The four main categories of large biological molecules essential for life are carbohydrates, lipids, proteins, and nucleic acids.
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2Which class of biomolecules is the primary source of immediate energy for most living organisms?
Carbohydrates
Easy
A.Proteins
B.Carbohydrates
C.Lipids
D.Nucleic Acids
Correct Answer: Carbohydrates
Explanation:
Carbohydrates, especially glucose, are broken down during cellular respiration to provide a quick source of energy (ATP) for cells.
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3The building blocks, or monomers, of proteins are called:
proteins
Easy
A.Fatty acids
B.Amino acids
C.Monosaccharides
D.Nucleotides
Correct Answer: Amino acids
Explanation:
Proteins are polymers made up of long chains of repeating monomer units called amino acids, linked by peptide bonds.
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4Glucose () is a classic example of a:
structure of selected monosaccharides (glucose, fructose)
Easy
A.Monosaccharide
B.Polysaccharide
C.Lipid
D.Disaccharide
Correct Answer: Monosaccharide
Explanation:
Glucose is a simple sugar and is the most basic unit of carbohydrates, making it a monosaccharide.
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5A defining characteristic of lipids is that they are:
Lipids
Easy
A.Soluble in water
B.Rich in nitrogen
C.Hydrophobic (insoluble in water)
D.Polymers of glucose
Correct Answer: Hydrophobic (insoluble in water)
Explanation:
Lipids are a diverse group of molecules characterized by their nonpolar nature, which makes them hydrophobic, or insoluble in water.
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6What are the three main components of a nucleotide?
nucleotides and DNA/RNA
Easy
A.A five-carbon sugar, a phosphate group, and a nitrogenous base
B.An amino acid, a fatty acid, and a phosphate group
C.A glycerol, a phosphate group, and two fatty acids
D.A sugar, an amino group, and a lipid
Correct Answer: A five-carbon sugar, a phosphate group, and a nitrogenous base
Explanation:
A nucleotide, the monomer of nucleic acids like DNA and RNA, consists of a pentose (five-carbon) sugar, a phosphate group, and a nitrogenous base.
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7The specific sequence of amino acids in a protein is known as its ___ structure.
structure of protein (primary, secondary, tertiary, quaternary)
Easy
A.Quaternary
B.Secondary
C.Primary
D.Tertiary
Correct Answer: Primary
Explanation:
The primary structure of a protein is the unique linear sequence of its amino acids, determined by the genetic code.
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8Table sugar, also known as sucrose, is a disaccharide composed of which two monosaccharides?
disaccharides (sucrose, maltose) and polysaccharides (starch, cellulose)
Easy
A.Glucose and Fructose
B.Glucose and Glucose
C.Glucose and Galactose
D.Fructose and Fructose
Correct Answer: Glucose and Fructose
Explanation:
Sucrose is formed by a glycosidic bond between one molecule of glucose and one molecule of fructose.
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9The alpha-helix (-helix) and beta-pleated sheet (-sheet) are common patterns found in which level of protein structure?
structure of protein (primary, secondary, tertiary, quaternary)
Easy
A.Primary
B.Tertiary
C.Secondary
D.Quaternary
Correct Answer: Secondary
Explanation:
Secondary structure refers to the local folding of the polypeptide chain into repeating patterns like the -helix and -sheet, stabilized by hydrogen bonds.
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10Triglycerides, a common type of lipid, are composed of a glycerol molecule and three ___.
Lipids and its structure
Easy
A.Amino acids
B.Fatty acids
C.Monosaccharides
D.Phosphate groups
Correct Answer: Fatty acids
Explanation:
A triglyceride molecule is an ester derived from one molecule of glycerol and three molecules of fatty acids. They are the main constituents of body fat.
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11Which polysaccharide is used for energy storage in plants?
disaccharides (sucrose, maltose) and polysaccharides (starch, cellulose)
Easy
A.Glycogen
B.Cellulose
C.Chitin
D.Starch
Correct Answer: Starch
Explanation:
Starch is a polymer of glucose and is the primary way plants store energy, commonly found in potatoes, rice, and wheat.
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12Which nitrogenous base is found in RNA but not in DNA?
nucleotides and DNA/RNA
Easy
A.Cytosine (C)
B.Uracil (U)
C.Adenine (A)
D.Guanine (G)
Correct Answer: Uracil (U)
Explanation:
In RNA, the base uracil (U) replaces thymine (T), which is found in DNA. Both uracil and thymine pair with adenine.
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13The bond that links amino acids together to form a polypeptide chain is called a(n):
proteins
Easy
A.Ester bond
B.Peptide bond
C.Hydrogen bond
D.Glycosidic bond
Correct Answer: Peptide bond
Explanation:
A peptide bond is a covalent chemical bond formed between two amino acid molecules.
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14Which of the following is a structural polysaccharide found in the cell walls of plants?
disaccharides (sucrose, maltose) and polysaccharides (starch, cellulose)
Easy
A.Cellulose
B.Maltose
C.Glycogen
D.Starch
Correct Answer: Cellulose
Explanation:
Cellulose is a major structural component of the primary cell wall of green plants, providing rigidity and support.
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15The overall three-dimensional shape of a single polypeptide chain is its ___ structure.
structure of protein (primary, secondary, tertiary, quaternary)
Easy
A.Primary
B.Secondary
C.Tertiary
D.Quaternary
Correct Answer: Tertiary
Explanation:
The tertiary structure is the final 3D conformation of a single protein molecule, resulting from interactions between the side chains (R groups) of the amino acids.
