Molecular Basis of Inheritance VBQs Class 12 Biology

VBQs for Class 12

VBQs Molecular Basis of Inheritance Class 12 Biology with solutions has been provided below for standard students. We have provided chapter wise VBQ for Class 12 Biology with solutions. The following Molecular Basis of Inheritance Class 12 Biology value based questions with answers will come in your exams. Students should understand the concepts and learn the solved cased based VBQs provided below. This will help you to get better marks in class 12 examinations.

Molecular Basis of Inheritance VBQs Class 12 Biology

Very Short Answer Type Questions

Question. Give an example of a codon having dual function.
Answer :
 AUG. 

Question. Name one amino acid, which is coded by only one codon. 
Answer :
Methionine / Tryptophan

Question. Retroviruses have no DNA. However the DNA of the infected host cell does possess viral DNA.
How is it possible ?
Answer :
RNA is the genetic material in retrovirus. This RNA forms DNA by the process of reverse transcription with the help of the enzyme called reversetranscriptase.

Question. Why is RNA more reactive in comparison to DNA? 
Answer :
2′ – OH Present in RNA (in every nucleotide) make it reactive.
Detailed Answer :
RNA is more reactive than DNA because :
(i) It is single stranded.
(ii) 2′ – OH group is present in every nucleotide.
(iii) It mutates faster. (Any two)

Question. Write the two specific codons that a translational unit of m-RNA is flanked by one on either sides.
Answer :
Start codon – AUG
Stop codon – UAA/UGA/UAG 

Question. What is a cistron ? 
Answer :
Cistron is a segment of gene which codes for a certain polypeptide or protein.

Question. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments ? 
Answer :
Density gradient centrifugation. 

Question. Which one out of Rho factor and Sigma factor acts as the initiation factor during transcription in a prokaryote ?
Answer :
Sigma factor (σ). 

Question. Name the enzyme that joins the small fragments of DNA of a lagging strand during DNA replication.
Answer :
DNA ligase. 

Question. Which one is tailed with adenylate residues between 3’ and 5’ end of hnRNA ?
Answer :
3’ end is tailed with polyadenylate residues. 

Question. Name the types of synthesis ‘a’ and ‘b’ occurring in the replication fork of DNA as shown below :

Molecular Basis of Inheritance VBQs Class 12 Biology

Answer : a— Leading strand (continuous)
b—Lagging strand (discontinuous) 

Question. Name the component a and b in the nucleotide with a purine given below. 

Molecular Basis of Inheritance VBQs Class 12 Biology

Answer : a— Phosphate group
b— Nitrogenous base 

Question. Write the conclusion Griffith arrived at the end of his experiment with Streptococcus pneumoniae.
Answer :
He concluded that the R-strain bacteria somehow been transformed by heat – killed S-strain bacteria, this must be due to transfer of genetic material 

Short Answer Type Questions

Question. Why does hnRNA need to undergo splicing?
Where does splicing occur in the cell ?
Answer :
 hnRNA has both exons and introns, Introns are non-coding regions, which are removed by the process called splicing, splicing occurs in the nucleus. 
Detailed Answer :
hnRNA is the primary transcript. It is a non functional. It contains both the coding region-exons and non coding regions called introns in RNA. It is called as heterogenous nuclear RNA or hnRNA. It gets functional only after processing by splicing.
During splicing the introns are removed and exons are joined. hnRNA also undergoes two additional processes called capping and tailing. During capping, the unusual nucleotide methyl guanosine triphosphate (mGPPP) is added to the 5′ end of hnRNA. In tailing, about 200-300 adenylate residues are added at 3′ end of mRNA. Now, this is the fully processed hnRNA which is called as mRNA. It is now functional and is ready for translation. The splicing of hnRNA occurs in the nucleus. 

Question. Protein synthesis machinery revolves around RNA but in the course of evolution it was replaced by DNA. Justify
Answer :
Since RNA was unstable and prone to mutations, DNA evolved from RNA with chemical modifications that make it more stable.
DNA has double stranded nature and has complementary strands. This further resist changes by evolving a process of repair. 

