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3 years ago

Oxidation of an amino acid via the citric acid cycle yields just as much ATP as oxidation of glucose.

1 answer:

7 0

I think it is true. If it is wrong, I am so sorry.

Please mark brainliest if the answer is correct. Hope this HELPS!!!!

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3 years ago

How do scientists use the differences in proteins compare Evolutionary Relationships between organisms?

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Explanation:

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3 years ago

Control-part of the body that determines what to do with the information about the change in setpoint (I will give brainliest)

Ann [662]

Answer:

Hypothalamus

Explanation:

The hypothalamus is involved in different daily activities like eating or drinking, in the control of the body's temperature and energy maintenance, and in the process of memorizing and in stress control. It also modulates the endocrine system through its connections with the pituitary gland.

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2 years ago

Dalton’s theory was identified using

Likurg_2 [28]

Answer:

Explanation:

Actually it was the experimental results of other scientists. The main stems of the theory was the indivisibility of small spheres (his image) that represented the elements he knew which reacted with other elements.

They seemed to give the same ratios in reactions (definite proportions) and multiple whole number ratios. Look up Lavoisier.

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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