Molecular Basis of Inheritance Class 12 Important Questions: CBSE Biology Chapter 5

Very Short Answer Questions                                                                             1 Mark

1. Name the factors for RNA polymerase enzymes which recognize the start and termination signals on DNA for the transcription process in Bacteria.

Ans: Sigma (s) factor and Rho(p)  factor)

2. Mention the function of the non-histone protein.

Ans: Packaging of chromatin

3. During translation what role is performed by tRNA?

Ans: (i) Structural role

(ii) Transfer of amino acid.

4. RNA viruses mutate and evolve faster than other viruses. Why?

Ans: -OH group is present on RNA, which is a reactive group so it is unstable and mutates faster.

5. Name the parts ‘X’ and ‘Y’ of the transcription unit given below.

Ans: X - Template strand, Y - Terminator.

6. Mention the dual functions of AUG.

Ans: (i) Acts as initiation codon for protein synthesis

(ii) It codes for methionine.

7. Write the segment of RNA transcribed from the given DNA 3´ -A T G C A G T A C G T C G T A ‘5´- Template Strand

5´ - T A C G T C A T G C A G C A T ‘3´ - Coding Strand.

Ans: 5’- UAACAAGGAGCAUCC – 3’ (In RNA ‘T’ is replaced by U)

8. Name the process in which unwanted mRNA regions are removed & wanted regions are joined.

Ans: RNA splicing.

9. Give the initiation codon for protein synthesis. Name the amino acid it codes for?

Ans: Initiation codon – AUG & its code for methionine.

10. In which direction, the new strand of DNA synthesized during DNA replication? 

Ans: 5’ to 3’

11. What is the function of aminoacyl tRNA synthetase?

Ans: Aminoacyl tRNA synthetase catalyses the activation of amino and attachment of activated amino acids to the 3-end of specific tRNA molecules.

12. What is point mutation?

Ans: Mutation due to change in a single base pair in a DNA sequence is called point mutation.

13. Name the enzyme that joins the short pieces in the lagging strand during the synthesis of DNA?

Ans: Ligase.

14. Name the enzyme which helps in the formation of peptide bonds. 

Ans: Peptidyl Transferase

15. Who experimentally proved that DNA replication is semi-conservative? 

Ans: Meselson & stahl.

16. What is a codon?

Ans: A triplet sequence of bases that codes for a single amino is called a codon.

17. Name the three nonsense codons.

Ans: UAA, UAG, UGA

18. What is the base-pairing pattern of DNA?

Ans: In DNA, adenine always binds with thymine & cytosine always binds with Guanine.

19. Mention the dual functions of AUG.

Ans: AUG codes for amino acid methionine & also acts as an initiator codon.

Short Answer Questions                                                                                    2 Marks

1. The process of termination during transcription in a prokaryotic cell is being represented here. Name the labels a, b, c, and d.

Ans: (a) DNA molecule

(b) mRNA transcript

(c) RNA polymers

(d) Rho factor

2. Complete the blanks a, b, c and d on the basis of the Frederick Griffith Experiment.

S Strain → inject into mice → (a)

R Strain → inject into mice →(b)

S strain (heat-killed) → inject into mice →(c)

S strain (heat-killed) + R strain (live)→inject into mice →(d)

Ans: (a) Mice die

  (b) mice live

  (c) mice live

  (d) mice die

3. Give two reasons why both the strands of DNA are not copied during transcription. 

Ans: (a) If both the strands of DNA are copied, two different RNAs which are complementary to each other and hence two different polypeptides will be produced; If  a segment of DNA produces two polypeptides, the genetic information machinery becomes complicated.

1. To identify criminals in the forensic laboratory.

2. To determine the real or biological parents in case of disputes.

3. To identify racial groups to rewrite biological evolution. (Any two)

(b) The two complementary RNA molecules (produced simultaneously)would form a double-stranded RNA rather than getting translated into polypeptides.

