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Molecular Basis of Inheritance | Unit Test - 4 Sets of PDF QPs in one page with answer key I Class 12


Set A (Max. score: 15)

  1. DNA: Thymine      RNA: ………………… (1)
  2. Semi-conservative model of DNA Replication was experimentally proved by ………………. (1)
  3. Point out the reasons for stability of DNA and mutability of RNA. (Any 2) (2)
  4. Construct the base sequence of mRNA transcribed from the following DNA sequence. (2)
    3’ CACGTGGACTGAGGACTCCTC 5’
  5. Mention the role of mRNA and tRNA in translation. (2)
  6. Identify the diagram and label A, B and C.
    7. PICTURENAME
  7. The given diagram represents Lac operon in absence of inducer. Redraw it in the presence of inducer. (2)
    8. PICTURENAME
  8. Write down the steps of DNA fingerprinting. (3)
Set A Answer key
  1. Uracil
  2. Matthew Meselson and Franklin Stahl
  3. Reasons for DNA stability:
    • Presence of thymine instead of uracil
    • Absence of reactive 2'-OH group in deoxyribose
    • Double-stranded.
    Reasons for RNA mutability:
    • Presence of uracil instead of thymine
    • Presence of reactive 2'-OH group
    • Single-stranded.
  4. 5'-GUGCACCUGACUCCUGAGGAG-3'
  5. mRNA: Carries the genetic information transcribed from DNA in the form of codons to direct the sequence of amino acids during protein synthesis.
    tRNA: Acts as an adapter molecule that reads the codon on mRNA via its anticodon loop and brings the corresponding specific amino acid to the ribosome.
  6. Identification: Nucleosome
    A: DNA
    B: H1 histone
    C: Histone octamer
  7. Lac operon in presence of Inducer (Lactose/ Allolactose):
    8. PICTURENAME
  8. Steps of DNA fingerprinting:
    1. Isolation of DNA.
    2. Digestion of DNA by restriction endonucleases.
    3. Separation of DNA fragments by electrophoresis.
    4. Transferring (blotting) of separated DNA fragments to synthetic membranes (nitrocellulose or nylon).
    5. Hybridisation using labeled VNTR probe.
    6. Detection of hybridised DNA fragments by autoradiography.

Set B (Max. score: 15)

  1. Loosely packed, light stained and transcriptionally active region of chromatin is called ………………. (1)
  2. UGG: Tryptophan      AUG: ………………… (1)
  3. Write down the steps of Griffith’s transforming principle experiment. (2)
  4. Analyze the diagram of a replication fork given below. (2)
    PICTURENAME
    1. What is meant by replication fork?
    2. Redraw the diagram correctly and label it.
  5. Briefly explain the following: (2)
    1. Repeated (repetitive) sequences.
    2. SNPs – single nucleotide polymorphism
  6. What are UTRs? Mention its significance. (2)
    1. What is the role of Regulatory or inhibitory gene in Lac operon?
    2. Name any two enzymes produced by the structural genes in Lac operon. (2)
  8. Briefly explain the steps of Processing of hnRNA to mRNA. (3)
Set B Answer key
  1. Euchromatin
  2. Methionine
  3. Steps of Griffith's experiment:
    S strain (virulent) → Inject into mice → Mice die.
    R strain (non-virulent) → Inject into mice → Mice live.
    Heat-killed S strain → Inject into mice → Mice live.
    Heat-killed S strain + Live R strain → Inject into mice → Mice die.
    1. A replication fork is the small Y-shaped open region of the DNA double helix where DNA unwinding and replication occurs.
    2. Correctly labeled diagram:
      3. PICTURENAME
    1. Repetitive DNA: DNA sequences consisting of small stretches of DNA repeated many times; make up a large portion of the eukaryotic genome and show high polymorphism.
    2. SNPs (Single Nucleotide Polymorphisms): Single-base DNA differences occurring among individuals at specific chromosomal locations; useful for finding chromosomal locations associated with disease and tracing human history.
  6. UTRs (Untranslated Regions): Additional sequences present on mRNA that are not translated into protein; located at both 5'-end (before start codon) and 3'-end (after stop codon).
    Significance: Required for efficient translation process and mRNA stability.
    1. The regulatory gene (i gene) codes for the repressor protein, which controls the switching on/off of the operon.
    2. Any 2: β-galactosidase, Permease, or Transacetylase.
  8. Processing of hnRNA to mRNA in eukaryotes:
    Splicing: Removal of non-coding introns and joining of coding exons in a defined order.
    Capping: Addition of an unusual nucleotide (methyl guanosine triphosphate) to the 5'-end of hnRNA.
    Tailing: Addition of adenylate residues (200–300 poly-A tail) to the 3'-end in a template-independent manner.

