= Solution
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 $15\,\mathrm{\mu g\,mL^{-1}}$ in the assay mixture. Therefore
$$
\begin{aligned}
c_{\mathrm{TEM1}}&\simeq200(4.3)=860\,\mathrm{\mu g\,mL^{-1}}=0.86\,\mathrm{mg\,mL^{-1}},\\
c_{\mathrm{TEM52}}&\simeq200(15)=3000\,\mathrm{\mu g\,mL^{-1}}=3.0\,\mathrm{mg\,mL^{-1}}.
\end{aligned}
$$
Multiplying by the $0.500\,\mathrm{mL}$ eluate volume gives \b[about $0.43\,\mathrm{mg}$ of <TEM-1> and $1.5\,\mathrm{mg}$ of <TEM-52>]. Only <TEM-52> meets the requirement.
Assuming the same recovery per culture volume, the required volumes are
$$
V_{\mathrm{culture,1}}\simeq50\frac{1}{0.43}\simeq116\,\mathrm{mL},\qquad V_{\mathrm{culture,52}}\simeq50\frac{1}{1.5}\simeq33\,\mathrm{mL}.
$$
\b[Use approximately $120\,\mathrm{mL}$ or more for <TEM-1>; the existing $50\,\mathrm{mL}$ is sufficient for <TEM-52>.] Rounding the former up to $125\,\mathrm{mL}$ is sensible given graphical precision. Reading the lower standard as $4$–$4.5\,\mathrm{\mu g\,mL^{-1}}$ gives a corresponding requirement of roughly $110$–$125\,\mathrm{mL}$, 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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