19. A spherical mammalian cell of radius ‘R’ is infected by a single coccus bacterium having 100 times smaller
radius. Given that the host cell will lyse when 1/2 of the cell volume is taken up by the bacterium,
approximately how many times will the bacterium divide before the host cell is lysed?
1. 19
2. 3X105
3. 106
4. 40

Calculating the Number of Times a Bacterium Divides Before Cell Lysis

In this problem, a spherical mammalian cell is infected by a single coccus bacterium. The host cell will lyse when half of its volume is taken up by the bacterium. We are asked to determine how many times the bacterium will divide before the host cell is lysed.

Step-by-Step Solution:

1. Understanding the Sizes of the Cell and Bacterium:

   • The radius of the mammalian cell is R, and the radius of the bacterium is R/100 (since the bacterium is 100 times smaller than the host cell).

   • The volume of a sphere is given by the formula V=43πr3V = \frac{4}{3} \pi r^3V=34​πr3.

2. Volume of the Host Cell:

   • The volume of the host cell is:

     Vcell=43πR3V_{\text{cell}} = \frac{4}{3} \pi R^3Vcell​=34​πR3

3. Volume of the Bacterium:

   • The volume of the bacterium is:

     Vbacterium=43π(R100)3=43πR31003V_{\text{bacterium}} = \frac{4}{3} \pi \left(\frac{R}{100}\right)^3 = \frac{4}{3} \pi \frac{R^3}{100^3}Vbacterium​=34​π(100R​)3=34​π1003R3​

4. Determining How Much of the Cell’s Volume is Occupied by the Bacterium:

   • The cell will lyse when half of its volume is taken up by the bacterium:

     12Vcell=12×43πR3\frac{1}{2} V_{\text{cell}} = \frac{1}{2} \times \frac{4}{3} \pi R^321​Vcell​=21​×34​πR3

5. How Many Times the Bacterium Can Divide:

   • Each time the bacterium divides, its volume increases by a factor of 2. The question asks how many times the bacterium will divide before its total volume occupies half the volume of the host cell.

   • Let the bacterium divide $n$ times. After $n$ divisions, the volume of the bacterium will be 2n×Vbacterium2^n \times V_{\text{bacterium}}2n×Vbacterium​.

   • Set this equal to half the volume of the cell:

     2n×Vbacterium=12Vcell2^n \times V_{\text{bacterium}} = \frac{1}{2} V_{\text{cell}}2n×Vbacterium​=21​Vcell​

   • Substitute the volume expressions for the bacterium and the cell:

     2n×43πR31003=12×43πR32^n \times \frac{4}{3} \pi \frac{R^3}{100^3} = \frac{1}{2} \times \frac{4}{3} \pi R^32n×34​π1003R3​=21​×34​πR3

   • Simplifying the equation:

     2n×R31003=R322^n \times \frac{R^3}{100^3} = \frac{R^3}{2}2n×1003R3​=2R3​ 2n=100322^n = \frac{100^3}{2}2n=21003​ 2n=1062=5×1052^n = \frac{10^6}{2} = 5 \times 10^52n=2106​=5×105

   • Taking the logarithm of both sides:

     nlog⁡(2)=log⁡(5×105)n \log(2) = \log(5 \times 10^5)nlog(2)=log(5×105) nlog⁡(2)=log⁡(5)+log⁡(105)n \log(2) = \log(5) + \log(10^5)nlog(2)=log(5)+log(105) $$n\log (2)=\log (5)+5$$ n≈5+log⁡(5)log⁡(2)≈5+0.70.3≈19n \approx \frac{5 + \log(5)}{\log(2)} \approx \frac{5 + 0.7}{0.3} \approx 19n≈log(2)5+log(5)​≈0.35+0.7​≈19

✅ Correct Answer:

(1) 19

Conclusion:

The bacterium will divide approximately 19 times before the host cell is lysed. This problem involves calculating the growth of the bacterium in relation to the volume of the host cell and requires knowledge of exponential growth and logarithms.