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Wittaler [7]
3 years ago
6

How did the carbon, hydrogen and oxygen atoms in glucose and fructose combine to form sucrose? Include in your description which

atoms bonded together from fructose and glucose to form sucrose and which atoms reformed to create a water molecule.
Biology
2 answers:
I am Lyosha [343]3 years ago
5 0

Answer:

1 oxygen atom and 1 hydrogen atom from the glucose molecule, and 1 oxygen atom from the fructose molecule were taken to make the water molecule. The rest of the atoms in the glucose and fructose molecules (glucose: 6 carbon, 11 hydrogen, 5 oxygen. Fructose: 6 carbon, 12 hydrogen, 5 oxygen.) combine to make sucrose(12 carbon, 22 hydrogen, 11 oxygen).

Lapatulllka [165]3 years ago
5 0

Answer:

A glycosidic linkage connects the carbon 1 in glucose and carbon 2 in fructose.

Also OH is removed from carbon 2 of fructose and carbon 1 of glucose to release water

Explanation:

Disaccharides such as Sucrose are formed by the combination of two or more monosaccharide.

A sucrose is formed when glucose (C6H12O6) combines with another monomer named fructose (C6H12O6) though a dehydration process where water is released and a glycosidic bond (-- O--) is formed

Both these monosaccharides have numbering on their carbon atoms starting from the terminal carbon closest to the carbonyl group. A glycosidic linkage connects the carbon 1 in glucose and carbon 2 in fructose.

Also OH is removed from carbon 2 of fructose and carbon 1 of glucose to release water

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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.

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