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 partial protein sequence is = MHLFNYLLHVADKALF

Since the mRNA sequence wll be the complementary of the c-DNA sequnce obtained with A binding to U, the mRNA sequence corresponding to the given cDNA clone will be as follows :

AUAUUUGAUUUGUUGGUCAAGCUCGACCACAAGCACCAGCCAAAAGUCTTCUAGGATTGCGACUGCAUGCAUCUGUUCAACUAUCUACUACACGUAGCCGACAAAGCUCUGUUCAU

Protein synthesis (translation) will always begin with an AUG (coding for a methionine). The coding frame would begin and appear as under :-

AUAUUUGAUUUGUUGGUCAAGCUCGACCACAAGCACCAGCCAAAAGUCTTCUAGGATTGCGACUGCAUGCAUCUGUUCAACUAUCUACUACACGUAGCCGACAAAGCUCUGUUCAU

The underlined sequence is the one which will code for the partial protein.

Hence the protein sequence would be :

AUG -> M

CAU -> H

CUG -> L

UUC -> F

AAC -> N

UAU -> Y

CUA -> L

CUA -> L

CAC -> H

GUA -> V

GCC -> A

GAC -> D

AAA -> K

GCU -> A

CUG -> L

UUC-> F

AU

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