Question

The amino acid sequences of peptide fragments obtained from a normal protein were compared with those obtained from the same protein synthesized by a defective gene. They were found to differ in only one peptide fragment. Their amino acid sequences are shown here:

Normal: Gln-Tyr-Gly-Thr-Arg-Tyr-Val

Mutant: Gln-Ser-Glu-Pro-Gly-Thr

a. What is the defect in DNA?

b. It was later determined that the normal peptide fragment is an octapeptide with a C-terminal Val-Leu. What is the C-terminal amino acid of the mutant peptide?

Short answer

1. The middle nucleotide (A) in the second triplet was deleted.
2. -Terminal two-letter code U (UCAG).

Step-by-step solution

Bases sequence in the normal and mutant peptides

The normal and mutant peptides would have the following base sequence in their mRNA;

Normal: CA (AG) UA (UC) GG (UCAG) AC (UCAG) CG (UCAG) UA (UC) GU (UCAG)

Mutant: CA (AG) UC (UCAG) GA (AG) CC (UCGA) GG (UCGA) AC(UCAG)

Defect in DNA (a)

(a)

The middle nucleotide (A) in the second triplet was deleted. This means that an A was deleted in the sense strand of DNA or a T was deleted in the template strand of DNA.

C-terminal in mutant peptide

(b)

The mRNA for the mutant peptide has an unused -terminal two-letter code. U (UCAG). The last amino acid in the octapeptide of the normal fragment is leucine, so its last triplet is UU (AG) or CU (UCAG). This means that the triplet for the last amino acid in the mutant is U (UCAG)(UC) and that the last amino acid in the mutant is one of the following: Phen, Ser, Tyr, or Cys.