Question. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer :
The two factors responsible for conferring stability to double helix structure of DNA are:
1. Stacking of one base over another
2. H-bonding between the nitrogenous base
Detailed Answer :
Factors conferring stability to Double Helical Structure of DNA are :
(i) Complementarity of the two strands of DNA due to complementary nitrogenous bases which form strong hydrogen bonds with each other. Adenine forms two hydrogen bonds with thymine and cytosine form 3 hydrogen bonds with guanine.
(ii) The base pairs are stacked with their planes one over the other in the double helical structure which provides extra stability.
Also, DNA is less reactive due to absence of reactive
—OH group at 2’ carbon. Evolution of a process of repairs which prevents their degradation and presence of thymine instead of uracil being more stable as nitrogenous base also provide stability to double helical structure of DNA. 
Question. Draw a neat labelled sketch of a replicating fork of DNA. 
Answer :

Molecular Basis of Inheritance VBQs Class 12 Biology

Polarity of the two strands of the fork to be shown and polarity as well as arrow mark of the lagging and leading strands to be shown with correct labellings.

Question. Differentiate between a cistron and an exon.
Answer : Difference between cistron and an exon :
Cistron :
A segment of DNA coding for a polypeptide chain, the structural gene in a transcription unit could be monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).
Exon : The coding sequences or expressed sequences are defined as exons. Exons are said to be those sequences that appear in mature or processed mRNA. 

Question. How do histones acquire positive charge ?
Answer :
Histones are rich in basic amino acids Lysine, Arginine (present as residues in their side chains), which are positively charged. 
Question. Describe the structure of a nucleosome.
Answer :
A unit of eight molecules of positively charged histones, negatively charged DNA, wrapped around the histones octamer, contains 200 bp of DNA helix
OR
In lieu of the above explanation the following diagram along with the following statement can be considered

Molecular Basis of Inheritance VBQs Class 12 Biology

DNA is negatively charged, histone is positively charged, 200 bp of DNA helix 2
Detailed Answer :
(i) A typical nucleosome contains 200 bp of DNA helix.
(ii) Negatively charged DNA is wrapped around positively charged histone octomer.
(iii) It constitute the repeating unit to form chromatin.
(iv) The chromatin appear as ’’beads on string’’.
(v) The chromatin is packed to form a solenoid structure and further supercoiling constitute looped structure called chromatin fibre.
(vi) Higher level packaging of chromatin requires nonhistone chromosomal (NHC) proteins.

Transcription

Very Short Answer Type Questions

Question. Write the function of RNA polymerase II.
Answer.In eukaryotes, RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hn RNA). 

Question. Which one out of Rho factor and Sigma factor act as initiation factor during transcription in a prokaryote? 
Answer.Sigma (s) factor acts as initiation factor during transcription in a prokaryote.

Question. At which ends do ‘capping’ and ‘tailing of hnRNA occur respectively? 
Answer.Capping and tailing of hnRNA occur respectively at 5′ end and 3′ end. 

Question. Which one of an intron and an exon is the reminiscent of antiquity? 
Answer.Intron is considered to be as the reminiscent of antiquity.

Question. Mention the two additional processing which hnRNA needs to undergo after splicing so as to to become functional. 
Answer.Capping and tailing.

Question. Differentiate between exons and introns.
Answer. Exons are the segments in genes which contain coding nucleotide sequences. These sequences are ultimately translated into polypeptide. Thus, exons carry genetic information. Introns are the segments in genes which contain non-coding nucleotide sequences. These do not form part of mRNA and are removed during the processing of hnRNA.

Question. Name the parts ‘A’ and ‘B’ of the transcription given below:

Molecular Basis of Inheritance VBQs Class 12 Biology

Answer.A – Promoter
B – Coding strand

Question. Why hnRNA is required to undergo splicing?
Answer.hnRNA undergo splicing in order to remove introns, which are intervening or non-coding sequences, and exons are joined in a defened order to form functional mRNA.

Question. What is a cistron? 
Answer. Cistron is a segment of DNA consisting of a stretch of base sequences that codes for one polypeptide chain, one transfer RNA (tRNA), ribosomal RNA (rRNA) molecule or performs any other specific function in connection with transcription, including controlling the functioning of other cistrons.

Question. When and at what end does the ‘tailing’ of hnRNA take place? 
Answer.‘Tailing’ of hnRNA take place during modification of hnRNA into functional mRNA. It takes place at 3′-end.

