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yanalaym [24]
3 years ago
15

What process in animal cells required oxygen

Biology
1 answer:
Debora [2.8K]3 years ago
6 0
I think you are referring to the Cellular respiration where you take in glucose and oxygen and make atp, lactic acid, sugar, and alcohol(basically fermentation)

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Barnacles live on whales. whales carry their barnacles around, which helps the barnacles feed. the whales are unaffected. this i
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The endosymbiotic theory states that eukaryotes evolved from prokaryotes. Which statement is part of the endosymbiotic
tatyana61 [14]

Note about the question:

Statements are missing, and I failed to find the complete question. So here I give you some details about the theory that will help you to indetify the correct statement.  

Answer and Explanation:                            

The endosymbiotic theory essentially states that some organelles of the eukaryotic cells, such as mitochondria and chloroplasts, were once free-living bacteria. Probably, these organisms must have been phagocytosed but not digested by another cell. On the contrary, these bacteria were able to adapt to their host in such a way that the two cells established a dependent relationship with each other.

It is speculated that chloroplasts derivate from cyanobacteria and that mitochondria derivate from rickettsias.

Cells would be beneficiated from this new bond, at the point that they could not survive by themselves anymore.

This theory is supported by a few characteristics of the chloroplasts and mitochondria that suggest that they once were a free cell. For example,

  • Both organelles present their genetic material. This DNI is independent of the cell´s DNA, is bi-catenary and circular, identical to the bacterial DNA, and very different from the one of the eukaryotic cells.
  • These organelles do not divide by mitosis. Instead, they do it by binary fission and are capable of synthesizing their ribosomes and organelles.
  • Both organelles present a double membrane, a characteristic that reinforces the idea of being phagocyted. The internal membrane looks identical to the bacteria membrane, while the external membrane looks like the eukaryotic one.
  • In fact, in this internal membrane are placed the energy centers, exactly as it occurs in bacterias membrane.
  • Finally, the sizes of the organelles are similar to the size of some procaryotes.
8 0
3 years ago
Pseudomonas putida is used for fermentation of the lactose present in cheese whey. The bacteria are cultivated in a steady-state
fgiga [73]

Answer:

Explanation:

Given that:

The dilution rate D = 0.28 h⁻¹

The concentration of lactose in the feed S_o = 2.0 \ g/L

The effluent S = 0.10 g/L

Also;

Y_{X/S} = 0.45 g\  X/g  \ S , \\ \\  Y_{X/O2  }= 0.25 g  \ X/g  \ O2,  \\ \\

Saturation C* = 8 mg/l

To calculate the steady-state biomass, we use the formula:

X = Y_{X/S}(S_o-S_e) \\ \\ X = 0.45(2.0 -0.10) \ g/L \\ \\ X= 0.45 (1.9) \ g/L \\ \\  X = 0.855\ g/L \\ \\  X = 855 \ mg/L

The biomass is 0.855 g/L

For a steady-state condition, the oxygen uptake rate can be illustrated by using the formula:

q_{o_2}X =\dfrac{\mu_X}{Y_{X/O_2}}

where;

\mu = dilution rate (D)

Thus, the steady-state can be expressed as:

q_{o_2}X =\dfrac{D}{Y_{X/O_2}}

q_{o_2}X =\dfrac{0.28}{0.25}

q_{o_2}X =1.12 \ h^{-1}

The specific rate of oxygen consumption q_{o_2}X =1.12 \ h^{-1}

b)

In the fermentation medium, if the desired DO concentration C_L = 2 mg/L

Here, the oxygen transfer is regarded as the rate-limiting step.

As such, the oxygen transfer rate(OTR) is equivalent to the oxygen uptake rate.

In this scenario, let's determine the oxygen transfer coefficient (K_{La}) by using the formula:

OTR =  K_{La}(C^* - C_L)

where;

K_{La}= coefficient of oxygen transfer

C* = saturation

Since OTR = q_{O_2}X

q_{o2}X = K_{La}(C^*-C_L) \\\\ (1.12 )(855) = K_{La}(8-2) \\ \\  957.6 = K_La (6) \\ \\ K_{La}= \dfrac{957.6}{6} \\ \\ K_{La} = 159.6 \ h^{-1}

Thus, the oxygen transfer coefficient K_{La} = 159.6 h⁻¹

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