Protein synthesis and processing MCQ Quiz in मराठी - Objective Question with Answer for Protein synthesis and processing - मोफत PDF डाउनलोड करा

Key PointsRelease factors in Translation:

• Protein synthesis ends when one of the three termination (stop) codons (UAG, UAA, and UGA) is reached.
• The UAG triplet is called the amber codon; UAA is the ochre codon; and UGA is sometimes called the opal codon.
• The A site is now entered not by a tRNA, but by a protein release factor.
• Eubacteria have three types of release factors. Release factors RF-1 and RF-2 recognize the stop codons.
• RF-1 recognizes the termination codons - 5'UAA3' and 5'UAG3'; and RF-2 recognizes 5'UAA3' and 5'UGA3.
• RF-1 and RF-2 after recognizing the termination codons activate the ribosome to hydrolyze the bond between tRNA and polypeptide.
• RF-1 and RF-2 recognize the stop codon and stimulate polypeptide release through a GGQ (Gly-Gly-Gin) motif which acts as a peptide anticodon.
• Release factor, RF-3 (related to the elongation factor EF-G) stimulates release of RF-1 and RF-2 from the ribosome after termination, in a reaction requiring energy from the hydrolysis of GTP.
• Eukaryotes have only two release factors: eRF-1, which recognizes all three termination codons and eRF-3.
• The role of eRF-3 is not clear.
• The release factors terminate translation, but they do not appear to be responsible for the dissociation of the ribosomal subunits.
• This is the function of an additional protein called ribosome recycling factor (RRF).
• This acts together with EF-G in a reaction that uses hydrolysis of GTP.
• RRF probably enters the P or A site and 'unlocks' the ribosome.
• Dissociation requires energy, which is released from GTP by EF-G, one of the elongation factors and also requires the initiation factor IF-3 to prevent the subunits from attaching together again.

Explanation:

• RF-1 and RF-2 recognize the stop codon and stimulate polypeptide release through a GGQ (Gly-Gly-Gin) motif which acts as a peptide anticodon.
• Reaction requiring energy from the hydrolysis of GTP.

Hence the correct answer is option 3

The correct answer is Option 3 i.e. Three.

Explanation-

• The six codons for arginine are CGU, CGC, CGA, CGG, AGA, and AGG.
• The genetic code is redundant, meaning that multiple codons can code for the same amino acid.
• In this case, all six codons for arginine share the same first two nucleotides (CG) and differ only in the third nucleotide. The third nucleotide, known as the wobble position, allows for some flexibility in base pairing.
• To decode the six arginine codons, only three tRNAs are needed, each recognizing a specific set of codons based on the wobble position.
• For example, a tRNA with the anticodon sequence 5'-CGU-3' can recognize codons CGU and CGC, while a tRNA with the anticodon sequence 5'-CGA-3' can recognize codons CGA and CGG. The third tRNA would recognize codons AGA and AGG.

Conclusion-In summary, the minimum number of tRNAs required to read the six codons for arginine is three, taking advantage of the redundancy in the genetic code and the wobble position.

Key Points

• Protein synthesis is the process of creating protein molecules.
• In biological systems, it involves amino acid synthesis, transcription, translation, and post-translational events.
• In amino acid synthesis, there is a set of biochemical processes that produce amino acids from carbon sources like glucose. 

Important Points 

• During protein synthesis, the formation of one peptide bond requires the participation of two amino acids and a molecule of ATP, as well as at least one molecule of each of the four nucleoside triphosphates (NTPs): ATP, GTP, CTP, and UTP.
• The process of protein synthesis begins with the formation of an initiation complex consisting of the small ribosomal subunit, mRNA, and initiator tRNA charged with methionine.
• The initiation complex then binds to the large ribosomal subunit, forming the functional ribosome.
• Next, the ribosome moves along the mRNA molecule, reading the genetic code and matching each codon with the appropriate tRNA molecule carrying the corresponding amino acid.
• When two amino acids are in close proximity, a peptide bond can form between them, releasing a molecule of water.
• This process is catalyzed by the ribosome, which also requires energy in the form of GTP hydrolysis.
• During this process, the nucleotide triphosphates (NTPs) are used as substrates for the formation of peptide bonds.
• The NTPs are hydrolyzed by the ribosome to release energy, which is used to drive the formation of the peptide bond between the two amino acids.