Short Answer Type Questions

Question. State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes. 
Answer.Difference between structural gene in prokaryotes and structural gene in eukaryotes.

Molecular Basis of Inheritance VBQs Class 12 Biology
Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Differentiate between a template strand and coding strand of DNA.
Answer.Differences between template strand and coding strand.

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. A template strand is given below. Write down the corresponding coding strand and the mRNA strand that can be formed, along with their polarity. 3′ ATGCATGCATGCATGCATGCATGC 5′
Answer.The corresponding coding strand is :
5′ TACGTACGTACGTACGTACGTACG 3′
The corresponding mRNA strand is
5′ UACGUACGUACGUACGUACGUACG 3′

Molecular Basis of Inheritance 

Question. Differentiate between a cistron and an exon.
Answer. Cistron is segment of DNA consisting of a stretch of base sequences that codes for one polypeptide chain, one transfer RNA (tRNA), ribosomal RNA (rRNA) molecule or performs any other specific function in connection with transcription, including controlling the functioning of other cistrons. Exons are the regions of a gene, which become part of mRNA and code for the different proteins.

Question. State the functions of the following in a prokaryote:
(a) tRNA
(b) rRNA 
Answer.
(a) tRNA helps in transferring amino acids to ribosome for synthesis of polypeptide chain. tRNA reads the genetic codes, carries amino acids to the site of protein synthesis and acts as an adapter molecule.
(b) rRNA is the most abundant RNA. Prokaryotic ribosomes are of three types 23S, 16S and 5S. 23S and 5S occur in large subunit of ribosome while 16S is found in smaller subunit. It plays structural and catalytic role during translation.

Question. Differentiate between exons and introns.
Answer.Differences between Introns and Exons.

Molecular Basis of Inheritance VBQs Class 12 Biology

Short Answer Type Questions

Question. Monocistronic structural genes in eukaryotes have interrupted coding sequences. Explain. How are they dierent in prokaryotes?
Answer.In eukaryotes, monocistronic structural genes could be said as interrupted, as both introns and exons are present.
The corresponding coding strand is :
5′ TACGTACGTACGTACGTACGTACG 3′
The corresponding mRNA strand is
5′ UACGUACGUACGUACGUACGUACG 3′

Question. (a) Construct a complete transcription unit with promotor and terminator on the basis of the hypothetical template strand given below:

Molecular Basis of Inheritance VBQs Class 12 Biology

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity. 
Answer.

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Describe with the help of a schematic representation the structure of a transcription unit. 
Answer.A transcription unit in DNA is defined by these regions in DNA :

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Write the help of a schematic diagram, explain the location and the role of the following in a transcription unit: Promoter, Structural gene, Terminator.
Answer.Schematic representation of transcription unit is as follows:

Molecular Basis of Inheritance VBQs Class 12 Biology

(i) Promoter – The promoter is located towards 5′-end of the coding strand. It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that defines the template and coding strands.
(ii) Structural gene – The structural genes code for the enzymes and proteins. It transcribe the mRNA for the same.
(iii) Terminator – The terminator is located towards 3′-end of the coding strand and it defines the end of the process of transcription.

Question. (a) Draw a schematic representation of the structure of a transcription unit and show the following in it:
(i) Direction in which the transcription occurs
(ii) Polarity of the two strands involved
(iii) Template strand
(iv) Terminator gene
(b) Mention the function of promoter gene in transcription. 
Answer.A transcription unit in DNA is defined by these regions in DNA :

Molecular Basis of Inheritance VBQs Class 12 Biology

(b) Promoter – The promoter is located towards 5′-end of the coding strand. It is a DNA sequence that has RNA polymerase recognition site, it also provide binding site for RNA polymerase. Presence of a promoter in a transcription unit defines the template and coding strands.

Question. (a) What are the transcriptional products of RNA polymerase III?
(b) Differentiate between ‘Capping’ and ‘Tailing’.
(c) Expand hnRNA. 
Answer.(a) The transcriptional products of RNA polymerase III are tRNA, 5SrRNA and snRNA.
(b) In capping, additional nucleotides (methyl guanosine triphosphate) are added to the 5′-end of hnRNA. In tailing, adenylate residues (200 – 300) are added at the 3′-end in a template independent manner.
(c) Heterogeneous nuclear RNA.

