Within a cell, proteins are synthesized on particles called ribosomes.Assuming ribosomes are generally spherical, what is the volume (in dm' and μL) of a ribosome whose average diameter is 21.4 nm (V of a sphere=4/3 πr³)?
V=43πr3r=d2=21.4 nm2=10.7 nmV=43×3.14×(10.7)3=43×3.14×1225.043 nm3=5128.846 nm3V = \frac{4}{3} \pi r^3 \\ r = \frac{d}{2} = \frac{21.4 \;nm}{2} = 10.7 \;nm \\ V = \frac{4}{3} \times 3.14 \times (10.7)^3 \\ = \frac{4}{3} \times 3.14 \times 1225.043 \;nm^3 \\ = 5128.846 \;nm^3V=34πr3r=2d=221.4nm=10.7nmV=34×3.14×(10.7)3=34×3.14×1225.043nm3=5128.846nm3
Convert nm3 to dm3 (L):
5128.846 nm3×(1 m109 nm)3×(102 cm1 m)3×(1 L103 cm3)=5.128×10−21dm3 or L5128.846 \;nm^3 \times (\frac{1 \;m}{10^9 \;nm})^3 \times (\frac{10^2 \;cm}{1 \;m})^3 \times (\frac{1 \;L}{10^3 \;cm^3}) = 5.128 \times 10^{-21} dm^3 \; or \;L5128.846nm3×(109nm1m)3×(1m102cm)3×(103cm31L)=5.128×10−21dm3orL
Convert to μL\mu LμL :
5.128×10−21 L×106 μL1 L=5.128×10−15 μL5.128 \times 10^{-21} \;L \times \frac{10^6 \; \mu L}{1 \;L} = 5.128 \times 10^{-15} \; \mu L5.128×10−21L×1L106μL=5.128×10−15μL
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