Protein Synthesis MCQ Quiz in मराठी - Objective Question with Answer for Protein Synthesis - मोफत PDF डाउनलोड करा

The correct answer is Both (2) and (3)

Concept:

• Gene expression is the process by which information from a gene is used to synthesize functional gene products such as proteins. This process is tightly regulated to ensure that the correct proteins are produced in the right amounts and at the right times.
• The regulation of gene expression can occur at multiple levels within the cell, including DNA replication, transcription, and translation.
• Transcription is the process where the DNA sequence of a gene is copied into RNA. Translation is the process where the RNA is used as a template to build proteins.

Explanation:

• DNA Replication: This is the process of copying the entire DNA molecule before cell division. While it is crucial for cell division and inheritance, DNA replication itself is not a primary control point for gene expression. Thus, this option is incorrect.
• Transcription: This is a critical control point for gene expression. During transcription, regulatory proteins and other factors can enhance or suppress the transcription of specific genes, thereby controlling which genes are expressed and to what extent. This is one part of the correct answer.
• Translation: This is another key control point for gene expression. During translation, the cell can regulate the efficiency and rate at which proteins are synthesized from mRNA. Factors such as the availability of ribosomes and tRNAs, as well as regulatory proteins that interact with the mRNA, can influence translation. This is the second part of the correct answer.
• Since both transcription and translation are crucial control points for gene expression.Therefore, the correct answer is Both 2 and 3

Concept:

• Transcription is the process of copying the genetic information from one strand of DNA onto the RNA.
• It takes place with the help of the enzyme DNA-dependent RNA polymerase.
• It takes place at a defined region of DNA that is known as the transcription unit.
• A transcriptional unit can be recognized by the following:
  • Template strand - It acts as the template for RNA pol to act on and has 3'→5' polarity.
  • Coding strand - It has 5'→3' polarity and is used as a reference strand for denoting upstream or downstream of a gene.
  • Structural gene - It contains the nucleotide sequence that actually codes for the required RNA.
  • Promoter - It is the nucleotide sequence that acts as a binding site for RNA pol enzyme.
  • Terminator - It is the nucleotide sequence that defines the end of transcription.
• The template strand is also referred to as the non-coding strand.

Explanation:

• The promoter is located upstream of the gene, while the terminator is located downstream of the gene.
• Upstream and downstream indicate towards the 5'-end and the 3'-end of coding strand respectively.
• The location of the promoter defines the coding and template strands.
• For instance, if the position of promoter is switched with that of the terminator, the coding strand becomes the template strand and vice versa.
• This ensures that the promoter is always upstream of the gene, which is towards the 5'-end of the coding strand.

Hence, the correct answer is option (3).

The correct answer is Only 2

Explanation-

RNA polymerase is an enzyme responsible for catalyzing the synthesis of RNA from a DNA template during transcription. Transcription involves three main stages: initiation, elongation, and termination.

Initiation: In this stage, RNA polymerase binds to the promoter region of the DNA, forming a transcription initiation complex. In prokaryotes, this process often involves the transient association of a sigma factor (σ) with the RNA polymerase core enzyme, which helps in recognizing and binding to the promoter sequence.

Elongation: Once the initiation complex is formed, RNA polymerase begins moving along the DNA template, synthesizing RNA in the 5' to 3' direction. During elongation, RNA polymerase does not require the transient association of initiation or termination factors. It simply catalyzes the addition of ribonucleotides to the growing RNA chain based on the complementary base pairing with the DNA template.

Termination: In the termination stage, RNA polymerase recognizes specific sequences in the DNA template that signal the end of the gene or transcription unit. In prokaryotes, termination often involves the transient association of a termination factor (ρ) with the RNA polymerase complex, leading to the release of the newly synthesized RNA molecule and dissociation of RNA polymerase from the DNA template.

Conclusion-RNA polymerase can perform elongation on its own without the need for additional factors such as the initiation factor (σ) or termination factor (ρ).

During elongation, RNA polymerase moves along the DNA template, catalyzing the addition of ribonucleotides to the growing RNA chain based on the complementary base pairing with the DNA template

Concept:

• Every cell in the body contains the same set of genes that on translation gives rise to specific proteins.
• However, not all enzymes (proteins) are required at all times.
• Hence there is a need to regulate the expression of genes.
• Gene regulation is the process by which the timing, location and amount of genes expressed are controlled.
• In prokaryotes, gene regulation takes place at the level of transcription.
• In eukaryotes, gene regulation can take place at,
  • Transcription level
  • Processing level
  • Transport of mRNA from the nucleus to the cytoplasm
  • Translational level
• Operon is a gene regulatory system that allows for controlling of the genes that need to be transcribed according to the needs of the cell.
• An operon comprises of 

1. Promotor - The site at which RNA polymerase binds for initiating transcription.
2. Operator - The site at which repressor or activator binds that affects transcription.
3. Structural gene - The gene that needs to be transcribed.

