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2 years ago

2. Which of the following can survive either with oxygen or without it?

2 answers:

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2. Which of the following can survive either with oxygen or without it?
facultative anaerobes

3. Where are you likely to find a photoautotroph?

near the surfaces of lakes and streams

4. During what process do prokaryotes exchange genetic information?

conjugation

5. What would be a direct consequence of the disappearance of nitrogen-fixing prokaryotes?

Organisms would not be able to get nutrients they need to make proteins

Mila [183]2 years ago

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2. The answer is; facultative anaerobes. Facultative means optionally while obligate means obligatory. Therefore obligate anaerobes only thrive in the presence of oxygen while obligate anaerobes thrive in the absence of oxygen. Examples of facultative anaerobes are Staphylococci and Escherichia bacterias

3. The answer is; near the surfaces of lakes and streams. Photoautotrophs are organisms that synthesize biomolecules and harness their energy from light. They use the process of photosynthesis to capture the light photons. The organism will be found on the surface of lakes and stream where sunlight can reach and to depths that light can penetrate.

4. The answer is; conjugation. In bacteria, this process occurs through a structure called pili that forms in the ‘male’ bacteria. It is through the exchange of the genetic material that bacteria confer traits such as drug resistance to other bacteria. It’s a horizontal method of gene transfer.

5. The answer is; Organisms would not be able to get nutrients they need to make proteins. Nitrogen-fixing bacteria fix nitrogen in the soil. This means that they convert the abiotic form of nitrogen from the atmosphere to organic molecules such as ammonia. Ammonia is absorbed by organisms such as plants and becomes a constituent of amino acids that make up proteins.

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The brain signals the body of an animal to move by transmitting electrical impulses through the neurons. An animal is then able

Fed [463]

Answer:

The muscles of the muscular system work with the bones of the skeletal system.

Explanation: The musculoskeletal system is what enables movement.

(I hope this helps, you can also search up the same question maybe someone also answered it.)

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1 year ago

As they flow over rotten logs as a fluid sheet, slime molds appear to lack any partitioning into cell units; however, slime mold

Brilliant_brown [7]

Answer:

Since cells are utilized at a point or the other in the organism life we can say the general concept of life-is-cellular holds and abandon the cell membrane partitions notion

Explanation:

where the metabolic processes occur that keep the organism alive is the cellular level of an organism (the general concept of life-is-cellular)

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2 years ago

Characteristics of the early u.s. navy "super frigates" included ______________. select all that apply.

Stolb23 [73]

The original six frigates of the United States Navy came up with the Naval Act of 1794. Two characteristics are distinct of these said early US. Navy "super frigates. They are:

· timbers put close together to be able to withstand heavy cannon fire

· hull reinforced to resist the pressures of heavier ordnance on deck

4 0

2 years ago

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

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

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

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 0.8 $D_{max}$

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

(d) Researchers discovered a strain of C. parvum that expresses a functional variation of the lactate dehydrogenase gene. A DNA

Gelneren [198K]

Answer:

Changing the allosteric site would definitely impact the sensitivity of the blocker, and we can not understand precisely how it is owing to our lack of awareness of the specific adjustments and the FX11 layout.

Explanation:

The move would most likely reduce affinity, and FX11 will no longer be as successful as inhibiting C. Growth of parvum. An inhibitor may reach an allosteric site since the site has some sizes and operational classes that precisely match the shape and operational categories of the inhibitor, which is how the association is obtained if the shape is modified and the inclination is affected.

Such chemicals can be used as human drugs because the mechanism we 're disrupting isn't that normal in human cells, we 're talking about lactic fermentation. C.parvum is a parasite that is present in the digestive tract, and these areas do not appear to experience aerobic glycolysis. The material that undergoes this process under other conditions is muscle tissue. It is possible that the absorbed drug can penetrate the bloodstream and touch other organs, and we would recommend that clinicians avoid exercise during this drug therapy.

3 0

2 years ago