# DBT JRF 2020 — Question 38

Exam: DBT JRF 2020 · Section: A · Topic: Biochemistry · Difficulty: Medium

## Question

How many peptide fragments can be generated from the complete digestion of the polypeptide AGRCDKCQANRSLMNF with trypsin?

## Options

- **A.** 2
- **B.** 4
- **C.** 6
- **D.** 3

## Answer

Option B

## Explanation

Trypsin cleaves specifically after Arg (R) and Lys (K). Mapping the sequence AGRCDKCQANRSLMNF: Position 3 = R → cleaves after R: fragment 1 = AGR. Position 7 = K → cleaves after K: fragment 2 = CDKC. Wait — C is at pos 4, D at 5, K at 7. So: AGR | CDK | CQANR | SLMNF. Actually: A-G-R(3)-C-D-K(6)-C-Q-A-N-R(11)-S-L-M-N-F(16). Cuts after R(3), K(6), R(11) → 4 fragments: AGR | CDK | CQANR | SLMNF. Total = 4 peptide fragments.

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