IndietroGeneral Chemistry and Biochemistry Practice Exam Guidance
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Q1. Which of the following is NOT a functional group?
Background
Topic: Organic Chemistry – Functional Groups
This question tests your ability to identify functional groups, which are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules.
Key Terms:
Functional group: A specific group of atoms within a molecule that is responsible for certain properties and reactions.
Examples: Amino group (–NH2), Alcohol group (–OH), Carbonyl group (C=O).
Step-by-Step Guidance
Review the definition of a functional group in organic chemistry.
Look at each answer choice and determine if it is a specific group of atoms (functional group) or something else (e.g., a macromolecule).
Recall that proteins are large biomolecules, not a specific functional group.
Try solving on your own before revealing the answer!
Final Answer: B) Protein
Protein is a macromolecule, not a functional group. The other options are all specific functional groups found in organic molecules.
Q2. If atoms with greatly differing electronegativities form a bond, that bond will be:
Background
Topic: Chemical Bonding – Electronegativity and Bond Polarity
This question tests your understanding of how differences in electronegativity between atoms affect the type of bond formed.
Key Terms:
Electronegativity: The tendency of an atom to attract electrons in a chemical bond.
Polar bond: A bond where electrons are shared unequally due to a difference in electronegativity.
Nonpolar bond: A bond where electrons are shared equally.
Step-by-Step Guidance
Recall the definition of electronegativity and how it affects electron sharing in bonds.
Consider what happens when two atoms have a large difference in electronegativity.
Think about whether the bond will be polar, nonpolar, or have other characteristics.
Try solving on your own before revealing the answer!
Final Answer: A) Polar
A large difference in electronegativity leads to a polar bond, where electrons are shared unequally.
Q3. The water molecule is polar because:
Background
Topic: Molecular Polarity and Structure
This question tests your understanding of why water is a polar molecule, considering both its shape and the electronegativity of its atoms.
Key Terms:
Polarity: A molecule is polar if it has a net dipole moment due to uneven electron distribution.
Electronegativity: Oxygen is more electronegative than hydrogen.
Molecular geometry: Water has a bent shape.
Step-by-Step Guidance
Recall the structure of a water molecule (H2O) and its bent geometry.
Consider the electronegativity difference between hydrogen and oxygen.
Think about how the shape and electron distribution contribute to water's polarity.
Try solving on your own before revealing the answer!
Final Answer: E) All of these are correct
All the listed reasons contribute to the polarity of water: asymmetric electron distribution, hydrogens on one side, and oxygen's higher electronegativity.
Q4. Ionic compounds and polar covalent compounds tend to dissolve in water because of:
Background
Topic: Intermolecular Forces and Solubility
This question tests your knowledge of the types of interactions that allow ionic and polar compounds to dissolve in water.
Key Terms:
Ion-dipole interaction: Attraction between an ion and a polar molecule.
Dipole-dipole interaction: Attraction between two polar molecules.
Hydrophobic interaction: Tendency of nonpolar substances to aggregate in water.
Step-by-Step Guidance
Recall how water interacts with ions and polar molecules.
Identify which types of intermolecular forces are involved in dissolving ionic and polar compounds.
Eliminate options that do not involve water's polarity or ionic interactions.
Try solving on your own before revealing the answer!
Final Answer: A) Ion-dipole and dipole-dipole interactions
These interactions are responsible for the solubility of ionic and polar compounds in water.
Q5. Which of the following characteristics makes for a good hydrogen bond donor?
Background
Topic: Hydrogen Bonding
This question tests your understanding of what makes a molecule or atom a good hydrogen bond donor.
Key Terms:
Hydrogen bond donor: An atom bonded to hydrogen that can participate in hydrogen bonding.
Electronegativity: High electronegativity increases hydrogen bond strength.
Nonbonding pair: Lone pairs on the acceptor, not the donor.
Step-by-Step Guidance
Recall the requirements for hydrogen bonding: a hydrogen atom attached to a highly electronegative atom (like O, N, or F).
Consider whether a nonbonding pair of electrons is needed on the donor or acceptor.
Determine which characteristics are necessary for a good donor.
Try solving on your own before revealing the answer!
Final Answer: A) A high electronegativity
A good hydrogen bond donor must have a hydrogen attached to a highly electronegative atom.
Q6. (Short answer) The pOH of a solution of 0.04 M HCl is:
Background
Topic: Acids and Bases – pH and pOH Calculations
This question tests your ability to calculate pOH from the concentration of a strong acid.