(c)  RNA polymerase carries out the polymerization in 53 direction and hence the DNA  strand with 35 polarity acts as the template strand. (Any two)

4. Mention any two applications of DNA fingerprinting.

Ans: (i) To identify criminals in the forensic laboratory.

(ii) To identify the real or biological parents in case of disputes.

(iii) To identify racial groups to rewrite the biological evolution.

5. State the 4 criteria which a molecule must fulfill to act as genetic material.

Ans: (i) Replica should be generated.

(ii) It should be chemically and structurally stable.

(iii) It should be able to express itself in the form of Mendelian characters.

(iv) It should provide the scope for slow changes (mutations) that are necessary for evolution.

6. “DNA polymerase plays a dual function during DNA replication” comment on the statement?

Ans: DNA polymerase plays a dual function –it helps in the synthesis of new strands & also helps in proofreading i.e replacement of RNA strands lay DNA fragments.

7. Three codons on mRNA are not recognized by tRNA. What are they? What is the general term used for them and what is their significance in protein synthesis?

Ans: UAG UAA & UGA are the three codons that are not recognized by tRNA; these are known as stop codons or nonsense codons. Since these three codons are not recognized by any tRNA they help in the termination of the protein chain during translation.

8. Give two reasons why both the strands of DNA are not copied during DNA transcription.

Ans: I)If both the strands code for RNA two different RNA molecules & two different proteins would be formed hence genetic machinery would become complicated

II) Since the two RNA molecules would be complementary to each other, they would wind together to form dsRNA without carrying out translation which means the process of transcription would be futile.

9. Why is it essential that tRNA binds to both amino acids & mRNA codons during protein synthesis?

Ans: It is essential that tRNA binds to both amino acids & mRNA codon because tRNA acts as an adapter molecule with picks up a specifically activated amino acid from the cytoplasm & further it is transferred to the ribosomal within the cytoplasm where proteins are synthesized. It attracts itself to the ribosome with the sequence specified by mRNA & finally, it transmits its amino acid to a new polypeptide chain.

10. What is a peptide bond? How is it formed?

Ans: A peptide bond is formed between the carboxylic group (COOH) of the first amino acid & the amino group (-NH₂) of the second amino acid. This reaction is catalyzed by peptidyltransferase.

11. Explain what happens in a frameshift mutation. Name one disease caused by the disorder.

Ans: A frameshift mutation is a type of mutation where the addition or deletion of one or two bases changes the reading from the site of mutation, resulting in a protein with different sets of amino acids.

12. What do you mean by “Central Dogma of Molecular genetics?”

Ans: The central dogma of molecular genetics is the flow of genetic information from DNA to DNA through replication, DNA to mRNA through transcription & mRNA to proteins through translation.

Central Dogma of Biology

13. Give two reasons why both the strands are not copied during transcription?

Ans: i) If both the strands code for RNA, two different RNA molecules & two different proteins are formed hence genetic machinery would be complicated.

ii) The two RNA molecules which are complementary to each other would bind together to form ds-RNA.

14. Why is the human Genome project considered a megaproject? 

Ans: The Human Genome project was called a mega project for the following facts.

1. The human genome has approximately 3.3 x 109bp if the cost of sequencing is the US g billion.

2. If the sequence which was obtained was to be stored in a typed form in books & if each page contained 1000 letters & each book contained 1000 pages then 3300 such books would be needed to store complete information.

3. The enormous quantity of data expected to be generated also necessitates the use of high-speed computational devices for data storage, retrieval & analysis.

15. Why is DNA & not RNA the genetic material in the majority of organisms?

Ans: The -OH group in the nucleotides of RNA is much more reactive & makes RNA labile & easily degradable thus, DNA and not RNA act as genetic material in the majority of organisms.

