4 ms·

85.8 grams of DNA is enough to overflow 64bits. Not that we're anywhere close to being able to use or process information on those scales, but it at least stor

85.8 grams of DNA is enough to overflow 64bits. Not that we're anywhere close to being able to use or process information on those scales, but it at least storage on that scale doesn't outright violate the laws of physics/information theory.

littlestymaar 5y ago
Storage needs don't affect pointer size though. To need bigger pointers, you must need an addressable space this big. That said, how to you make your calculation about DNA's information density? Without counting water, I've arrived at 720g for one “mole of bits”, which is a bunch of orders of magnitude higher than your figure: 2 amino-acids (262g/mole) [1], 2 Deoxyribose (135g/mole each) and 2 Phosphate (94g/mole each). [1]: and it's the same value be it A-T or G-C, that's interesting
- spqr0a1 5y ago
  The information density is even higher than that (double), because one pair of bases has 4 possible states. The pair A-T is symbolically distinct from T-A, for example.
jamal-kumar 5y ago
Not anywhere close? I've heard things about using DNA itself for the memory storage format which is super interesting. I guess it just has bandwidth limitations and is more for long term storage? [1] [1] https://www.scientificamerican.com/article/dna-the-ultimate-data-storage-solution/ https://www.scientificamerican.com/article/dna-the-ultimate-...
- d_tr 5y ago
  85 grams of DNA contain around 8 * 10^22 base pairs, 2 bits per pair, so around 2 * 10^22 bytes. So you need about 75 bits to address all these bytes. I do not know why Laremere chose 85.8 grams though as these are quite above the 64 bit limit, unless I got my multiplications wrong :P
  - sdenton4 5y ago
    Not that anything uses that much DNA, anyway: "The male nuclear diploid genome extends for 6.27 Gigabase pairs (Gbp), is 205.00 cm (cm) long and weighs 6.41 picograms (pg)." Or, many many orders of magnitude less than 85 grams...

littlestymaar 5y ago

Storage needs don't affect pointer size though. To need bigger pointers, you must need an addressable space this big. That said, how to you make your calculation about DNA's information density? Without counting water, I've arrived at 720g for one “mole of bits”, which is a bunch of orders of magnitude higher than your figure: 2 amino-acids (262g/mole) [1], 2 Deoxyribose (135g/mole each) and 2 Phosphate (94g/mole each). [1]: and it's the same value be it A-T or G-C, that's interesting

spqr0a1 5y ago
The information density is even higher than that (double), because one pair of bases has 4 possible states. The pair A-T is symbolically distinct from T-A, for example.

jamal-kumar 5y ago

Not anywhere close? I've heard things about using DNA itself for the memory storage format which is super interesting. I guess it just has bandwidth limitations and is more for long term storage? [1] [1] https://www.scientificamerican.com/article/dna-the-ultimate-data-storage-solution/ https://www.scientificamerican.com/article/dna-the-ultimate-...

d_tr 5y ago
85 grams of DNA contain around 8 * 10^22 base pairs, 2 bits per pair, so around 2 * 10^22 bytes. So you need about 75 bits to address all these bytes. I do not know why Laremere chose 85.8 grams though as these are quite above the 64 bit limit, unless I got my multiplications wrong :P
- sdenton4 5y ago
  Not that anything uses that much DNA, anyway: "The male nuclear diploid genome extends for 6.27 Gigabase pairs (Gbp), is 205.00 cm (cm) long and weighs 6.41 picograms (pg)." Or, many many orders of magnitude less than 85 grams...