Question

1.When GTP hydrolysis is inhibited in the translation initiation, what would be the most possible outcome?

Select one:

A.
IF1 cannot associate with 30S ribosomal subunit.

B.
The binding affinity of IF2 with initiator tRNA would be stronger.

C.
Premature binding of 50S subunit to 30S subunit.

D.
IF3 cannot bind to 30S subunit because the binding requires energy.

E.
Poor binding occurs between the first aminoacyl-tRNA and mRNA.

2.

The sequence of a DNA template fragment encoding for protein is 5’ CAG CTA ATG 3’. What is the amino acid sequence of the peptide produced (from N-terminal to C-terminal)?

Select one:

A. fMet

B. His

C. Gln-Leu-Met

D. fMet-Leu-Gln

E. Val-Ile-Asp

3.

An antibiotic has been developed to target the 23S RNA of 50S subunit in E. coli. What would be the possible outcome(s)?

(1)    Binding of 50S to 30S would not occur.

(2)    30S subunit would not bind to mRNA at a correct position.

(3)    Polypeptide would not be synthesized.

Select one:

A. (1) only

B. (2) only

C. (3) only

D. (1) and (2) only

E. (1), (2) and (3)

4.

The coding sequence on an mRNA contains 137 codons including start and stop codons. How many nucleotides are required to produce this mRNA? How many residues are encoded by this mRNA?

Select one:

A. 405, 135

B. 408, 136

C. 408, 137

D. 411, 136

E. 411, 137

Answer 1

1.When GTP hydrolysis is inhibited in the translation initiation, what would be the most possible outcome?

Answer - C. Premature binding of 50S subunit to 30S subunit

IF - 3 and IF - 1 first bidn to 30 S subunit. A GTP is bound to IF-2. IF - 2 now places the initiator tRNA into P site. Now it binds to 50S subunit of ribosome. Binding of 50S subunit leads to hydrolysis of GTP . When it is not hydrolysed, premature binding of 50S and 30S subunit occurs.

IF-3 binding to 30S subunit doesn' require energy. IF-1 association with 30 S subunit doesn't require GTP.

Binding affinity of IF-2 to aminoacyl tRNA is not dictated by GTP binding.

2. The sequence of a DNA template fragment encoding for protein is 5’ CAG CTA ATG 3’. What is the amino acid sequence of the peptide produced (from N-terminal to C-terminal)?

D. fMet-Leu-Gln

The mRNA is read in a 3' to 5' direction and the proteins are accordingly produced in a N to C direction with the first amino acid of protein being the N termini. So according to this, the first amino acid coded by ATG is fMet. ATG is the initiation codon, The first amino acid of a protein is generally fMet which acts as a marker of translation initiation. The next amino acid will be Leucine and the last will be Glutamine.

3.

An antibiotic has been developed to target the 23S RNA of 50S subunit in E. coli. What would be the possible outcome(s)?

(1)    Binding of 50S to 30S would not occur.

(2)    30S subunit would not bind to mRNA at a correct position.

(3)    Polypeptide would not be synthesized.

Answer - C. (3) only

23S rRNA of 50S subunit is responsible for the peptidyl transferase activity which joins the amino acids through peptide bonds. Due to this protein synthesis would stop and polypeptide would not be synthesized.

Binding of 50S and 30S doesn't involve the 23S rRNA and thus it would not be affected.

30S subunit is unaffected by this antibiotic since 23S rRNA is not a constituent of 30S subunit.

4.

The coding sequence on an mRNA contains 137 codons including start and stop codons. How many nucleotides are required to produce this mRNA? How many residues are encoded by this mRNA?

Answer -D. 411, 136

The total number of nucleotides in a codon =3

Number of codons = 137

So, net nucleotide number = 137 x 3 = 411

We must know that the amino acid coded by start codon is included in the polypeptide but the amino acid coded by stop codon is not included.

Thus total number of amino acids will be one less than the given since stop codon is also included  = 137 - 1 = 136.