Key Terms and Formulas:
HCl is a strong acid and dissociates completely:
Step-by-Step Guidance
Recognize that HCl is a strong acid, so M.
Calculate the pH using .
Use the relationship to set up the calculation for pOH.
Try solving on your own before revealing the answer!
Final Answer: pOH = 12.6
Since HCl is a strong acid, the calculation is straightforward.
Q7. (Short answer) If a solution has a pH = 6, the [H+] is:
Background
Topic: Acids and Bases – pH and [H+] Relationship
This question tests your ability to convert pH to hydrogen ion concentration.
Key Formula:
Step-by-Step Guidance
Recall the formula relating pH and [H+]: .
Substitute pH = 6 into the formula to set up the calculation.
Try solving on your own before revealing the answer!
Final Answer: M
Using the formula, M for pH 6.
Q8. The amino and carboxyl groups of amino acids are bonded to which carbon?
Background
Topic: Amino Acid Structure
This question tests your knowledge of the general structure of amino acids.
Key Terms:
Alpha-carbon: The central carbon atom in an amino acid, bonded to the amino group, carboxyl group, hydrogen, and side chain (R group).
Step-by-Step Guidance
Recall the general structure of an amino acid.
Identify which carbon is the alpha-carbon and what is attached to it.
Compare the answer choices to your knowledge of amino acid structure.
Try solving on your own before revealing the answer!
Final Answer: A) Both are bonded to the alpha-carbon
In all standard amino acids, both the amino and carboxyl groups are attached to the alpha-carbon.
Q9. Which group consists only of amino acids with polar side chains?
Background
Topic: Amino Acid Classification
This question tests your ability to identify amino acids with polar side chains.
Key Terms:
Polar side chain: Side chains that can form hydrogen bonds with water.
Examples: Serine, threonine, cysteine (all polar); leucine, valine, isoleucine (nonpolar).
Step-by-Step Guidance
Recall which amino acids have polar side chains.
Go through each answer choice and identify the polarity of each amino acid listed.
Eliminate any group that contains a nonpolar amino acid.
Try solving on your own before revealing the answer!
Final Answer: B) Serine, threonine, and cysteine
All three have polar side chains, while the others include nonpolar amino acids.
Q10. The amount of energy released during a reaction tells nothing about the rate at which that reaction will occur.
Background
Topic: Thermodynamics vs. Kinetics
This question tests your understanding of the difference between thermodynamic favorability and reaction rate.
Key Terms:
Thermodynamics: Concerns energy changes (ΔG, ΔH) in reactions.
Kinetics: Concerns the rate (speed) of a reaction.
Step-by-Step Guidance
Recall the difference between thermodynamic and kinetic properties of reactions.
Consider whether energy release (ΔG) affects the rate of reaction.
Decide if the statement is true or false based on your understanding.
Try solving on your own before revealing the answer!
Final Answer: A) True
Thermodynamics (energy released) does not determine the rate; kinetics does.
Q11. What does amphiphatic mean?
Background
Topic: Molecular Properties – Amphiphatic Molecules
This question tests your understanding of the term "amphiphatic" as it applies to molecules.
Key Terms:
Amphiphatic: Molecules that have both hydrophilic (water-loving) and hydrophobic (water-fearing) regions.
Step-by-Step Guidance
Recall the definition of amphiphatic and examples (e.g., phospholipids).
Compare each answer choice to the definition.
Eliminate options that refer to charge or acid/base properties.
Try solving on your own before revealing the answer!
Final Answer: C) Having both hydrophilic and hydrophobic regions
Amphiphatic molecules have both water-attracting and water-repelling parts.
Q12. Biochemists use a modified value for ΔG values because:
Background
Topic: Biochemical Thermodynamics
This question tests your understanding of why standard free energy changes (ΔG°) are modified in biochemistry.
Key Terms:
ΔG: Gibbs free energy change.
Standard state: 1 M concentration, 1 atm, 25°C.
Biochemical standard state: Adjusted for pH 7 and other biological conditions.
Step-by-Step Guidance
Recall the standard conditions for ΔG° in chemistry.
Consider why these conditions are not realistic for biological systems.
Identify which answer choices reflect these differences.
Try solving on your own before revealing the answer!
Final Answer: D) All of the above
All listed reasons are why biochemists use a modified standard state for ΔG.
