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Cloud [144]
3 years ago
13

I don't get this. This is apart of my assignment and I don't get it.​

Biology
1 answer:
nika2105 [10]3 years ago
3 0

Answer:

Producer means an organism that can produce it's own food. Plants are prime examples of producers

Primary Consumer means organisms that primarily eat plants and have low energy levels. Squirrels fit into that category. They're almost always herbivores.

Secondary Consumers eat both plants and animals in the Primary Consumer. Crows are technically secondary consumers but I've never seen one eat a squirrel before.

Tertiary Consumers are predators that have the highest energy levels and are at the top of the food chain. Coyotes eat both squirrels and crows.

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Answer:

Qui-Gon Jinn, to Anakin Skywalker. Midi-chlorians were intelligent microscopic life forms that lived symbiotically inside the cells of all living things. When present in sufficient numbers, they could allow their host to detect the pervasive energy field known as the Force.

Explanation:

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3 years ago
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What process is used by the cell to make mRNA?
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RNA makes a copy of a gene from DNA to mRNA .
6 0
4 years ago
assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

6 0
3 years ago
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Thepotemich [5.8K]

Answer:

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3 0
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Select the correct answer.
GrogVix [38]

Answer:

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