Chemistry Of Life Chapter 2 Answer Key

Chemistry of Life – Chapter 2 Answer Key

The Chemistry of Life chapter 2 explores the fundamental building blocks of biological molecules, the role of water, and the principles governing macromolecular structure. Below is a complete, step‑by‑step answer key that follows the typical textbook layout, providing clear explanations, calculations, and diagrams where needed. Use this guide to verify your solutions, deepen your understanding, and prepare for exams or laboratory work.

1. Introduction to Biological Molecules

Answer example:
The six most abundant elements are carbon (C), hydrogen (H), oxygen (O), nitrogen (N), phosphorus (P), and sulfur (S). They constitute roughly 99 % of the dry weight of living organisms because they form the backbone of proteins, nucleic acids, carbohydrates, and lipids.

2. Water – The Solvent of Life

2.1 Physical Properties of Water

1. Polarity – The O–H bonds are polar; the molecule has a bent shape, giving a dipole moment of 1.85 D.
2. Hydrogen bonding – Each water molecule can form up to four hydrogen bonds, leading to high cohesion, surface tension, and specific heat.
3. High specific heat (4.18 J g⁻¹ °C⁻¹) – Stabilizes temperature in organisms.
4. Density anomaly – Ice is less dense than liquid water because hydrogen bonds create an open lattice.

Sample calculation (specific heat):
A 50 g water sample absorbs 209 J of heat and its temperature rises from 20 °C to 25 °C. Verify the specific heat.

[ q = m \times c \times \Delta T \ c = \frac{q}{m\Delta T} = \frac{209\ \text{J}}{50\ \text{g}\times5\ \text{°C}} = 0.836\ \text{J g}^{-1}\text{°C}^{-1} ]

The result (0.On top of that, g. 84 J g⁻¹ °C⁻¹) is lower than the accepted value, indicating experimental error (e., heat loss to the container).

2.2 Water as a Reactant

• Hydrolysis – Water splits into H⁺ and OH⁻, breaking bonds in polymers (e.g., disaccharides → monosaccharides).
• Condensation (dehydration synthesis) – Removal of water joins monomers (e.g., amino acids → dipeptide).

Equation example:

[ \text{Glucose} + \text{Fructose} \xrightarrow{\text{hydrolysis}} \text{Maltose} + \text{H}_2\text{O} ]

3. Carbohydrates

3.1 Monosaccharide Structure

• General formula: C(n)H({2n+2})O(_n) (n = 3–7).
• Aldoses have an aldehyde group; ketoses have a ketone group.
• D- and L- configurations are defined by the chiral carbon farthest from the carbonyl.

Typical question: Draw the Fischer projection of D‑glucose and label the C‑2 to C‑5 hydroxyl groups.

Answer summary:

• C‑1: aldehyde (CHO)
• C‑2: OH on the right (R)
• C‑3: OH on the left (L)
• C‑4: OH on the right (R)
• C‑5: OH on the right (R) → D‑configuration.

3.2 Disaccharides and Polysaccharides

Polysaccharide classification:

• Storage – Starch (amylose + amylopectin) in plants; glycogen in animals.
• Structural – Cellulose (β‑1→4 linkages) in plant cell walls; chitin (N‑acetylglucosamine) in arthropod exoskeletons.

Problem: Calculate the number of glucose units in a glycogen particle that contains 10 % protein by mass, assuming the total mass is 100 mg and the protein is negligible in volume.

Solution:

• Protein mass = 10 mg → carbohydrate mass = 90 mg.
• Molar mass of glucose = 180 g mol⁻¹.

[ \text{Moles of glucose} = \frac{0.090\ \text{g}}{180\ \text{g mol}^{-1}} = 5.0\times10^{-4}\ \text{mol} ]

[ \text{Number of units} = 5.In practice, 0\times10^{-4}\ \text{mol} \times 6. 022\times10^{23}\ \text{units mol}^{-1} \approx 3.

4. Lipids

4.1 Fatty Acid Nomenclature

• Saturated – No C=C double bonds; e.g., stearic acid (C₁₈:0).
• Unsaturated – One or more C=C bonds; notation C₁₈:1 (one double bond).
• cis vs. trans – Geometry around double bond influences membrane fluidity.

Example question: Identify the number of double bonds and the configuration in oleic acid (C₁₈:1 Δ⁹ cis).

Answer: One double bond at carbon 9 in the cis configuration, creating a kink that prevents tight packing.

4.2 Triglyceride Formation

[ \text{Glycerol} + 3\ \text{fatty acids} \xrightarrow{\text{esterification}} \text{Triglyceride} + 3\ \text{H}_2\text{O} ]

Energy content: Each gram of fat yields ≈ 9 kcal, about twice the energy of carbohydrates or proteins (≈ 4 kcal g⁻¹).

4.3 Phospholipids and Membrane Structure

• Amphipathic: polar head (phosphate + choline) + two non‑polar fatty tails.
• Form bilayers because hydrophobic tails avoid water while heads interact with aqueous environments.

Diagram description (textual):
A cross‑section shows parallel rows of phospholipids; the hydrophilic heads face outward toward the extracellular fluid and cytosol, while the tails create an interior hydrophobic core.

For more on this topic, read our article on x 2 x 8 0 or check out words that start with the letter o.

