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DaniilM [7]
3 years ago
11

All of the following statements about photosynthesis are true EXCEPT chlorophyll is located in the thylakoid membrane and is the

light-absorbing pigment that drives photosynthesis in chemiosmosis, electron transport chains pump hydrogen ions across a membrane from areas of low to high concentration the Calvin cycle is a series of reactions that uses energy from ATP and NADPH to convert CO2 to glucose the light reactions convert solar energy to chemical energy stored in the form of ADP and FAD+
Biology
1 answer:
Korolek [52]3 years ago
7 0
The last one: light reactions don’t make ADP and FAD+, they make ATP and NADPH
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The lifetime of an intracellular signaling molecule must be short for the cell to be able to cause a change in the concentration
ivann1987 [24]

Answer:

Statement B is false.

Explanation:

PKA and PKG is both have binding domains to attach with nucleoside monophosphate and help in modulating these enzymes that explains that PKA is more related to the PKG than PKC at the level of amino acid.

At the level of  polypeptide chain level PKG has single polypeptide chain whereas in PKA modulatory domains are located on different poly peptide chain. Both are activated by the nucluoside triphosphate, more precisely PKA by cAMP and PKG is by cGMP respectively, whereas PKC is activated by DAG or Ca or both depending on the isofom.

Thus, the correct answer is - option B.

4 0
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
What are the four body openings in which HIV can be transferred through?
pav-90 [236]
Saliva,unprotected intercourse,contact with infected blood (blood transfusion) 
and women with an unborn child.
5 0
3 years ago
Which of the following is found in both cellular respiration and in the light reactions of photosynthesis?
alexdok [17]
Glycolysis.......hi....
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What provides the energy for the cell to move?
Yuri [45]
I believe the answer is glucose

5 0
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Read 2 more answers
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