• Gratuitous inducer:
  • It is a molecule that is analog (structurally similar) to an inducer that initiates transcription.
  • In the absence of the actual inducer, a gratuitous inducer can initiate transcription.

​

Explanation:

• Option 1: It can bind with the promoter region of a gene but cannot facilitate gene expression - INCORRECT
  • ​A gratuitous inducer binds to a repressor and not the gene itself.
  • By binding to the repressor it causes conformational changes in the repressor.
  • As a result, the repressor can no longer bind to the operator region and thus transcription can take place.
• Option 2: Isopropyl thiogalactoside (IPTG) resembles lactose - CORRECT
  • ​IPTG is a chemical that mimics allolactose.
  • Allolactose is an isomer of lactose. It acts as an inducer in the lac operon.
  • In the presence of allolactose, the repressor does not bind to the operator site.
  • As a result of which transcription of genes that are required for the metabolism of lactose takes place.
  • In the absence of allolactose, IPTG acts as a gratuitous inducer of the lac operon. It enables the transcription of genes.
  • Thus, Isopropyl thiogalactoside (IPTG) resembles lactose.
• Option 3: It is after the binding of gratuitous inducer with the main inducer molecule a gene can be expressed - INCORRECT
  • ​An inducer on its own can enable transcription to take place. It does not require the binding of a gratuitous inducer.
  • A gratuitous inducer on other hand is employed only when the original inducer is not present to regulate the gene transcription.
• Option 4: It resembles the inducer molecule but has no role in gene expression - INCORRECT
  • ​A gratuitous inducer is an analog of the inducer that can initiate transcription in the absence of an inducer.
  • Thus a gratuitous inducer resembles an inducer molecule and it also plays a role in gene expression.

So the correct answer is option 2 (Isopropyl thiogalactoside (IPTG) resembles lactose).

Correct answer:2

Concept:

• The process of copying genetic information from one strand of DNA into RNA is termed transcription or mRNA synthesis.
• The mRNA formed after transcription is non-functional, containing both the coding region, exon, and the non-coding part, intron. This type of RNA is called heterogenous RNA or hnRNA.
• This hnRNA is unstable and cannot be directly translated into proteins. It has to undergo some post-transcriptional modifications to become stable for translation.
• The post-transcriptional modification includes Capping, Tailing, and Splicing.
• Capping: The process of addition of an unusual nucleotide, 7 methyl guanosine to the 5' end of the hnRNA is called capping. It forms a cap-like structure on the 5' end of hnRNA.
• Tailing: The process of addition of 200-300 nucleotides of adenylic acid to the 3' end of hnRNA is called tailing. This tail is called poly-A tail.
• Splicing: After capping and tailing now the hnRNA undergoes a process where the non-coding regions and introns are removed and the coding region exons are joined together to form the functional mRNA by the process called splicing.
• The removal of introns is done with the help of the ribonuclease enzyme and the exons are joined with the help of the enzyme DNA ligase.

Explanation:

Option:1-  Exons and Introns do not appear in the mature RNA- INCORRECT

• Introns do not appear, but exons are present.
• So, this option is incorrect

Option  2)  Exons appear but introns do not appear in the mature RNA-CORRECT

• This option is correct as exons are present in the mature RNA.
• So, this option is correct

Option  3)  Introns appear but exons do not appear in the mature RNA-INCORRECT

• Introns do not appear in the mature RNA
• So, this option is incorrect

Option  4)  Both exons and introns appear in the mature RNA-INCORRECT

• Both are not present, only the coding part is present.
• So, this option is incorrect.

​So, the correct option is Option 2: Exons appear but introns do not appear in the mature RNA.

The correct answer is option 2.

Solution

Concept:-

• A set of three-letter combinations of bases that code for a particular amino acid is called a codon.
• There are 64 codons out of which 61 code for amino acids and three are stop codons.

Explanation:

Let us look at the salient features of genetic code:

• The three stop codons (UAA, UAG, UGA) do not code for any amino acid. They are also referred to as terminator codons.
• Each codon, out of the 61 codons, codes for only one amino acid. Hence it is unambiguous and non-overlapping.
• The codons are nearly universal that is UUU codes for phenylalanine in humans as well as bacteria.
• One amino acid can be coded by more than one codon. Hence, code is degenerate.
• The codon in mRNA is read in a contiguous manner and there are no punctuations.
• AUG act as an initiator codon and codes for amino acid methionine.

So, the correct answer is option 2.

Concept:

• The translation is a process of the formation of protein or polypeptide from messenger RNA.
• This process takes place in the cytoplasm of the cell.
• There are three steps in which translation occurs- Initiation, elongation, and termination.
• The process of translation, where the genetic code carried by mRNA is translated into a sequence of amino acids to form a protein, begins with the binding of mRNA to the ribosome.