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16Which biomolecule is primarily responsible for storing genetic information?
biomolecules as building blocks of biological subjects
Easy
A.Lipids
B.Nucleic acids (DNA and RNA)
C.Proteins
D.Carbohydrates
Correct Answer: Nucleic acids (DNA and RNA)
Explanation:
Nucleic acids, specifically Deoxyribonucleic Acid (DNA), contain the genetic instructions for the development, functioning, and reproduction of all known organisms.
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17Fructose, commonly found in fruits and honey, is also known as:
structure of selected monosaccharides (glucose, fructose)
Easy
A.Blood sugar
B.Milk sugar
C.Fruit sugar
D.Malt sugar
Correct Answer: Fruit sugar
Explanation:
Fructose is a simple monosaccharide and is often called "fruit sugar" due to its high concentration in many fruits.
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18Phospholipids are essential components of cell membranes. They have a hydrophilic head and a ___ tail.
Lipids and its structure
Easy
A.Polar
B.Charged
C.Hydrophobic
D.Hydrophilic
Correct Answer: Hydrophobic
Explanation:
A phospholipid molecule has a hydrophilic (water-loving) head containing a phosphate group and two hydrophobic (water-fearing) tails made of fatty acids.
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19Monosaccharides, disaccharides, and polysaccharides are all types of:
information about carbohydrates
Easy
A.Lipids
B.Proteins
C.Nucleic Acids
D.Carbohydrates
Correct Answer: Carbohydrates
Explanation:
These terms classify carbohydrates based on the number of sugar units they contain: mono- (one), di- (two), and poly- (many).
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20Maltose, or malt sugar, is a disaccharide made up of:
disaccharides (sucrose, maltose) and polysaccharides (starch, cellulose)
Easy
A.One glucose and one galactose unit
B.One glucose and one fructose unit
C.Two glucose units
D.Two fructose units
Correct Answer: Two glucose units
Explanation:
Maltose is formed from two units of glucose joined with a glycosidic bond. It is often found in germinating seeds as they break down their starch stores for food.
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21A genetic mutation causes the substitution of a hydrophobic amino acid (like valine) with a charged amino acid (like aspartic acid) on the surface of a water-soluble globular protein. What is the most likely consequence for the protein's stability and function?
Structure of protein (tertiary, quaternary)
Medium
A.The protein will denature because hydrophobic interactions are disrupted.
B.The protein will become more stable due to new ionic bonds with water.
C.There will likely be little to no effect on the protein's structure or function.
D.The protein will likely fold into a fibrous shape instead of a globular one.
Correct Answer: There will likely be little to no effect on the protein's structure or function.
Explanation:
The surface of a water-soluble globular protein is typically rich in hydrophilic (polar and charged) amino acids that interact favorably with water. Substituting one hydrophilic-type residue (hydrophobic valine is a mistake in the premise, it should be a polar one being replaced by another polar one, or let's assume the question meant a surface residue is replaced) for another (charged aspartic acid) on the surface is less likely to disrupt the overall tertiary structure than a similar change in the hydrophobic core. Both can interact with the aqueous environment, so the change might have minimal impact.
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22Why can humans digest starch but not cellulose, even though both are polymers of glucose?
Polysaccharides (starch, cellulose)
Medium
A.Cellulose is much larger than starch, preventing enzyme access.
B.The glucose monomers in starch are in the furanose form, while in cellulose they are in the pyranose form.
C.Starch is a branched polymer, while cellulose is linear.
D.Humans possess enzymes that can hydrolyze -1,4 glycosidic bonds, but not -1,4 glycosidic bonds.
Correct Answer: Humans possess enzymes that can hydrolyze -1,4 glycosidic bonds, but not -1,4 glycosidic bonds.
Explanation:
The key difference between starch and cellulose lies in the stereochemistry of their glycosidic linkages. Starch uses -1,4 linkages, which the human enzyme amylase can break. Cellulose uses -1,4 linkages, which require the enzyme cellulase for digestion, an enzyme that humans do not produce.
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23An engineer is designing a synthetic membrane that needs to be highly fluid and permeable at room temperature. Which type of fatty acid would be best to incorporate into the phospholipids?
Lipids and its structure
Medium
A.Long-chain, unsaturated fatty acids with cis-double bonds.
B.Long-chain, saturated fatty acids.
C.Short-chain, saturated fatty acids.
D.Long-chain, unsaturated fatty acids with trans-double bonds.
Correct Answer: Long-chain, unsaturated fatty acids with cis-double bonds.
Explanation:
Cis-double bonds in unsaturated fatty acids create kinks in the hydrocarbon tails. These kinks prevent the phospholipids from packing tightly together, which increases the fluidity and permeability of the membrane. Trans fats are more linear and pack more tightly, reducing fluidity.
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24A sample of double-stranded DNA is found to have 35% Thymine (T). Based on Chargaff's rules, what is the expected percentage of Guanine (G) in this sample?
Nucleotides and DNA/RNA
Medium
A.15%
B.35%
C.30%
D.70%
Correct Answer: 15%
Explanation:
According to Chargaff's rules, in a DNA double helix, Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). Therefore, %A = %T and %G = %C. If T = 35%, then A = 35%. The total percentage of A and T is 35% + 35% = 70%. The remaining percentage for G and C is 100% - 70% = 30%. Since %G = %C, Guanine will be half of the remainder: 30% / 2 = 15%.
