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Correct choice is (c) Shine and DALGARNO sequence

To explain: Eukaryotes are not polycistronic and the 5’ cap corresponds to the 5’ end of the mRNA near which it MUST START translating. Although there is not much defined sequence but all the same there is a consensus.

Correct choice is (B) By binding to A site

Easy explanation: IF1 binds to the A site of the 30S SUBUNIT so that the 1^st aminoacyl-tRNA can’t sit at the A site. It then sits on the P site i.e. correct position.

Right choice is (d) Colicin E3

To EXPLAIN I would say: Colicin e# effects initiation at the Shine and DALGARNO SEQUENCE, however, it has no effect on elongation of termination. Thus, it will prevent NEW initiation but will not halt already translating ribosomes.

The CORRECT option is (a) Before initiation

The explanation: IF2 guides the 1^st tRNA to the P site of the ribosome. Then it is GTP bound. When PROPERLY PLACED the GTP hydrolyses to GDP and the IF2 is released.

The CORRECT CHOICE is (B) RNA

The explanation: RNA was the 1^st to arise which served both as information storage and as enzyme NAMELY ribozyme. Later it transferred the activities to more specific molecules namely DNA and protein respectively.

The CORRECT choice is (d) 4

To explain I WOULD say: You will OBTAIN 4 BANDS corresponding to heavy 50 and 30S SUBUNIT, heavy 50S and light 30S subunit, heavy 30S and light 50S subunit and light 50S and 30S subunit respectively.

Right OPTION is (B) 50S ribosomal RRNA

The best explanation: The peptidyl transferase activity of the ribosome is located in the 50S subunit of the ribosome in the rRNA and not in the protein. This has been supported with sound proof.

The CORRECT choice is (a) P

To ELABORATE: fMettRNA-fMet BINDS to the P site of the ribosome in TRANSLATIONAL INITIATION and then the site is blocked by initiation factor.

The correct answer is (a) Glycine

Best explanation: The N terminal amino acid can be REMOVED only if the N+1 amino acid is small, i.e. Gly, ALA, SER, THR, Pro, Val or Cys. Here all other than Glycine are larger amino acid. So, only glycine in the N+1 position will SUFFICE the result.

Correct ANSWER is (B) The subunits will dissociate

Explanation: Low Mg^2+ concentration FACILITATES dissociation of the subunits of RNA polymerases. On the contrary increasing Mg^2+ concentration will promote association in VITRO.

Correct option is (c) 2‘-OH of the A base

For EXPLANATION I would say: Class II aminoacyl tRNA SYNTHETASE transfers the AMINO acid to the 2’-OH of the tRNA. There is a quick equilibrium between the 2’ and the 3’-OH. On the other hand class I amino acyl tRNA synthetase transfers it to the 3’-OH.

The correct ANSWER is (d) Variable ARM

The best explanation: The variable arm, as its name suggest, differs for different TRA. Some amino acyltRNA SYNTHETASE can use them as recognition feature.

Correct OPTION is (a) True

The best I can explain: The TRNA against all the degenerate codons for an amino acid are not present in the same concentration in the cell. Thus, it can sometimes stall the POLYMERASE as in trp operon. Even in eukaryotes this stalling of polymerase could LEAD to domain formation.

The correct choice is (c) Transition of one base

For explanation I would say: In all possibilities transition of a base would have least harmful effect as there are DEGENERATE CODONS for same amino acid. The NEW codon COULD thus be a silent mutation. Also if it is the 3rd base due to wooble HYPOTHESIS it will have negligible effect.

The correct CHOICE is (d) Two codons can code for same amino acid

The best explanation: Degeneracy means that there can be two or more codons that can code for same amino acid with there being TRIPLET CODON and 4 BASES there are 64 possible combinations and only 20 amino acids. This was then observed that different codon COULD be degenerate i.e. code for same amino acid.

Right option is (b) 50S subunit has peptidy transferase activity

Easiest explanation: In ROBERT Traut and MUNRO EXPERIMENT puromycin was used as an assay of peptidyl transfer activity. They observed that rRNA was ABLE to transfer the amino acid chain from tRNA on P SITE to the aminoacyl tRNA on A site.

The correct CHOICE is (d) It stands upto PROTEINASE K treatment

The explanation is: Ribosomes treated by SDS and proteinase K shows a remarkable ability to still carryout peptidyl transfer activity. HOWEVER, extraction with PHENOL destroys this activity. This may be because it DISTURBS some higher order structure.

Correct choice is (c) Co^2+

For explanation: The Methionyl aminopeptidase which removes the terminal METHIONINE residue in the peptide chain uses Co^2+ as a cofactor. While on the other hand many polymerase USE Mg^2+and DNA binding DOMAIN uses Zn^2+.

Correct option is (c) 16S rRNA ~ 10 nucleotide upstream to START site

For explanation: The 16S rRNA forms the smaller SUBUNIT which recognizes the mRNA via its Shine and DALGARNO SEQUENCE. This is a purine rich sequence about 10 amino ACIDS upstream to translation start site.

