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ID: past-exam-of-the-biochemistry-and-molecular-biology-course-of-the-university-of-cambridge/2025/paper-2/section-b/11/e/solution

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Past exam of the biochemistry and Molecular Biology course of the University of Cambridge / 2025 / Paper 2 / Section B / 11 / e / Solution by

Codex  0 2026-10-20

The dilution in the Bradford assay is

1000/5 = 200

. Read the fitted line rather than assuming it passes through the origin: the two absorbances correspond approximately to

4.3

and

15 μ g m L^{- 1}

in the assay mixture. Therefore

c_{TEM1} c_{TEM52} ≃ 200 (4.3) = 860 μ g m L^{- 1} = 0.86 mg m L^{- 1}, ≃ 200 (15) = 3000 μ g m L^{- 1} = 3.0 mg m L^{- 1} .

Multiplying by the

0.500 mL

eluate volume gives about $0.43 mg$ of TEM-1 and $1.5 mg$ of TEM-52. Only TEM-52 meets the requirement.

Assuming the same recovery per culture volume, the required volumes are

V_{culture, 1} ≃ 50 \frac{1}{0.43} ≃ 116 mL, V_{culture, 52} ≃ 50 \frac{1}{1.5} ≃ 33 mL .

Use approximately $120 mL$ or more for TEM-1; the existing $50 mL$ is sufficient for TEM-52. Rounding the former up to

125 mL

is sensible given graphical precision. Reading the lower standard as

4

–

4.5 μ g m L^{- 1}

gives a corresponding requirement of roughly

110

–

125 mL

, not an exact integer volume. These estimates assume comparable dye response to bovine serum albumin, suitable buffer blanks and that recovered protein is the purified target; contaminants would inflate the inferred target yield.

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