IChO 2022 · Theory problem 1

Rapid and Visual Nucleic Acid Testing for COVID-19

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Fast and simple methods for early detection of COVID-19 are urgently needed. Gold nanoparticles-aided detection is one of the promising methods. Gold nanoparticles are widely used as a visible readout for test strips due to their high molar extinction coefficients (molar absorptivity). The color appearance of gold nanoparticles is closely associated with their sizes and dispersity. Generally, the larger the gold nanoparticles, the more reddish the color is. Once the nanoparticles are aggregated, the color shifts from red to blue.

When the surface of gold nanoparticles is modified with two kinds of single-stranded nucleic acid fragments \(a\) and \(b\), the gold nanoparticles would aggregate in the presence of the target nucleic acid (\(a’b’\)), causing the color change of the solution from red to blue in several minutes (as indicated below). Based on this principle, the target nucleic acids collected from the coronavirus in a sample can be detected.

Figure 1

1.1 Indicate the absorption band region of the dispersed gold nanoparticles with an appropriate letter (a-f in the color wheel of Figure 0.1 in General Instructions).

The Color Wheel

Figure 2

Figure 0.1

1.2 Indicate the change in the absorption spectrum of the gold nanoparticles once they are aggregated. Compared to the absorption wavelength of the dispersed gold nanoparticles, the wavelength of the aggregated nanoparticles

(a) becomes longer

(b) becomes shorter

(c) does not change

Gold nanoparticles are composed of gold atoms that are closely packed as solid gold (density \(\rho = 19.3\ \mathrm{g\,cm^{-3}}\)).

1.3 Calculate how many gold atoms (\(N\)) are there in a spherical gold nanoparticle with a diameter of 30.0 nm.

The synthesis of gold nanoparticles depends on the redox reaction between hydrogen tetrachloroaurate(III) (\(\ce{HAuCl4 * 3H2O}\), \(MW = 394\)) and reducing substances (such as sodium citrate). 5.2 mg of \(\ce{HAuCl4 * 3H2O}\) was completely converted into uniform spherical gold nanoparticles with a diameter of 30.0 nm in a 100.0 mL reaction solution. The absorbance of the resulting red solution was 0.800 measured at 530 nm by a UV-Vis spectrometer.

1.4 Calculate the molar extinction coefficient (molar absorptivity) of the resulting gold nanoparticles solution at 530 nm per mol of gold nanoparticles. The optical path of the cuvette used is 1 cm. If you failed to get the number of gold atoms (\(N\)) in 1.3 , use \(N = 1.00 \times 10^5\).

The standard addition method was used for colorimetric detection of the target nucleic acid. The original throat swab sample was collected and divided equally into two portions. After addition of the probe solutions and water (as indicated below), the absorbance of the two resulting solutions was measured at 600 nm separately.

No. Volume of throat swab sample (mL) Volume of solution of gold nanoparticles modified with nucleic acid fragments (mL) Volume of standard solution containing \(2.0\ \mu\mathrm{g\,mL^{-1}}\) target nucleic acid (mL) Volume of water (mL) Absorbance
1 0.10 0.80 0.00 0.10 0.400
2 0.10 0.80 0.10 0.00 0.900

1.5 Calculate the concentration of the viral nucleic acids in the original throat swab sample.

Official English text of the problem.

Answers

  • 1.1: d, the green region, approximately \(490\text{–}560\ \mathrm{nm}\).
  • 1.2: (a) becomes longer: the absorption band undergoes a red shift.
  • 1.3: \(\boxed{N = 8.34\times10^5\ \text{gold atoms per nanoparticle}}\).
  • 1.4: \(\boxed{\varepsilon_{530}=5.1\times10^9\ \mathrm{L\,mol^{-1}\,cm^{-1}}}\), per mole of nanoparticles.
  • 1.5: \(\boxed{C_{\text{original}}=1.6\ \mu\mathrm{g\,mL^{-1}}}\).

Solution

1.1

The observed color is complementary to the absorbed color. On the supplied color wheel, red lies opposite green. Therefore, the dispersed nanoparticles, which appear red, absorb in the green region, d, approximately \(490\text{–}560\ \mathrm{nm}\). The stated measurement wavelength of \(530\ \mathrm{nm}\) lies in this region.

1.2

The aggregated nanoparticles appear blue. On the supplied wheel, blue is complementary to orange, corresponding to approximately \(580\text{–}620\ \mathrm{nm}\).

