. Because RNA molecule with different sequences will be formed
  2. Because RNA molecule with same sequences will be formed
  3. Because RNA molecule with identical sequences will be formed
  4. Because DNA molecule with different sequences will be formed

Answer : A (Because RNA molecule with different sequences will be formed)

Question 33: The strand which do not code for anything is called

  1. Coding strand
  2. Non – coding strand
  3. Template strand
  4. Antisense strand

Answer : A (Coding strand )

Question 34: If the coding strand has the sequence

5′ — ATCGATCG — 3′

then find out the sequence of non – coding strand.

  1. 3′ — TAGCTAGC — 5′
  2. 5′ — TACGTACG — 3′
  3. 5′ —UAGGUACG — 3′
  4. 5′ — UACFUACG — 3′

Answer : A (3′ — TAGCTAGC — 5′)

Question 35: Identify the correct pair of mRNA type and its function.

  1. Messanger RNA — Provides the template
  2. Transfer RNA — brings amino acids and reads genetic code
  3. Ribosomal RNA — Plays catalytic role during translation
  4. All of the above

Answer : D (All of the above)

 Question 36: In bacteria, which enzyme catalyses the transcription of all types of RNA (mRNA, RNA and rRNA)?

  1. DNA dependent RNA polymerase
  2. DNA dependent DNA polymerase
  3. RNA dependent RNA polymerase
  4. RNA dependent DNA polymerase

Answer : A (DNA dependent RNA polymerase)

Question 37: Splicing is the removal of

  1. Introns and exons are joined
  2. Exons end introns are joined
  3. Exons only
  4. Exons and introns

Answer : A (Introns and exons are joined)

Question 38: The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.

  1. A – transcription, B – replication, C – James Watson
  2. A – translation, B – transcription, C – Erwin Chargaff
  3. A – transcription, B – translation, C – Francis Crick
  4. A – translation, B – extension, C – Rosalind Franklin

Answer : C ( A – transcription, B – translation, C – Francis Crick)

Question 39: Who developed the technique of synthesizing RNA molecules with well defined combination of bases to develop genetic code?

  1. Crick et. al
  2. Har Gobind khorana
  3. Matthaei
  4. Nirenberg

Answer: : B (Har Gobind khorana)

Question 40: The one aspect, which is not a salient feature of genetic code is, its being

  1. Degenerate
  2. Ambiguous
  3. Universal
  4. Specific

Answer : B (Ambiguous)

 Question 41: Codons are unambiguous, means one codon code for

  1. More than one amino acid
  2. Two amino acids
  3. Only one amino acid
  4. Non-sense amino acid

Answer : C (Only one amino acid)

Question 42: The codon AUG codes for (in eukaryotes)

  1. Methionine
  2. Histidine
  3. Tryptophan
  4. Alanine

Answer : A (Methionine)

Question 43: Identify the stop codons in given options.

  1. UAA, UAG, UGA
  2. UCA, UCC, UCA
  3. UGC, UCG, UCC
  4. UUU, UAT, UTA

Answer : A (UAA, UAG, UGA)

Question 44: Which mutation of the genetic bases gives the proof that codon is triplet and reads in a contagious manner?

  1. Frameshift mutation
  2. Point mutation
  3. Both A and B
  4. Inversion mutation

Answer : A (Frameshift mutation)

Question 45: tRNA is a compact molecule which looks like

  1. M – shaped
  2. P – shaped
  3. L – shaped
  4. K – shaped

Answer : C (L – shaped)

 Question 46: Which enzyme will be produced in a cell in which there is a non – sense mutation in the lac Y – gene ?

  1. B – galactosidase
  2. Lactose permease
  3. Transacetylase
  4. Lactose permease and transacetylase

Answer : A (B – galactosidase)

Question 47: The accessibility of the promoter regions of prokaryotic DNA is (in many cases) regulated by the interaction of proteins with the sequences termed as

  1. Regulator
  2. Promoter
  3. Operator
  4. Structural genes

Answer : C (Operator)

Question 48: An operon is considered to regulate a unit

  1. Translational unit
  2. Genetic unit
  3. Protein unit
  4. Enzymatic unit

Answer : B (Genetic unit)

Question 49: Gene regulation governing lactose operon of E.coli the lac I gene products is

  1. Positive and inducible because it can be induced by lactose
  2. Negative and inducible because it repressor protein prevents transcription
  3. Negative and repressible because repressor proteins prevents transcription
  4. Feedback inhibition because excess of B – galactosidase can switch off transcription

Answer : B (Negative and inducible because it repressor protein prevents transcription)

Question 50: Lactose is transported into cells through

  1. B-galactosidase
  2. Permease
  3. Transacetylase
  4. Transferase

Answer : B (Permease)

 Question 51: Given the diagram of the lac operon showing an operon of inducible enzymes. Identify components and enzymes (A, B, C, D and E).

  1. A – Galactosidase, B – Permease, C – Transacetylase, D – Repressor protein, E – Inducer (lactose)
  2. A – Galactosidase, B – Permease, C – Transacetylase, D – Inducer (lactose), E – Repressor protein
  3. A – Galactosidase, B – Transacetylase, C – Permease, D – Repressor protein, E – Inducer (lactose)
  4. A – Permease, B – Transacetylase, C – Galactosidase, D – Repressor protein, E – Inducer (lactose)

Answer : A (A – Galactosidase, B – Permease, C – Transacetylase, D – Repressor protein, E – Inducer (lactose))

Question 52: Repressor proteins of lac operon binds to

  1. Exons
  2. Introns
  3. Operator
  4. Structural genes

Answer : C (Operator)

Question 53: Why glucose and galactose cannot act as an inducer for lac operon ?

