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2 years ago

What element is the most abundant in the earth's crust?

1 answer:

3 0

The element that is the most abundant in the Earth's crust is oxygen. It makes up 46.6% of the crust. It is followed by silicon, aluminum, iron, etc.

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A purple flower with an unknown genotype is crossed with a white flower. Determine the genotype of the purple flower if purple (

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What initiates an action potential on a muscle cell?

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A change in membrane potential

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Uma ameba que tivesse seu habitat no meio marinho, sendo isotônica em relação a esse meio, com certeza poderia dispensar em sua

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2 years ago

A researcher is examining the response of corn to herbivory by white-tailed deer. She sets up a corn field with a very high fenc

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Answer:

The herbivory by the white-tailed deer

Explanation:

The independent variable can be defined as the variable that who's values do not depend on those of another.

The independent variable in this case is the herbivory by the white-tailed deer which is divided into two categories.

The dependent variables is the response of the corn to the independent variable which are measurements to be taken as shown in the question; the height of the corn plants and the number of ears or corn each plant produces.

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2 years ago

Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l

erastova [34]

Answer:

a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d) the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

$D = \frac {D_{max}[S] }{ks+[S]}$

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

$0.5 = \frac{S}{1.3+[S]}$

$\\\0.65=0.5[S]$

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

Substrate concentration when the dilution rate is 0.8 $D_{max}$ is calculated as:

$D = \frac {D_{max}[S] }{ks+[S]}$

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore , the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

Maximum dilution rate is calculated using the expression $D_{max} = \frac{1}{time}$

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

The cell productivity at 0.8 $D_{max}$ can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8 $D_{max}$

Cell yield = $\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}$

Cell productivity at 0.8 $D_{max}$ = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

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2 years ago