Exam (elaborations)
XII_Chem_New_Chap-10 Biomolecules (143 AR Items)
XII_Chem_New_Chap-10 Biomolecules (143 AR Items)
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Spread the Happiness. Assertion and Reason Statements Feedback: 9008271516XII_CHEMISTRY_NEW_CHAPTER-10: BIOMOLECULES _ A&R TEST ITEMS
# Correct Assertion Correct Reason
10.1 Carbohydrates; 10.1.1 Classification of Carbohydrates; 10.1.2 Monosaccharides
Carbohydrates are polyhydroxy aldehydes or ketones, The presence of specific functional groups (polyhydroxy aldehyde or
1 or compounds yielding such units on hydrolysis. ketone) defines carbohydrates.
Carbohydrates are classified into monosaccharides, The extent of hydrolysis determines the classification of
oligosaccharides, and polysaccharides based on the carbohydrates (mono-, oligo-, or polysaccharides).
2
number of monosaccharide units yielded upon
hydrolysis.
Monosaccharides are further classified as aldoses or The functional group present differentiates aldoses from ketoses
3 ketoses based on the functional group present (aldehyde vs. keto).
(aldehyde or keto).
Disaccharides are a specific category of Disaccharides are distinguished from other oligosaccharides by
4 oligosaccharides yielding two monosaccharide units consistently yielding two monosaccharide units.
on hydrolysis.
Monosaccharides are the fundamental building blocks Monosaccharides cannot be further hydrolyzed into simpler units.
5 of carbohydrates as they cannot be broken down
further.
Carbohydrates are also called saccharides, derived The term "saccharides" is synonymous with carbohydrates.
6
from the Greek word "sakcharon" meaning sugar.
Carbohydrates are classified as reducing or non- Reducing sugars can reduce Fehling's solution and Tollens' reagent.
7
reducing sugars based on their reducing properties.
The classification of carbohydrates is based on their Their hydrolysis behavior determines how carbohydrates are
8
behavior upon hydrolysis. classified.
Carbohydrates are primarily produced by plants and The definition of carbohydrates encompasses their primary
are a very large group of naturally occurring organic production by plants, their classification as naturally occurring
9
compounds with a general formula (Cn(H2O)y) that organic compounds, and acknowledges limitations of the general
doesn't always hold true. formula.
Disaccharides yield two monosaccharide units on The number of monosaccharide units obtained classifies
10 hydrolysis, and the units can be the same or different. disaccharides, and these units can be identical or distinct.
Monosaccharide nomenclature includes the number The inclusion of the carbon atom count in monosaccharide
11
of carbon atoms for precise structural identification. nomenclature facilitates identification.
Monosaccharides are further classified based on both The classification of monosaccharides depends on both their carbon
12 the number of carbon atoms and the functional group atom count and the specific functional group they contain.
they contain (aldose or ketose).
10.1.2.1 Glucose; Structure of Glucose; Cyclic Structure of Glucose
Glucose reacts with hydroxylamine and HCN Confirms a carbonyl group (C=O) in glucose, essential for these
13
(functional groups with C=O). reactions.
Glucose oxidation with Br2 water yields gluconic acid Indicates an aldehydic group (adjacent to the primary alcohol)
14
(carboxylic acid). susceptible to oxidation to carboxylic acid.
Acetylation of glucose with acetic anhydride forms Confirms five hydroxyl groups in glucose, as each reacts with acetic
15
glucose pentaacetate. anhydride to form an ester linkage.
Glucose oxidation with HNO3 yields saccharic acid Indicates a primary alcohol group (-OH bonded to a primary carbon)
16
(dicarboxylic acid). susceptible to further oxidation to a dicarboxylic acid.
Molecular formula (C6H12O6) of glucose suggests a The formula depicts six carbons, twelve hydrogens, and six oxygen
17
straight chain structure. atoms, consistent with a straight chain.
Open-chain structure cannot explain all properties Inconsistency between the structure's functional groups and
18 (Schiff's test, NaHSO3 addition, anomers). observed behaviors necessitates a more comprehensive
explanation.
Cyclic hemiacetal structures explain the absence of a Ring formation involving the C-5 hydroxyl group masks the aldehyde
19
free aldehyde group. functionality, hindering reactions typical of aldehydes.
Cyclic hemiacetal structures explain the existence of The configuration of the hydroxyl group at C1 (anomeric carbon)
20
anomers. differs between anomers due to the cyclic structure.
Cyclic and open-chain structures coexist in Explains the presence of multiple crystalline forms and properties
21
equilibrium. not fully explained by either structure alone.
Specific cyclic structure with C-5 hydroxyl The specific ring closure involving C-5 hydroxyl can result in
22 involvement explains the two crystalline forms. different spatial arrangements, leading to distinct crystalline forms.
