IChO 2000 · Theory problem 5

Protein and DNA

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Figure 1

DNA is composed of 2’-deoxy-nucleotides carrying the bases adenine (A), guanine (G), cytosine (C) and thymine (T). The molar mass of the 2’-deoxy-nucleotide-5’-triphosphates is given in table 2:

Table 2

dNTP Molar mass /g mol\(^{-1}\)
dATP 487
dGTP 503
dCTP 464
dTTP 478

5.1 Calculate the molar mass of a double stranded DNA fragment consisting of 1000 base pairs with a uniform distribution of the four bases.

This DNA fragment can be isolated and cloned by using the PCR method (polymerase chain reaction), in which a heat stable DNA polymerase enzyme multiplies the number of molecules of a specific piece of DNA in a cyclic process. Under optimal conditions the number of double-stranded DNA copies doubles in each cycle. Using the PCR method you perform 30 cycles starting from a single double stranded DNA molecule.

5.2 Calculate the approximate mass of the DNA you obtain from this experiment.

The bacteria-virus T4 enzyme - polynucleotide kinase (PNK) catalyzes the transfer of the terminal phosphate of ATP (\(\gamma\)-orthophosphate) to the 5'-hydroxyl termini of ribo- and deoxyribonucleotides:

Figure 2

PNK is commonly used to label DNA at the 5’-end with the radioactive phosphorus isotope \(\ce{^{32}P}\) using ATP in which the \(\gamma\)-P (the outermost of the phosphorus atoms) is replaced with \(\ce{^{32}P}\). The amount of \(\ce{^{32}P}\) and thus the amount of labeled DNA can be measured.

A 10 µL solution containing double stranded DNA is labelled 100 % with \([\gamma\text{-}\ce{^{32}P}]\mathrm{ATP}\) by PNK. 37 days ago, the specific activity of \([\gamma\text{-}\ce{^{32}P}]\mathrm{ATP}\) was 10 Ci/mmol or \(370\cdot10^9\) Bq/mmol. \(\ce{^{32}P}\) has a half-life of 14.2 days, and during the decay a \(\beta\)-particle is emitted. Now the labelled DNA emits 40000 \(\beta\)-particles/s.

5-3 Calculate the concentration of the DNA solution.

In an experiment in which PNK is incubated with \([\gamma\text{-}\ce{^{32}P}]\mathrm{ATP}\) and single stranded DNA, the reaction can be monitored by isolating labeled DNA and measuring the \(\beta\)-particle emission. Using this kind of measurements in a 1 cm\(^{3}\) experimental mixture, a labeling of 9 nmol DNA/min was calculated. PNK has a catalytic rate constant (turnover number) of 0.05 s\(^{-1}\) and molar mass of 34620 g mol\(^{-1}\).

5.4 Calculate the concentration (in mg/cm\(^{3}\)) of PNK in the experimental mixture.

Aromatic amino acids, tryptophan, tyrosine and phenylalanine absorb UV light of a wavelength between 240 nm and 300 nm. In a protein containing several aromatic amino acids, the sum of the molar absorptivity per amino acid \(\Sigma\varepsilon_{\text{amino acid}}\), is approximately equal to the molar absorptivity, \(\varepsilon_{\text{protein}}\), for the protein. The molar absorptivity, \(\varepsilon_{\text{amino acid}}\), at 280 nm for tyrosine, tryptophan and phenylalanine is 1400 M\(^{-1}\) cm\(^{-1}\), 5600 M\(^{-1}\) cm\(^{-1}\) and 5 M\(^{-1}\) cm\(^{-1}\), respectively. The absorbance of a 10 µM solution of PNK is 0.644 at 280 nm and with 1.00 cm light path. The amino acid sequence of PNK contains 14 tyrosines and 9 phenylalanines.

\((\mathrm{M}=\mathrm{mol\,dm^{-3}})\)

5.5 Calculate the number of tryptophan residues in a PNK molecule.

Official English text of the problem.

Answers

  • 5.1: \(M_{\mathrm{DNA}}\approx6.18\times10^5\ \mathrm{g\,mol^{-1}}\).
  • 5.2: \(m_{\mathrm{DNA}}\approx1.10\times10^{-9}\ \mathrm{g}=1.10\ \mathrm{ng}\).
  • 5.3: \(c_{\mathrm{DNA}}\approx6.58\times10^{-5}\ \mathrm{mol\,dm^{-3}}=65.8\ \mu\mathrm{M}\), counting each labelled 5′ end as one DNA molecule (amount of labelled DNA = amount of 32P, as the statement says).
  • 5.4: \(c_m(\mathrm{PNK})\approx0.104\ \mathrm{mg\,cm^{-3}}\), or about \(0.10\ \mathrm{mg\,cm^{-3}}\).
  • 5.5: 8 tryptophan residues per PNK molecule.

Solution

5.1 — Molar mass of the DNA fragment

There are two nucleotides per base pair, so the fragment contains

\[ N_{\mathrm{nt}}=2(1000)=2000 \]

nucleotide residues. A uniform distribution means 500 residues of each base.

