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Ganezh [65]
3 years ago
7

Question 5 please answer

Biology
1 answer:
otez555 [7]3 years ago
4 0

Answer:

C.

Explanation:

I believe it would be C.

The tree is just starting to get  flowers, nut its already come out of the ground and so its not in the seed stage, and it does not have bark yet  so it cant be an adult tree

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What are some examples of how a cell's structure relates to its function in the body?
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The structure and shape of each type of human cell depends on what function it will perform in the body. For example, red blood cells (RBCs) are very small, flat discs, which allows them to easily fit through narrow capillaries and around sharp corners in the circulatory system to deliver oxygen throughout the body.

The elongated shape of muscle cells allows the contraction proteins to line up in an overlapping pattern that makes muscle flexing possible.

And human sperm cells’ structures allow them to “swim” long distances to reach an egg for fertilization

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Vance understands why rainfall amounts affect different types of vegetation positively and others negatively. vance is demonstra
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Vance is demonstrating conceptual knowledge

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3 years ago
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

7 0
3 years ago
Cell is to tissue as
AfilCa [17]
D. tissue is to organ

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Read 2 more answers
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