Question. Explain the role of DNA-dependent RNA polymerase in transcription.
Answer. Transcription requires DNA-dependent RNA polymerase. RNA polymerase binds to promoter and initiates transcription. With the help of RNA polymerase the adjacent ribonucleotides held over DNA template join to form RNA chain. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription. In eukaryotes there are at least three DNAdependent RNA polymerases in the nucleus. The RNA polymerase I transcribes rRNAs (28S, 18S and 5.8S), whereas the RNA polymerase II transcribes precursor of mRNA, the heterogenous nuclear RNA (hnRNA). RNA polymerase III is responsible for transcription of tRNA, 5S rRNA and some snRNAs. Whereas prokaryotes have only single DNA dependent RNA polymerase.

Question. Differentiate between the following:
(a) Promoter and terminator in a transcription unit.
(b) Exon and intron in an unprocessed eukaryotic mRNA. 
Answer.(a) Differences between promoter and terminator.

Molecular Basis of Inheritance VBQs Class 12 Biology

Differences between Introns and Exons.

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Describe the elongation process of transcription in bacteria. 
Answer. With the help of RNA polymerase the adjacent ribonucleotides held over DNA template join to form RNA chain. As the RNA chain formation initiates, the sigma (s) factor of the RNA polymerase separates. RNA polymerase (core enzyme) moves along the DNA template causing elongation of RNA chain at the rate of some 30 nucleotides per second. RNA synthesis stops as soon as polymerase reaches the terminator region.

Question. (a) Name the enzyme that catalyses the transcription of hnRNA.
(b) Why does the hnRNA need to undergo changes? List the changes hnRNA undergoes and where in the cell such changes take place.
Answer.
(a) RNA polymerase II transcribes hnRNA.
(b) Post transcription processing of hnRNA is required to convert primary transcript of all types of RNA into functional RNAs. The eukaryotic transcription involves certain complexity, one of the complexity is that the primary transcript contain both the exons and the introns (which are non-functional). Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order. The transcribed eukaryotic RNA or heterogenous nuclear RNA (hnRNA), undergoes additional processing called as capping and tailing.
In capping, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5’ – end of hnRNA. In tailing, adenylate residues (200 – 300) are added at the 3′-end in a template independent manner. It is the fully processed hnRNA, now called mRNA. This process takes place in the nucleus of a cell and then mRNA is transported out of the nucleus for translation.

Question. Describe the initiation process of transcription in bacteria. 
Answer. A DNA transcription unit has a promoter region, initiation site, coding region and a terminator region. Transcription begins at the initiation site and ends at the terminator region. A promoter region has RNA polymerase recognition site and RNA polymerase binding site. Chain opening occurs in the region occupied by TATAATG nucleotides (TATA box) in most prokaryotes. Enzymes required for chain separation are unwindases, gyrases and single stranded binding proteins.
During initiation of transcription, RNA polymerase (common in prokaryotes and specific in eukaryotes) binds itself to the promoter region. The two strands of DNA uncoil progressively from the site of polymerase binding. One of the two strands of DNA (3′ → 5′) functions as template for transcription of RNA (template strand). Transcript formation occurs in 5′ → 3′ direction.
Ribonucleoside triphosphates present in the surrounding medium come to lie opposite the nitrogen bases of the DNA template (anti-sense strand).They form complementary pairs; U opposite A, A opposite T, C opposite G and G opposite C. A pyrophosphate is released from each ribonucleoside triphosphate to produce ribonucleotide.

Long Answer Type Questions

Question. Explain the process of transcription in prokaryotes. How is the process different in eukaryotes? 
Answer.
Mechanism of transcription in prokaryotes :
In bacteria/prokaryotes, transcription occurs in contact with cytoplasm as their DNA lies in the cytoplasm.

(a) Activation of ribonucleotides – The four types of ribonucleotides are adenosine monophosphate (AMP), guanosine monophosphate (GMP), uridine monophosphate (UMP) and cytidine monophosphate (CMP). They occur freely in the nucleoplasm. Prior to transcription the nucleotides are activated through phosphorylation. Enzyme phosphorylase is required alongwith energy. The activated or phosphorylated ribonucleotides are adenosine triphosphate (ATP), guanosine triphosphate (GTP), uridine triphosphate (UTP) and cytidine triphosphate (CTP).

