Access the latest CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set D. We have provided free printable Class 12 Biology worksheets in PDF format, specifically designed for Chapter 5 Molecular Basis of Inheritance. These practice sets are prepared by expert teachers following the 2025-26 syllabus and exam patterns issued by CBSE, NCERT, and KVS.
Chapter 5 Molecular Basis of Inheritance Biology Practice Worksheet for Class 12
Students should use these Class 12 Biology chapter-wise worksheets for daily practice to improve their conceptual understanding. This detailed test papers include important questions and solutions for Chapter 5 Molecular Basis of Inheritance, to help you prepare for school tests and final examination. Regular practice of these Class 12 Biology questions will help improve your problem-solving speed and exam accuracy for the 2026 session.
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Very Short Answer Questions
Question. Write the function of RNA polymerase II.
Answer. RNA polymerase II transcribes precursor of mRNA or hnRNA.
Question. Give an example of a codon having dual function.
Answer. AUG acts as an initiation codon and also codes for methionine.
Question. Name the transcriptionally active region of chromatin in a nucleus.
Answer. Euchromatin or exon.
Question. Mention the two additional processing which hnRNA needs to undergo after splicing so as to become functional.
Answer. Capping and tailing.
Question. Mention how does DNA polymorphism arise in a population.
Answer. DNA polymorphism in a population arise due to presence of inheritable mutations at high frequency.
Question. What do ‘X’ and ‘Y’ represent in the transcription unit of the DNA molecule shown?
Answer. X—Template Strand Y—Terminator.
Question. Why is the Human Genome Project called a mega project?
Answer. Human Genome Project is called a mega project because of following reasons:
(i) Sequencing of more than 3 × 109 bp.
(ii) Identification of all the genes present in human genome
(iii) High expenditure of more than 9 billion dollars.
(iv) Identification of all the alleles of genes and their functions.
(v) Storage of data for sequencing would require space equal to 3300 books of 1000 pages each if each page contains 1000 letters.
Question. Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer. (a) Transcription: It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. One strand of DNA is used as template strand for the synthesis of a complementary strand of RNA called mRNA.
(b) Polymorphism: Genetic polymorphism meAnswer occurrence of genetic material in more than one form. It is of three major types, i.e., allelic, SNP and RFLP.
Allelic polymorphism: Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.
SNP or single nucleotide polymorphism: Over 1.4 million single base DNA differences have been observed in human beings. According to SNP, every human being is unique. SNP is very useful for locating alleles, identifying disease-associated sequence and tracing human history.
(c) Translation: It is the process during which the genetic information which is stored in the sequence of nucleotides in an mRNA molecule is converted, following dictations of the genetic code, into the sequence of amino acids in the polypeptide. It takes place in cytoplasm in both eukaryotes and prokaryotes.
(d) Bioinformatics: The science which deals with handling, storing of huge information of genomics as databases, analysing, modelling and providing various aspects of biological information, especially the molecules connected with genomics and proteomics is called bioinformatics.
Following are some applications of bioinformatics:
(i) Phylogeny or evolutionary biology: By comparing gene, nucleotide, protein and amino acid sequence, the organisms are grouped and phylogeny is established.
(ii) Chemoinformatics: Information about chemical combination, their effects, side effects,pharmaceutical chemicals, etc., by databases is done.
(iii) Faster drug research: Analysis of drug–ligand complexes and assessment of binding interaction help in development of computer-aided drug design.
(iv) Proteomics: Protein sequence databases are useful in diagnostics, healthcare and drug research.
Question. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer. DNA template
(i) DNA polymerase for DNA replication.
(ii) RNA polymerase for RNA synthesis or transcription.
RNA template
(i) RNA-dependent RNA polymerase for synthesis of RNA in some RNA viruses.
(ii) Reverse transcriptase to synthesise cDNA (complementary DNA) over RNA template.
Short Answer Questions
Question. Write the full form of VNTR. How is VNTR different from a probe?
Answer. Full form of VNTR is Variable Number of Tandem Repeats
| VNTR | Probe |
| The segment of DNA which shows very high no. of repetitive nucleotide sequences which vary from person to person are called variable number tandem repeats. | The small fragments of DNA strands which are highly selective and specific to be complementary to VNTR sequences are called probe. |
Question. Following are the features of genetic codes. What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Answer. Stop codon does not code for any amino acid and terminates the synthesis of polypeptide chain.
Unambiguous codon: one codon codes for one amino acid only.
Degenerate codon: some amino acid are coded by more than one codon.
Universal codon: Genetic code is same for all organisms (bacteria to humAnswer).
Question. Name indicating their functions, a few additional enzymes, other than DNA polymerase and ligase, that are involved in the replication of DNA with high degree of processivity and accuracy.
