A) \[1/21\]
B) \[1/11\]
C) \[10/21\]
D) \[10/11\]
Correct Answer: B
Solution :
| Radius of histone octamer (r) =4.5 nm |
| Height of histone octamer (h) = 5 nm |
| Volume of one histone octamer = \[\pi {{\operatorname{r}}^{2}}h\] |
| \[=\frac{22}{7}\times 4.5\times 4.5\times 5\]\[=318.214n{{m}^{3}}\]\[=318.214\times 1{{0}^{-27}}{{m}^{3}}\] |
| Total volume of histone octamers present in human genome\[=318.214\times {{10}^{-27}}{{\operatorname{m}}^{3}}\times 32\times {{10}^{6}}\] |
| \[=10182.85\times 1{{0}^{-21}}{{m}^{3}}\] |
| Radius of nucleus \[=3\mu \operatorname{m}=3\times {{10}^{-6}}nm\] |
| Volume of nucleus \[=\frac{4}{3}\pi {{\operatorname{r}}^{3}}\]\[=\frac{4}{3}\times \frac{22}{7}\times 3\times {{10}^{-6}}\times 3\times {{10}^{-6}}\times 3\times {{10}^{-6}}\]\[=113.14\times {{10}^{-18}}\] |
| Fraction of the volume of nucleus occupied by histone octamers \[\frac{10182.85\times {{10}^{-21}}}{113.14\times {{10}^{-18}}}\]\[=0.09\approx 1/11\] |
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