(b) Binding of RNA polymerase to DNA duplex – On a signal from the cytoplasm, DNA segment become ready to transcribe. The RNA polymerase enzyme binds to a specific site, called promoter, in the DNA double helix. Prokaryotes have only one RNA polymerase that synthesise all types of RNA. The promoter also determines which DNA strand is to be transcribed. Thus, a promotor region has RNA polymerase recognition site and RNA polymerase binding site.

(c) Base pairing – Ribonucleoside triphosphates present in the surrounding medium come to lie opposite the nitrogen bases of the DNA template (anti-sense strand). They form complementary pairs; U opposite A, A opposite T, C opposite G and G opposite C. A pyrophosphate is released from each ribonucleoside triphosphate to produce ribonucleotide.

(d) Formation of RNA chain – With the help of RNA polymerase the adjacent ribonucleotides held over DNA template join to form RNA chain. A single RNA polymerase recognise promoter and initiation region is prokaryotes. As the RNA chain formation initiates, the sigma (s) factor of the RNA polymerase separates. RNA polymerase (core enzyme) moves along the DNA template causing elongation of RNA chain at the rate of some 30 nucleotides per second. RNA synthesis stops as soon as polymerase reaches the terminator region. Rho factor (r) has ATP-ase activity and also possesses 4-8 adenine ribonucleotides.

(e) Separation of RNA chain – With the help of rho factor, the fully formed RNA chain is now released. One gene forms several molecules of RNA, which are released from the DNA template one after the other on completion. The released RNA is called primary transcript.

(f) Duplex formation – As the RNA chain is released, the transcribed region of the DNA molecule gets hydrogen bonded to the sense strand and the two are spirally coiled to assume the original double helical form. The protective protein coat is added again to the DNA duplex. Gyrases, helicases and helix stabilizing proteins are released.
Table : Differences between prokaryotic and eukaryotic transcription

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Draw a labelled schematic structure of a transcription unit. Explain the function of each component of the unit in the process of transcription. 
Answer.

Molecular Basis of Inheritance VBQs Class 12 Biology

(i) Promoter – The promoter is located towards 5′-end of the coding strand. It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that defines the template and coding strands.
(ii) Structural gene – The structural genes code for the enzymes and proteins. It transcribe the mRNA for the same.
(iii) Terminator – The terminator is located towards 3′-end of the coding strand and defines the end of the process of transcription.
(iv) Coding strand – The strand of DNA with 5′ → 3′ polarity is the coding strand, it does not code for RNA.
(v) Template strand – The strand of DNA with 3′ → 5′ polarity acts as the template for transcription of mRNA.

Question. Explain the process of transcription in eukaryotes.
Answer.
In eukaryotes, transcription occurs throughout I-phase in differentiated cells but more so in G1 and G2 phases of cell cycle inside the nucleus. Depending upon the requirement, a structural gene may transcribe one to numerous RNA molecules. The transcription products move out into cytoplasm for translation. Transcription requires a DNA dependent RNA polymerase. Eukaryotes have three RNA polymerase, Pol I (Pol A) (for ribosomal or rRNAs except 5S rRNA). Pol II (for mRNA, snRNAs) and Pol III (for transfer or tRNA, 5S rRNA., and some snRNAs). Eukaryotic RNA polymerases also require transcription factors for initiation.
Prior to transcription, the nucleotides are activated through phosphorylation. Enzyme phosphorylase is required alongwith energy. Each DNA transcription segment has a promoter region, initiation site, coding region and a termintor region. RNA polymerase (common in prokaryotes and specific in eukaryotes) binds itself to the promoter region. The two strands of DNA uncoil progressively from the site of polymerase binding. One of the two strands of DNA (3′ → 5′) functions as a template for transcription of RNA. Transcript formation occurs in 5′ → 3′ direction.
Ribonucleoside triphosphate present in the surrounding medium form complementary pairs. With the help of RNA polymerase the adjacent ribonucleotides held over DNA template join to form RNA chain. In eukaryotes, there are separate transcription factor and RNA polymerase for activation of transcription. RNA polymerase (core enzyme) moves along the DNA template causing elongation of RNA chain at the rate of some 30 nucleotides per second. RNA synthesis stops as soon as polymerase reaches the terminator region. In eukaryotes, the transcription unit yields a monocistronic mRNA. Diagrammatic representation of transcription in eukaryotes is as follows:

