After four successive mitotic divisions of a | vedclass

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(D) In arithmetic growth,the rate of growth is constant.After each mitotic division,one daughter cell remains meristematic (continues to divide),while

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$4$

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(D) In arithmetic growth,the rate of growth is constant.After each mitotic division,one daughter cell remains meristematic (continues to divide),while the other daughter cell undergoes differentiation and maturation,thereby losing its ability to divide.Therefore,after $1$st division,$1$ cell differentiates.After $2$nd division,another $1$ cell differentiates (total $2$).After $3$rd division,another $1$ cell differentiates (total $3$).After $4$th division,another $1$ cell differentiates (total $4$).Thus,after $4$ successive mitotic divisions,the total number of cells that have lost the ability of cell division is $4$.

After four successive mitotic divisions of a cell,in an arithmetic growth pattern,the total number of cells that have lost the ability of cell division is . . . . . . .

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