Thursday, 29 June 2006

homework - What is the total number of rounds of cleavage during mammalian embryonic development?

It's not a totally answerable question, since some types of cells are going to divide more times than others. But for an estimate, take as a starting proposition that there are 1 trillion cells in the adult human body. [1] The average weight for a human is 62kg. [2] Average birth weight is about 3.4 kg. [3]



So that implies roughtly (3.4/62)* 1 trillion = 55 billion cells in a newborn.



You then take the log base 2 of 55 billion, which gives you the exponent you have to hang on 2 in order to get 55 billion, which is about 35. Then add one for that additional cell division to get from one to two cells == 36 divisions.



Of course I'm just using math, not biology, so your actual reality may vary. Certainly some cells will reproduce more often than others, maybe cells actually grow in mass instead of dividing (i.e., baby cells might have less mass than adult cells) so the baby-cell-count could be off, lots of possible sources of error.



[1] http://www.nichd.nih.gov/publications/pubs/fragileX/sub3.cfm



[2] http://en.wikipedia.org/wiki/Body_weight#Average_weight_around_the_world



[3] http://en.wikipedia.org/wiki/Infant#Weight

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