Important Questions for Class 12 Chapter 6: Molecular Basis of Inheritance

Genes are the basic unit of heredity. Most genes are made of strands of genetic material called DNA. DNA contains all the hereditary information of an individual. This information is passed on from one generation to the other in the form of homologous chromosomes. The DNA is transcribed into mRNA and translated into proteins. This is known as central dogma.

Explore important questions on Molecular Basis of Inheritance for better understanding of the concept.

Very Short Answer Type Questions

Q.1. State the role of histones in DNA packaging.


  • The basic histone proteins neutralize the acid nature of DNA.
  • They play a role in gene regulation.
  • They help the DNA to wind around itself.

Also Read: DNA packaging.

Q.2. What is the difference between heterochromatin and euchromatin?

A.2. Heterochromatin is that part of the chromosomes which is tightly packed and is genetically inactive, whereas euchromatin is the part that of chromosomes which is uncoiled or loosely packed and is transcriptionally active.

More Details: Difference between heterochromatin and euchromatin

Q.3. Name any three viruses with RNA as the genetic material.

A.3. Human Immunodeficiency Virus, Influenza Virus, Tobacco Mozaic Virus.

Explore More: Viruses

Q.4. What is the reason for the discontinuous synthesis of DNA on one of the parental strands?

A.4. The DNA synthesis occurs in 5’-3’ direction. The DNA strands act as templates. One of the strands is synthesized in the 3’-5’ direction. The other strand is synthesized in the opposite direction producing small stretches of DNA known as Okazaki fragments. That is why DNA synthesis is discontinuous.

Q.5. The sequence of the coding strand of DNA in a transcription unit is mentioned below.


Write the sequence for:

  1. Its complementary strand
  2. It’s mRNA



Q.6. What is DNA polymorphism?

A.6. DNA polymorphism is the variation in the DNA sequence arising due to mutation at non-coding sequences.

Q.7. Comment on the statement “Retroviruses do not follow central dogma.”

A.7. Retroviruses have RNA as genetic material. That is why they do not follow the central dogma. Instead, the RNA is converted into DNA by the enzyme reverse transcriptase.

Q.8. Sometimes the young ones born have an extremely different set of eyes or limbs. Give a relevant explanation for the abnormality.

A.8. This happens due to non-coordination in the regulation of expression in the set of genes associated with the development of organs.

Q.9. Discuss the dual polymerase present in E.coli.

A.9. DNA polymerase III is found in E.coli that helps in the replication process. It performs the 5’-3’ polymerase activity as well as 3’-5’ exonuclease activity. It has the ability to proofread the wrong nucleotides and substitutes it with the correct one.

Q.10. Mention the functions of:

  1. Methylated guanosine cap
  2. poly-A tail


  1. Methylated guanosine cap- It attaches the mRNA to the smaller sub-units of the ribosome during translation initiation.

  2. Poly-A tail- It increases the length of the mRNA and provides longevity to the mRNA.

Q.11. Mention two functions of AUG.

A.11. The two important functions of AUG include:

  • It acts as initiation codon for protein synthesis
  • It codes for methionine

Q.12. State the function of amino acyl t-RNA synthetase.

A.12. Amino acyl t-RNA synthetase attaches appropriate amino acid on the t-RNA molecule.

Short Answer Type Questions

Q.1. Define the following:

  1. Promoter
  2. tRNA
  3. Exons


  1. Promoter– The promoter serves as the binding site for RNA polymerase for transcription. It helps in transcription initiation.

  2. tRNA– tRNA or transfer RNA is involved in the translation of proteins. It reads the genetic code on the RNA and transfers specific amino acids to mRNA on the ribosomes.

  3. Exons– Exons are the coding regions present in a DNA sequence in eukaryotes.

Q.2. Describe the role of the ribosome in translation.

A.2. Ribosomes are made up of rRNA. It performs the following functions during translation.

  • The two ribosomal sub-units sandwich around the mRNA.

  • It provides bing site for tRNA.

  • With each triplet codon of mRNA moving through the ribosome, a specific tRNA with its own anticodon is recruited and a polypeptide is synthesized.

  • When the ribosome encounters a stop codon, it dissociates and no more polypeptides are synthesized.

Q.3. What is cistron? Differentiate between monocistronic and polycistronic transcription units.

A.3. Cistron refers to a DNA sequence that code for specific polypeptide during translation. Differences:

Monocistronic Polycistronic
It codes for only one protein. It codes for several proteins.
Eukaryotic mRNAs are monocistronic Prokaryotic mRNAs are polycistronic

Q.4. Name and state the functions of the enzymes involved in DNA replication.