16. Mention any four important characteristics of the genetic code. 

Ans: Genetic codon has the following important features:-

1. Each codon is a triplet consisting of three bases.

2. Each codon codes for only one amino acid i.e. – unambiguous.

3. Some amino acids are coded and more than one codon is said to be degenerative.

4. Codons are read in a continuous manner in direction & have no punctuation.

17. Why is it that transcription & translation could be coupled in the prokaryotic cells but not in eukaryotic cells?

Ans: In prokaryotes the mRNA synthesized does not require any processing to become active both transcription & translation occurs in the same cytosol but In Eukaryotes, the primary transcript contains both exon & intron & is subjected to a process called splicing where introns are removed or exons are joined in a definite order to form mRNA.

Short Answer Questions                                                                                    3 Marks

1. Give six points of difference between DNA and RNA in their structure/chemistry and function.

Ans: Below given are the six points of difference between DNA and RNA in their structure/chemistry and function:

2. Explain how the hnRNA becomes the mRNA. OR Explain the process of splicing, capping, and tailing which occur during transcription in Eukaryotes.

Ans: hnRNA is the precursor of mRNA. It undergoes:

1. Splicing: Introns are removed and exons are joined together.

2. Capping: an unusual nucleotide (methyl guanosine triphosphate is added to the 5´ end of hnRNA.

3. Adenylate residues (200-300) are added at the 3´ end of hnRNA.

3.  Name the three major types of RNAs, specifying the function of each in the synthesis of the polypeptide.

Ans: (i) mRNA-(Messenger RNA): decides the sequence of amino acids.

(ii) tRNA-(Transfer RNA) : (a) Recognises the codon on mRNA (b) transport the amino acid to the site of protein synthesis.

(iii) rRNA (Ribosomal RNA): Plays the structural and catalytic role during translation.

4. Enlist the goals of the Human genome project.

Ans: The Human Genome Project (HGP) is an international scientific research project whose goal is to determine the sequence of chemical base pairs that make up human DNA, as well as to discover and map all of the human genome's genes, both physically and functionally.

5. A tRNA is charged with the amino acid methionine.

i. Give the anticodon of this tRNA.

Ans: UAC

ii. Write the Codon for methionine.

Ans: AUG

iii.  Name the enzyme responsible for the binding amino acid to tRNA. 

Ans: Amino-acyl tRNA synthetase

6. Illustrate schematically the process of initiation, elongation, and termination during transcription of a gene in a bacterium.

Ans: During translation in bacteria, mRNA serves as a template, tRNA transports amino acids and reads the genetic information, and rRNAs serve as structural and catalytic components.

There is a single DNA-dependent RNA polymerase that catalyzes the transcription of all types of RNA in bacteria.

RNA polymerase connects to the promoter and starts transcription (Initiation). It also helps the helix open and continues elongation in some way.

The nascent RNA, as well as the RNA polymerase, falls off once the polymerases approach the terminator region. This results in termination.

Transcription of a Bacterial Gene

7. What is transformation? Describe Griffith’s experiment to show transformation? What did he prove from his experiment?

Ans: Transformation means a change in the genetic makeup of an individual. Fredrick Griffith conducted a series of experiments on streptococcus pneumoniae. He observed two strains of this bacterium –one forming smooth colonies with capsule (s-type) & the other forming rough colonies without capsule (R-type).

1. When live s-type cells are injected into mice, they produce pneumonia & mice die.

2. When liveR-type cells are injected into mice, the disease was not produced and did not appear.

3. When heat-killed S-type cells were injected into mice, the disease did not appear.

4. When heat-killed S-type cells were mixed with live R-cells & injected into mice, the mice died.

He concluded that R-strain bacteria had somehow been transformed by heat-killed S-strain bacteria which must be due to the transfer of genetic material.

8. The base sequence on one strand of DNA is ATGTCTATA

(i) Give the base sequence of its complementary strand.

Ans: TAC AGATAT

(ii) If an RNA strand is transcribed from this strand what would be the base sequence of RNA?

Ans: UACA GUAU

(iii) What holds these base pairs together?

Ans: Hydrogen bonds hold these base pairs together. Adenine & thymine are bounded by two hydrogen bonds & cytosine & Guanine are bonded by three hydrogen bonds.

9. Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled by identifying the real biological father?

Ans: This case to identify the real biological father could have settled lay DNA – fingerprinting technique. In this technique:-

1. First of all, the DNA of the two claimants who have to be tested is isolated.

2. Isolated DNA is then digested with a suitable restriction enzyme & digest is subjected to gel electrophoresis.

3. The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.

4. The electrophoresed DNA is then transferred from getting into a nitrocellulose filter paper where it is fixed.

5. A known sequence of DNA is prepared called probe – DNA & is labeled with radioactive isotope 32p & then the probe is added to nitrocellulose paper.

6. The nitrocellulose paper is photographed on X-ray film through autoradiography. The film is analyzed to determine the presence of hybrid nucleic acid.

Then, the DNA fingerprints of the two claimants are compared with the DNA fingerprint of the lady & her daughter, whosoever matches with each other would be declared as the biological father of her daughter.

10. The length of DNA in a eukaryotic cell is N 2.2 m How can such a huge DNA be packaged in a nucleus of micrometer in diameter.

Ans: In eukaryotes, the DNA is wrapped around a positively charged histone octamer into a structure called a nucleosome. An atypical nucleosome consists of 200bp of DNA helix.

The repeating units of nucleosomes form chromatin fibers.

These chromatin fibers condense at the metaphase stage of cell division to form chromosomes. The packaging of chromatin at a higher level requires an additional set of proteins called non-histone chromosomal proteins thus in the nucleus, certain regions of the chromatin are loosely packed & they Stain lighter than the other region, these are called euchromatin. The other regions are lightly packed & they stain darker & is called heterochromatin.

11. A tRNA is charged with amino acid methionine.

i. At what site in the ribosome will the tRNA bind?

Ans: P- site

ii. Give the anticodon of this tRNA?

Ans: UAC

iii. What is the mRNA codon for methionine?

Ans: AUG

iv. Name the enzyme responsible for this binding?

Ans: Aminoacyl tRNA synthetase

12. Describe the continuous & discontinuous Synthesis of DNA?

Ans: Synthesis of a new strand of DNA takes place to lay the addition of fresh nucleotides to the 3 – OH group of the last nucleotide of the primer. This synthesis takes place in 5 directions enzymes that catalyze this DNA DNA-polymerase synthesis of a strand called leading strands is continuous. The replication of the second strand of the DNA molecule is discontinuous on a strand called a lagging strand.

Primase initiates primer synthesis on the strand near the fork. The RNA – primer thus formed provides free for replication of single-stranded regions on the lagging strand; the new complementary strand is formed in small fragments of DNA called Okazaki fragments. It is called discontinuous because it has to be initiated several times & every time an Okazaki fragment is produced.

13. What are the three types of RNA & Mention their role in Protein Synthesis? 

Ans: There are three types of RNA :

1. Messenger RNA (mRNA):- It is a single-stranded RNA that brings the genetic information of DNA transcribed on it for protein synthesis.

2. Transfer RNA (tRNA):- It has a clover leaf-like structure that acts as an adapter a molecule that contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end that binds to the specific amino acid on the other hand.

3. Ribosomal RNA (rRNA):- Ribosomes provide the site for the synthesis of protein & catalyze the formation of the peptide bond.

14. Define bacterial transformation. Who proved it experimentally & how?

Ans: The transformation is a mode of exchange or transfer of genetic information between organisms or from one organism to another.

Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in the following steps:-

1. When S-III strains of bacteria are injected into mice. It developed pneumonia & died.

2. When R-II strains are injected into mice, they do not develop pneumonia & survive.

3. When heat-killed S-III strains of bacteria are injected into mice, No symptoms of pneumonia develop & mice remain healthy.

