Very Short Answer Questions

Question. Mention the carbon positions to which the nitrogenous base and the phosphate molecule are respectively linked in the nucleotide given below:

Answer. Nitrogenous base is linked to first carbon.
Phosphate is linked to fifth carbon.

Question. Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer. Gel electrophoresis is used to separate DNA fragments.

Question. Mention one difference to distinguish an exon from an intron. 
Answer. Exon is the coded or expressed sequence of nucleotides in mRNA.
Intron is the intervening sequence of nucleotides not appearing in processed mRNA.

Question. In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates, still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why? 
Answer. DNA polymerase is highly specific to recognise only deoxyribonucleoside triphosphates.
Therefore it cannot hold RNA nucleotides.

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

Question. What are ‘a’ and ‘b’ in the nucleotide with purine represented below?

Answer. ‘a’ is phosphate group and ‘b’ is purine (adenine/guanine).

Question. Name the negatively charged and positively charged components of a nucleosome.
Answer. In a nucleosome, the negatively charged component is DNA and positively charged component is histone octamer.

Short Answer Questions

Question. State the dual role of deoxyribonucleoside triphosphates during DNA replication.
OR
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.
Answer. (i) Deoxyribonucleoside triphosphates act as substrates for polymerisation.
(ii) These provide energy from its two terminal phosphates for polymerisation reaction.

Question. (a) Name the molecule ‘M’ that binds with the operator.
(b) Mention the consequences of such binding.

Answer. (a) M is the repressor.
(b) When repressor binds with the operator, transcription stops.

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

Question. Discuss the role of enzyme DNA ligase plays during DNA replication.
Answer. DNA ligase joins or seals the discontinuous DNA fragments.

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. Coding strand: 5′ TACGTACGTACGTACGTACGTACG 3′mRNA strand: 5′ UACGUACGUACGUACGUACGUACG 3′

Question. Draw the structure of a tRNA charged with methionine. 
Answer. 

Question. Retrovirus do not follow central dogma. Comment.
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is reverse transcribed to its complementary DNA first, then transcriped to RNA and proteins. Hence, retrovirus are not known to follow central dogma.

Question. If the sequence of the coding strand in a transcription unit is written as follows:
5′—ATGCATGCATGCATGCATGCATGCATG—3′
Write down the sequence of mRNA.
Answer. 5′—AUGCAUGCAUGCAUGCAUGCAUGCAUG—3′

Question. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer. The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest semi-conservative mechanism of DNA replication in which one strand of parent is conserved while the other complementary strand formed is new.

Question. Differentiate between the following:
(a) Repetitive DNA and satellite DNA
(b) mRNA and tRNA
(c) Template strand and coding strand

Answer. (a) Differences between repetitive DNA and satellite DNA

(b) Differences between mRNA and tRNA

(c) Differences between template strand and coding strand

Question. Draw a labelled schematic diagram of a transcription unit.
Answer. 

Question. Differentiate between the genetic codes given below:
(a) Unambiguous and Universal.
(b) Degenerate and Initiator
Answer. 

Question. Explain the process of transcription in a bacterium.
Answer. • In prokaryotes, the structural gene is polycistronic and continuous.
• In bacteria, the transcription of all the three types of RNA (mRNA, tRNA and rRNA) is catalysed by single DNA-dependent enzyme, called the RNA polymerase.
• All three RNA’s are needed to synthesize a protein in cell. mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNA plays structural and catalytic role durinng translation.
• The transcription is completed in three steps: initiation, elongation and termination.
• Initiation: s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion following the rule of complementarity.

Question. (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion. 
Answer. (i) George Gamow.
(ii) He proposed that there are four bases and 20 amino acids So, there should be atleast 20 different genetic codes for these 20 amino acids.
The only possible combinations that would meet the requirement is combinations of 3 bases that will give 64 codons.

Question. Name the category of codons UGA belongs to. Mention another codon of the same category.Explain their role in protein synthesis.
Answer. UGA is a stop or termination codon.
UAA, UAG are the other stop codons of the category.
They prevent the elongation of the polypeptide chain by terminating translation.