Question: Explain why cholesterol increases membrane stability at low temperature.

Answer: Cholesterol inserts between phospholipid tails, reducing the packing defects caused by unsaturated kinks. At low temperature, it prevents the fatty‑acid chains from packing too tightly, maintaining fluidity and preventing the membrane from becoming brittle.

5. Proteins

5.1 Amino Acid Structure

• General formula: NH₂–CH(R)–COOH.
• R‑group determines side‑chain properties (non‑polar, polar, acidic, basic).

Key concept: Peptide bond is a planar amide linkage formed by a condensation reaction between the carboxyl group of one amino acid and the amino group of the next, releasing water.

5.2 Levels of Protein Structure

1. Primary – Linear sequence of amino acids (determined by DNA).
2. Secondary – α‑helix (hydrogen bonds every i→i+4) and β‑sheet (hydrogen bonds between adjacent strands).
3. Tertiary – Overall 3‑D folding driven by hydrophobic interactions, disulfide bridges, ionic bonds.
4. Quaternary – Assembly of multiple polypeptide subunits (e.g., hemoglobin).

Sample problem: Predict the effect of substituting a hydrophobic leucine with a polar serine at position 45 in the core of a globular protein.

Answer: The substitution introduces a polar side chain into a hydrophobic core, destabilizing the protein by disrupting hydrophobic interactions. This may lead to partial unfolding or loss of function.

5.3 Enzyme Kinetics (Chapter 2 often includes a brief intro)

• Michaelis–Menten equation:

[ v = \frac{V_{\max}[S]}{K_m + [S]} ]

• Interpretation: (K_m) = substrate concentration at half‑maximal velocity; lower (K_m) = higher affinity.

Calculation example:
Given (V_{\max}=120\ \mu\text{mol min}^{-1}) and (K_m=5\ \text{mM}), find the rate when ([S]=10\ \text{mM}).

[ v = \frac{120 \times 10}{5 + 10} = \frac{1200}{15} = 80\ \mu\text{mol min}^{-1} ]

6. Nucleic Acids

6.1 Nucleotide Composition

• Phosphate group – Provides negative charge.
• Pentose sugar – Ribose (RNA) or deoxyribose (DNA).
• Nitrogenous base – Purines (A, G) or pyrimidines (C, T/U).

Key reaction: Polymerization via phosphodiester bonds between the 3′‑OH of one nucleotide and the 5′‑phosphate of the next, releasing pyrophosphate.

6.2 DNA vs. RNA

Typical question: Write the complementary DNA strand for 5′‑ATG CCT‑3′.

Answer: 3′‑TAC GGA‑5′ (reading 5′→3′ gives 5′‑AGG CAT‑3′).

7. Integrated Metabolism Overview (Connecting Chapter 2 to later topics)

• Catabolism of carbohydrates → glycolysis → pyruvate → acetyl‑CoA.
• Anabolism of fatty acids uses acetyl‑CoA and NADPH from the pentose phosphate pathway.
• Protein turnover recycles amino acids for gluconeogenesis or urea cycle.

Conceptual diagram (described): A flow chart starts with glucose, branches to glycolysis, TCA cycle, and oxidative phosphorylation; another branch shows fatty acid synthesis from citrate; a third branch illustrates amino‑acid deamination feeding into the TCA cycle.

8. Frequently Asked Questions (FAQ)

Q1. Why is water considered a “universal solvent”?
A: Its polarity and ability to form hydrogen bonds allow it to dissolve ionic compounds (e.g., salts) and polar molecules (e.g., sugars), facilitating biochemical reactions.

Q2. How do unsaturated fatty acids affect membrane fluidity?
A: The cis double bonds introduce kinks, preventing tight packing of phospholipid tails, thus increasing fluidity, especially at lower temperatures.

Q3. What distinguishes a protein’s primary structure from its tertiary structure?
A: Primary structure is the linear amino‑acid sequence; tertiary structure is the three‑dimensional folding of that chain, driven by interactions among side chains.

Q4. Can you convert a polysaccharide into a nucleic acid?
A: Direct conversion is not possible; however, carbon skeletons from carbohydrates can be redirected through metabolic pathways (e.g., the pentose phosphate pathway) to generate ribose‑5‑phosphate, a precursor for nucleotides.

Q5. Why do enzymes often require a cofactor?
A: Cofactors (metal ions or organic molecules) can participate directly in the chemical transformation, stabilize transition states, or assist in substrate binding.

9. Summary and Study Tips

• Master the properties of water; they underlie every biochemical process.
• Memorize the basic structures of the four major macromolecule classes and be able to draw representative diagrams (e.g., Fischer projection of glucose, phospholipid bilayer).
• Practice calculations involving molar masses, specific heat, and enzyme kinetics; they appear frequently in exam questions.
• Connect concepts: understand how carbohydrate catabolism supplies acetyl‑CoA for fatty‑acid synthesis, and how amino‑acid degradation feeds the TCA cycle.
• Use visual aids: sketching structures and pathways reinforces memory and clarifies relationships.

By reviewing each section of this answer key, you will reinforce the core concepts of Chapter 2, develop problem‑solving skills, and be well prepared for quizzes, lab reports, and cumulative examinations in the Chemistry of Life course.