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. (a) Describe the process of transcription in bacteria.
(b) Explain the processing the hnRNA needs to undergo before becoming functional mRNA of eukaryotes.
Answer.
(a) Transcription is the process of copying genetic information from one strands of DNA into RNA. A transcription unit of a DNA has three regions a promoter, a structural gene and a terminator. Bacterial structural gene in a transcription unit is polycistronic. Transcription requires a DNA dependent RNA polymerase. Prokaryotes have only one DNA dependent RNA polymerase which synthesises all types of RNA.
Three major types of RNAs in a bacteria are mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA). All three RNAs are needed to synthesise a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation. In bacteria/prokaryotes, transcription occurs in contact with cytoplasm as their DNA lies in the cytoplasm.
RNA polymerase binds to promoter and initiates transcription (Initiation). It uses nucleoside triphosphates as substrate and polymerises the mRNA strand in a template depended fashion following the rule of complementarity. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls offf, so also the RNA polymerase. This results in termination of transcription.

(b) In eukaryotes, the primary transcript which is often larger than the functional RNA is called heterogeneous nuclear RNA or hnRNA. Posttranscription processing is required to convert primary transcript of all types of RNAs into functional RNAs. It is of four types.
(i) Cleavage : Larger RNA precursors are cleaved to form smaller RNAs.
(ii) Splicing : Eukaryotic transcripts possess extra segments called introns or intervening sequences or noncoding sequences. They do not appear in mature or processed RNA. The functional coding sequences are called exons. Splicing is removal of introns and fusion of exons to form functional RNAs.
(iii) Terminal additions (capping and tailing) : Additional nucleotides are added to the ends of RNAs for specific functions, e.g., CCA segment in tRNA, cap nucleotides at 5′ end of mRNA or poly-A segments (200-300 residues) at 3′ end of mRNA. Cap is formed by modification of GTP into 7-methyl guanosine or 7 mG.
(iv) Nucleotide modifications :They are most common in tRNA-methylation (e.g., methyl cytosine, methyl guanosine), deamination (e.g., inosine from adenine), dihydrouracil, pseudouracil, etc. 

Genetic Code

Very Short Answer Type Questions

Question. How does a degenerate code differ from an unambiguous one? 
Answer. 
The difference between unambiguous and degenerate codons is :

Molecular Basis of Inheritance VBQs Class 12 Biology

Question. Give an example of a codon having dual function.
Answer.AUG codon has dual functions. It codes for methionine (met) and also acts as an initiation codon for polypeptide synthesis

Question. Mention the role of the codons AUG and UGA during protein synthesis. 
Answer.AUG has dual functions. It functions as initiation codon during protein synthesis and also codes for methionine. UGA does not specify any amino acid hence it functions as terminator codon.

Short Answer Type Questions

Question. Explain the structure of a tRNA and state why it is known as an adaptor molecule.
Answer. Structure of tRNA can be explained by means of L-form model (Given by Klug 1974) and by means of clover leaf model (given by Holley 1965).
In tRNA molecule, about half of the nucleotides are base paired to produce paired stems. Five regions are unpaired or single stranded – AA-binding site, T ψ C loop, DHU loop, extra arm and anticodon loop.
(i) Anticodon Loop. It has 7 bases out of which three bases form anticodon (nodoc) for recognising and attaching to the codon of mRNA.
(ii) AA-Binding Site. It is amino acid binding site.The site lies at the 3′ end opposite the anticodon and has CCA – OH group. The 5′ end bears G. Amino acid or AA binding site and anticodon are the two recognition sites of tRNA.
(iii) T ψ C Loop. It has 7 bases out of which ψ (pseudouridine) and rT (ribothymidine) are unusual bases. The loop is the site for attaching to ribosome.
(iv) DHU Loop. The loop contain 8–12 bases. It is largest loop and has dihydrouridine. It is binding site for aminoacyl synthetase enzyme.
(v) Extra Arm. It is a variable side arm or loop which lies between T ψ C loop and anticodon. It is not present in all tRNAs. tRNA is known as an adapter molecule because it transfers amino acids to ribosomes during protein synthesis for synthesis of polypeptides