Question. Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer. Tandemness in repeats provides many copies of the sequence for fingerprinting and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

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

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.

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′

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

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. Differentiate between the following:
(a) Promoter and terminator in a transcription unit.
Answer. (a) Differences between promoter and terminator.

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.

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. 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 DNA- dependent 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. (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 off, 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. Post- transcription 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.

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)andcytidinemonophosphate (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 () 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 () 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.

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.

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. 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,  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. 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. 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  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  C Loop. It has 7 bases out of which 
(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  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. 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. 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 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.

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.
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. 

Question. Correlate the codons of mRNA strand with amino acids of a polypeptide translated.
5′ AUGACCUUUCAUUCGUGUAA 3′
→ mRNA
Met-Thr-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. 

Question. (a) Why is tRNA called an adaptor?
Answer. (a) tRNA is also known as an adapter molecule as it is meant for transffering amino acids to ribosome during 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.

Question.State the functions of ribozyme and release factor in protein synthesis respectively.
Answer. Ribozyme (catalytic RNA) is present in ribosome and joins the amino acids together by peptide bond formation to form protein chains. Release factor (RF) is GTP dependent. It binds to the stop codon, terminates translation and release the complete polypeptide from the ribosome. 

Question. Where does peptide bond formation occur in a bacterial ribosome and how? (Foreign 2014)
Answer. Peptide bond formation occurs in the peptidyl transferase centre present in the larger sub-unit of ribosome. Peptide bond formation occurs between the nascent polypeptide chain and the new amino acid resulting in the elongation of polypeptide chain. An aminoacyl tRNA complex reaches the A-site and attaches to mRNA codon next to initiation codon with the help of its anticodon. The step requires GTP and an elongation factor. A peptide bond (—CO —NH—) is established between the carboxyl group (—COOH) of amino acid attached to tRNA at P-site and amino group (—NH2) of amino acid attached to tRNA at A-site. The reaction is catalysed by enzyme peptidyl transferase which is an RNA- enzyme.
A lot of energy is consumed in protein synthesis. For every single amino acid incorporated the peptide chain one ATP and two GTP molecules are used.

Question. Mention the role of ribosomes in peptide-bond formation. How does ATP facilitate it?
Answer. When a small sub-unit of ribosome encounters an mRNA, the process of translation of mRNA to protein begins. There are two sites in the large sub- unit for the subsequent amino acids to bind to and thus, be close enough to each other for the formation of peptide bond. The ribosome also acts as a catalyst (23 S rRNA in bacteria is ribozyme) for the formation of peptide bond.
Formation of a peptide bond requires energy. Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA – a process called charging of tRNA.

Question. How is the translation of mRNA terminated?
Explain. 
Answer. Polypeptide chain synthesis is terminated when a nonsense codon of mRNA reaches the A-site of ribosome during protein synthesis. There are three nonsense codons-UAA, UAG and UGA. These codons do not specify any amino acid. So, translation is stopped. GTP dependent release factor, cleaves the polypeptide from the terminal tRNA releasing the product from the ribosome.

Question. How do the initiation and termination of translation process occur in bacteria? Where are untranslated regions located in an mRNA?
Mention their role.
Answer. Initiation : When the small subunit of ribosome encounters the mRNA, translation begins. The mRNA binds to the small subunit of ribosome catalysed by initiation factor. There are two sites on the larger subunit of ribosome, the P-site and the A-site. The small subunit attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P-site. The initiator tRNA (methionine-tRNA) binds to the P-site.
Termination : The synthesis of polypeptide chain is terminated when stop/terminator codons UAA, UAG or UGA comes at the A-site of ribosome. These codons do not code for any amino acid.
Guanosine triphosphate (GTP) dependent release factors cleaves the polypeptide chain from the terminal tRNA, releasing the polypeptide from the ribosome.
An mRNA has some additional sequences that are not translated and are referred as untranslated regions (UTR). The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon). They are required for efficient translation process.