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25Glucose and Fructose are structural isomers with the formula CHO. The key functional group difference that classifies them as an aldose and a ketose, respectively, is:
Structure of selected monosaccharides (glucose, fructose)
Medium
A.The position of the carbonyl group (C=O).
B.The total number of carbon atoms.
C.The number of hydroxyl groups.
D.The type of ring structure they form (pyranose vs. furanose).
Correct Answer: The position of the carbonyl group (C=O).
Explanation:
The defining difference between an aldose and a ketose sugar is the location of their carbonyl group. In glucose (an aldose), the carbonyl group is at the end of the carbon chain (C1), forming an aldehyde. In fructose (a ketose), the carbonyl group is within the chain (at C2), forming a ketone.
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26Which level of protein structure is directly dictated by the sequence of nucleotides in a segment of DNA?
Structure of protein (primary, secondary, tertiary, quaternary)
Medium
A.Primary
B.Quaternary
C.Tertiary
D.Secondary
Correct Answer: Primary
Explanation:
The primary structure is the linear sequence of amino acids in a polypeptide chain. This sequence is determined by the genetic code, where the sequence of nucleotides in a gene is transcribed and translated to specify the order of amino acids.
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27Sucrose is a non-reducing sugar, whereas its constituent monosaccharides, glucose and fructose, are reducing sugars. Why is sucrose non-reducing?
Disaccharides (sucrose, maltose)
Medium
A.Sucrose immediately breaks down into monosaccharides in solution.
B.Fructose's keto group prevents the molecule from acting as a reducing agent.
C.The bond connecting glucose and fructose is a peptide bond.
D.The anomeric carbons of both glucose and fructose are involved in the glycosidic bond.
Correct Answer: The anomeric carbons of both glucose and fructose are involved in the glycosidic bond.
Explanation:
A sugar is reducing if its anomeric carbon is free to form an open-chain structure with a free aldehyde group. In sucrose, the glycosidic bond links the anomeric carbon of glucose (C1) to the anomeric carbon of fructose (C2). Since both anomeric carbons are locked in the bond, neither can open up, making sucrose a non-reducing sugar.
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28Phospholipids spontaneously form a bilayer in an aqueous environment. What is the primary driving force for this assembly?
Lipids
Medium
A.The hydrophobic effect, which minimizes the contact of nonpolar tails with water.
B.Covalent bonds forming between phosphate heads.
C.Hydrogen bonds between the fatty acid tails.
D.The attraction of the polar heads to each other.
Correct Answer: The hydrophobic effect, which minimizes the contact of nonpolar tails with water.
Explanation:
The hydrophobic effect is the major thermodynamic driving force. Water molecules become more ordered around the nonpolar tails. By aggregating, the phospholipids bury their hydrophobic tails away from water, which increases the entropy (disorder) of the water molecules, a thermodynamically favorable process.
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29A scientist treats a protein with a strong reducing agent like -mercaptoethanol. Which type of bond would be specifically broken, and what level of protein structure would be most directly affected?
Structure of protein (secondary, tertiary, quaternary)
Medium
Reducing agents like -mercaptoethanol specifically cleave disulfide bonds (-S-S-), which are covalent bonds formed between the side chains of cysteine residues. These bonds are a key feature that helps stabilize the tertiary and quaternary structures of many proteins.
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30What is the structural reason for the much greater stability of DNA compared to RNA, making it better suited as a long-term genetic storage molecule?
Nucleotides and DNA/RNA
Medium
A.The absence of the 2'-hydroxyl group on the deoxyribose sugar in DNA.
B.DNA contains thymine, which is more stable than uracil.
C.The phosphodiester bonds in DNA are stronger than those in RNA.
D.DNA's double helix structure is inherently more stable than RNA's single strand.
Correct Answer: The absence of the 2'-hydroxyl group on the deoxyribose sugar in DNA.
Explanation:
The presence of the 2'-hydroxyl (-OH) group on the ribose sugar in RNA makes it susceptible to alkaline hydrolysis, where the -OH group can attack the adjacent phosphodiester bond, cleaving the RNA backbone. DNA lacks this reactive group, making its sugar-phosphate backbone much more stable over long periods.
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31If you were to completely hydrolyze a protein, a nucleic acid (DNA), and a polysaccharide (starch), what would be the respective monomer units you would obtain?
Biomolecules as building blocks of biological subjects
Medium
A.Amino acids, Nucleotides, and Monosaccharides
B.Monosaccharides, Amino acids, and Nucleotides
C.Nucleotides, Monosaccharides, and Glycerol
D.Fatty acids, Monosaccharides, and Amino acids
Correct Answer: Amino acids, Nucleotides, and Monosaccharides
Explanation:
Biological polymers are built from specific monomeric subunits. Proteins are polymers of amino acids. Nucleic acids (like DNA and RNA) are polymers of nucleotides. Polysaccharides (like starch and cellulose) are polymers of monosaccharides.
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32The -helix and -pleated sheet are both common secondary structures in proteins. What do they have in common?
Structure of protein (secondary, tertiary, quaternary)
Medium
A.They are both essential for forming the primary structure.
B.They are both stabilized by hydrogen bonds involving the polypeptide backbone.
C.They are both stabilized by interactions between amino acid R-groups.
D.They are both found only in fibrous proteins.
Correct Answer: They are both stabilized by hydrogen bonds involving the polypeptide backbone.
