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Ugo [173]
4 years ago
7

Unit Test

Biology
1 answer:
butalik [34]4 years ago
7 0

Answer:

I am pretty sure it is water

Explanation:

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d. cells

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Determine whether each phrase describes an enzyme, an active site, or a substrate. move each phrase to the category it describes
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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

4 0
3 years ago
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