Question. Name the major types of RNAs and explain their role in the process of protein synthesis in a prokaryote. 
Answer. The three type of RNA are ribosomal RNA, messenger RNA and transfer RNA 
(a) mRNA - Messenger RNA bring coded information from DNA and takes part in its translation by bringing amino acids in a particular sequence during the synthesis of polypeptide. However, the codons of mRNA are not recognised by amino acids but by anticodons of their adapter molecules (tRNAs → aa-tRNAs). Translation occurs over the ribosomes. The same mRNA may be reused time and again. In the form of polysome, it can help synthesise a number of copies simultaneously.
(b) tRNAs -They are transfer or soluble RNAs which pick up particular amino acids (at CCA or 3′ end) in the process called charging. The charged tRNAs take the same to mRNA over particular codons corresponding to their anticodons. A tRNA can pickup only a specific amino acid though an amino acid can be specified by 2-6 tRNAs. Each tRNA has an area for coming in contact with ribosome (T  C) and the enzyme amino acyl tRNA synthetase (DHU).
(c) Ribosomes - Protein synthesis occurs over the ribosomes, Ribosomes are, therefore, also called protein factories. Each ribosome has two unequal parts, small and large. The larger subunit of ribosome has a groove for pushing out newly. 

Question. Explain the process of translation.
Answer. The process of decoding of the message from mRNA to protein with the help of tRNA, ribosome and enzyme is called translation (protein synthesis). Protein synthesis occurs over ribosomes.
The 4 main steps in protein synthesis (translation) are
: activation, initiation, elongation and termination of polypeptide chain.
The newly synthesised mRNA joins the smaller subunit of ribosome at 5′ end. mRNAs carry the codon and tRNAs carry the anticodon for the same codon. Activation of amino acid is catalysed by the enzyme aminoacyl tRNA synthetase in the presence of ATP. In presence of ATP an amino acid combines with its specific amino acyl-tRNA synthetase to produce aminoacyl adenylate enzyme complex. This reacts with tRNA to form aminoacyl-tRNA complex. Activated tRNA is taken to ribosome mRNA complex for initiation of protein synthesis.
Initiation of protein synthesis is accomplished with the help of initiation factor which are 3 (IF3, IF2, IF1) in prokaryotes and 9 in eukaryotes (eIF2, eIF3, eIF1, eIF4A, eIF4B, eIF4C, eIF4D, eIF5, eIF6). The ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA. A polypeptide chain forms as tRNAs deliver amino acids to the ribosome. Large ribosomal subunit binds the initiation complex forming two (A and P) binding site for tRNA molecules. The first site is P site or peptidyl site which is occupied by tRNAmet. The second site is A or amino acyl site and is positioned over the second codon. The ribosome proceeds to the elongation phase of protein synthesis. During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. The enzyme peptidyl synthetase catalyses the formation of peptide bond between the carboxylic group of amino acid at P site and amino group of amino acid at A site. Enzyme translocase brings about the movement of mRNA by one codon.
The termination of protein synthesis occur when a non-sense codon reaches at A site of ribosome. The chain detaches from the ribosome. A release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome. Two subunits of ribosomes dissociate with the help of dissociation factor.

Question. Differentiate between the following: Inducer and repressor in lac operon.
Answer. Inducer is a chemical (substrate, hormone or some other metabolite) which after coming in contact with the repressor, changes the latter into non-DNA binding state so as to free the operator gene. The inducer for lac-operon of Escherichia coli is lactose (actually allolactose, or metabolite of lactose.)
Repressor is a regulator protein meant for blocking the operator gene so that the structural genes are unable to form mRNAs.

Question. (a) What do ‘Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project (HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP. 
Answer. (a) In human genome project, ‘Y’ stands for yeast in YAC (yeast artificial chromosomes) and
‘B’ stands for bacterial in BAC (bacterial artificial chromosomes). These are specialised vectors used during sequencing in human genome project.
(b) In human genome, less than 2 percent of the genome codes for proteins and functions of only
50% of discovered genes are known.
(c) Human genome has SNPs at 1.4 million locations. Expanded form of SNPs is Single Nucleotide Polymorphism.