Q13. By definition, a spontaneous reaction is one:
Background
Topic: Thermodynamics – Spontaneity
This question tests your understanding of what makes a reaction spontaneous.
Key Terms:
Spontaneous reaction: A reaction that occurs without outside intervention (ΔG < 0).
Exergonic: Energy is released.
Rate: Not related to spontaneity.
Step-by-Step Guidance
Recall the definition of spontaneity in thermodynamics.
Consider whether spontaneity is about energy change, rate, or both.
Eliminate options that confuse spontaneity with reaction speed.
Try solving on your own before revealing the answer!
Final Answer: A) Where energy is released
Spontaneous reactions have a negative ΔG, meaning energy is released, but this does not imply the reaction is fast.
Q14. The production of larger molecules from smaller ones is called:
Background
Topic: Metabolism – Anabolism vs. Catabolism
This question tests your knowledge of metabolic pathways and terminology.
Key Terms:
Anabolism: Building larger molecules from smaller ones.
Catabolism: Breaking down larger molecules into smaller ones.
Step-by-Step Guidance
Recall the definitions of anabolism and catabolism.
Identify which term refers to the synthesis of complex molecules.
Eliminate options that refer to breakdown or are too general.
Try solving on your own before revealing the answer!
Final Answer: C) Anabolism
Anabolism is the process of building larger molecules from smaller ones.
Q15. An oxidizing agent:
Background
Topic: Redox Reactions
This question tests your understanding of oxidation and reduction, specifically the role of the oxidizing agent.
Key Terms:
Oxidizing agent: The substance that is reduced (gains electrons) and causes another to be oxidized.
Reduction: Gain of electrons.
Step-by-Step Guidance
Recall the definitions of oxidation and reduction.
Remember the mnemonic "OIL RIG" (Oxidation Is Loss, Reduction Is Gain of electrons).
Determine what happens to the oxidizing agent during a redox reaction.
Try solving on your own before revealing the answer!
Final Answer: D) Gains electrons
The oxidizing agent gains electrons and is reduced in the process.
Q16. (Essay) Given the following information about Tyrosine, find its pI (isoelectric point). Alpha carboxyl group pKa=2.1, alpha-amino group pKa=9.1, side chain pKa=10.1. Show which pKa’s you are using and why.
Background
Topic: Amino Acids – Isoelectric Point (pI)
This question tests your ability to calculate the isoelectric point of an amino acid with an ionizable side chain.
Key Terms and Formulas:
pI (isoelectric point): The pH at which the amino acid has no net charge.
For amino acids with ionizable side chains, pI is the average of the two pKa values that surround the neutral (zwitterion) form.
Step-by-Step Guidance
Write out the three relevant pKa values: carboxyl (2.1), amino (9.1), and side chain (10.1).
Determine the charge states of tyrosine at different pH values.
Identify which two pKa values to average for the pI (those that bracket the neutral form).
Set up the formula for pI as the average of the appropriate pKa values.
Try solving on your own before revealing the answer!
Final Answer: pI = (9.1 + 10.1)/2 = 9.6
For tyrosine, the pI is the average of the pKa values for the amino group and the side chain, since the side chain is the next group to lose a proton after the zwitterion form.
Q17. (Essay) You are given the task of making a buffer solution for a laboratory setting, with a desired pH of 6.7. Can you use the phosphate buffer system to make this buffer? Use the Henderson-Hasselbalch equation mathematically to answer this question.
Background
Topic: Buffers and the Henderson-Hasselbalch Equation
This question tests your ability to determine if a buffer system is appropriate for a given pH using the Henderson-Hasselbalch equation.
Key Terms and Formulas:
Buffer: A solution that resists changes in pH upon addition of acid or base.
Henderson-Hasselbalch equation:
Phosphate buffer system: , with
Step-by-Step Guidance
Recall the pKa of the phosphate buffer system (about 7.2).
Set up the Henderson-Hasselbalch equation for the desired pH (6.7).
Calculate the ratio of base to acid required to achieve pH 6.7.
Determine if this ratio is practical (within a factor of 10) for effective buffering.
Try solving on your own before revealing the answer!
Final Answer: Yes, you can use the phosphate buffer system.
Using the Henderson-Hasselbalch equation: , so , which means . This is within the effective buffering range (ratio between 0.1 and 10), so the phosphate buffer system is suitable for pH 6.7.
Q18. (Essay) Draw the structure for the amino acid, Proline. What class of amino acids does Proline fall under?
