Biology by Bradford
Unit 1: Chemistry of Life. Practice questions with scoring guidelines.
96 original AP-style questions on Unit 1, written to the CED learning objectives: 74 multiple-choice, 22 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 1.1 Structure of Water and Hydrogen Bonding 20 questions
- 1.2 Elements of Life 13 questions
- 1.3 Introduction to Biological Macromolecules 15 questions
- 1.4 Properties of Biological Macromolecules 32 questions
- 1.5 Structure and Function of Biological Macromolecules 17 questions
- 1.6 Nucleic Acids 18 questions
In the bank for Unit 1
- 74 multiple-choice
- 22 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.
A single-celled organism has cytoplasm that is roughly 70% water by mass. Compared with an otherwise identical cell whose interior is mostly a nonpolar oil, how will the water-rich cell respond to a sudden burst of heat, and why?
- More slowly, because water has a high specific heat
- More slowly, because water is a good solvent
- More quickly, because water has a low specific heat
- At the same rate, since specific heat is irrelevant
Both a triglyceride and a polysaccharide are composed of only carbon, hydrogen, and oxygen, yet the triglyceride releases far more energy per gram when oxidized. Which difference in composition best accounts for this?
- The triglyceride contains energy-storing nitrogen
- The triglyceride has more oxygen and fewer C–H bonds
- The triglyceride has less oxygen and more C–H bonds
- The polysaccharide contains energy-rich phosphorus
An enzyme normally has a positively binding, charged amino acid in its active site that attracts a negatively charged substrate. A mutation replaces this amino acid with one that has a nonpolar side chain, and the mutant enzyme's reaction rate falls to nearly zero even though the rest of the protein folds normally. Which conclusion is best supported?
- Nonpolar side chains always improve binding between an enzyme and its substrate
- Enzyme function depends only on the total number of amino acids, not their identity
- The mutation must have increased the enzyme's affinity for its substrate
- The identity of a single amino acid can be critical to enzyme function
A student is given two nucleic acid samples. