We have a unknown peptide .. One aliquot is treat with cyanogen bromide. Given t

Question

We have a unknown peptide .. One aliquot is treat with cyanogen bromide. Given the (N-terminal to C-terminal) make the original peptide.

Trypsin treatment

Cys—Tyr—Asp—Met—Trp—Ile—Lys

Arg—Val—Phe—Arg

Asn—Gln

Asn—Met—Tyr—Ala—Ser—Phe—Lys

Cyanogen bromide treatment

Arg—Val—Phe—Arg—Asn—Met

Trp—Ile—Lys—Asn—Gln

Tyr—Ala—Ser—Phe—Lys—Cys—Tyr—Asp—Met

1)Trp—Ile—Lys—Asn—Gln—Arg—Val—Phe—Arg—Asn—Met—Tyr—Ala—Ser—Phe—Lys—Cys—Tyr—Asp—Met

2)Arg—Val—Phe—Arg—Asn—Met—Tyr—Ala—Ser—Phe—Lys—Cys—Tyr—Asp—Met—Trp—Ile—Lys—Asn—Gln

3)Asn—Met—Tyr—Ala—Ser—Phe—Lys—Cys—Tyr—Asp—Met—Trp—Ile—Lys—Arg—Val—Phe—Arg—Asn—Gln

4)Cys—Tyr—Asp—Met—Trp—Ile—Lys—Arg—Val—Phe—Arg—Asn—Gln—Asn—Met—Tyr—Ala—Ser—Phe—Lys

5)Asn—Gln—Cys—Tyr—Asp—Met—Trp—Ile—Lys—Asn—Met—Tyr—Ala—Ser—Phe—Lys—Arg—Val—Phe—Arg

6)Tyr—Ala—Ser—Phe—Lys—Cys—Tyr—Asp—Met—Arg—Val—Phe—Arg—Asn—Met—Trp—Ile—Lys—Asn—Gln

None of the Above

Explanation / Answer

In trypsin treatment, trypsin cleaves COOH side of arginine and lysine whereas in Cyanogen bromide treatment, CyanoBr cleave COOH side of methonine. Hence in trypsin and cyanogen bromide treatment fragments, Lys or Arg and Met must be present at the end of the fragments respectively, unless it is a terminal fragment.

Hence the sequence is,

2) Arg- val- phe- Arg- Asn- Met- Tyr- Ala- Ser- Phe- Lys- Cys- Tyr- Asp- Met- Trp- Ile- Lys- Asn- Glu.

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