  1. Because they cannot bind with the repressor
  2. Because they can bind with the repressor
  3. Because they can bind with the operator
  4. Because they can bind with the regulator

Answer : A (Because they cannot bind with the repressor)

Question 54: Identify the statement not concerned with human genome project.

  1. It was completed in 2003
  2. It aims to determine the sequence of 3 billion chemical base pairs and store it in data bases
  3. It associated ethical legal and social issues arising from the project
  4. It is not associated with non-human organisms DNA sequences

Answer : D (It is not associated with non-human organisms DNA sequences)

Question 55: How genetic and physical maps were generated in HGP?

  1. By using DNase
  2. By using RNase
  3. By using restriction endonuclease
  4. By using automated DNA sequences

Answer : C (By using restriction endonuclease )

Question 56: SNP-Single Nucleotide Polymorphisms is

  1. Location on RNA where the single base differs
  2. Location on proteins where the single base differs
  3. Location on genome where the single base of DNA differs
  4. Location on genome where many bases of DNA differs

Answer : C (Location on genome where the single base of DNA differs)

Question 57: SNPs can be used for

  1. Finding chromosome locations for disease associated sequences
  2. Tracing human history
  3. Evolution
  4. All of the above

Answer : D (All of the above)

Question 58: Satellite DNA or repetitive DNA

  1. Do not code for any protein
  2. Forms a large portion of human genome
  3. Shows high degree of polymorphism
  4. All of the above

Answer : D (All of the above)

Question 59: VNTR belongs to the class of satellite DNA referred to as

  1. Microsatellite DNA
  2. Minisatellite DNA
  3. Megasatellite DNA
  4. Repititive DNA

Answer : B (Minisatellite DNA)

Question 60: Steps in DNA fingerprinting are

 

                                       Isolation of DNA

                                                    ↓

                                  Digestion of DNA by (A)

                                                    ↓

                                Separation of DNA by (B)

                                                     ↓

                                Transfering of DNA to (C)

                                                     ↓

                               DNA hybridisation using (D)

                                                      ↓

                            Detecting of hybridised DNA by (E)

 

 Complete the accompanying flowchart. A, B, C, D and E in the flowchart are

  1. A – Restriction endonuclease, B – Electrophoresis, C – Nitrocellulose or nylon, D – Labelled VNTR probe, E – Autoradiography
  2. A – Electrophoresis, B – Restriction endonuclease, C – Nitrocellulose or nylon, D – Labelled VNTR probe, E – Autoradiography
  3. A – Restriction endonuclease, B – Electrophoresis, C – Labelled VNTR probe, D – Nitrocellulose or nylon, E – Autoradiography ,
  4. A – Restriction endonuclease, B – Electrophoresis C – Nitrocellulose or nylon, D – Autoradiography, E – Labelled VNTR probe

Answer : A (A – Restriction endonuclease, B – Electrophoresis, C – Nitrocellulose or nylon, D – Labelled VNTR probe, E – Autoradiography)

Question 61: While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaffs rule it can be concluded that

  1. It is a double-stranded circular DNA
  2. It is single-stranded DNA
  3. It is a double-stranded linear DNA
  4. No conclusion can be drawn

Answer : B (It is single-stranded DNA)

Question 62: The first genetic material could be

  1. Protein
  2. Carbohydrates
  3. DNA
  4. RNA

Answer : D (RNA)

Question 63: DNA is a polymer of nucleotides which are linked to each other by 3 – 5′ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?

  1. Replace purine with pyrimindines
  2. Remove/Replace 3′ OH group in deoxyribose
  3. Remove/Replace 2′ OH group with some other group in deoxyribose
  4. Both B and C

Answer : B ( Remove/Replace 3′ OH group in deoxyribose)

Question 64: Which one of the following steps in transcription is catalysed by RNA polymerase?

  1. Initiation
  2. Elongation
  3. Termination
  4. All of these

Answer : C (Termination)

Question 65: With regard to mature mRNA in eukaryotes

  1. Exons and introns do not appear in the mature RNA
  2. Exons appear but introns do not appear in the mature RNA
  3. Introns appear but exons do not appear in the mature RNA
  4. Both exons and introns appear in the mature RNA

Answer : B (Exons appear but introns do not appear in the mature RNA)

Question 66: The amino acid attaches to the tRNA at its

  1. 5′ – end
  2. 3′ – end
  3. Anticodon site
  4. DHU loop

Answer : B (3′ – end)

Question 67: In E. coli, the lac operon gets switched on when

  1. Lactose is present and it binds to the repressor
  2. Repressor binds to operator
  3. RNA polymerase binds to the operator
  4. Lactose is present and it binds to RNA polymerase

Answer : A (Lactose is present and it binds to the repressor)

Question 68: Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory protein?

  1. They only increase expression
  2. They only decrease expression
  3. They interact with RNA polymerase but do not affect the expression
  4. They can act both as activators and as repressors

Answer : D (They can act both as activators and as repressors)

Question 69: Which was the last human chromosome to be completely sequenced?

  1. Chromosome 1
  2. Chromosome 11
  3. Chromosome 21
  4. Chromosome – X

Answer : A (Chromosome 1)

Question 70: The human chromosome with the highest and least number of genes in them are respectively

  1. Chromosome 21 and Y
  2. Chromosome 1 and X
  3. Chromosome 1 and Y
  4. Chromosome X and Y

Answer :  C (Chromosome 1 and Y)

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