Each DNA base can be represented in 4 bits because there’s only 4 values: A, T, G or C. (This is a simplification because of epigenetic base modifications, but works for now). So just take the length of DNA and multiply by 4 for the amount of bits the data represents. You do have to add up the lengths of all the chromosomes considering that sperm only have one copy of each chromosome, not two. Also factor in that there is no mitochondria in sperm. Lastly, about half of the sperm will be carrying an X and half will be carrying a Y chromosome so just take the length of the (X + Y) / 2 for the sex chromosome. Now just multiply by the amount of sperm in an average ejaculation and you’ll get a good estimate.
How do they calculate that?
Each DNA base can be represented in 4 bits because there’s only 4 values: A, T, G or C. (This is a simplification because of epigenetic base modifications, but works for now). So just take the length of DNA and multiply by 4 for the amount of bits the data represents. You do have to add up the lengths of all the chromosomes considering that sperm only have one copy of each chromosome, not two. Also factor in that there is no mitochondria in sperm. Lastly, about half of the sperm will be carrying an X and half will be carrying a Y chromosome so just take the length of the (X + Y) / 2 for the sex chromosome. Now just multiply by the amount of sperm in an average ejaculation and you’ll get a good estimate.
How do they ejaculate that?
Each base-pair requires two bits, not four. Picking an arbitrary convention: 00 = A, 01 = T, 10 = G, 11 = C.
To represent 4 values, you only need 2 bits