4. When a mixture of heat-killed S-III strain & live R-II strain is injected into mice, they develop pneumonia & die.

From these results, Griffith concluded that the presence of heat-killed S-III bacteria must convert living R-II type bacteria to type S-III to restore the capacity for capsule formation. This was called “BACTERIAL TRANSFORMATION”

S strain →Inject into mice R strain→Mice die

 R strain→Inject into mice →Mice live

S strain (heat-killed)→Inject into mice→Mice live

S strain (heat-killed) + R strain (live)→Inject into mice→Mice die

Long Answer Questions                                                                                    5 Marks

1. What is meant by semi-conservative replication? How did Meselson and Stahl prove it experimentally?

Ans: An experiment was performed by  Meselson and Stahl using E.coli to prove that DNA replication is semi-conservative in nature. They grew E.coli in a medium. Then heavy DNA was separated from normal (14N) by centrifugation in CsCl density gradient. The DNA extracted, after one generation of transfer from 15N medium to 14N medium, had an intermediate density. The extracted DNA, after two generations, consisted of equal amounts of light and hybrid DNA. Thus, it was proved that DNA replicates in a semiconservative manner.

2. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain.

Ans: Lac Operon consists of the following :

1. Structural genes: z, y, a which transcribe a polycistronic mRNA. gene ‘z’ codes for b-galactosidase gene' codes for permease. gene‘a’ codes for transacetylase.

2. Promotor: The site where RNA polymerase binds for transcription.

3. Operator: acts as a switch for the operon.

4. Repressor: It binds to the operator and prevents the RNA polymerase from transcribing.

5. Inducer: Lactose is the inducer that inactivates the repressor by binding to it. Allows access for the RNA polymerase to the structural gene and transcription.

3. What is an operon? Describe the major steps involved in an operon?

Ans: An operon is a group of controllers & structural genes which control the catabolism of the cell genetically lactose operon/lac operon.

1. When inducer or lactose is absent:-

The lac regulator gene synthesizes a repressor protein by transcription & translation. This repressor protein binds with the operator site of lac operon & blocks RNA polymerase. Thus, RNA polymerase is unable to transcribe mRNA & structural genes are unable to translate enzyme B- B- galactosidase.

2. When inducer or lactose is present:

The lac regulator gene transcribes mRNA & synthesizes active lac repressor protein & at the same time lactose is converted into isomer allolactose. Allolactose binds to an active lac repressor due to which it is converted to an inactive repressor. This inactive repressor is released from the operator site of lac operon & RNA polymerase binds to promoter & starts to transcribe mRNA & forms β-galactosidase which converts lactose into glucose vs galactose.

Thus, the presence of lactose determines whether or not lac. The repressor is bound to the operator & genes are expressed on not.

4. What do you mean semi-conservative nature of DNA replication? Who proved it & how?

Ans: The semiconservative nature of DNA replication suggested that during replication two strands would separate & each acts as a template for the synthesis of a new complementary strand so, that after complete replication, each DNA molecule would have one parental & one newly synthesized strand thus, half the information is conserved over a generation. Mathew Messselson& Franklin Stahl have performed an experiment using Escherichia coli to prove that DNA replication is semiconservative. They grew E.coli in a medium containing 15NH4Cl until 15N was incorporated in the two strands of newly synthesized DNA this heavy DNA can be separated from normal DNA lay centrifugation in CsCl density gradient. Then they transferred the cells into a medium with normal X14NH4Cl & took samples at various time intervals & extracted DNA & centrifuged them to measure their densities. The DNA is extracted from the cells after one generation to transfer from X15N medium to X14N

5. Where do transcription & translation take place in a prokaryotic cell? Describe the three steps involved in translation?

Ans: In a prokaryotic cell, both transcription & translation occurs in the cytoplasm. It consists of the following steps:-

1. Activation of Amino Acids:- amino acids are activated in the presence of ATP lay enzaminoacy tRNA Synthetase.

2. Binding of Activated Amino Acid With tRNA:- Activated amino acids bind with specific tRNA to form charged tRNA.

3. Initiation of Polypeptide Chain:- Initiation codon is AUG which codes for methionine. The initiation codon of mRNA binds to the p-site of the ribosome with the help of initiation factors.

