Question

Write an account of cleavage during embryonic development in humans.

Answer 1

1. Cleavage:
   Cleavage is the process of early mitotic division of the zygote into a hollow multicellular blastula. It does not involve the growth of the daughter cells. The cells formed by cleavage are called blastomeres. Since there is no growth phase between the cleavages, the size of blastomeres will be reduced with every successive cleavage. As the size reduces, the metabolic rate increases. Subsequent cleavages are thus faster than the earlier ones. This requires rapid replication of DNA and high consumption of oxygen.
2. Process of cleavage:
   In humans, cleavage is holoblastic i.e. the whole zygote gets divided. The cleavage planes may be longitudinal or meridional and equatorial or horizontal. It is radial and indeterminate i.e. fate of each blastomere is not predetermined. The 1st cleavage in the zygote is meridional and occurs about 30 hours after fertilization. It divides longitudinally into two blastomeres, one slightly larger than the other. The 2nd cleavage is also longitudinal but at the right angle to the 1st one and occurs after 30 hours of 1st cleavage. The 3rd cleavage is horizontal. After 3rd cleavage, the embryo is in the 8-cell stage. While the cleavages occur, the young embryo is gradually being pushed towards the uterus. By the end of the 4th day after fertilization, the embryo is a solid ball of 16-32 cells and externally looking like mulberry. This stage is thus called morula.
3. Morula:
   The morula shows cells of two types:

1. smaller, clearer cells towards the outer side.
2. inner cell mass of larger cells. Cells are compactly arranged. Till the formation of morula, the zona pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula. The morula reaches the isthmus and gains entry into the uterus by the end of day 4.