Answer. (i) Helicase: opens the helix
(ii) Topoisomerases: removes the supercoiling of DNA relieves the tension due to unwinding
(iii) Primase: synthesises RNA primer
(iv) Telomerase: to synthesises the DNA of telomeric end of chromosomes.
Question. Mention the role of ribosomes in peptide bond formation. How does ATP facilitate it?
Answer. There are two sites in the large subunit of the ribosome, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst for the formation of peptide bond 23S rRNA in bacteria is a ribozyme. Amino acids become activated by binding with its tRNA in the presence of aminoacyl tRNA synthetase and ATP.
Question. (a) List the two methodologies which were involved in human genome project. Mention how they were used.
(b) Expand ‘YAC’ and mention what was it used for.
Answer. The two methodologies involved in human genome project are:
(a) (i) Expressed Sequence Tags: Identifying all the genes that are expressed as RNA
(ii) Sequence Annotation: Sequencing the whole set of genome coding or non coding sequences and later assigning different region with functions.
(b) ‘YAC’ stands for Yeast Artificial Chromosome. It is used as a cloning vectors.
Question. State the functions of the following in a prokaryote:
(i) tRNA (ii) rRNA
Answer. (i) tRNA reads the genetic codes, carries amino acids to the site of protein synthesis and act as an adaptor molecule.
(ii) rRNA plays structural and catalytic role during translation.
Question. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer. Factors responsible for conferring stability to double helix structure are presence of hydrogenbonds,the plane of one base pair stacks over the other, complementary presence of thymine in place of uracil.
Question. (a)
Look at the above sequence and mention the events (A), (B) and (C).
(b) What does Central Dogma state in Molecular Biology? How does it differ in some viruses?
Answer. (a) A—DNA replication, B—Transcription, C—Translation
(b) Central Dogma in Molecular Biology states that information flows in the order
DNA →RNA → Proteins
It differs in some viruses as the flow of information is in reverse direction, that is, from DNA to RNA.
Question. What are satellite DNA in a genome? Explain their role in DNA fingerprinting.
OR
Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer. A small stretch of DNA sequence that repeats many a time, shows a high degree of polymorphism and forms a bulk of DNA in a genome called satellite DNA.
(i) They do not code for any proteins.
(ii) They form large part of the human genome.
(iii) They show high degree of polymorphism and are specific to each individual.
Question. During in vitro synthesis of DNA, a researcher used 2′, 3′-dideoxycytidine triphosphate as raw nucleotide in place of 2′-deoxy cytidine triphosphate, other conditions remaining as standard. Will further polymerisation of DNA continue up to the end or not? Explain.
Answer. Further polymerisation up to the end would not occur as the 3 –OH on sugar molecule is not for forming ester bond, which is required to add another nucleotide.
Question. What is aminoacylation? State its significance.
Answer. Aminoacylation of tRNA or charging of tRNA is the activation of amino acids in the presence of ATP and their linkage to their cognate tRNA.
If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically.
Question. Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having all the four nitrogenous bases and showing the correct polarity.
Answer.
Question. The base sequence of one strand of DNA is TACTAGGAT.
(i) Write the base sequence of the RNA got after transcription of the given sequence.
(ii) What is the distance maintained between the two consecutive pairs of bases in the DNA molecule?
(iii) Who contributed the base complementary rule?
Answer. (i) AUGAUCCUA.
(ii) 3.4 Å or 0.34 nm.
(iii) Chargaff contributed the base complementary rule.
Question. Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain.
Answer. DNA is negatively charged and has positively charged histone proteins. These help DNA to be held in a place in large loops called nucleoid.
Please click on below link to download CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set D
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Important Practice Resources for Class 12 Biology
Chapter 5 Molecular Basis of Inheritance CBSE Class 12 Biology Worksheet
Students can use the Chapter 5 Molecular Basis of Inheritance practice sheet provided above to prepare for their upcoming school tests. This solved questions and answers follow the latest CBSE syllabus for Class 12 Biology. You can easily download the PDF format and solve these questions every day to improve your marks. Our expert teachers have made these from the most important topics that are always asked in your exams to help you get more marks in exams.
NCERT Based Questions and Solutions for Chapter 5 Molecular Basis of Inheritance
Our expert team has used the official NCERT book for Class 12 Biology to create this practice material for students. After solving the questions our teachers have also suggested to study the NCERT solutions which will help you to understand the best way to solve problems in Biology. You can get all this study material for free on studiestoday.com.
Extra Practice for Biology
To get the best results in Class 12, students should try the Biology MCQ Test for this chapter. We have also provided printable assignments for Class 12 Biology on our website. Regular practice will help you feel more confident and get higher marks in CBSE examinations.
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