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3 years ago

The concentration of pathogen needed to kill 50% of an infected group of test animals is the __________.

Biology

1 answer:

natulia [17]3 years ago

6 0

Answer:

LD50 (median lethal dose)

Explanation:

Whenever one needs to quantify the virulence of a pathogen then this is done by using a set of laboratory animals.
LD50 (median lethal dose) is an important indicator of virulence.
LD50 is defined as the number of pathogens that are required to kill 50% of the total infected animals.
For the calculation of LD50, the animals are infected with a known number of pathogens and then a graph is plotted for the percentage of animals killed against the concentration of the pathogen used for infection.<em> The point at which the graph indicates the death of 50% animals, the corresponding concentration of inoculation of a pathogen is taken as the LD50 of that particular pathogen. </em>

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A spheroidal bacterium with a diameter of 1 mm contains 2 molecules of a particular protein. What is the molar concentration?

creativ13 [48]

Answer : The molar concentration is, $6.35\times 10^{-18}mol/L$

Explanation :

First we have to calculate the moles of protein.

$\text{Moles of protein}=\frac{\text{Number of molecules of protein}}{\text{Avogadro's number}}$

Number of molecules of protein = 2 molecules

Avogadro's number = $6.022\times 10^{23}molecules/mol$

$\text{Moles of protein}=\frac{2\text{ molecules}}{6.022\times 10^{23}molecules/mol}=3.32\times 10^{-24}mol$

Now we have to calculate the radius.

Radius = $\frac{Diameter}{2}=\frac{1mm}{2}=0.5mm=5.0\times 10^{-4}m$

Conversion used : (1 mm = 0.001 m)

Now we have to calculate the volume.

$V=\frac{4}{3}\pi r^3$

$V=\frac{4}{3}\times 3.14\times (5.0\times 10^{-4}m)^3$

$V=5.23\times 10^{-10}m^3=5.23\times 10^{-7}L$

Conversion used : (1 m³ = 1000 L)

Now we have to calculate the molar concentration.

$Concentration=\frac{\text{Moles of protein}}{\text{Volume}}$

$Concentration=\frac{3.32\times 10^{-24}mol}{5.23\times 10^{-7}L}$

$Concentration=6.35\times 10^{-18}mol/L$

Thus, the molar concentration is, $6.35\times 10^{-18}mol/L$

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Explanation:

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