During incorporation into DNA, the two terminal phosphate groups of a nucleotide triphosphate leave as pyrophosphate. Using the ionic mass convention of the table, the removed pyrophosphate group is \(\ce{P2O7^{4-}}\), with

\[ M_{\mathrm{PP_i}} =2(31)+7(16) =174\ \mathrm{g\,mol^{-1}}. \]

Pyrophosphate ion has SMILES [O-]P(=O)([O-])OP(=O)([O-])[O-].

The average molar mass of an incorporated nucleotide residue is therefore

\[ \overline M_{\mathrm{residue}} =\frac{487+503+464+478}{4}-174 =309\ \mathrm{g\,mol^{-1}}. \]

Neglecting the small contribution from terminal groups,

\[ \boxed{ M_{\mathrm{DNA}}\approx2000(309) =6.18\times10^5\ \mathrm{g\,mol^{-1}} }. \]

5.2 — Mass after 30 PCR cycles

Each cycle doubles the number of double-stranded DNA molecules. Starting from one molecule,

\[ N=2^{30}=1\,073\,741\,824. \]

Using \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\),

\[ n_{\mathrm{DNA}} =\frac{2^{30}}{N_{\mathrm A}} =1.783\times10^{-15}\ \mathrm{mol}. \]

Thus,

\[ m_{\mathrm{DNA}} =n_{\mathrm{DNA}}M_{\mathrm{DNA}} =(1.783\times10^{-15})(6.18\times10^5)\ \mathrm{g}, \]

\[ \boxed{ m_{\mathrm{DNA}}\approx1.10\times10^{-9}\ \mathrm{g} =1.10\ \mathrm{ng} }. \]

5.3 — DNA concentration from radioactive labelling

The specific activity decreases according to

\[ a(t)=a_0\,2^{-t/t_{1/2}}. \]

After 37 days,

\[ a(37) =(370\times10^9)\,2^{-37/14.2} =6.079\times10^{10}\ \mathrm{Bq\,mmol^{-1}}. \]

One emitted \(\beta\)-particle corresponds to one decay, so the measured activity is

\[ A=40000\ \mathrm{Bq}. \]

Since the amount of labelled DNA equals the amount of 32P (each labelled 5′ end counts as one DNA molecule),

\[ n_{\mathrm{DNA}} =\frac{A}{a(37)} =\frac{40000}{6.079\times10^{10}} =6.580\times10^{-7}\ \mathrm{mmol} =6.580\times10^{-10}\ \mathrm{mol}. \]

The sample volume is

\[ V=10\ \mu\mathrm{L}=1.0\times10^{-5}\ \mathrm{dm^3}. \]

Consequently,

\[ \boxed{ c_{\mathrm{DNA}} =\frac{6.580\times10^{-10}}{1.0\times10^{-5}} =6.58\times10^{-5}\ \mathrm{mol\,dm^{-3}} =65.8\ \mu\mathrm{M} }. \]

No factor of two is applied, because each labelled 5′ end counts as one DNA molecule.

5.4 — PNK mass concentration

The observed labelling rate is

\[ \dot n_{\mathrm{DNA}} =\frac{9\ \mathrm{nmol}}{60\ \mathrm{s}} =0.15\ \mathrm{nmol\,s^{-1}}. \]

Using the stated turnover number as the reaction rate per enzyme molecule,

\[ \dot n_{\mathrm{DNA}} =k_{\mathrm{cat}}n_{\mathrm{PNK}}. \]

Therefore,

\[ n_{\mathrm{PNK}} =\frac{0.15\ \mathrm{nmol\,s^{-1}}}{0.05\ \mathrm{s^{-1}}} =3.0\ \mathrm{nmol}. \]

The corresponding enzyme mass is

\[ m_{\mathrm{PNK}} =(3.0\times10^{-9}\ \mathrm{mol}) (34620\ \mathrm{g\,mol^{-1}}) =1.0386\times10^{-4}\ \mathrm{g} =0.10386\ \mathrm{mg}. \]

Since the experimental volume is \(1\ \mathrm{cm^3}\),

\[ \boxed{ c_m(\mathrm{PNK}) =0.10386\ \mathrm{mg\,cm^{-3}} \approx0.10\ \mathrm{mg\,cm^{-3}} }. \]

5.5 — Number of tryptophan residues

The Beer–Lambert law gives

\[ A=\varepsilon_{\mathrm{PNK}}cl. \]

Hence,

\[ \varepsilon_{\mathrm{PNK}} =\frac{0.644} {(10\times10^{-6}\ \mathrm{mol\,dm^{-3}})(1.00\ \mathrm{cm})} =64400\ \mathrm{M^{-1}\,cm^{-1}}. \]

Let \(x\) be the number of tryptophan residues. Adding the contributions from the three aromatic amino acids,

\[ 64400=14(1400)+x(5600)+9(5). \]

Thus,

\[ x=\frac{64400-19600-45}{5600} =7.99196. \]

The residue count must be an integer, and the absorptivity relation is approximate:

\[ \boxed{x=8}. \]

The residue is L-tryptophan; its neutral free-amino-acid SMILES is N[C@@H](Cc1c[nH]c2ccccc12)C(=O)O. Within PNK, its amino and carboxyl groups participate in the peptide backbone.

The solution was checked against the official solution.