Explanation:

• Initiation begins with the assembly of ribosomes on mRNA.
• This is followed by activation or aminoacylation of tRNA.
• The binding of amino acids with their tRNA in presence of enzyme tRNA synthetase is known as aminoacylation of tRNA.
• This step requires ATP.
• In the elongation step, amino acids are joined to form a polypeptide or protein.
• Each tRNA brings amino acid for a specific codon on mRNA.
• In termination, the polypeptide is released as ribosomes reach to stop codon.​

​

​So, the correct answer is option 3.

Concept:

• Transcription is the process of copying the genetic information from one strand of DNA onto the RNA.
• Transcription involves 3 basic steps:
  • Initiation
  • Elongation
  • Termination
• In eukaryotes, the mRNA is formed is the form of heterogenous RNA (hnRNA).
• This hnRNA needs to go through some post-transcriptional modification processes to form the mature mRNA that can be translated into polypeptide.

Important Points

• Initiation Factor - 
  • In prokaryotic transcription, RNA polymerase enzyme binds at the promoter site of DNA with the help of sigma (σ) factor and initiates transcription.
  • The sigma factor is also known as the initiation factor.
• Introns - 
  • In eukaryotes, the primary transcribed RNA has both exons and introns and is known as the hn-RNA or heterogeneous nuclear RNA. 
  • Introns are the non-coding sequence of RNA that are removed before RNA is translated.
  • The exons are the part of RNA that translates to a protein.
  • Splicing is the process of removing introns and the joining together of exons in the primary RNA transcript.
  • This forms the mature RNA from the hn-RNA.
• Termination Factor - 
  • Transcription termination is carried out by the Rho factor (ρ) that binds to the terminator site.
  • When the Rho factor comes across the RNA polymerase enzyme, the RNA is unwound from the DNA.
  • The nascent RNA is released along with the enzyme.
  • The DNA double helix reforms and Rho protein dissociates, terminating the transcription process.
• Adenylate residue - 
  • Polyadenylation is the process of adding about 200-300 adenylate residues to the 3'-end of the newly formed RNA.
  • These residues form a poly-A tail at the 3'-end of the mRNA and thus, this process is also known as tailing.
  • In an eukaryotic cell, the RNA molecules have the risk of being degraded by cytoplasmic enzymes when the RNA is transported from nucleus to cytoplasm.
  • Polyadenylation helps in protecting the 3'-end of the coding sequence of the RNA from digestive enzymes.

Corrected Table:

Hence, the correct answer is option 4.

Key Points

• Lactose is used as an energy source in E. coli in absence of glucose.
• Lactose can be hydrolyzed into its monomeric subunits glucose and galactose with the help of β-galactosidase enzyme.
• This enzyme is produced in E. coli by a polycistronic gene that is regulated by the lac operon.

Explanation:

The lac operon consists of the following parts:

• Structural Gene - It consists of 3 genes:
  • z - codes for β-galactosidase
  • y - codes for permease
  • a - codes for transacetylase
• Operator -
  • It is present adjacent to the structural gene.
  • It is the site for the binding of repressor protein.
• Regulator -
  • It comprises of the i-gene, which codes for the repressor protein.
  • The repressor protein is synthesized all the time constitutively.
• Promoter -
  • It is the transcription initiation site where RNA polymerase binds.
• Inducer -
  • It is the molecule that determines whether the repressor will bind to the operator or not.
  • Hence, it regulates the operon.
  • Example - Lactose in lac operon.

Additional Information

Lac Operon Regulation - 

• In the absence of lactose, the repressor protein binds at the operator site and prevents RNA polymerase to transcribe the structural genes.
• But in the presence of lactose, transcription of the genes take place.
• Inducer binds to the repressor and inactivates it by forming the repressor inducer complex.
• This complex prevents the repressor from binding with the operator.
• Thus, the RNA polymerase slides along the DNA without any hindrance and thus, the genes get expressed.

Key Points

• DNA fingerprinting is a process that identifies differences in some specific regions of DNA sequence called repetitive DNA.
• In these sequences, a small stretch of DNA is repeated many times.
• Major peaks are formed by the bulk of repetitive DNA forms and other small peaks are called satellite DNA.
• The technique of DNA fingerprinting was initiated by Alec Jeffreys.
• He used satellite DNA as a probe that showed a very high degree of polymorphism called Variable Number of Tandem Repeats(VNTR).

Steps of DNA Fingerprinting:

1. Isolation of DNA
2. Digestion of DNA by restriction endonuclease
3. Separation of DNA fragments by gel electrophoresis
4. Transferring of separated DNA fragments to synthetic membranes like nylon and nitrocellulose - Southern Blotting technique
5. Hybridization by using radio-labelled VNTR probes.
6. Detection of hybridized DNA fragments by autoradiography.

So the correct option is option 2: C – A – E – B – D.

Additional Information

• It is used in forensic science to solve criminal issues.
• DNA fingerprinting has been widely used to solve paternity issues.