4. Elongation of Polypeptide Chain:-

1.  Second, activated amino acid along its tRNA reaches the ‘A’ site & binds to mRNA codon next to AUG.

2.  A peptide bond is formed between two amino acids by a peptidyl transferase.

3. Ribosomes translate mRNA in -direction due to which free tRNA slips away &peptidyltRNA reaches P – site. Now the third amino acid reaches the A – site & the process continues.

4.  Termination of Polypeptide Chain:- When a termination codon (UAA, UAG, UGA) reaches at A- site translation terminates Since there is no specific tRNA for these codons.

6. Who performed the blender experiment? What does this experiment prove? Describe the steps followed in this experiment?

Ans: The proof for DNA as the genetic material came from the experiments of Hershey & chase who worked with bacteriophages.

The bacteriophage on infection injects only the DNA into the bacterial cell & not the protein coat.

Bacterial cells treat the viral DNA as their own & subsequently manufacture more virus particles.

They grew some viruses on a medium that contained radioactive Phosphorus & some others on a medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not proteins because DNA contains phosphorus. Similarly, viruses grown on radioactive sulfur contain radioactive protein because DNA does not contain sulfur.

Radioactive phages are allowed to infect E. coli bacteria & soon after infection the cultures were gently agitated in a blender to separate the adhering protein coat of the virus from the bacterial cells. It was found that when a phage containing radioactive DNA was used to infect the bacteria its radioactivity was found in bacterial cells indicating that DNA had been injected into the bacterial cells. So, the DNA is the genetic material & not proteins.

Points to Remember from Class 12 Biology Chapter 5: Molecular Basis of Inheritance

• DNA (Deoxyribonucleic acid) is a double-stranded helical molecule composed of nucleotides. Each nucleotide consists of a phosphate group, a deoxyribose sugar, and a nitrogenous base (Adenine - A, Thymine - T, Cytosine - C, and Guanine - G). 

• DNA replication is a semi-conservative process where each strand of the parent DNA molecule serves as a template for the synthesis of a new complementary strand. 

• The process involves enzymes like DNA helicase, DNA polymerase, and DNA ligase, ensuring the accurate copying of genetic material.

• The central dogma explains the flow of genetic information in a cell: DNA → RNA → Protein. 

• Transcription is the process where an mRNA molecule is synthesised using one strand of DNA as a template. 

• Translation is the process by which mRNA is used as a template to synthesise proteins in the ribosomes. 

• Mutations are changes in the genetic material (DNA) that can occur due to errors in replication, exposure to mutagens, or other factors.

• This chapter provides the foundation for understanding how genetic information is passed from one generation to the next and how it influences an organism’s traits. 

• It is essential for understanding genetic diseases, inheritance patterns, and advancements in biotechnology.

Benefits of Important Questions for Class 12 Biology Chapter 5

• Important questions help reinforce key concepts of genetics, such as DNA structure, replication, transcription, and translation. 

• These questions cover all crucial topics from the chapter, helping students develop a holistic understanding.

• Important questions often include problem-solving scenarios related to genetic mutations, genetic codes, and recombinant DNA, which help students strengthen their analytical and reasoning abilities.

• Practising with important questions helps students manage their time more effectively during exams.

• These questions help students relate theoretical knowledge to real-world applications in genetics, biotechnology, and medical sciences, providing context and relevance to the subject matter.

Conclusion

Understanding the concepts of the Molecular Basis of Inheritance is crucial for securing top marks in your Class 12 Biology exams. With Vedantu’s resources, including important questions and detailed explanations, students can deepen their understanding of complex topics like DNA replication, transcription, translation, and genetic code. Whether you're preparing for your board exams or competitive exams like NEET, our comprehensive study materials ensure you are well-equipped to excel.

Related Study Materials for Class 12 Biology Chapter 5

Chapter-wise Important Questions for Biology Class 12

Important Study Material for Biology Class 12