Question. In human genome which one of the chromosomes has the most genes and which has the fewest? 
Answer. Chromosome 1 has 2968 genes while Y-chromosome has 231 genes. They are maximum and minimum genes for human chromosomes respectively. 

Question. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments? 
Answer. Repetitive/satellite DNA can be separated from bulk genomic DNA by using density gradient centrifugation.

Question. How does DNA polymorphism arise in a population?
Answer. DNA polymorphism in a population arise due to mutations.

Question. Write the full form of VNTR. How is VNTR different from ‘Probe’?
Answer. VNTR stands for Variable Number of Tandem Repeats.
VNTRs are short nucleotide repeats in DNA that are specific to each individual and vary in number from person to person. DNA probes, are radioactive, have repeated base sequence complementary to VNTRs.

Very Short Answer

Question. How RNA acts?
Answer. RNA used to act as a genetic material as well as a catalyst.

Question. How DNA has evolved?
Answer. DNA has evolved from RNA.

Question. Who witnessed a miraculous transformation in the bacteria?
Answer. Frederick Griffith witnessed a miraculous transformation in the bacteria.

Question. Who proposed a scheme for replication of DNA?
Answer. Watson and Crick had proposed the scheme for replication of DNA.

Question. Who discovered that proteases and RNases did not affect transformation?
Answer. Oswald Avery, Colin Macleod and Maclyn McCarty discovered protease and RNasesthat did not affect transformation.

Short Answer

Question. What are the criteria a molecule can follow to act as a genetic?
Answer. 
A molecule that can act as a genetic material criteria are:
1. It should be able to generate its replica.
2. It should be stable chemically and structurally.
3. It should provide the scope for the slow changes that are required for evolution.
4. It should be able to express itself in the form of Mendelian characters.

Question. What do you understand by the Griffith transforming principle?
Answer. Griffith states that the R-strain bacteria must have taken up what he called a transforming principle from the heat-killed S bacteria, which allowed them to transform into smooth-coated bacteria and become virulent. It must be due to the transfer of the genetic material.

Question. What are the properties of genetic material?
Answer. 
The properties of genetic materials are:
1. It must be stable.
2. It must be capable of being expressed when needed.
3. It must be capable of accurate replication.
4. It must be transmitted from parent to progeny without change.

Question. What do you understand by RNA World?
Answer.  The RNA world is a hypothetical stage in the evolutionary history of life on earth, in which self-replicating RNA molecules proliferated before the evolution of DNA and proteins. The term also refers to the hypothesis that posits the existence of this stage.

Question. What do you mean by the semiconservative replication?
Answer. Semiconservative replication describes the mechanism of DNA replication in all known cells. This process is known as semi-conservative replication because two copies of the original DNA molecule are produced. Each copy contains one original strand and one newly-synthesized strand.

 Long Answer

Question. State the difference between DNA polymerase and RNA polymerase?
Answer. 
1. DNA polymerase is the enzyme which synthesizes new DNA molecules from DNA nucleotides in a process called DNA replication whereas RNA polymerase is the enzyme responsible for the synthesis of RNA molecules from DNA in a process called transcription.
2. DNA polymerase is used in DNA replication whereas RNA polymerase is used in transcription.
3. DNA polymerase uses DNA nucleotides to synthesize a new strand whereas RNA polymerase uses RNA nucleotides to synthesize a new strand.
4. DNA polymerase synthesizes a double-stranded DNA molecule whereas RNA polymerase synthesizes a single-stranded RNA molecule.
5. DNA polymerase requires a primer for the initiation of replication whereas RNA polymerase does not require a primer for the initiation of transcription.

