carbs Flashcards

(21 cards)

1
Q

what are monosaccharides?

A

monosaccharides are the simplest forms of carbohydrates, with a general formula of (CH2O)n, where n is 3 and more. they have a carbonyl group and a hydroxyl group on the remaining carbons. the presence of a carbonyl group and multiple OH groups provides monosaccharides with hydrophilic functional groups, allowing them to be soluble

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2
Q

orientation of alpha and beta glucose

A

alpha glucose : H on top for C1, H on top of C2, alternate until C4, H below on C5, H20H on C6
beta glucose : OH on top for C1, alternate until C4, H below on C5, H20H on C6

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3
Q

solubility of monosaccharides

A

due to presence of OH groups that can form hydrogen bonds with water, they are soluble in water
since they are soluble, they affect osmotic pressure

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4
Q

formation of alpha and beta glucose

A

in solution, almost all glucose molecules spontaneously react to form one of two ring structures (alpha and beta), where both forms are isomers and exist in equilibrium

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5
Q

reducing ability of monosaccharides

A

all monosaccharides are reducing sugars, where they can carry out reduction, reducing blue copper sulfate containing Cu2+ ions to brick red CuO precipitate containing Cu+ ions

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6
Q

describe advantages of plants condensing a-glucose molecules into starch

A

large amounts of glucose can be stored without affecting the water potential of the cell. it is compact and more glucose can be stored, and released as a substrate for respiration. a-1,6 glycosidic bond provides free end that can be hydrolysed easily

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7
Q

formation of disaccharide

A

formed from condensation reaction joint 2 molecules together

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8
Q

formation and breaking of glycosidic bonds

A

two hydroxyl groups line alongside each other, one combines with a hydrogen atom from another to form a water molecule. this allows the formation of an oxygen bridge (C-O-C) between the two monosaccharides

reverse of condensation is hydrolysis which involves addition of water to break the glycosidic bonds

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9
Q

type of glycosidic bonds in amylose, amylopectin, cellulose and glycogen

A

amylose : a-1,4
amylopectin : a-1,4 and a-1,6
cellulose : b-1,4
glycogen a-1,4 and a-1,6

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10
Q

polysaccharides

A

polymers with a few hundred to few thousand monosaccharides joined by glycosidic linkages. the structure and function are determined by its sugar monomers and by the position of the glycosidic linkages

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11
Q

amylose

A

chains of alpha glucose joined together by a-1,4 glycosidic bonds. it is unbranded consisting of 250 to 4000 a glucose units. the polymer forms a helical structure which makes it compact.

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12
Q

solubility of amylose

A

it is poorly soluble in water as most of its OH groups of a glucose are projected into the helix and are involved in intramolecular hydrogen bonding within the helix, thus few OH groups are available for hydrogen bonding with the water.

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13
Q

amylopectin

A

similar structure as amylose with helical chains of a glucose join together by a-1,4 glycosidic bonds, except that it is branched with the branches linked by a-1,6 glycosidic bonds
the branches provide multiple sites for enzymes to bind and hydrolyse the molecule to release more glucose per unit time

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14
Q

solubility of amylopectin

A

it is more soluble that amylose as branching decreases the ability of chains to get together with one another and increases the binding of water molecules to the chains

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15
Q

glycogen

A

similar structure to amylopectin consisting of helical chains of a glucose joined together by a-1,4 glycosidic bonds, and a-1,6 glycosidic bonds between branched points, but glycogen has more extensive branching (every 8-14 residues) than amylopectin (every 25 residues)

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16
Q

cellulose

A

linear unbranched polymer of b-glucose linked by b-1,4 glycosidic bonds. the alternated b glucose molecules are rotated 180º with respect to each other, where the flipped orientation allows for the formation of interchain hydrogen bonds.

17
Q

cellulose structure

A

there is a large number of hydrogen cross-links formed between the linear cellulose chains due to the large number beta glucose units with protruding OH groups in each chain.

hydrogen bonds are individually weak, but the large number of OH groups allows extensive cross-linking by hydrogen bonding, forming microfibrils that combine to form larger bundles which associate to form cellulose fibres, giving cellulose high tensile strength.

since the OH groups are involved in cross-linking between the cellulose chains, the OH groups aren not free to form hydrogen bonds with water, and therefore cellulose is insoluble in water.

criss-crossing of many microfibrils creates a mesh work resulting in a porous structure that allows of the passage of water and solute molecules. it also forms the cell wall, which helps to distribute stress in all directions.

18
Q

iodine test

A

test for presence of starch
starch present, yellowish-brown solution turns blue-black
glycogen present, yellowish-brown solution turns reddish-brown
both polysaccharides absent, solution remains yellowish-brown

19
Q

benedict’s test

A

presence of reducing sugars

equal amounts of benedict’s reagent (2cm3) to test sample, mix well and place into boiling water bath
reducing sugar : in increasing concentration, blue solution turns from green, yellow-orange, orange to brick red suspension
no reducing sugar : solution remains clear blue

20
Q

non-reducing sugars

A

equal volumes of benedict’s solution to test sample, mix well and place into boiling water bath. (only continue if solution remains clear blue - no reducing sugar)
add equal volumes of HCL to test sample, mix well and place into boiling water bath. allow to cool then neutralise with sodium hydrogen carbonate, until no effervescence is observed. add double the volume of benedict’s solution to test tube, mix well and place in boiling water bath
non-reducing sugar (sucrose) present : sample forms green, yellow, orange, brick-red precipitate (in varying concentrations)
non-reducing sugar absent : solution remains clear blue

21
Q

why is cellulose synthesised in the cell surface membrane and not inside the cell?

A

it is a large and insoluble molecule and cannot pass through the cell surface membrane