# Does the DNA of one sperm contain 37.5 MB of information?

[Source]

In an episode of the BBC show QI - Quite Interesting (Series J, Episode 1) Stephen Fry said:

How much information do you think is in the DNA of one little sperm...?

It's 37.5 MB...

...a normal male ejaculation, if there is such a thing, is equivalent of 15,875 GB. That's about 7500 laptops worth of information...

The shows Twitter page summarized it:

A sperm has 37.5 MB of DNA info.
One ejaculation transfers 15,875 GB of data, equivalent to that held on 7,500 laptops.

(with "200 million sperm per ejaculation" one would actually get 7150 TB;
but I'm more interested in where the 37.5 MB number comes from)

My Question:

• Does the DNA of one sperm contain 37.5 MB of information?
• I was very annoyed when this was broadcast, as the claim of 7500TB is clearly false. Each sperm is approximately a random shuffle of 1/2 the DNA of the parent and so 200 million selections of 1/2 the parental DNA is not going to multiply the information contained by 200 million! The 37.5MB sounds reasonable order of magnitude, exact values will depend on how you encode the information etc. I would crunch the numbers myself but would that be acceptable as an answer?
– Nick
Commented Sep 18, 2012 at 10:39
• Wikipedia seems to claim a figure between 2MB (haploid difference from standard reference) and 700-800MB (complete haploid genome). Not sure what set of approximations QI used to get 37.5MB.
– Nick
Commented Sep 18, 2012 at 10:55
• "equivalent of 15,875 GB" is total BS, exact copy of same information is not extra information. It's like saying that copying "lorem ipsum" few thousand times is "equivalent to contents of the Library of Congress" Commented Sep 18, 2012 at 11:47
• @RobZ: Depends if they really talk about information or data. In coding theory the information content won't be changed by a lossless encoding scheme. It just determines how much data you need to represent it. Of course, 200 million times the same messages gives you 200 million times the data size, but no more additional information. Commented Sep 18, 2012 at 18:12
• @Vartec - Actually its like saying that you could copy Lorum Ipsum enough times to fill a datastore with the same volume of information as contained in the LOC. There is no claim that the data being transferred is not redundant at all.
Commented Sep 19, 2012 at 15:39

I am not sure where these numbers come from and the answer depends on how you encode the genome data and if you define all the redundancy (unnecessary, repetitive data) as "information".

First of all, the humane genome contains somewhere around 3.1 (men) to 3.2 (women) billion base pairs. Since the X chromosome is three times longer than the Y chromosome, women have a higher total genome length than men.

A base pair is made of two of the four nucleobases adenine, cytosine, guanine and thymine, but only the four combinations AT, TA, CG and GC are possible as the A and T nucleobases won't bond with the C and G nucleobases and vice versa. These four combinations can be encoded with two bits, so that 6.2-6.4 gigabits or about 750 megabytes are required to store an exact copy of the genome.

Now, even if you need 750 megabytes to store the "raw data" from a human genome, at least a computer scientist will have a hard time defining all of this as "information". E.g. if you record 74 Minutes of complete silence on a CD, the disc contains roughly 750 megabytes of "data" as well, but actually no "information". Large parts of the human genome are repetitive, only a very small part actually differ between different individuals and from the difference, several base pair sequences only occur in a few well-defined varieties.

There is actually some research in the field "how to store a human genome as compact as possible", since genome databases most likely are going to expand rapidly and scientists need efficient ways to share data. Some tools are available for this purpose, e.g. DNAzip, which using a ~5 gigabyte dictionary (permanent data) can compress a human genome down to roughly 4 megabytes.

• CAG and T are nucleotides, not proteins. Proteins are long strings of amino acids; nucleotides are small cyclic molecules. Commented Sep 19, 2012 at 15:09
• @matt_black: Aren't they actually nucleobases, to be very precise? Commented Sep 19, 2012 at 16:32
• @Tor-EinarJarnbjo: A, C, G and T can be used to identify both the nucleobase (for instance adenine) and the nucleoside (for instance adenosine).
– nico
Commented Sep 19, 2012 at 17:37
• The second number is interesting but not really an answer to the question: the information content is certainly more than 4 MB since you can’t just ignore the dictionary size. Commented Sep 19, 2012 at 22:51
• I have to say that I’m unhappy that this is the accepted answer. The 37 MB number is in the ballpark of often-quoted numbers in bioinformatics. Whether or not it’s correct it requires some explanation, and this is entirely lacking here. Unfortunately, I can’t for the life of me remember how the number was derived. Commented Sep 20, 2012 at 12:08

For a simpler answer, you can just look at the size of an ASCI encoded text file containing the human genome's information. This, of course, is not the information content of the genome which, as you can see from the answer above and the comments in this thread, is not that easy to define.

In any case, when biologists work on the genome sequence, it tends to be in the form of FASTA sequences. The human genome as a multi fasta file is ~3Gb. See, for example, the file `UCSC/hg19/Sequence/WholeGenomeFasta/genome.fa` obtained when extracting this archive.

Again, I stress that this is not the information content of the genome. For those of us who are not information theorists though, it gives an easy way of picturing the genome's size in a format we are familiar with: text.

• A good compression algorithm should be able to compress the human genome file and produce something with minimal output size. That might give a better indication as there one would have a file containing a dictionary plus the data to expand the dictionary, which would be the (minimal?) amount of "information" needed.
– Raf
Commented Jan 15 at 6:47