Question. One of the salient features of the genetic code is that it is nearly universal from bacteria to humans. Mention two exceptions to this rule.Why are some codes said to be degenerate?
Answer.
Exceptions to the universality of genetic code are:
(i) UAA and UGA are termination codons and do not code for any amino acid. But in Paramecium and some other ciliates, these codons code for glutamine.
(ii) Genetic code is non-overlapping is most organisms. But, f × 174 has 5375 nucleotides that code for 10 proteins which require more than 6000 bases. Three of its genes E, B and K overlap other genes. Nucleotide sequence at the beginning of E gene is contained within gene D. Likewise gene K overlaps with genes A and C. A similar condition is found in SV-40.
More than one codons code for a single amino acid, thus are called degenerate codons. In degenerate codes, the first two nitrogen bases are similar while the third one is different, e.g., UUU and UUC are the degenerate codes that code for amino acid phenylalanine.

Question. Name the category of codons UGA belongs to.Mention another codon of the same category.Explain their role in protein synthesis.
Answer. UGA belongs to the category of termination codons. Other codons of same category are UAG and UAA. The termination codons do not code for any amino acid and therefore terminate the process of protein synthesis.

Question. Genetic codes can be universal and degenerate.Write about them, giving one example of each.
Answer.
Genetic code is universal i.e., a codon specifies the same amino acid from a virus to a tree or human being. Example : mRNA from chick oviduct introduced in Escherichia coli produces ovalbumen in the bacterium exactly similar to one formed in chick.
Genetic code is degenerative i.e., all other amino acids, except tryptophan and methionine, are specified by two (e.g. phenylalanine – UUU, UUC) to six (e.g., arginine–CGU, CGC, CGA, CGG, AGA, AGG) codons. They are therefore called degenerate or redundant codon. In degenerate codons, generally the first two nitrogen bases are similar while the third one is different.

Question. Following are the features of genetic codes.What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon. 
Answer.
(a) Stop codon :
 Codons that do not code for any amino acids and signal polypeptide chain termination. E.g., UAA, UAG, UGA.
(b) Unambiguous codon : Codons that specify only one amino acid and not any other. E.g., AUG codes for methionine.
(c) Degenerate codon : More than one codons codes for a single amino acid. In degenerate codons, generally the first two nitrogen bases are similar while the third one is different. E.g., UUU and UUC codes for phenylalanine.
(d) Universal codon : A codon that is applicable universally i.e., specifies the same amino acid from a virus to a tree or human being.

Question. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its anticodon. 
Answer. AUG acts as an initiation codon and it also codes for amino acid methionine. The sequence of bases from which it is transcribed is TAC. Its anticodon is UAC.

Question. (a) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(b) Explain the basis on which he arrived at this conclusion. 
Answer.(a) George Gamow suggested that the genetic code should be made up of three nucleotides.
(b) He proposed that since there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases made up of three nucleotides. Combination of 43 (4 × 4 × 4) would generate 64 codons; generating many more codons than required.

Question. Genetic code is specific and nearly universal. Justify. 
Answer.Genetic code is specific as one codon codes for only one amino acid and it is nearly universal as the same codon would code for same amino acid from bacteria to human. But some exceptions to this rule have been found in mitochondrial codons and in some protozoans. 

Short Answer Type Questions

Question. (a) Name the scientist who called tRNA as an adaptor molecule.
(b) Draw a clover leaf structure of tRNA showing the following:
(i) Tyrosine attached to its amino acid site.
(ii) anticodon for this amino acid in its correct site (codon for tyrosine is UAC).
(c) What does the actual structure of tRNA look like?
Answer.
(a) Francis Crick
(b)

Molecular Basis of Inheritance VBQs Class 12 Biology

(c) Actual structure of tRNA is compact and looks like an inverted L.

Question. One of the codons on mRNA is AUG. Draw the structure of tRNA adaptor molecule for this codon. Explain the uniqueness of this tRNA?
Answer.

Molecular Basis of Inheritance VBQs Class 12 Biology

This tRNA is specific for methionine and can act as initiator tRNA.