Concept:

• Elongation is the phase of the protein-synthesis pathway that is responsible for the growth of nascent polypeptide chains.
• The two-site model postulates that the ribosome has two sites for transfer RNA (tRNA) binding: one that binds aminoacyl-tRNAs preferentially and a second that is specific for peptidyl-tRNAs.
• Study of the effects of tRNA and elongation factor binding on the reactivity of ribosomal RNA bases has generated the findings on which the hybrid-sites model is based.
• The 16-S-rRNA-reactivity changes observed when tRNAs bind to ribosomes depend entirely on the interactions involving the anticodon stem/loop of tRNA; the 23-S-rRNA alterations seen reflect interactions with the CCA end of tRNA only. Elongation factor Tu (EF-Tu) is the most abundant protein in bacterial cells.
• There are about 10 copies per ribosome, about as much as there is aminoacyl-tRNA, and most of the EF-Tu in the cell is found complexed with aminoacyl-tRNA.
• In addition to catalyzing the delivery of aminoacyltRNAs to messenger RNA-programmed ribosomes, the complexation of aminoacyltRNA with EF-Tu protects it from deacylation.

Explanation:

Statement A. the translocation activity is independent of GTP hydrolysis

• ​Efficient translocation of mRNA-tRNAs, a process required to move a new codon into the A site, requires the binding of elongation factor G (EF-G) to the ribosome and subsequent guanosine triphosphate (GTP) hydrolysis.

• Thus this option is not true

Statement B. translocation activity is completely dependent on GTP and EF-G

• During protein synthesis, tRNAs and their associated codons on the mRNA are translocated from the A (aminoacyl) to the P (peptidyl) and the E (exit) sites of the ribosome.
• This process is induced by the binding of a universally conserved elongation factor G (EF-G) in bacteria and elongation factor 2 (EF-2) in eukaryotes.
• EF-G accelerates translocation by ~50,000 fold
• As it is possible to observe translocation in vitro in the absence of EF-G and GTP this statement is not true

Statement C. translocation activity is inherent in ribosomes; however the rate of translocation in vivo is enhanced significantly in presence of GTP and EF-G.​

• During protein synthesis, mRNA and tRNAs are moved through the ribosome by the dynamic process of translocation.
• Sequential movement of tRNAs from the A (aminoacyl) site to the P (peptidyl) site to the E (exit) site is coupled with movement of their associated codons in the mRNA.
• As it is possible to observe translocation in vitro in the absence of EF-G and GTP, the translocation process appears to be an inherent property of the ribosome that is enhanced by the presence of EF-G and GTP.
• The classical model suggests that EF-G bound to GTP drives the translocation movement and that the subsequent GTP hydrolysis results in the dissociation of EF-G.
• Thus this option is true

Statement D. the molecular mechanism of translocation invitro is completely different from that in vivo

• Consider  the explanation above this statement is not true

Hence the correct answer is option 4

Concept:

• Leishmania's kinetoplast DNA is made up of catenated minicircle molecules (20 to 50 per associate) and maxicircle molecules (20 to 104 per associate).
• It was demonstrated that the maxicircle cryptogenic guide RNA (gRNA) molecules are encoded by the minicircles.
• Minicircles have only ever been shown to have this one significant cellular function, but their redundancy is still a mystery.
• More than 60 minicircle classes are expected to be held in Leishmania.

Explanation:

• The mitochondrion of Leishmania tarentolae has a family of tiny RNA molecules that may be involved in RNA editing.
• These "guide" RNA (gRNA) molecules are encoded in the mitochondrial maxicircle DNA's intergenic regions and have sequences that precisely correspond to complementary forms of the mature mRNAs found in the edited areas.
• Additionally, several gRNAs' 5' regions can hybridize with nearby mRNAs barely 3' of the preedited region.
• A model is described in which a partially hybridized gRNA and preedited mRNA serve as the substrate for numerous cycles of cleavage, uridylate addition or deletion, and religation, ultimately leading to a fully hybridized gRNA and mature edited mRNA.

Therefore, the correct answer is option 1.

