a) Describe the elongation process of translation in E.coli with suitable diagram. State the function of Aminoacyl-synthetase in translation. (4+1)
Ans→ The elongation process of translation in E. coli involves the addition of amino acids to a growing polypeptide chain. It takes place at the ribosome, which has three main sites: the A site (aminoacyl site), the P site (peptidyl site), and the E site (exit site).
Steps of Elongation:
- Binding of Aminoacyl-tRNA to the A site:
The elongation cycle begins when an aminoacyl-tRNA is charged with its corresponding amino acid, binds to the A site of the ribosome. This binding is facilitated by the elongation factor EF-Tu (in E. coli) and GTP is hydrolyzed during this step, and EF-Tu is released.
- Peptide Bond Formation:
– Once the aminoacyl-tRNA is in the A site, the ribosome catalyzes the formation of a peptide bond between the amino acid in the A site and the growing polypeptide chain in the P site.
- Translocation:
– After the peptide bond is formed, the ribosome moves (or translocates) along the mRNA by one codon. This step is facilitated by the elongation factor EF-G and the hydrolysis of GTP. The ribosome shifts so that the tRNA in the A site moves to the P site, and the tRNA in the P site moves to the E site, where it will exit the ribosome.
- Release of tRNA:
– The empty tRNA in the E site is released from the ribosome after translocation. This step is also driven by GTP hydrolysis, and the elongation cycle repeats until a stop codon is there.
Significance: Aminoacyl-tRNA synthetase catalyzes the attachment of a specific amino acid to its corresponding tRNA molecule.
b) What type of products would you expect with the following partial heterozygotes of E.coli for lac operon in presence or in absence of lactose and why? The partial heterozygotes are
- i) i+ O- Z+ Y+ / i- O+ Z- Y-
- ii) i+ Oc Z+ Y- / i- O+ Z- Y+
Ans→ (i) i+ O- Z+ Y+ / i+ O+ Z- Y-
In the absence of lactose:
- The i⁺ allele produces a functional repressor that binds to O⁺ on the functional operon in the second strand, keeping it inactive. However, it cannot bind O⁻ on the first strand, allowing the genes (Z⁺ Y⁺) to be constitutively expressed.
- Product: β-galactosidase and permease will be produced.
In the presence of lactose:
- Lactose binds to the repressor, inactivating it. This allows the first strand (O⁻ Z⁺ Y⁺) to continue expressing Z⁺ Y⁺, while the second strand remains non-functional due to Z⁻ Y⁻.
- Product: β-galactosidase and permease will be produced.
ii) i+ Oc Z+ Y- / i- O+ Z- Y+
In the absence of lactose
- The constitutive operator (Oc) allows the expression of Z⁺on the first strand regardless of the repressor’s activity. The second strand (Z⁻ Y⁺) remains inactive because of the i⁻ O⁺ configuration.
- Product: β-galactosidase will be produced, but permease will not.
In the presence of lactose:
- Lactose inactivates the repressor, but the first strand (Oc Z⁺ Y⁻) continues expressing Z⁺ constitutively. The second strand (O⁺ Z⁻ Y⁺) becomes active, producing permease (Y⁺).
- Product: Both β-galactosidase and permease will be produced.
(c) What is PCR? Briefly describe the procedure of PCR. State the significance of PCR. (1+3+1)
Ans→ PCR (Polymerase Chain Reaction) is a technique to amplify specific DNA sequences, creating millions of copies in a short time.
The Polymerase Chain Reaction (PCR) procedure involves three main steps, repeated in cycles:
- Denaturation (90-95°C): DNA is heated to separate the double strands into single strands.
- Annealing (50-65°C): Short DNA primers bind to complementary sequences on the single-stranded DNA.
- Extension (72°C): Taq DNA polymerase synthesizes new DNA strands by adding nucleotides to the primers.
Significance of PCR:
- Disease Diagnosis: Detects genetic mutations, pathogens, and diseases like COVID-19, HIV, and cancer.
- Forensic Applications: Amplifies DNA from crime scenes for identification.
- Genetic Research: Facilitates cloning, gene sequencing, and mutation analysis.
Molecular Biology 2020
a) Describe the salient features of DNA according to ‘Watson and Crick’ model, with suitable diagram. (3+2=5)
Ans→The salient features of DNA according to the Watson and Crick model are:
- Double Helix Structure: DNA consists of two polynucleotide strands coiled around each other in a right-handed helix.
- Complementary Base Pairing: Adenine (A) pairs with Thymine (T) via two hydrogen bonds. Guanine (G) pairs with Cytosine (C) via three hydrogen bonds.
