IChO 1984 · Theory problem 7

DNA: mass, coding capacity and base composition

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Deoxyribonucleic acid (DNA) represents the genetic program of all living beings. The human genetic program is subdivided into 23 chromosomes.

7.1 Calculate the mass of a DNA thread in grams, which reaches form earth to the moon (340,000 km). A mass of 1 g represents 1,000 nucleotide pairs.

One nucleotide pair (base pair) has a length of 0.34 nm.

7.2 Give estimation on how many nucleotid pairs are stored in the chromosome set of a human being. Human cells can synthesize 50,000 different proteins, which are on the average 300 amino acids long. Only 2 % of the DNA code for proteins.

7.3 The DNA of the bacteriophage M13 shows the following base composition:

A: 23 %, T: 36 %, G: 21 %, C: 20 % (mole %)

What does the base composition tell about the structure of the DNA?

Official English text of the problem.

Answers

  • 7.1: \(\boxed{1\times10^{-3}\ \mathrm{g}=1\ \mathrm{mg}}\), taking 1,000 nucleotide pairs to have a mass of \(10^{-18}\ \mathrm{g}\).
  • 7.2: \(\boxed{\approx2.3\times10^{9}\text{ nucleotide pairs}}\) in the 23-chromosome set; the unrounded estimate is \(2.25\times10^{9}\).
  • 7.3: M13 DNA is single-stranded: its base proportions are incompatible with a fully complementary double-stranded molecule.

Solution

7.1

The printed mass datum ("1 g") is a misprint; the intended value, also used in the official solution, is \[ m(1{,}000\text{ nucleotide pairs})=10^{-18}\ \mathrm{g}. \]

Convert the Earth–Moon distance into nanometres: \[ L=340{,}000\ \mathrm{km} \times10^{12}\ \frac{\mathrm{nm}}{\mathrm{km}} =3.4\times10^{17}\ \mathrm{nm}. \]

Since each nucleotide pair contributes \(0.34\ \mathrm{nm}\) to the length, the number of pairs is \[ N=\frac{3.4\times10^{17}\ \mathrm{nm}} {0.34\ \mathrm{nm/pair}} =1.0\times10^{18}\ \text{pairs}. \]

Therefore, \[ m=(1.0\times10^{18}\ \text{pairs}) \left(\frac{10^{-18}\ \mathrm{g}}{10^{3}\ \text{pairs}}\right) =\boxed{1\times10^{-3}\ \mathrm{g}}. \]

7.2

Each amino acid is encoded by a codon of three nucleotides. Thus, a protein containing 300 amino acid residues requires approximately \[ 300\times3=900 \] nucleotides on its coding strand, corresponding to 900 base pairs of double-stranded DNA.

For 50,000 different proteins, the total coding requirement is \[ N_{\mathrm{coding}} =50{,}000\times300\times3 =4.5\times10^{7}\ \text{base pairs}. \]

This represents \(2\%=0.02\) of the total DNA: \[ 0.02N_{\mathrm{total}}=4.5\times10^{7}. \] Hence, \[ N_{\mathrm{total}} =\frac{4.5\times10^{7}}{0.02} =2.25\times10^{9} \approx\boxed{2.3\times10^{9}\ \text{base pairs}}. \]

This is the total across the stated 23 chromosomes. No additional factor of two is required: a base pair already includes the complementary nucleotides of both strands. Stop codons are neglected at the precision of this estimate.

7.3

The four base identities are listed below; the SMILES describe the free bases.

Symbol Base SMILES
A Adenine Nc1ncnc2[nH]cnc12
T Thymine Cc1c[nH]c(=O)[nH]c1=O
G Guanine Nc1nc2[nH]cnc2c(=O)[nH]1
C Cytosine Nc1cc[nH]c(=O)n1

In fully complementary double-stranded DNA, adenine pairs with thymine, and guanine pairs with cytosine. Counting both strands therefore requires \[ n_{\mathrm A}=n_{\mathrm T}, \qquad n_{\mathrm G}=n_{\mathrm C}. \]

The supplied composition instead gives \[ x_{\mathrm A}=23\%\ne36\%=x_{\mathrm T}, \qquad x_{\mathrm G}=21\%\ne20\%=x_{\mathrm C}. \]

In particular, the large difference between adenine and thymine is incompatible with a fully complementary duplex. The intended structural conclusion is therefore \[ \boxed{\text{M13 DNA is single-stranded.}} \]

The composition alone does not determine whether the strand is circular or linear. A complete DNA SMILES cannot be specified from these percentages: its nucleotide sequence and connectivity are missing. SMILES also does not specify its three-dimensional conformation.

The solution was checked against the official solution.