The correct answer is Option 2 i.e.Sigma factor

Key Points

• RNA polymerase is the enzyme responsible for transcription, which is the process of synthesizing RNA from a DNA template.
• However, RNA polymerase cannot initiate transcription on its own.
• It requires the help of other proteins called sigma factors to recognize and bind to the promoter sequence on DNA.
• Promoters are specific DNA sequences located upstream of the transcription start site that contain information for transcription initiation.
• Sigma factors recognize and bind to the promoter sequence, which results in the formation of a stable RNA polymerase-promoter complex.
• This complex then undergoes a series of conformational changes, leading to the opening of the DNA double helix and the initiation of RNA synthesis.
• Different sigma factors recognize different promoter sequences, which allows RNA polymerase to selectively initiate the transcription of specific genes in response to environmental cues.
• For example, the sigma factor σ-70 recognizes promoters of genes involved in housekeeping functions, while the sigma factor σ-54 recognizes promoters of genes involved in nitrogen metabolism.
• Other factors are also present, which make up the core enzyme, and then the sigma factor binds to the core enzyme and makes up the holoenzyme.
  • 2α: Assembly of the core enzyme
  • β: catalytic activity, catalyzes the synthesis of RNA by polymerization of rNTP
  • β': stabilizes the single strand
  • ω: assembly of holoenzyme
  • σ: recognition of promoter and binding to core enzyme

Explanation:

Option 1: DNA photolyase

• DNA photolyase is an enzyme that repairs DNA damage caused by exposure to ultraviolet (UV) light.
• It recognizes and reverses the damage caused by the formation of pyrimidine dimers in the DNA strand.
• However, DNA photolyase is not involved in transcription or ensuring stable binding of RNA polymerase at the promoter site.

Option 2: Sigma factor

• Sigma factors are protein subunits of RNA polymerase that recognize the promoter sequence on DNA and ensure stable binding of RNA polymerase to the promoter site.
• Sigma factors are essential for transcription initiation in bacteria and are involved in recognition of specific promoter sequences, melting of DNA strands, and recruitment of the RNA polymerase holoenzyme to the promoter site.
• Once the RNA polymerase holoenzyme is stably bound to the promoter site, transcription can begin.

Option 3: DNA glycosylase

• DNA glycosylase is an enzyme that repairs DNA damage caused by oxidative stress.
• It recognizes and removes damaged bases from the DNA strand.
• However, DNA glycosylase is not involved in transcription or ensuring stable binding of RNA polymerase at the promoter site.

Option 4: Rec A

• Rec A is a protein involved in DNA repair and recombination. It plays a role in repairing DNA damage caused by double-strand breaks and promoting genetic exchange between homologous DNA molecules.
• However, Rec A is not involved in transcription or ensuring stable binding of RNA polymerase at the promoter site.

​Therefore, the correct answer is option 2.

The correct answer is Nucleotide addition resulting in incorporation of a stem-loop structure in the mRNA upstream of the Shine-Dalgarno sequence.

Concept:

• Translation is the process where ribosomes synthesize proteins using mRNA as a template.
• Several factors can influence translation efficiency, including the structure of the mRNA, availability of initiation factors, and modifications to the translation machinery.
• The Shine-Dalgarno sequence is a ribosomal binding site in bacterial mRNA, located upstream of the start codon, which helps recruit the ribosome to initiate translation.
• Disruptions in mRNA structure, ribosomal binding sites, or initiation factors can lead to translation inhibition.

Explanation:

Nucleotide addition resulting in incorporation of a stem-loop structure in the mRNA upstream of the Shine-Dalgarno sequence does NOT lead to translation inhibition.

• The Shine-Dalgarno sequence remains intact and accessible for ribosome binding in this scenario.
• The stem-loop structure formed upstream of the Shine-Dalgarno sequence does not interfere with ribosomal binding or translation initiation.
• Translation will proceed normally as the ribosome can still recognize and bind to the Shine-Dalgarno sequence effectively.

Other Options:

• Option 2: Nucleotide addition resulting in the Shine-Dalgarno sequence being a part of a stem-loop structure in the mRNA.
  • This modification impairs ribosome binding as the Shine-Dalgarno sequence becomes inaccessible due to its incorporation into a stem-loop structure.
  • Translation initiation is hindered, leading to translation inhibition.
• Option 3: Expression of an eIF2 mutant that mimics its phosphorylated state.
  • eIF2 (eukaryotic initiation factor 2) plays a crucial role in translation initiation in eukaryotes by delivering the initiator tRNA to the ribosome.
  • When eIF2 is phosphorylated, it prevents the recycling of eIF2-GDP to eIF2-GTP, resulting in translation inhibition.
  • A mutant mimicking the phosphorylated state of eIF2 will inhibit translation initiation.
• Option 4: Mutation that leads to a decrease in the processivity of the capping enzyme that leaves numerous mRNAs devoid of a CAP structure.
  • In eukaryotes, the 5' CAP structure is essential for ribosome recognition and translation initiation.
  • A deficiency in capping enzymes results in uncapped mRNAs, which cannot be efficiently translated, leading to translation inhibition.

