You have determined the sequence of a cDNA clone to be: TATAAACTGGACAACCAGTTCGAG

Question

You have determined the sequence of a cDNA clone to be: TATAAACTGGACAACCAGTTCGAGCTGGTGTTCGTGGTCGGTTTTCAGAAGATCCTAACGCTGACGTACGTAGACAAGTTGATAGATGATGTGCATCGGCTGTTTCGAGACAAGTA Since the cDNA sequence is so short, you suspect it contains only a portion of the protein coding sequence. Using the table below (and the single letter codes), what is the partial protein sequence? Remember, there are 3 possible reading frames. How did you decide which reading frame to use?

(A) (B) The Genetic Code UUU Phe (F) UCU Ser (S) UAU Tyr (Y) UGU Cys (C) UUC UAC UGC UCC UUA Leu (L UCA UAA Stop UGA Stop UUG UCG UAG Stop UGG Trp (W) CUU Leu (L) CCU Pro (P) CAU His (H) CGU Arg (R) CUC CGC CAC CAA Gln (o) CGA CUA CCA CUG CCG CGG CAG AUU lle (l) ACU Thr (T) AAU Asn (N) AGU Ser (S) AUC II ACC AAC AGC AAA Lys (K) AGA Arg (R) AUA ACA AUG Met (M) ACG II AAG AGG GUU Val (V) GCU Ala (A) GAU Asp (D) GGU Gly (G) GUC GCC GGC GAC GUA GCA GAA Glu (E) GGA GCG GUG GAG tb 10-06

Explanation / Answer

The protein sequence always starts with starting codon (AUG) and ends with stop codon (UAA, UAG, UGA). So, the sequence which starts with AUG or ends with stop codons can considered as coding sequence.

The analysis is

DNA:
TATAAACTGGACAACCAGTTCGAGCTGGTGTTCGTGGTCGGTTTTCAGAAGATCCTAACGCTGACGTACGTAGACAAGTTGATAGATGATGTGCATCGGCTGTTTCGAGACAAGTA

mRNA:
AUAUUUGACCUGUUGGUCAAGCUCGACCACAAGCACCAGCCAAAAGUCUUCUAGGAUUGCGACUGCAUGCAUCUGUUCAACUAUCUACUACACGUAGCCGACAAAGCUCUGUUCAU

Frame 1:

1 AUA UUU GAC CUG UUG GUC AAG CUC GAC CAC AAG CAC CAG CCA AAA 45
1 Ile Phe Asp Leu Leu Val Lys Leu Asp His Lys His Gln Pro Lys 15

46 GUC UUC UAG GAU UGC GAC UGC AUG CAU CUG UUC AAC UAU CUA CUA 90
16 Val Phe End Asp Cys Asp Cys Met His Leu Phe Asn Tyr Leu Leu 30

91 CAC GUA GCC GAC AAA GCU CUG UUC AU- 117
31 His Val Ala Asp Lys Ala Leu Phe XXX 38

Frame 2:

2 UAU UUG ACC UGU UGG UCA AGC UCG ACC ACA AGC ACC AGC CAA AAG 46
1 Tyr Leu Thr Cys Trp Ser Ser Ser Thr Thr Ser Thr Ser Gln Lys 15

47 UCU UCU AGG AUU GCG ACU GCA UGC AUC UGU UCA ACU AUC UAC UAC 91
16 Ser Ser Arg Ile Ala Thr Ala Cys Ile Cys Ser Thr Ile Tyr Tyr 30

92 ACG UAG CCG ACA AAG CUC UGU UCA U-- 118
31 Thr End Pro Thr Lys Leu Cys Ser XXX 38

Frame 3:

3 AUU UGA CCU GUU GGU CAA GCU CGA CCA CAA GCA CCA GCC AAA AGU 47
1 Ile End Pro Val Gly Gln Ala Arg Pro Gln Ala Pro Ala Lys Ser 15

48 CUU CUA GGA UUG CGA CUG CAU GCA UCU GUU CAA CUA UCU ACU ACA 92
16 Leu Leu Gly Leu Arg Leu His Ala Ser Val Gln Leu Ser Thr Thr 30

93 CGU AGC CGA CAA AGC UCU GUU CAU --- 119
31 Arg Ser Arg Gln Ser Ser Val His XXX 38

THE PARTIAL SEQUENCE IS

TACGTAGACAAGTTGATAGATGATGTGCATCGGCTGTTTCGAGACAAGTA

THE ANSWER IS FRAME 1. YOU CAN FIND THE MET AA (AUG CODON) SO IT IS THE PART OF PROTEIN SEQUENCE.

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