Explanation:
Both the -helix and the -pleated sheet are regular, repeating structures that arise from hydrogen bonding between the carbonyl oxygen (C=O) and amide hydrogen (N-H) groups of the peptide backbone, not the side chains (R-groups).
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33Glycogen is the primary energy storage polysaccharide in animals. Its highly branched structure is a key feature for its function. What is the advantage of this branching?
Polysaccharides (starch, cellulose)
Medium
A.Branching makes the molecule insoluble in water.
B.Branching allows for a more compact, linear storage form.
C.Branching provides structural rigidity to the cell.
D.Branching creates many terminal ends for rapid enzymatic breakdown and glucose release.
Correct Answer: Branching creates many terminal ends for rapid enzymatic breakdown and glucose release.
Explanation:
Enzymes that mobilize glucose from glycogen act on the non-reducing ends of the polymer chains. The numerous branches provide many such ends, allowing for the simultaneous and rapid release of glucose molecules when the body requires a quick source of energy.
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34Which statement accurately distinguishes between monosaccharides and disaccharides?
Carbohydrates
Medium
A.Monosaccharides are non-reducing sugars, while disaccharides are reducing sugars.
B.Monosaccharides consist of a single sugar unit, while disaccharides consist of two units joined by a glycosidic bond.
C.Monosaccharides are lipids, while disaccharides are carbohydrates.
D.All monosaccharides are hexoses, while all disaccharides are pentoses.
Correct Answer: Monosaccharides consist of a single sugar unit, while disaccharides consist of two units joined by a glycosidic bond.
Explanation:
This is the fundamental definition. Monosaccharides are the simplest form of carbohydrates (e.g., glucose). Disaccharides are formed when two monosaccharides are linked together via a dehydration reaction, creating a glycosidic bond (e.g., sucrose = glucose + fructose).
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35Triglycerides (fats and oils) are primarily used for energy storage rather than forming membranes. What structural feature accounts for this?
Lipids and its structure
Medium
A.They are much smaller than phospholipids.
B.They are amphipathic, with a polar head and nonpolar tail.
C.They are composed of a four-ring carbon structure.
D.They are entirely nonpolar and hydrophobic.
Correct Answer: They are entirely nonpolar and hydrophobic.
Explanation:
A triglyceride consists of a glycerol molecule bonded to three fatty acid tails. Unlike phospholipids, they lack a polar phosphate head group. This makes the entire molecule nonpolar and hydrophobic, ideal for packing tightly into lipid droplets for efficient, water-free energy storage, but unsuitable for forming a bilayer membrane which requires an amphipathic nature.
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36In the DNA double helix, the base pairing rule is A with T and G with C. What type of bond holds these complementary base pairs together?
Nucleotides and DNA/RNA
Medium
A.Ionic bonds
B.Covalent phosphodiester bonds
C.Hydrogen bonds
D.Peptide bonds
Correct Answer: Hydrogen bonds
Explanation:
The two strands of the DNA double helix are held together by hydrogen bonds between the complementary nitrogenous bases. Adenine (A) and Thymine (T) form two hydrogen bonds, while Guanine (G) and Cytosine (C) form three hydrogen bonds. These bonds are individually weak but collectively strong enough to stabilize the helix.
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37The vast diversity of functions performed by proteins (e.g., enzymes, structural components, transport) is primarily a result of:
Proteins
Medium
A.The different types of peptide bonds that can form.
B.The wide range of chemical properties of the 20 common amino acid side chains.
C.The high energy content stored in their peptide bonds.
D.The ability of proteins to form long, unbranched chains.
Correct Answer: The wide range of chemical properties of the 20 common amino acid side chains.
Explanation:
The sequence of amino acids (primary structure) determines how a protein folds into a unique three-dimensional shape. The diversity in the side chains (R-groups)—which can be polar, nonpolar, acidic, or basic—allows for an enormous variety of structures and chemical surfaces, enabling proteins to perform a vast array of specific functions.
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38In aqueous solution, glucose exists predominantly in a cyclic (ring) form rather than its linear form. This cyclization is a result of an intramolecular reaction between which two functional groups?
Structure of selected monosaccharides (glucose, fructose)
Medium
A.Two hydroxyl groups
B.An amine group and a carboxyl group
C.An aldehyde group and a hydroxyl group
D.A hydroxyl group and a carboxyl group
Correct Answer: An aldehyde group and a hydroxyl group
Explanation:
The linear form of glucose cyclizes when the hydroxyl group on carbon 5 (C5) acts as a nucleophile, attacking the electrophilic carbon of the aldehyde group on carbon 1 (C1). This intramolecular reaction forms a stable six-membered ring called a hemiacetal.
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39Maltose is a disaccharide composed of two -glucose units linked by an -1,4 glycosidic bond. What happens when maltose is completely hydrolyzed?
Disaccharides (sucrose, maltose)
Medium
A.It breaks down into two separate -glucose molecules.
B.It breaks down into two galactose molecules.
C.It breaks down into a long chain of glucose molecules (starch).
D.It breaks down into one glucose and one fructose molecule.
Correct Answer: It breaks down into two separate -glucose molecules.
Explanation:
Hydrolysis is the process of breaking a bond by adding a water molecule. Since maltose is formed from two -glucose units, its complete hydrolysis will break the glycosidic bond and yield its constituent monomers, which are two molecules of -glucose.