Thus, aggregation shifts the absorption from the green region toward longer wavelengths:

\[ \lambda_{\mathrm{aggregated}}>\lambda_{\mathrm{dispersed}}. \]

The answer is (a) becomes longer.

1.3

Use the standard constants

\[ M(\ce{Au})=197.0\ \mathrm{g\,mol^{-1}}, \qquad N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}. \]

The nanoparticle radius is

\[ r=\frac{30.0}{2}\ \mathrm{nm} =15.0\ \mathrm{nm} =1.50\times10^{-6}\ \mathrm{cm}. \]

Its volume and mass are

\[ V_{\mathrm p} =\frac43\pi r^3 =\frac43\pi(1.50\times10^{-6}\ \mathrm{cm})^3 =1.4137\times10^{-17}\ \mathrm{cm^3}, \]

\[ m_{\mathrm p} =\rho V_{\mathrm p} =(19.3\ \mathrm{g\,cm^{-3}}) (1.4137\times10^{-17}\ \mathrm{cm^3}) =2.7285\times10^{-16}\ \mathrm g. \]

Therefore, the number of gold atoms in one nanoparticle is

\[ N=\frac{m_{\mathrm p}}{M(\ce{Au})}N_{\mathrm A} =\frac{2.7285\times10^{-16}\ \mathrm g} {197.0\ \mathrm{g\,mol^{-1}}} (6.022\times10^{23}\ \mathrm{mol^{-1}}). \]

\[ \boxed{N=8.34\times10^5\ \text{atoms per nanoparticle}}. \]

1.4

Each formula unit of the precursor contains one gold atom. Complete conversion therefore gives

\[ n_{\ce{Au}} =\frac{5.2\times10^{-3}\ \mathrm g} {394\ \mathrm{g\,mol^{-1}}} =1.3198\times10^{-5}\ \mathrm{mol}. \]

Because each nanoparticle contains \(N\) gold atoms, the amount of nanoparticles is

\[ n_{\mathrm{NP}}=\frac{n_{\ce{Au}}}{N} =\frac{1.3198\times10^{-5}}{8.34\times10^5} \approx1.582\times10^{-11}\ \mathrm{mol}. \]

Their molar concentration in \(100.0\ \mathrm{mL}\) is consequently

\[ c_{\mathrm{NP}} =\frac{1.582\times10^{-11}\ \mathrm{mol}} {0.1000\ \mathrm L} =1.582\times10^{-10}\ \mathrm{mol\,L^{-1}}. \]

Using the Beer–Lambert law,

\[ A=\varepsilon c_{\mathrm{NP}}l, \]

\[ \varepsilon_{530} =\frac{0.800} {(1.582\times10^{-10}\ \mathrm{mol\,L^{-1}}) (1\ \mathrm{cm})} \approx5.06\times10^9\ \mathrm{L\,mol^{-1}\,cm^{-1}}. \]

Rounding to two significant figures, consistent with the supplied precursor mass,

\[ \boxed{\varepsilon_{530} =5.1\times10^9\ \mathrm{L\,mol^{-1}\,cm^{-1}}}. \]

This coefficient is expressed per mole of nanoparticles, so the concentration used must count particles.

1.5

Both measured solutions have the same final volume:

\[ V_f=1.00\ \mathrm{mL}. \]

Let \(C_x\) be the concentration in the original sample. Each solution contains \(V_x=0.10\ \mathrm{mL}\) of that sample. The second solution additionally contains \(V_s=0.10\ \mathrm{mL}\) of standard with concentration \(C_s=2.0\ \mu\mathrm{g\,mL^{-1}}\).

Using the linear absorbance response assumed in this standard-addition determination, with a common proportionality constant \(k\),

\[ A_1=k\frac{C_xV_x}{V_f}, \qquad A_2=k\frac{C_xV_x+C_sV_s}{V_f}. \]

Subtracting and taking a ratio eliminates both \(k\) and \(V_f\):

\[ \frac{A_1}{A_2-A_1} =\frac{C_xV_x}{C_sV_s}. \]

Hence,

\[ C_x =\frac{A_1}{A_2-A_1}\frac{C_sV_s}{V_x} =\frac{0.400}{0.900-0.400} \frac{(2.0\ \mu\mathrm{g\,mL^{-1}})(0.10\ \mathrm{mL})} {0.10\ \mathrm{mL}}. \]

\[ \boxed{C_x=1.6\ \mu\mathrm{g\,mL^{-1}}}. \]

The solution was checked against the official solution.