Right choice is (c) Wrong amino acid incorporation

Best EXPLANATION: STREPTOMYCIN interacts with the smaller subunit of the ribosome and prevents the initiation at HIGHER concentration. At lower concentration, it leads to the incorporation of more ERRORS.

The correct option is (c) TRNA BINDING site

For EXPLANATION: Puromycinis a tRNA analogue that has an amide bond in place of ester bond. This is why when the RIBOSOME transfers the PEPTIDE over this; it flees with the peptide and stops the translation.

The correct answer is (a) True

Explanation: PDF co-Translationally removes the formyl group from the formyl-METHIONINE amino acid. In absence of this, the N terminal methionine would pose a problem in the structure and function of the PROTEIN.

Right choice is (d) OxoG base

Easiest explanation: UNLIKE most RNA TRNA has some special bases like pseudo URIDINE, dihydroxy uridine etc. But oxoG is a base PRODUCED due to oxidative damage, it is not normally found in RNA.

Right CHOICE is (C) 50 angestron

Easy explanation: There is a gap of 50 angestron between the CATALYTIC site of the ribosome and the nearest protein to that point. This ensures that the catalytic activity is due to the RRNA and not any protein as proteins can’t influence a reaction 50 angestron away from it.

The correct ANSWER is (c) Tetracyclin

Explanation: While CHLORAMPHENICOL also BLOCKS the peptidyl transferase ACTIVITY, tetracycline does so by blocking the A site. Then the A site amino acid can’t interact with the P site one for the REACTION to take place.

Right option is (a) Use of low Mg^2+ concentration

The BEST I can explain: The condition of 33% methanol is not usual for a celluar interior. Thus Robert Traut and MUNRO’s using 33% methanol (or ethanol) to carry out peptidyl transfer ACTIVITY was criticized.

Right choice is (d) C/T rich

To explain: As the Shina and Dalgarno region is PURINE rich i.e. A/G rich, its CORRESPONDING DNA sequence should be PYRIMIDINE rich or C/T rich. This is because a purine base PAIRS with a pyrimine so a Pyrimidine will code for a purine in RNA.

Right answer is (b) False

The EXPLANATION: The ribosome is a MULTI protein complex while the MRNA is much smaller. In reality, the mRNA slides PAST the ribosome in place of the ribosome sliding on it.

Right OPTION is (a) Fusidic acid

To ELABORATE: Fusidic acid inhibits elongation in prokaryotes in a way that it BINDS to Ef-G-GDP permanently and doesn’t allow the ribosome to FALL off.

The CORRECT answer is (b) S34

Explanation: While naming the RIBOSOMAL proteins L and S stand for LARGER and smaller subunits respectively. The larger subunit has 34 proteins while the smaller ONE has 21, so there could not be S34 protein in the ribosome.

The correct option is (d) 18S

To ELABORATE: The 18S rRNA of a smaller SUBUNIT for eukaryotes and the 16S rRNA of the smaller subunit for prokaryotes recognizes the MRNA initiation site.

Correct OPTION is (c) CCA

Easiest explanation: The CCA SEQUENCE of the AMINOACYL arm of tRNA is responsible for binding the AMINO acid. While AUG is a start codon and T ψ C is another arm of the tRNA.

Correct CHOICE is (b) Phe

For explanation I WOULD say: While poly A oligo-rNTP would incorporate lysine, due to repeated AAA CODON, oligo-dNTP actually makes poly U oligo-rNTP, having codon UUU that codes for PHENYL alanine.

Right option is (B) Alanine

Easiest explanation: The ANTICODON loop plays no part in recognizing the amino acid. It can simply recognize the CODON. So, it will call in the Ala CHARGED lysine-tRNA to alysine AAA codon and add the alanine to the peptide.

Right option is (B) A peptide chain to an amino acid

Easiest explanation: Ribosomal peptidyl transferase TRANSFERS a peptide chain over the TRNA in P site of the ribosome to the tRNA’s amino acid in the A site of the same ribosome.

The correct CHOICE is (d) Gln

To elaborate: The 3’-OH of the aminoacyl arm of the tRNA enters the BINDING pocket of the synthetase to interact with a Gln AMINO acid correctly placed. This is USED to distinguish it from ASP and Asn.

Right CHOICE is (b) Asp

The best I can EXPLAIN: Asp is charged by class II amino acyl tRNA synthetase. It recognized the structural FEATURES and not the anticodon LOOP, there is also no bending of the 3’-OH end of the amino acyl arm.

The correct option is (c) L SHAPED

For EXPLANATION: The tertiary structure of the tRNA is L shaped while the secondary structure is cloverleaf shaped. The tRNA has three stem loop structures.

Right choice is (d) 6

To explain I would say: While the smaller 30S SUBUNIT of the ribosome is folded into 4 major DOMAINS the LARGER 50S subunit had 6 major domains with its 23S and 5S rRNA.

Right OPTION is (c) 256

For explanation: There are 4 DNA bases and TRIPLET coodon so number of possible COMBINATIONS id 4^3=64. Now if the codon was QUADRUPLET the combinations would be 4^4=256.