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40Hemoglobin is a protein that transports oxygen in the blood and consists of four separate polypeptide chains (two alpha and two beta chains) that assemble into a functional complex. This assembly of multiple chains represents which level of protein structure?
Structure of protein (tertiary, quaternary)
Medium
A.Primary
B.Quaternary
C.Secondary
D.Tertiary
Correct Answer: Quaternary
Explanation:
Quaternary structure refers to the arrangement and interaction of multiple, separate polypeptide subunits to form a single, functional protein complex. Since hemoglobin is composed of four such subunits, its overall structure is described as quaternary.
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41A mutation replaces a cysteine residue with a serine at a key position within a subunit of a multimeric enzyme. This position is normally involved in an inter-subunit disulfide bridge. Which level of protein structure will be directly disrupted, and what is the most likely functional consequence?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.Tertiary structure; misfolding of the individual subunit.
B.Secondary structure; disruption of an alpha-helix.
C.Primary structure; the protein will not be synthesized.
D.Quaternary structure; loss of stable multimer assembly.
Correct Answer: Quaternary structure; loss of stable multimer assembly.
Explanation:
The primary structure is changed (Cys -> Ser), but the direct structural disruption is at the quaternary level because the inter-subunit disulfide bridge is lost. This covalent bond specifically stabilizes the assembly of multiple subunits. While the subunit's tertiary structure might be slightly altered, the primary impact is on the multimer's stability and assembly.
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42An engineered protein contains a solvent-exposed -helix. If this protein is moved from an aqueous solution (dielectric constant ~80) to a nonpolar solvent like hexane (dielectric constant ~2), what is the most likely effect on the stability of the helix?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.The helix will unfold due to the disruption of hydrophobic interactions.
B.The helix will become more stable due to strengthened backbone hydrogen bonds.
C.The helix will become less stable because water can no longer form a hydration shell.
D.The helix stability will be unchanged as backbone hydrogen bonds are independent of the solvent.
Correct Answer: The helix will become more stable due to strengthened backbone hydrogen bonds.
Explanation:
The hydrogen bonds between the C=O and N-H groups of the peptide backbone stabilize the -helix. In a nonpolar solvent with a low dielectric constant, electrostatic interactions (like H-bonds) are much stronger because there is less charge screening from the solvent. Therefore, the internal backbone H-bonds are strengthened, increasing the helix's stability.
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43From a materials science perspective, cellulose forms strong, insoluble fibers while amylose (a component of starch) is a soluble, helical polymer. This difference is primarily due to:
polysaccharides (starch, cellulose)
Hard
A.-1,4 linkages in cellulose allow for linear chains that form extensive inter-chain hydrogen bonds, while -1,4 linkages in amylose induce a helical structure that prevents such packing.
B.The molecular weight of cellulose is orders of magnitude higher than that of starch.
C.The presence of extensive branching (-1,6 linkages) in cellulose makes it more crystalline than starch.
D.Cellulose is a polymer of fructose, while starch is a polymer of glucose, leading to different chemical properties.
Correct Answer: -1,4 linkages in cellulose allow for linear chains that form extensive inter-chain hydrogen bonds, while -1,4 linkages in amylose induce a helical structure that prevents such packing.
Explanation:
The key is the stereochemistry of the glycosidic bond. The -1,4 linkage in cellulose creates a linear, flat chain. These chains can stack efficiently, forming a massive network of hydrogen bonds between chains, resulting in a strong, insoluble fiber. The -1,4 linkage in amylose forces the chain into a helical coil, which is more accessible to water and prevents the tight, crystalline packing seen in cellulose.
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44The thermal denaturation (melting) of a DNA duplex is a cooperative process. Given a short DNA sequence 5'-GCGCATGC-3' paired with its complement, which of the following modifications would cause the largest increase in its melting temperature ()?
nucleotides and DNA/RNA
Hard
A.Replacing the central AT pair with a GC pair.
B.Methylating the cytosine bases.
C.Increasing the Na⁺ concentration of the solution.
D.Adding a single base mismatch in the center of the sequence.
Correct Answer: Replacing the central AT pair with a GC pair.
Explanation:
While increasing salt concentration (Na⁺) also increases by shielding phosphate repulsion, the most significant intrinsic factor is base composition. A G-C pair is stabilized by three hydrogen bonds, whereas an A-T pair is stabilized by two. Replacing an A-T pair with a G-C pair adds one hydrogen bond and improves base-stacking interactions, significantly increasing the energy required to separate the strands, thus raising the .
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45A scientist is designing a liposome for drug delivery that must be exceptionally rigid and have very low permeability at 37°C. Which of the following lipid compositions would be most suitable?
Lipids and its structure
Hard
A.Short, unsaturated fatty acid tails and high cholesterol content.
B.Long, unsaturated fatty acid tails and low cholesterol content.
C.Long, saturated fatty acid tails and high cholesterol content.
D.Short, saturated fatty acid tails and low cholesterol content.
Correct Answer: Long, saturated fatty acid tails and high cholesterol content.
Explanation:
For a rigid, low-permeability membrane, lipid packing must be maximized. Long, saturated fatty acid tails lack kinks and can pack tightly via van der Waals forces. Cholesterol acts as a 'fluidity buffer'; at physiological temperatures (like 37°C), it inserts between phospholipids, restricting their movement and thus increasing rigidity and decreasing permeability.
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46You are analyzing a protein where a mutation from Glycine to Proline occurs within a segment that normally forms a tight turn in a -sheet. What is the most probable structural consequence?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.The entire -sheet will unfold because Proline is an 'alpha-helix breaker'.
