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4vir4ik [10]
1 year ago
5

. In the video, the domain Eukarya includes all eukaryote organisms. In a six kingdom system, these eukaryotes can make up four

of the six kingdoms. Name these separate eukaryotic kingdoms and list descriptive words and examples for each. In your descriptions, use the bolded vocabulary words correctly from #1-3 on the previous page.
Biology
1 answer:
guajiro [1.7K]1 year ago
5 0

Answer:

Fungi, Animalia, Plantae, and Protista

Explanation:

Fungi: A cluster of organisms that produce spores and feed on organic matter, examples are Mushrooms, mold and yeast.

Animals: a classification of living organisms that almost always have organs, and feed on other organic matter, they also usually have a nervous system. Examples are, Pigeons, Lions, Whales, etc.

Plants: organisms such as grass, trees, moss, flowers, and grow in the same spot. They absorb water, and inorganic matter through their roots, and usually have green coloring to do photosynthesis.

Protists: These organisms are usually microscopic and unicellular. They contain a nucleus and other specialized organelles, and some can move around. They vary widely of what they do, with some like Amoebas moving and eating other cells, while algae stay put and perform photosynthesis.

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The equation y = 8x + 12, where x is the number of hours and y is the total cost, represents what the surf instructor charges fo
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Not researched this answer was from common knowledge so if you have a correction or better answer please inform me I will be happy to learn more.

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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

b)

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

c)

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

d)

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 0.8D_{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
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