- Antiparallel Strands: The two strands run in opposite directions, one in the 5′ to 3′ direction and the other in the 3′ to 5′ direction.
- Sugar-Phosphate Backbone: Each strand has a backbone made of alternating deoxyribose sugar and phosphate groups.
- Specific Base Sequence: The sequence of bases on one strand determines the sequence on the other strand.
- 10 Base Pairs Per Turn: The helix completes one turn approximately every 10 base pairs (34 Å or 3.4 nm).
- Diameter: The diameter of the helix is uniform at about 2 nanometers (20 Å).
- Stability: The double helix is stabilized by hydrogen bonds between base pairs and hydrophobic interactions between stacked bases.
b) Define transcription. Describe the ‘RNA chain elongation process’ in E.coli, with suitable diagram. (1+3+1)
Ans→Transcription is the process in which DNA is copied into messenger RNA (mRNA) by the enzyme RNA polymerase.
1) Initiation Transition: After RNA polymerase binds to the promoter and synthesizes a short RNA, it transitions from initiation to elongation.
2) RNA Polymerase Movement: The RNA polymerase unwinds the DNA double helix as it moves along the template strand in the 5′ to 3′ direction.
3) Nucleotide Addition: Ribonucleotides complementary to the DNA template are added to the growing RNA chain at the 3′ end.
4) RNA-DNA Hybrid Formation: A transient RNA-DNA hybrid is formed during elongation, stabilized within the active site of RNA polymerase.
5) Topological Adjustments: DNA gyrase and topoisomerase relax supercoils generated ahead of and behind the polymerase.
6) Proofreading: RNA polymerase exhibits some proofreading capability to correct mismatches or remove misincorporated nucleotides.
c) Make a list of enzymes/factors and their function that are involved in the ‘DNA replication’ process in prokaryotes. Define ‘OKAZAKI’ fragment. (3+2=5)
Ans→
- DNA Helicase: Unwinds the double-stranded DNA by breaking hydrogen bonds between complementary bases.
2. Single-Strand Binding Proteins (SSBs): Stabilize the unwound single-stranded DNA and prevent it from reannealing.
3. DNA Gyrase (Topoisomerase II): Relieves supercoiling ahead of the replication fork by cutting and rejoining DNA strands.
4. Primase: Synthesizes short RNA primers to provide a starting point for DNA polymerase.
5. DNA Polymerase III: The main enzyme that synthesizes new DNA strands by adding nucleotides complementary to the template strand.
6. DNA Polymerase I: Removes RNA primers and replaces them with DNA.
7. DNA Ligase: Joins Okazaki fragments on the lagging strand by forming phosphodiester bonds.
Okazaki fragments are short DNA segments synthesized on the lagging strand during DNA replication. They are later joined by DNA ligase to form a continuous strand.
a) Give an account of double strand break model of DNA recombination. (5)
b) Write short notes on the following:
- i) Wobble hypothesis
- ii) Exon shuffling
Ans→The Wobble Hypothesis, proposed by Francis Crick in 1966, explains how a single tRNA can recognize multiple codons during protein synthesis. According to this hypothesis:
- Base Pairing Flexibility: The third base of a codon (at the 3′ end of mRNA) and the first base of the anticodon (at the 5′ end of tRNA) exhibit non-standard pairing, allowing “wobble.”
- Efficiency: This flexibility reduces the number of tRNA molecules required for translation, as fewer tRNAs can pair with multiple codons.
- Base Pairing Rules: In the wobble position:
– Guanine (G) can pair with Uracil (U). Inosine (I) in the tRNA can pair with Adenine (A), Cytosine (C), or Uracil (U).
Exon shuffling is a molecular mechanism through which new genes are formed by rearranging or recombining exons from different genes. This process creates novel proteins with unique functions.
Mechanism: Exons are transferred between genes through mechanisms like recombination, transposons, or retrotransposition.
Significance: It drives evolutionary innovation by generating genetic diversity.
Example: Domains of immunoglobulins in antibodies are a result of exon shuffling.
c) The table below gives the genotypes (lac operon) of several partial diploid E.coli strain. Fill the phenotypes using ‘+’ for B-galactosidase synthesis and ‘-’ for no B-galactosidase synthesis. Glucose is absent in all cases.
