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vladimir2022 [97]
1 year ago
6

When Charles Darwin visited the Galapagos Islands in the 1800's, he observed many types of organisms that were similar but lived

on different islands. The four species of mockingbirds found on the Galapagos Islands are shown below. Each species lives on a different island, These species are very similar, but the Hood mockingbird has a longer beak than the other three species. Which of the following best explains the difference?​
Biology
1 answer:
anastassius [24]1 year ago
8 0

Answer:

C

Explanation:

It is right trust me. It needs a longer beak to reach food in smaller areas and it has adapted to the enviroment in many years

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Black mice and tan mice are native species that are observed in a particular region. Human habitation in the region caused an in
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Explanation:

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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

5 0
2 years ago
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