Question

. A double-stranded DNA molecule with the sequence shown here produces, in vivo, a polypeptide that is five amino acids long. TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA ATG TAC TAG TAA AGT GCC TTA AAG ATC GTA CAT a. Which strand of DNA is transcribed and in which direction? Support your response by indicating the 5 amino acid sequence that results when this transcript is translated. (2 pts) b. Label the 5’ and the 3’ ends of each strand in the original DNA molecule. (1 pt) c. If an inversion occurs between the second and third triplets, from the left and right ends, respectively, and the same strand of DNA is transcribed, how long will the resultant polypeptide be and what is its sequence? (Hint: rewrite the DNA molecule AFTER the inversion.) (3 pts) d. Assume that the original molecule is intact and that the bottom strand is transcribed from left to right. Give the base sequence, and label the 5’ and 3’ ends of the anticodon that inserts the fourth amino acid into the nascent polypeptide. What is this amino acid?

Answer 1

Aand B. 3’ TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA 5’

              5’ ATG TAC TAG TAA AGT GCC TTA AAG ATC GTA CAT 3’

Lower strand is transcribed from left to right direction..

Lower strand Mrna = AUG UAC GAU CUU UAA

N- Met – Tyr-Asp-Leu – Stop codon

C. If second and third triplets inversion

3’ TAC ATG ATC ATT TCA CGG AAT TTC TAC GAT GTA 5’

5’ ATG TAC TAG TAA AGT GCC TTA AAG ATG CTA CAT 3’

MRNA = 5’- AUG UAG – 3’

N- MET – STOP CODON

D. Bottom strand is transcribed from left to right.

mRNA =              5’ AUG AUC AUU UCA CGG AAU UUC UAG 3’

ANTICODON      3’                                                                    5’

MET – ILE – Ile – ser - Arg