Genotype
- i) I⁺ P⁻ O⁺ Z⁺ / I⁺ P⁺ O⁺ Z⁺
- ii) I⁺ Oc Z⁻ / I⁺ O⁺ Z⁺
- iii) I⁺ O’ Z⁺ / I⁺ O⁺ Z⁺
- iv) Is Oc Z⁺ / I⁺ O⁺ Z⁺
- v) Is O⁺ Z⁻ / i⁺ O⁺ Z⁻
Ans – Here is the table filled with phenotypes based on the given genotypes and conditions. The lac operon phenotypes are determined by the functionality of the regulatory elements and structural genes.
| Genotype | Non-Inducer (Lactose absent) | Inducer (Lactose present) | Explanation |
| i) I⁺ P⁻ O⁺ Z⁺ / I⁺ P⁺ O⁺ Z⁺ | _ | + | In the first copy, the promoter (P⁻) is non-functional. The second copy has a functional promoter, so lacZ is inducible. |
| ii) I⁺ Oc Z⁻ / I⁺ O⁺ Z⁺ | + | + | The Oc mutation leads to constitutive expression of the Z⁻ copy (non-functional). The second copy (O⁺ Z⁺) is inducible. |
| iii) I⁺ O’ Z⁺ / I⁺ O⁺ Z⁺ | _ | + | Both copies are inducible and functional. They require lactose for β-galactosidase expression. |
| iv) Is Oc Z⁺ / I⁺ O⁺ Z⁺ | + | + | The Oc mutation leads to constitutive expression of the Z⁺ gene. The Is mutation prevents induction of the second copy. |
| v) Is O⁺ Z⁻ / i⁺ O⁺ Z⁻ | _ | _ | Both copies have non-functional lacZ. The Is mutation prevents induction, but this does not matter due to Z⁻. |
Molecular Biology 2022
3. Answer any two of the following questions:
a) Briefly describe any DNA sequencing method that you know. What is the application of northern blotting? (3+2)
Ans→ Northern Blotting
① Preparation of RNA from Sample: All the target mRNA’s molecules should be extracted from sample.
② Separated by Gel electrophoresis: In gel electrophoresis, the mixture of mRNA molecules are separated according to their size, molecular weight and charge using an electric field.
③ Blotting: In next step the mRNA from the gel electrophoresis, gel will be transferred onto a nylon membrane by the process of blotting.
④ Hybridization: The RNA bound to membrane is treated with labelled DNA probe for hybridization.
⑤ Washing: The blot membrane is washed to remove unwanted probe and RNA.
⑥ Visualization by Autoradiogram: Visualize the labelled DNA RNA hybrid under audiogram which give patterns of band.
Application of Northern Blotting:
- 1) Detecting a specific mRNA in a sample.
- 2) Studying gene expression.
- 3) Construct the cDNA library.
- 4) Detection or diagnosis of a disease.
b) What is nucleotide excision repair? Describe the nucleotide excision repair mechanism in E.coli. (1+4=5)
Ans→ Nucleotide excision repair (NER) is a DNA repair mechanism that removes bulky DNA lesions, such as those caused by UV radiation or chemical damage.
The NER mechanism removes the thymine dimer (T=T) and other damaged bases. In E. coli NER mechanism involves several key steps.
1) The damaged DNA is recognized by a trimeric protein containing Uvr-A dimer and Uvr-B protein.
2) Uvr-AB Complex (trimeric protein) uses energy from ATP to bend the damaged site.
3) Uvr-A dimer is released, and the Uvr-C protein bind to the Uvr-B DNA complex.
4) Uvr-C protein (endonuclease) cleaves the fourth / fifth phosphodiester bond from the damaged nucleotide on the 3′ side and eighth phosphodiester bond for 5’ side.
5) Uvr-D or DNA helicase breaks the H-bond on the damaged DNA complex part.
6) After that, DNA polymerase I synthesises the complementary DNA nucleotide sequence.
7) DNA ligase seals the remaining nick portion in the DNA molecule.
- c) i) Write a short note on micro RNA. (2)
- ii) Explain different steps of Northern Blot using a flow diagram only. (2)
- iii) What is exon shuffling? (1)
Ans→ (i) microRNAs (miRNAs) are small non-coding RNAs that regulate gene expression by binding to target mRNAs, causing their degradation or inhibiting translation. They are vital in development, cell function, and disease processes.
Different steps of Northern Blotting
RNA Extraction
↓
RNA Quantification & Quality Check
↓
RNA Denaturation
↓
Gel Electrophoresis
↓
Transfer to Membrane (Nitrocellulose/ Nylon)
↓
RNA Fixation (UV Crosslinking/ Heat)
↓
Hybridization with Labeled Probe
↓
Washing to Remove Unbound Probes
↓
Detection of Signal (Autoradiography/ Chemiluminescence)
↓
Analysis of Results
Exon shuffling is a process where exons are rearranged or combined through recombination events, creating new genes with novel functions.