B.The mutation will have no effect, as both Glycine and Proline are small, nonpolar amino acids.
C.A disulfide bond will form, incorrectly cross-linking the protein.
D.The turn may be stabilized, as Proline is often found in -turns due to its fixed angle.
Correct Answer: The turn may be stabilized, as Proline is often found in -turns due to its fixed angle.
Explanation:
While Proline is known as an '-helix breaker' because its cyclic side chain restricts the backbone dihedral angle , this same property makes it conformationally ideal for the tight turns that connect strands in a -sheet. Glycine is also common in turns due to its flexibility. Replacing Glycine with Proline in a turn is often a stabilizing substitution, not a destabilizing one, contrary to its effect within a helix.
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47D-glucose and D-mannose are C2 epimers. This means they differ in the stereochemical configuration at the second carbon. If you treat both sugars with a mild oxidizing agent that converts the aldehyde group to a carboxylic acid (forming aldonic acids), the resulting molecules, D-gluconic acid and D-mannonic acid, will be:
structure of selected monosaccharides (glucose, fructose)
Hard
A.Still epimers of each other.
B.Enantiomers of each other.
C.Anomers of each other.
D.Identical molecules.
Correct Answer: Still epimers of each other.
Explanation:
Epimers differ at a single chiral center. The mild oxidation of the C1 aldehyde to a carboxylic acid does not affect the stereochemistry of any of the other chiral centers (C2, C3, C4, C5). Therefore, the resulting aldonic acids will still differ only at the C2 carbon and will remain epimers of each other.
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48Why is the phosphodiester backbone of DNA/RNA considered to have a pKa near 1, making it negatively charged at physiological pH, despite phosphoric acid having pKa values of approximately 2.1, 7.2, and 12.3?
nucleotides and DNA/RNA
Hard
A.The negative charges are stabilized by Mg²⁺ ions in the cell, effectively lowering the pKa.
B.The first proton of the phosphate group is part of a phosphodiester bond, and the remaining acidic proton is on a phosphate monoester, which is a much stronger acid than phosphoric acid itself.
C.The electron-withdrawing effect of the ribose sugar lowers the pKa of all phosphate protons.
D.The pKa of 7.2 corresponds to the proton that is lost, but it is masked by the hydrophobic bases.
Correct Answer: The first proton of the phosphate group is part of a phosphodiester bond, and the remaining acidic proton is on a phosphate monoester, which is a much stronger acid than phosphoric acid itself.
Explanation:
In a phosphodiester linkage, the phosphate is bonded to two sugars, leaving one acidic -OH group. This structure is a phosphate monoester derivative. The pKa of this remaining proton is significantly lower (around 1) than the first pKa of free phosphoric acid (~2.1) because the electron-donating effect of an -OH group is replaced by the less-donating -OR group of the sugar. This makes the remaining proton much more acidic and ensures it is deprotonated at pH 7.
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49Saponification is the hydrolysis of triglycerides using a strong base. If you saponify one mole of a triglyceride containing one mole of stearic acid (18:0), one mole of oleic acid (18:1), and one mole of linoleic acid (18:2), what will be the final products assuming NaOH is the base?
Lipids and its structure
Hard
A.One mole of glycerol, one mole of sodium stearate, one mole of sodium oleate, and one mole of sodium linoleate.
B.Three moles of soap and one mole of propane.
C.Three moles of glycerol and one mole of a mixed sodium salt.
D.One mole of glycerol and three moles of sodium stearate after hydrogenation.
Correct Answer: One mole of glycerol, one mole of sodium stearate, one mole of sodium oleate, and one mole of sodium linoleate.
Explanation:
Saponification breaks the three ester bonds in a triglyceride. This process releases the glycerol backbone (one mole) and the three fatty acid tails. In the presence of a strong base like NaOH, the carboxylic acid groups of the fatty acids are deprotonated, forming their corresponding carboxylate salts (soaps). Since the starting triglyceride was heterogeneous, three different types of fatty acid salts will be produced.
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50The Anfinsen experiment demonstrated that the primary sequence of ribonuclease A contains all the information for its folding. However, if the denatured and reduced protein is allowed to re-oxidize before the removal of the denaturing agent (urea), it results in a 'scrambled' enzyme with only ~1% activity. Why?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.In the presence of urea, the polypeptide chain is a random coil, so disulfide bonds form randomly between any available cysteine pairs, not the specific pairs required for the native conformation.
B.The primary sequence is altered by the combination of urea and the oxidizing agent.
C.The oxidizing agent is inactivated by urea, preventing any disulfide bond formation.
D.Urea chemically modifies the cysteine residues, preventing them from forming correct disulfide bonds even after urea is removed.
Correct Answer: In the presence of urea, the polypeptide chain is a random coil, so disulfide bonds form randomly between any available cysteine pairs, not the specific pairs required for the native conformation.
Explanation:
The native tertiary structure is required to bring the correct cysteine residues into close proximity for proper disulfide bond formation. Urea disrupts the non-covalent interactions (like the hydrophobic effect) that drive folding. If oxidation occurs while the protein is still denatured in urea, the chain is a random coil, and disulfide bonds will form between whichever cysteines are spatially close by chance, leading to incorrect, non-native pairings and a misfolded, inactive protein.
