How Many Atoms to Encode the Human Genome?

We often hear about how the deoxyribonucleic acid (DNA) macro-molecule that contains the human genome has about 3 billion base pairs, but we rarely hear about the atoms. I was curious to know how many such a complex structure required…

Warning: Back of the envelope calculations.

Here we go:

3 billion base pairs equals 6 billion nucleotides.

The human genome uses 4 types of nucleotides:

  • Adenine (A)
  • Guanine (G)
  • Cytosine (C)
  • Thymine (T)

‘T’ is always associated with ‘A’, and ‘G’ with ‘C’.

Each of these nucleotides is composed of a nitrogen-containing base, a five-carbon sugar, and a phosphate group.

To simplify, we’ll only look at Thymine, which is pretty representative of the others in size.

Thymine’s base is a pyrimidine ring compound and it contains 15 atoms. Its pentose sugar has 15 atoms also (the 5-carbon sugar used for ribonucleic acid (RNA) is very similar, but it has an extra oxygen atom). The phosphate group contains 4 atoms.

Total for a thymine nucleotide: 34 atoms.

Multiply that by 6 billion nucleotides and you get 204 billion atoms.

Of course, that’s just an estimate since human DNA isn’t composed of only ‘T’ nucleotides, and I think some of them can have more than one phosphate group (not sure). But it should be in the right ballpark.

The enzymes of DNA polymerase can copy human DNA, shuffling hundreds of billions of atoms with great precision, including proofreading and sometimes error correction, in a very short amount of time (if you know the precise timing, please let me know).

Now that is nanoscale molecular manufacturing that works! If that’s not impressive, I don’t know what is.


4 Responses to “How Many Atoms to Encode the Human Genome?”

  1. Is There a DNA Puzzle in Alberts’ Molecular Biology of the Cell? « Michael Graham Richard Says:

    […] Michael Graham Richard Stay Curious. « How Many Atoms to Encode the Human Genome? […]

  2. Yegor Says:

    Thank you very much for this answer!

  3. Salim Says:

    Ammmazing !!!
    Thanks so much for the info Sir, this is what I was looking for a long time.
    My calculations were a bit close to yours… but anyway, you followed a good methot for it.

    Salim D.

  4. austyn perkins Says:


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