The correct answer is RNA polymerase III

Concept:

• In eukaryotic cells, different types of RNA are synthesized by different RNA polymerases.
• RNA polymerase III is specifically responsible for synthesizing certain small RNAs, including the 5S ribosomal RNA (rRNA).
• 5S rRNA is an essential component of the large subunit of the ribosome and plays a crucial role in the protein synthesis process.

Explanation:

• RNA polymerase I: This enzyme is primarily responsible for the synthesis of most ribosomal RNA (rRNA), except for 5S rRNA. It synthesizes the 18S, 5.8S, and 28S rRNAs which are components of the ribosome.
• RNA polymerase II: This enzyme is responsible for transcribing messenger RNA (mRNA), which carries the genetic information from DNA to the ribosome for protein synthesis. It also transcribes some small nuclear RNAs (snRNAs) involved in RNA splicing.
• RNA polymerase III: This enzyme synthesizes 5S rRNA, tRNAs (transfer RNAs), and other small RNAs. The 5S rRNA is a structural and functional component of the large ribosomal subunit.
• RNA polymerase σ (sigma) subunit: This is a part of the bacterial RNA polymerase, not eukaryotic. The sigma factor is essential for initiating transcription in bacteria, but it does not relate to the synthesis of 5S rRNA in eukaryotic cells.

The correct answer is: (A) and (D).

Explanation:

• Eukaryotic 80S Ribosome: The eukaryotic 80S ribosome consists of two subunits: the large 60S subunit and the small 40S subunit.
  • The 60S subunit contains:
    • 28S rRNA
    • 5.8S rRNA
    • 5S rRNA
    • 49 proteins (not 34 proteins)
  • The 40S subunit contains:
    • 18S rRNA
    • 33 proteins.

Options:

• (A). 28S rRNA, 5.8S rRNA: These are part of the 60S subunit.
• (B). 5S rRNA and 34 proteins: The 5S rRNA is part of the 60S subunit, but 34 proteins is incorrect. The correct number is 49 proteins.
• (C). 23S rRNA: This is incorrect for eukaryotic ribosomes; the 23S rRNA is found in prokaryotic ribosomes, not eukaryotic ones.
• (D). 5S rRNA and 49 proteins: 5S rRNA is indeed part of the 60S subunit, and there are 49 proteins associated with the 60S subunit.

Key Points

•  Prokaryotic ribosomes are slightly smaller than eukaryotic ribosomes, with a 70S designation (compared to the 80S ribosomes in eukaryotes). The prokaryotic ribosome consists of two subunits:
  • Large Subunit (50S)
  • Small Subunit (30S)

• 50S Subunit:

  • 23S rRNA: The primary rRNA molecule in the large subunit.
  • 5S rRNA: Smaller rRNA molecule in the large subunit.
  • 34 proteins: These are the proteins associated with the 50S subunit.

• 30S Subunit:

  • 16S rRNA: The rRNA molecule in the small subunit.
  • 21 proteins: These are the proteins associated with the 30S subunit.

• The 16S rRNA (30S subunit) and 23S rRNA (50S subunit) are critical for ribosome function, particularly in the binding of tRNA and mRNA during protein synthesis.
• The 5S rRNA is also present in the large subunit, playing a role in ribosome structure and function.

The correct answer is Shine-Dalgarno sequences

Explanation:

• Prokaryotic mRNAs contain a conserved polypurine tract, consensus AGGAG, located about seven nucleotides upstream from the AUG initiation codon.
• This conserved hexamer is called the Shine-Dalgarno sequence.
• The Shine-Dalgarno sequence plays a critical role in the initiation of protein synthesis in prokaryotes by aligning the mRNA on the ribosome for proper translation initiation.
• This sequence helps the ribosome to recognize the start codon (AUG) on the mRNA, ensuring that translation begins at the correct location.

Other Options:

• Kozak's sequences: These are conserved sequences found in the mRNA of eukaryotes, not prokaryotes.
• Consensus sequences: This is a general term for sequences that are conserved across different molecules or species, but it does not specifically refer to the Shine-Dalgarno sequence in prokaryotic mRNA.
• TATA box sequences: These are DNA sequences found in the promoter region of genes in eukaryotes, not prokaryotes. They play a role in the initiation of transcription, not translation.