Question

5 pts Question 33 The following prokaryotic DNA (same as above) is transcribed into RNA and the MRNA transcript is translated into protein. 31 21 11 1 ATGGGTTACT ATGAGGAGTT GACACACAAG AGGAGGTAGC 71 61 51 41 AGTATGGGTA TAATCTAATG CGTAATTGAG GAGGTAGTTG 101 111 91 81 ACGTATGCAT AGATAACGTA CGGGGGGGAA ACCCCCCCTT 141 131 121 TTTTTTTTTC GAGCAATAAA AGGGTTACAG Useful sequences: -35: 5' TTGACAT 3', -10: 5' TATAAT 3 (Pribnow box), Shine Dalgarno sequence: 5' AGGAGGU 3 THE GENETIC CODE THE GENETIC CODE SECOND LETTER A G C U ou UGU Cys UGC) UCU UAU Туг UAC) Phe Uuc UCC Ser UCA Opal UAA (terminator) UGA terminator UGG Trp Ochre UUA Leu UUG UAG Amber (terminator) UCG CGU CAU CUU CCU CGC Arg CGA CAC ccc CUC Pro Leu C CAA Gin CAG CUA CCA CGG cCG CUG AGU Ser AAU Asn AAC ACU AUU AGC AUC leu A AUA ACC Thr ACA AGA Arg AGG AAA Lys AAG AUG Met (initiator) ACG) GUU GGU GGC Gly GGA GCU GAU Asp GAC GCC Ala GCA GUC Val GUA GAA Glu GAG GGG GUG (initiator) GCG THIRD (3') LETTER UCAC UCAG UCAG UCAG FIRST (5') LETTER What will the sequence be for the protein translated from this mRNA? N-Met-Asn-Ser-C N-Met-Gly-lle-lle-C N-Met-His-Arg-C N-Met-Glu-Ser-C 5 pts Question 34 Babies in a hospital were switched at birth. Identify each mother's baby Baby 2 Baby 3 Baby 1 Mother 2 Mother 1 Ladder 1 2 3 4 6 + Mother 1 and Baby 2, Mother 2 and Baby 1 Mother 1 and Baby 3, Mother 2 and Baby 2 Mother 1 and Baby 1, Mother 2 and Baby 3 Mother 1 and Baby 2, Mother 2 and Baby 3

Answer 1

33. B. N-met-gly-ile-ile-C

34. D. Mother 1 and baby 2 , mother 2 baby 3.

It is because the band match in lane 2 to lane 4 and lane 3 to lane 6.