Background
Topic: Amino Acid Structure and Classification
This question tests your ability to recall the structure and classification of proline.
Key Terms:
Proline: An amino acid with a unique cyclic structure.
Nonpolar: Proline is classified as a nonpolar, aliphatic amino acid.
Step-by-Step Guidance
Recall the general structure of amino acids and how proline is different (its side chain forms a ring with the amino group).
Draw the structure, showing the ring connecting the alpha-amino group to the side chain.
Identify the class of amino acids proline belongs to (nonpolar, aliphatic).
Try drawing and classifying before revealing the answer!
Final Answer:
Proline's structure is a five-membered ring connecting the alpha-amino group to the side chain. It is classified as a nonpolar, aliphatic amino acid.
Structure: 
Q19. (Essay) Draw out the structure for the amino acid, Histidine. What class of amino acids does Histidine fall under?
Background
Topic: Amino Acid Structure and Classification
This question tests your ability to recall the structure and classification of histidine.
Key Terms:
Histidine: An amino acid with an imidazole side chain.
Polar, basic: Histidine is classified as a polar, basic amino acid.
Step-by-Step Guidance
Recall the general structure of amino acids and the unique imidazole ring in histidine's side chain.
Draw the structure, showing the imidazole ring attached to the alpha-carbon.
Identify the class of amino acids histidine belongs to (polar, basic).
Try drawing and classifying before revealing the answer!
Final Answer:
Histidine's structure includes an imidazole ring as its side chain. It is classified as a polar, basic amino acid.
Structure: 
Q20. (Essay) Draw out the structure of the dipeptide between Alanine and Valine, starting with the N-terminal residue and ending with the C-terminal residue.
Background
Topic: Peptide Bond Formation
This question tests your ability to draw a dipeptide, showing the peptide bond between two amino acids.
Key Terms:
Dipeptide: Two amino acids joined by a peptide bond.
N-terminal: The end with a free amino group.
C-terminal: The end with a free carboxyl group.
Step-by-Step Guidance
Draw the structure of alanine (N-terminal) and valine (C-terminal).
Show the peptide bond formed between the carboxyl group of alanine and the amino group of valine.
Label the N-terminal and C-terminal ends of the dipeptide.
Try drawing the dipeptide before revealing the answer!
Final Answer:
The dipeptide structure is: 
The N-terminal is alanine, and the C-terminal is valine.
Q21. (Essay) What is the effective pH range of a buffer? Using the Henderson-Hasselbalch equation, prove this mathematically. You can use the pKa for the acetic acid buffer, 4.76.
Background
Topic: Buffers and Buffer Range
This question tests your understanding of the effective pH range of a buffer and how to use the Henderson-Hasselbalch equation to justify it.
Key Terms and Formulas:
Buffer range: The pH range over which a buffer effectively resists changes in pH, typically pKa ± 1.
Henderson-Hasselbalch equation:
Step-by-Step Guidance
Recall the general rule for buffer range (pKa ± 1).
Set up the Henderson-Hasselbalch equation for the extremes of the buffer range (when and $10$).
Calculate the pH at these ratios using the given pKa (4.76).
Interpret the results to determine the effective buffer range.
Try solving on your own before revealing the answer!
Final Answer:
The effective buffer range is pH 3.76 to 5.76 (pKa ± 1). This is shown mathematically by plugging and $10$ into the Henderson-Hasselbalch equation:
Q22. (Essay) List and explain two ways in which biochemical pathways can compensate for nonspontaneous reactions. Be specific.
Background
Topic: Metabolic Pathways and Thermodynamics
This question tests your understanding of how cells drive nonspontaneous (endergonic) reactions forward.
Key Terms:
Coupling: Linking a nonspontaneous reaction to a spontaneous one (e.g., ATP hydrolysis).
Product removal: Shifting equilibrium by removing products.
Step-by-Step Guidance
Recall that nonspontaneous reactions have a positive ΔG and require energy input.
Think of ways cells can make these reactions proceed (e.g., coupling to ATP hydrolysis).
Consider how changing concentrations (removing products) can drive reactions forward.
Try listing and explaining two methods before revealing the answer!
Final Answer:
Reaction Coupling: Cells couple nonspontaneous reactions to highly exergonic reactions (like ATP hydrolysis) to make the overall process spontaneous.
Product Removal: By continuously removing products (e.g., via subsequent reactions), the equilibrium shifts to favor product formation, driving the reaction forward.