Question

Lecture Homework Assignment BIO 2010 Microbiology Print Name: hanan Bonelt Section LOOS Sequences of single strands of DNA are shown below for two different pieces of ONA Assume that the DNA sequences are not at the beginning of a gene and that the spaces between each DNA triplet represent the correct reading frame. Label all sequences and 3', where appropriate. Keep everything aligned properly. A. Synthesize the complementary DNA sequence. B. Transcribe the mRNA sequence starting with the 5 end on the left-hand side C. Write the complementary tRNA sequence for each reading frame. D. Use the genetic code table from our lecture notes or from our textbook (Bauman, 4 ed. p. 208; Bauman, 3"ed, p. 209; Bauman, 2nd ed., p. 211) to translate the amino acid sequence of the encoded polypeptide, using the three-letter code for each amino acid Note: Use only the 20 amino acids that are most commonly found in all living organisms 3 1. AAC CTC AAA TAC GTT CAG GGT GCG GTA CAC ATT CCG TTA TCA ATC TAG TAG GGA AAA CCC GAC CIT

Answer 1

Ans 1. A) the complementary DNA sequence will be:-

5'- TTG GAG TTT ATG CAA GTC CCA CGC CAT GTG TAA-3'

B) the mRNA sequence of the above strand will be :-

3'- AAC CUC AAA UAC GUU CAG GGU GCG GUA CAC AUU-5'

C) for the above codons on mRNA tRNA will have anticodons on it which will be complementary to the above strand:-

UUG GAG UUU AUG CAA GUC CCA CGC CAU GUG UAA

D) the amino acid sequence will be according to the codons present on the mRNA:-

Asn-Leu-Lys-Tyr-Val-Gln-Gly-Ala-Val-His-Ile

Ans 2. A)the complementary DNA sequence will be:-

3'- GGC AAT AGT TAG ATC ATC CCT TTT GGG CTG GAA-5'

B) the mRNA sequence of the strand starting with 5' end on the left hand side(in the question) :-

3'-GGC AAU AGU UAG AUC AUC CCU UUU GGG CUG GAA-5'

C)for the above codons on mRNA tRNA will have anticodons on it which will be complementary to the above strand:-

CCG UUA UCA AUC UAG UAG GGA AAA CCC GAC CUU

D)the amino acid sequence will be according to the codons present on the mRNA, translation occurs from the 5' end of the mRNA to the 3' end of the mRNA therefore for convenience let us rotate the mRNA strand by 180 degree, we get:-

5'-GAA CUG GGG UUU CCU AUC AUC UAG AGU AAU GGC-3'

the amino acid sequence will be-

Glu-Leu-Gly-Phe-Pro-Ile-Ile-stop

due to presence of stop codon on mRNA that is UAG further translation would not take place.