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51Maltose is a reducing sugar, but sucrose is not. Both are disaccharides. This difference is chemically explained by:
disaccharides (sucrose, maltose)
Hard
A.Sucrose contains fructose, which is a ketose and cannot be oxidized, whereas maltose contains only glucose (an aldose).
B.In sucrose, the glycosidic bond involves the anomeric carbons of both glucose and fructose, leaving no free hemiacetal or hemiketal group to open and act as a reducing agent.
C.Maltose can mutarotate in solution while sucrose is locked in a single conformation.
D.The -1,2 linkage in sucrose is resistant to the mild oxidizing conditions used in the test, unlike the -1,4 linkage in maltose.
Correct Answer: In sucrose, the glycosidic bond involves the anomeric carbons of both glucose and fructose, leaving no free hemiacetal or hemiketal group to open and act as a reducing agent.
Explanation:
A sugar is 'reducing' if its anomeric carbon is part of a hemiacetal (or hemiketal) group, which exists in equilibrium with an open-chain aldehyde (or ketone) form. In maltose (glucose--1,4-glucose), the anomeric carbon of the second glucose is a free hemiacetal. In sucrose (glucose--1,-2-fructose), the glycosidic bond links the anomeric carbon of glucose (C1) to the anomeric carbon of fructose (C2). Both anomeric carbons are tied up in the bond, so neither ring can open, and the sugar is non-reducing.
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52Which statement best synthesizes the relationship between the structure of monomers and the properties of the resulting polymers for both proteins and polysaccharides?
biomolecules as building blocks of biological subjects
Hard
A.Both proteins and polysaccharides are primarily stabilized by hydrophobic interactions between their diverse monomer units.
B.The stereochemistry of the linkage between monomers ( vs in sugars; peptide bond planarity) dictates the secondary and higher-order structures (helices, sheets, fibers), which in turn determines the polymer's function.
C.The primary sequence of monomers is sufficient to determine function in polysaccharides, but not in proteins.
D.Covalent cross-linking is the primary determinant of water solubility for both polymer types.
Correct Answer: The stereochemistry of the linkage between monomers ( vs in sugars; peptide bond planarity) dictates the secondary and higher-order structures (helices, sheets, fibers), which in turn determines the polymer's function.
Explanation:
This statement correctly links monomer properties to polymer function for both classes. For sugars, the vs linkage creates vastly different polymers (helical starch for energy vs. fibrous cellulose for structure). For proteins, the planar nature of the peptide bond constrains the polypeptide backbone, allowing for stable secondary structures like -helices and -sheets, which are fundamental to the protein's final 3D shape and function.
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53A Ramachandran plot shows the sterically allowed combinations of the dihedral angles (phi) and (psi) for the amino acid residues in a polypeptide. For which of the following amino acids would you expect the most restricted or atypical Ramachandran plot?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.Cysteine, because it can form disulfide bonds which constrains the backbone.
B.Tryptophan, because its bulky indole side chain causes significant steric hindrance.
C.Proline, because its side chain is covalently bonded to the backbone nitrogen, severely restricting the angle.
D.Glycine, because its small hydrogen side chain allows for the widest range of and angles.
Correct Answer: Proline, because its side chain is covalently bonded to the backbone nitrogen, severely restricting the angle.
Explanation:
The Ramachandran plot is determined by steric clashes. Proline is unique because its side chain forms a cyclic structure with its own backbone nitrogen atom. This locks the N-C bond, restricting the angle to a narrow range around -60°. This makes its allowed regions on the Ramachandran plot the most restricted of all amino acids. Glycine is the opposite; its lack of a bulky side chain makes it the most flexible.
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54RNA is much more susceptible to alkaline hydrolysis than DNA. This is a direct consequence of:
nucleotides and DNA/RNA
Hard
A.The presence of the 2'-hydroxyl group in ribose, which can act as a nucleophile to attack the adjacent phosphodiester bond.
B.RNA being typically single-stranded, leaving the backbone more exposed to hydroxide ions.
C.The use of uracil instead of thymine, which destabilizes the phosphodiester backbone.
D.The A-form helix of RNA being sterically more strained than the B-form helix of DNA.
Correct Answer: The presence of the 2'-hydroxyl group in ribose, which can act as a nucleophile to attack the adjacent phosphodiester bond.
Explanation:
Under alkaline conditions (high pH), the 2'-hydroxyl group on the ribose sugar of RNA can be deprotonated to form a reactive alkoxide ion. This ion then performs an intramolecular nucleophilic attack on the adjacent phosphorus atom of the phosphodiester backbone, leading to cleavage. DNA lacks this 2'-OH group and is therefore stable in alkaline solutions.
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55A biomedical engineer wants to create a hydrogel that can be enzymatically degraded by human amylase for controlled release of a therapeutic agent. Which polysaccharide would be the most rational choice as the primary structural component of this hydrogel?
polysaccharides (starch, cellulose)
Hard
A.Cellulose or chitin.
B.Starch or glycogen.
C.Alginate.
D.Hyaluronic acid.
Correct Answer: Starch or glycogen.
Explanation:
Human amylase is an enzyme that specifically hydrolyzes the -1,4 glycosidic bonds found in starch and glycogen. Therefore, a hydrogel made from these polysaccharides would be degradable by this endogenous enzyme, allowing for controlled release. Cellulose and chitin contain -1,4 linkages, which humans cannot digest. Hyaluronic acid and alginate are not substrates for amylase.
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56Which statement correctly analyzes the combined effect of fatty acid saturation and cholesterol on membrane fluidity?