Question. Explain the experimental proof performed by the Matthew Meseslon and Franklin Stahl?
Answer. The experimental proof performed by the Matthew Meseslon and Franklin Stahl on Escherichia coli and subsequently in higher organisms like plants and human cells are:
1. They grew Escherichia coli in a medium containing 15NH4Cl (15 N is the heavy isotopes of the nitrogen, as the only nitrogen source for many generations. The result was that 15N was incorporated into newly synthesized DNA as well as other nitrogen containing compounds.
2. The heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient. We also have to take care of that 15N is not a radioactive isotope and it can be separated from 14N only based on the densities.
3. Then they transferred the cells into a medium with the normal 14NH4Cl and took the samples at the various definite time intervals as the cells multiplied and extracted the DNA that remained as the double-strained helices. The various samples were separated independently on CsCl gradients to measure the densities of the DNA.
4. The DNA that was extracted from the culture one generation after the transfer from 15N to 14N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of the light DNA.

Question.How enzymes are involved in DNA replication?
Answer. DNA molecule and both strands in that molecule are referred to as parent strands. The goal of DNA replication is to make a second DNA molecule, using the parent strands as a template to create two new daughter strands. The term semi-conservative refers to the fact that both parent strands are conserved, or saved, in each of the new molecules. So, in semi-conservative replication, the parent strands split apart, but each remains whole, while new daughter strands are added onto them. The end result is two DNA molecules, each containing one parent strand and one daughter strand. Enzymes involved in DNA replication are: Helicase (unwinds the DNA double helix), Gyrase (relieves the build-up of torque during unwinding), Primase (lays down RNA primers), DNA polymerase III (main DNA synthesis enzyme), DNA polymerase I (replaces RNA primers with DNA), Ligase (fills in the gaps). Enzyme replication has to follow the three activities:
1. The 5′ → 3′ polymerase activity is responsible for primer extension or DNA synthesis.
2. The 5′ → 3′ exonuclease activity is involved in excision of DNA strands during DNA repair; it removes ~ 10 bases at a time. An exonuclease digests nucleic acids (here DNA) from one end, and it does not cut DNA internally.
3. The 3′ → 5′ exonuclease activity is responsible for proof-reading.

Question. State the difference between DNA and DNase?
Answer. 
1. DNA is a nucleic acid whereas DNase is an enzyme or protein.
2. DNA occurs inside the nucleus whereas DNase occurs in the cytoplasm.
3. DNA contains the genetic information required by the growth, development, and reproduction of organisms whereas DNase catalyses the hydrolytic cleavage of phosphodiester bonds.
4. DNA serves as the hereditary material of most organisms whereas DNase cleaves DNA into oligosaccharides.
5. DNA contains genes with relevant information whereas DNase is involved in the purification of RNA.
6. DNA refers to a self-replicating material present in nearly all living organisms as the main constituent of chromosomes, serving as the carrier of genetic information whereas DNase refers to an enzyme which catalyses the hydrolysis of DNA into oligonucleotides and smaller molecules.

Question. Write short note on transcription?
Answer. The process of copying genetic information form one strand f the DNA into RNA is termed as transcription. Transcription takes place in the nucleus. It uses DNA as a template to make an RNA molecule. RNA then leaves the nucleus and goes to a ribosome in the cytoplasm, where translation occurs. Translation reads the genetic code in mRNA and makes a protein. Transcription uses one of the two exposed DNA strands as a template; this strand is called the template strand. The RNA product is complementary to the template strand and is almost identical to the other DNA strand, called the non-template or coding strand. A transcription unit in DNA is defined primarily by the three regions in the DNA: a promoter, the structural gene, a terminator. Sequence of nucleotides in DNA that codes for a single RNA molecule, along with the sequences necessary for its transcription; normally contains a promoter, an RNA-coding sequence, and a terminator. RNA polymerase is the main transcription enzyme. Transcription begins when RNA polymerase binds to a promoter sequence near the beginning of a gene (directly or through helper proteins). RNA polymerase uses one of the DNA strands (the template strand) as a template to make a new, complementary RNA molecule.