Question. Correlate the codons of mRNA strand with amino acids of a polypeptide translated.
5′ AUGACCUUUCAUUCGUGUAA 3′
→ mRNA
Met-r-Phe-His-Ser-Cys.
→ Translated polypeptide
Infer any 3 properties of genetic code with examples from the above information.
Answer. From the given example it is inferred that genetic code is degenerate, universal and unambiguous. 
(a) Stop codon : Codons that do not code for any amino acids and signal polypeptide chain termination. E.g., UAA, UAG, UGA.
(b) Unambiguous codon : Codons that specify only one amino acid and not any other. E.g., AUG codes for methionine.
(c) Degenerate codon : More than one codons codes for a single amino acid. In degenerate codons,generally the first two nitrogen bases are similar while the third one is different. E.g., UUU and UUC codes for phenylalanine.
(d) Universal codon : A codon that is applicable universally i.e., specifies the same amino acid from a virus to a tree or human being.

Question. (a) Name the scientist who postulated the presence of an adapter molecule that can assist in protein synthesis.
(b) Describe its structure with the help of a diagram. Mention its role in protein synthesis.
Answer.
(a) Crick postulated the presence of tRNA as an adapter molecule.
(b) Following is the clover leaf model of tRNA.

Molecular Basis of Inheritance VBQs Class 12 Biology

Role of tRNA in protein synthesis:
(i) tRNA is an adapter molecule meant for transferring amino acids during protein synthesis. tRNA binds to a particular amino acid at 3′ end. The charged tRNA take the same amino acid to mRNA over particular codons corresponding to their anticodons.
(ii) tRNA holds peptidyl chains over the mRNAs.
(iii) The initiator tRNA has the dual function of initiation of protein synthesis as well as bringing in of the first amino acid.

Question. (a) Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis. 
Answer.(a) Enzyme RNA polymerase III is responsible for the transcription of tRNA. Methionine is the amino acid with which the initiator tRNA gets linked.
(b) Anticodon of initiator tRNA i.e., tRNA linked with methoinine carries this amino acid to mRNA, establishes temporary hydrogen bonds with the initiation codon (AUG) of mRNA to start the process of protein synthesis.

Question. Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated form it.
AUG UUU UCU UUU UUU UCU UAG
Met – Phe – Ser – Phe – Phe – Ser
Answer.
Salient features of genetic code are:
(a) Triplet : Three adjacent nitrogen bases constitute a codon which specifies one particular amino acid in a polypeptide chain, e.g., in the given polypeptide, codon AUG specifies methionine, codon UUU specifies phenylalanine and so on.
(b) Initiation codon AUG (codes for methionine) acts as an initiation codon for polypeptide synthesis.
(c) Termination codons : Three termination codons (UAA, UAG and UGA) do not code for any amino acid, e.g., in the given polypeptide, UAG does not code for any amino acid.
(d) Unambiguous codons : One codon specifies only one amino acid and not any other, e.g., UCU codes for serine (Ser) only and UUU codes for phenylalanine (Phe) only.
(e) Commaless : The genetic code is continuous and does not possess pauses after the triplets.
(f) Degeneracy of Code : Since there are 64 triplet codons and only 20 amino acids, the incorporation of some amino acids must be influenced by more than one codon. e.g., amino acid Phe is specified by two codons i.e. UUU and UUC.
(g) Colinearity : There is a colinearity between the sequence of nitrogen bases on DNA or mRNA and the sequence of amino acids in a polypeptide chain. 

Long Answer Type Questions

Question. (a) What is a genetic code ?
(b) Explain the following :
Degenerate code; Unambiguous code; Initiator code.
Answer.
(a) The relationship between the sequence of amino acids in a polypeptide and nucleotide sequence of DNA or mRNA is called genetic code.
(a) Stop codon : Codons that do not code for any amino acids and signal polypeptide chain termination. E.g., UAA, UAG, UGA.
(b) Unambiguous codon : Codons that specify only one amino acid and not any other. E.g., AUG codes for methionine.
(c) Degenerate codon : More than one codons codes for a single amino acid. In degenerate codons, generally the first two nitrogen bases are similar while the third one is different. E.g., UUU and UUC codes for phenylalanine.
(d) Universal codon : A codon that is applicable universally i.e., specifies the same amino acid from a virus to a tree or human being.
Initiator codons are AUG and rarely GUG,which code respectively for methionine and valine.
They initiate the process of translation.