10.1.2.2 Fructose; Structure of Fructose; 10.1.3 Disaccharides; 10.1.4 Polysaccharides; 10.1.5 Importance of Carbohydrates
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XII_CHEMISTRY_NEW_CHAPTER-10: BIOMOLECULES _ A&R TEST ITEMS
# Correct Assertion Correct Reason
Disaccharides like maltose and lactose are reducing Unbonded aldehyde group on a monosaccharide unit allows
23
sugars. participation in reduction reactions.
Sucrose, a disaccharide, is non-reducing. Glycosidic bond formation involves the reducing ends of glucose and
24
fructose, leaving no free aldehyde group.
Cyclic structures of fructose can affect its reactivity Ring formation restricts accessibility of the ketone group,
25
compared to the open-chain form. potentially hindering reactions requiring a free carbonyl group.
Fructose, a ketohexose, has a different spatial The presence of a ketone (fructose) versus an aldehyde (glucose)
26 arrangement compared to glucose, an aldohexose. influences how the molecules fold and form cyclic structures,
leading to distinct spatial configurations.
Starch serves as a readily available energy source. α-glycosidic linkages and a mixed linear/branched structure allow
27
for efficient enzymatic breakdown to release glucose for energy.
Humans cannot digest cellulose as readily as starch. Humans lack enzymes to break β-glycosidic linkages (C1-C4) in
28 cellulose, while human enzymes cleave α-glycosidic linkages (C1-C4
and C1-C6) in starch.
The positions of glycosidic bonds influence spatial Glycosidic bond locations can affect the 3D structure and enzyme
29 arrangements and potentially digestibility in interaction, potentially impacting digestibility. (Requires further
disaccharides (e.g., maltose/lactose vs. sucrose). investigation for definitive cause-effect).
The absence of an aldehyde group in the Acetic acid reacts with hydroxyl groups to form esters, indicating
30 pentaacetate of D-glucose suggests chemical conversion of the aldehyde group into an acetate group by all five
modification with acetic acid. available hydroxyls on the glucose molecule.
Glucose and sucrose are soluble in water, while Multiple hydroxyl groups in glucose and sucrose form hydrogen
cyclohexane and benzene are not. bonds with water molecules, enabling them to dissolve.
31
Cyclohexane and benzene lack polar groups for hydrogen bonding
with water, leading to insolubility.
Lactose hydrolysis yields glucose and galactose. The glycosidic linkage in lactose connects glucose and galactose, and
32 hydrolysis breaks this bond to produce these two monosaccharides.
10.2 Proteins; 10.2.1 Amino Acids; 10.2.2 Classification of Amino Acids; 10.2.3 Structure of Proteins; 10.2.4 Denaturation
Amino acid side chains influence protein folding and The interactions between side chains (ionic interactions, hydrogen
33 function. bonding, hydrophobic interactions) dictate protein folding and
shape, affecting function.
Amino acids with charged side chains can influence Charged side chains can form ionic interactions with water
34
protein solubility in water. molecules, enhancing protein solubility.
Changes in temperature or pH can denature proteins. Physical or chemical alterations disrupt hydrogen bonds and other
35 stabilizing interactions, leading to protein unfolding and loss of
function.
Disulfide bonds contribute to the stability of the Disulfide bonds covalently link cysteine side chains, creating a more
36
tertiary structure of proteins. rigid structure and preventing unfolding.
Fibrous proteins are generally insoluble in water. Fibrous proteins often have hydrophobic side chains on the exterior,
37
limiting hydrogen bond formation with water.
Globular proteins are generally soluble in water. Polar amino acid side chains on the outer surface of globular
38 proteins form hydrogen bonds with water molecules, enabling
solubility.
Peptide bonds form by condensation between a This reaction eliminates a water molecule and links amino acids in a
39 carboxyl group of one amino acid and the amino polypeptide chain.
group of another.
Smaller and more polar amino acids tend to form Smaller amino acids have a higher surface area-to-volume ratio,
40 more hydrogen bonds with water, leading to greater allowing for more hydrogen bonds with water molecules. Polar
water solubility. amino acids can also form hydrogen bonds with water.
The folding of polypeptide chains determines the Hydrogen bonding between peptide backbones causes folding into
41 secondary and tertiary structures of proteins. α-helix and β-pleated sheet structures (secondary), while various
interactions influence overall folding (tertiary).
The globular shape of many enzymes allows for a The folding of the polypeptide chain in globular enzymes creates a
42
specific substrate binding site. pocket or cleft that specifically binds to the substrate molecule.
Amino acids have zwitterionic character in aqueous Amino acids can gain a proton (amino group) and lose a proton
43
solutions. (carboxyl group), resulting in a dipolar ion.
The zwitterionic character of amino acids contributes Zwitterions can form hydrogen bonds with water molecules due to
44
to their solubility. the presence of polar groups.
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