Set C (Max. score: 15)

  1. A nitrogenous base is linked to the OH of 1' C pentose sugar through an ………………. to form nucleoside. (1)
  2. During translation ………… acts as adaptor molecule. (1)
    1. Mention the two major goals of HGP.
    2. Write any 2 applications of DNA fingerprinting. (2)
  4. What are the additional complexities in eukaryotes for the transcription process? (2)
  5. Mention any four salient features of genetic code. (2)
  6. Mention the steps of Hershey & Chase experiment. What was the conclusion of this experiment? (2)
  7. There are 4 levels of gene expression in eukaryotes. What are they? (2)
  8. Observe the diagram of Central Dogma given below. (3)
    PICTURENAME
    1. Name the processes a, b, c & d.
    2. In which organisms, the central dogma is reversed (RNA to DNA)?
    3. Name the enzyme responsible for reversal of central dogma.
Set C Answer key
  1. N-glycosidic linkage
  2. tRNA (transfer RNA / soluble RNA)
    1. Major goals of HGP (Any 2):
      • Identify all the approximately 20,000–25,000 genes in human DNA.
      • Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
      • Store this information in databases and improve tools for data analysis.
    2. Applications of DNA fingerprinting (Any 2):
      • Forensic tool to identify criminals/suspects.
      • Settling paternity or maternity disputes.
      • Determining genetic diversity and population relationships.
  4. • Presence of three types of RNA polymerases (RNA Pol I, II, and III) transcribing different types of RNA.
    • Primary transcript (hnRNA) contains both exons and introns, requiring processing to yield functional mRNA.
  5. Salient features of genetic code (Any 4):
    • Triplet code: 61 codons code for amino acids; 3 codons are stop codons.
    • Unambiguous and specific: One codon codes for only one particular amino acid.
    • Degenerate: Some amino acids are coded by more than one codon.
    • Universal: A given codon codes for the same amino acid from bacteria to humans (e.g., UUU codes for Phenylalanine).
    • Comma-less (contiguous): The code is read in mRNA in a continuous fashion without punctuation.
    • Initiator codon: AUG dual function (codes for Methionine and acts as initiator).
  6. Steps:
    • Infection: Bacteriophages with (35S)-labeled protein coat and (32P)-labeled DNA infected E. coli.
    • Blending: Agitation in a blender to detach viral coats from bacterial cells.
    • Centrifugation: Separation of viral particles (supernatant) from bacterial cells (pellet).
    Conclusion: DNA enters the bacterial cell from the virus, proving that DNA is the genetic material, not protein.
  7. Four levels of gene expression in eukaryotes:
    1. Transcriptional level (formation of primary transcript).
    2. Processing level (regulation of splicing).
    3. Transport of mRNA from nucleus to the cytoplasm.
    4. Translational level.
    1. a: Replication
      b: Transcription
      c: Translation
      d: Reverse transcription
    2. Retroviruses (e.g., HIV)
    3. Reverse transcriptase

Set D (Max. score: 15)

  1. In a DNA, Adenine is 26%. What would be the percentage of Cytosine? (1)
  2. Odd one out. Justify your answer. (1)
    UAA, AUG, UAG, UGA
    1. Name two enzymes involved in replication.
    2. Mention their role. (2)
  4. Explain the results of Biochemical characterization of transforming principle experiment. (2)
    1. What is Charging (aminoacylation) of tRNA?
    2. Name the site of translation. (2)
  6. Briefly describe how the Lac operon is switched off in the absence of inducer. (2)
  7. Expand: (a) HGP      (b) VNTR (2)
  8. Observe the diagram of transcription unit given below: (3)
    PICTURENAME
    1. Label a, b, c, d in the figure.
    2. During transcription, both DNA strands are not copied. Explain the reason.
Set D Answer key
  1. 24%.
  2. AUG
    Justification: AUG is a start/initiation codon. Others are stop (termination) codons.
  3. a & b. Enzymes and their roles:
    • DNA-dependent DNA polymerase: Catalyses the polymerisation of deoxynucleotides in the 5'→3' direction with high speed and fidelity.
    • DNA Ligase: Joins the discontinuously synthesised Okazaki fragments on the lagging strand.
  4. • Proteases and RNases did not affect transformation, proving proteins and RNA are not the transforming substance.
    • Digestion with DNase inhibited transformation, concluding that DNA is the hereditary material responsible for transformation.
    1. Charging of tRNA: The process in which an amino acid is activated in the presence of ATP and linked to its cognate tRNA at the 3'-end.
    2. Site of translation: Ribosome (in the cytoplasm).
  6. In the absence of an inducer (lactose), the regulatory gene (i gene) continuously transcribes repressor mRNA, producing an active repressor protein.
    The repressor binds to the operator region (o) of the operon, preventing RNA polymerase from transcribing the structural genes (z, y, a).
  7. (a) HGP: Human Genome Project
    (b) VNTR: Variable Number of Tandem Repeats
    1. a: Promoter
      b: Template strand (3'→5')
      c: Terminator
      d: Coding strand (5'→3')
    2. Why both strands are not copied during transcription:
      • The code for proteins is different in both strands. This complicates the translation.
      • If 2 RNA molecules are produced simultaneously, they would be complimentary. It forms a double stranded RNA and prevents translation.
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