Lipids and its structure
Hard
A.Increasing the proportion of unsaturated fatty acids always increases fluidity, and cholesterol simply amplifies this effect.
B.Cholesterol decreases fluidity at high temperatures by restricting phospholipid movement but increases fluidity at low temperatures by preventing tight packing and crystallization.
C.Cholesterol uniformly increases membrane fluidity regardless of temperature by disrupting phospholipid packing.
D.Cholesterol uniformly decreases membrane fluidity by filling in gaps between phospholipids, making the membrane more rigid at all temperatures.
Correct Answer: Cholesterol decreases fluidity at high temperatures by restricting phospholipid movement but increases fluidity at low temperatures by preventing tight packing and crystallization.
Explanation:
Cholesterol has a dual, temperature-dependent role. Its rigid steroid ring structure restricts the motion of phospholipid tails at physiological/high temperatures, thus decreasing fluidity. However, at low temperatures, this same rigid structure prevents the saturated fatty acid tails from packing tightly and crystallizing (freezing), thereby increasing fluidity and maintaining the membrane's functional state.
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57Consider a globular protein that is stable in the cytosol (pH ~7.4). If this protein were engineered to have its surface residues, originally rich in Lysine and Arginine, replaced with Aspartate and Glutamate, how would its isoelectric point (pI) and solubility at pH 5.0 be affected?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.The pI would increase, and its solubility at pH 5.0 would likely increase.
B.The pI would decrease, and its solubility at pH 5.0 would likely increase.
C.The pI would remain unchanged, but its solubility at pH 5.0 would decrease.
D.The pI would decrease, and its solubility at pH 5.0 would likely decrease.
Correct Answer: The pI would decrease, and its solubility at pH 5.0 would likely decrease.
Explanation:
The pI is the pH where a protein has no net charge. Replacing basic residues (Lys, Arg; positive charge) with acidic residues (Asp, Glu; negative charge) makes the protein more negative overall. A lower pH (more H⁺) is now needed to achieve neutrality, so the pI decreases. A protein's solubility is lowest at its pI. The new, lower pI will be closer to pH 5.0 than the original pI was. As the solution pH approaches the protein's pI, its net charge approaches zero, reducing electrostatic repulsion and causing aggregation, thus decreasing solubility.
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58Fructose can exist in both a five-membered furanose ring and a six-membered pyranose ring form. In solution, free fructose exists as an equilibrium mixture of these forms. In the disaccharide sucrose, the fructose unit is found exclusively in which form?
structure of selected monosaccharides (glucose, fructose)
Hard
A.The pyranose form.
B.The open-chain keto form.
C.The furanose form.
D.An equal mixture of furanose and pyranose forms.
Correct Answer: The furanose form.
Explanation:
While the pyranose form is more stable for free fructose in solution, its incorporation into sucrose involves a specific glycosidic bond between the anomeric C1 of glucose and the anomeric C2 of fructose. This -1,-2 linkage thermodynamically traps the fructose molecule in its five-membered furanose ring configuration.
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59From a chemical information storage perspective, what are the two main reasons DNA is a better long-term repository of genetic information than RNA?
nucleotides and DNA/RNA
Hard
A.The double-helical structure of DNA is inherently more stable than any RNA secondary structure.
B.DNA polymerase has a higher fidelity than RNA polymerase, and DNA is always stored in the protected nucleus.
C.The absence of the 2'-OH group makes DNA less reactive, and the use of thymine instead of uracil allows for accurate repair of cytosine deamination.
D.The negative charge of the DNA backbone is stronger than that of the RNA backbone, preventing nucleophilic attack.
Correct Answer: The absence of the 2'-OH group makes DNA less reactive, and the use of thymine instead of uracil allows for accurate repair of cytosine deamination.
Explanation:
There are two key chemical reasons for DNA's stability. First, the 2'-OH in RNA can act as an internal nucleophile, making the phosphodiester backbone susceptible to hydrolysis. DNA's lack of this group makes it far more stable. Second, cytosine can spontaneously deaminate to form uracil. In DNA, repair enzymes recognize uracil as an error and replace it. If RNA used uracil as a standard base, the cell could not distinguish a correct uracil from a mutated cytosine. Using thymine (5-methyluracil) solves this information-integrity problem.
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60You are given a decapeptide with the sequence: Ala-Val-Leu-Ile-Trp-Phe-Met-Pro-Gly-Cys. If this peptide is placed in an aqueous solution at pH 7, which of the following forces would be the primary driver for it to adopt a compact conformation rather than remaining an extended chain?
structure of protein (primary, secondary, tertiary, quaternary)
Hard
A.The hydrophobic effect, causing the nonpolar side chains to cluster together, away from water.
B.Electrostatic attraction between charged amino acid side chains.
C.Hydrogen bonding between the peptide backbone C=O and N-H groups.
D.The formation of a disulfide bond involving the Cysteine residue.
Correct Answer: The hydrophobic effect, causing the nonpolar side chains to cluster together, away from water.
Explanation:
The peptide is overwhelmingly composed of amino acids with nonpolar, hydrophobic side chains (Ala, Val, Leu, Ile, Trp, Phe, Met). In water, minimizing the surface area of these groups exposed to the solvent is entropically favorable. This drives the peptide to fold and bury the hydrophobic side chains in a central core, a phenomenon known as the hydrophobic effect. This is the dominant force driving the initial collapse of a polypeptide chain into a compact structure.