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A bust cell has intracellular bacteria symbionts. If the growth rate of bacterial symbiont is always `10%` higher than that of the host cell, after 10 generations of the host cell the density of bacteria in host cells will increase -A. by `10%`B. two-foldC. ten-foldD. hundred-fold |
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Answer» Correct Answer - BLet us consider, Mass of host cell = m, Mass of bacterial symbiont= n Now, for 1st generation, the host cell divides into two so total mass of host cell=2m Total mass of bacterial symbiont in two 1st generation cells = 2n+(2/10)*n Mass of bacterial symbiont in one of the two cells of 1st generation = n+(1/10)*n In general, mass of bacterial symbiont in one of the (2^k)th cells of kth generation = n+(k/10)*n Thus, for 10th generation, mass of bacterial symbiont in one of the (2^10)th cells = n+(10/10)*n=n+n=2n. Hence, density of bacterial symbiont in 10th generation increased by 2 folds (n to 2n).Option B correct. |
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