A.4. The enzymes involved in DNA replication are:

  • DNA Polymerase- They help in the synthesis of DNA molecules during replication.

  • Ligase- This enzyme joins two strands of DNA

  • Helicase- Unwinds the DNA strands during replication

  • Topoisomerase- Resolves the supercoiling of DNA

  • Primase- Helps in the synthesis of RNA primer during DNA replication

Q.5. What is the Human Genome Project? List any six of its features.

A.5. The Human Genome Project is an international scientific project started with the aim of identifying all the genes in the human DNA from both a physical and functional perspective. The features of the Human Genome Project are:

  • The human genome has 3164.7 million nucleotide bases.

  • The average gene consists of 3000 bases. The largest human being dystrophia has 2.4 million bases.

  • The functions of 50% of the genes are not discovered yet.

  • Less than 2% of the genes encode proteins.

  • There are a large number of repeated sequences in the human genome.

  • The maximum genes are present in the chromosome I whereas least are present in chromosome Y.

Q.6. How is an abnormal haemoglobin molecule formed? What are the consequences of such abnormality?

A.6. The abnormal haemoglobin molecule is formed due to point mutation. The p-globin chain of haemoglobin molecule is mutated which replaces glutamic acid with valine at the sixth position. As a result, the erythrocytes became sickle-shaped and the oxygen-carrying capacity of the cells is inhibited.

Q.7. Is it possible to use DNA probes such as VNTRs in the DNA fingerprinting of a bacteriophage?

A.7. The genome of a bacteriophage is very small and contains all the coding sequences. They do not have a repetitive sequence such as VNTRs. That is why no DNA fingerprinting can be done for bacteriophages.

Q.8. Why does the lac operon show a low level of expression all the time?

A.8. Lactose can be transferred from the medium into the bacterial cell only in the presence of permease. In the absence of permease, lactose is not transported into the cell and it fails to act as an inducer, thereby, showing a low level of expression.

Q.9. Can alternate splicing of exons enable a structural gene to code for several iso-proteins from one and the same gene? Give reasons.

A.9. The alternate splicing of exons is sex-specific, tissue-specific, and even developmental stage-specific. The alternate splicing helps in encoding a single gene for several iso-proteins. If this splicing is absent there would have been new genes for every protein or isoprotein.

Q.10. Mutation in a single base in a gene may not always result in a gain or loss of a function. Comment.

A.10. Due to the degeneracy of codons mutation at the third base of codons do not result in any phenotypic change.

Long Answer Type Questions

Q.1. Enumerate the post-transcriptional modifications in a eukaryotic mRNA.

A.1. Transcription is the process of conversion of DNA to mRNA. The post-transcriptional modifications include:

  • Capping at 5’-end
  • Poly-A tail at 3’-end
  • mRNA splicing

The 5’ cap protects the RNA from ribonucleases. The poly-A tail protects the mRNA from enzymatic degradation. The introns are spliced during mRNA splicing and the exons are joined together to form a continuous sequence that codes for a functional protein.

Q.2. Explain the process of translation.

A.2. Translation is the process of protein synthesis in which the mRNA is used to synthesize proteins. The mRNA sequence is decoded to specify the amino acid of a polypeptide. The process of translation is carried out in the following steps:

  • Initiation
  • Elongation
  • Termination

Q.3. Explain the process of DNA fingerprinting.

A.3. DNA fingerprinting is a technique that is used to analyze the genetic makeup of living beings. It is widely used in paternal disputes to identify the biological parents of the child, and also to identify the criminal during forensic investigations.

Extended Reading: DNA fingerprinting

Q.4. What is an operon? Explain an inducible operon.

A.4. An operon is the functional unit of DNA that contains a cluster of genes controlled by a single promoter. It consists of the following components:

  • The DNA fragment that transcribes the mRNA.

  • A promoter where the RNA polymerase binds and initiates the transcription.

  • An operator that is a DNA sequence adjacent to the promoter where the repressor protein binds.

  • Regulator gene that codes for a repressor protein

  • Inducer that prevents the repressor from binding to the operator.

The lac operon of E.coli is an inducible operon.

Q.5. Explain the process of DNA replication.

A.5. DNA replication is a biological process of producing two identical strands of DNA from the original strand. The original strand is known as the parent strand and the new strands are known as the daughter strands. This is achieved by a number of enzymes such as DNA polymerase, helicase, primase, topoisomerase, and ligase. For more information on Molecular Basis Of Inheritance and Biology related topics keep visiting BYJU’S website. You can also refer to the BYJU’S app for further details.

Read More: DNA replication

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