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2 years ago

14

I WILL MARK BRAINLYEST

1 answer:

Wittaler [7]2 years ago

3 0

The correct answer is D: stable and not growing or shrinking. I hope this helped:)

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Based on the table, which domains contain only one cell type?

sveticcg [70]

Answer:

Bacteria and Archaea

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3 years ago

A population crash occurs when

jeka94

The correct answer is that a population overshoots the carrying capacity and environmental pressures and shortfalls begin to exert their effects. A certain area of environment only has limited resources such as food, water, habitat, etc. These limited resources are called the carrying capacity of the environment. If the population exceeds the carrying capacity of the environment, there will be not enough resources to sustain the population therefore members of the population will die and the overall population number will crash hence the term <em>population crash. </em>

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3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST

Montano1993 [528]

The second one due to the fact that pioneer plants come in much later

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3 years ago

Read 2 more answers

g A conditional knockout mutant is created by inserting one ____________ site into each of the introns flanking an essential exo

Paraphin [41]

Answer:
loxP sites

Explanation:
Generally, two loxP sites are inserted into the introns flanking an essential exon of the target gene.

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

4 0

3 years ago