A nonapeptide was determined to have the following amino acid composition: 2 Lys, 2 Gly, 2 Phe, His, Leu, Met 1. The native peptide was incubated with 1-fluoro-2,4-dinitrobenzene (FDNB) and then hydrolyzed; 2,4-dinitrophenyl-histidine was identified by HPLC. 2. When the native peptide was exposed to cyanogen bromide (CNBr), an octapeptide and free glycine were recovered. 3. Incubation of the native peptide with trypsin gave a pentapeptide, a tripeptide, and free Lys. 2,4-dinitrophenyl-histidine was recovered from the pentapeptide, and 2,4-dinitrophenyl-phenylalanine was recovered from the tripeptide. 4. Digestion with the pepsin produced a dipeptide, a tripeptide, and a tetrapeptide. The tetrapeptide was composed of 2 Lys, Phe, and Gly. Protease Site of Cleavage Protease Site of Cleavage Trypsin Lys or Arg (C) Thermolysin Leu, Ile, or Val (N) Staph Protease Asp or Glu (C) Asp-N-Protease Asp (N) Chymotrypsin Phe, Trp, or Tyr (C) Pepsin Phe, Trp, or Tyr (N) (Assume these proteases cut only at the amino acids listed above) The native sequence was determined to be: NH2 - - - - - - - - - - COOH

The native sequence of the peptide is: His-Leu-Phe-Gly-Lys-Lys-Phe-Met-Gly

Hydrolysis with FDNB gives 2,4-dinitrophenylhistidine. This shows that the His amino acid is the amino terminal residue.

CNBr cleaves the peptide bond at C-terminus of Met. Hence, the 8th amino acid residue would be Met and 9th amino acid residue would be Gly as the resultant peptide was an octapeptide and free Gly.

Trypsin cleaves the peptide bond at Lys residues. Digestion with trypsin resulted in a pentapeptide. Hence, 5th amino acid would be Lys. As the 8th and 9th amino acid residues were already determined. So, the possiblity of other free Lys position resulted from this digestion would be 6th position. Hence, the 6th amino acid residue would be Lys.

2,4-dinitrophenylalanine was recovered from the tripeptide. This suggests that the amino terminus residue in the tripeptide obtained from trypsin digestion would be Phe. This indicates that 7th position of the nonapeptide would be Phe.

Till this step, the 1,5,6,7,8,9 amino acid residues were known to be as follows:

His-x-x-x-Lys-Lys-Phe-Met-Gly

Pepsin cleaves the peptide bond at Phe amino acid residues. So, from the above peptide sequence, one bond at which pepsin cleaves is at the 7th positioned Phe residue. This proves the production of dipeptide. In addition to this, a tripeptide and a tetrapeptide are formed. As the tetrapeptide sequence composed of (Lys) 2, Phe, and Gly. There is a possiblity of Gly being present at 4th position. This results in the sequence as:

His-x-x-Gly-Lys-Lys-Phe-Met-Gly

Hence, the tripeptide resulted from the pepsin cleavage would be from the first 3 amino acid residues. As pepsin cleaves at Phe, the 3rd amino acid residue would be Phe. This clearly demonstrates that the remaining Lys residue would be at 2nd position.

there are 35 white marbles

step-by-step explanation:

y- yellow marbles

4y + 5 = 45

subtract 5 from both sides

4y = 40

divide for from both sides

y = 10

plug in 10 to get 40 yellow marbles then subtract by 5 to get 35 marbles

thus there are only 10 yellow marbles

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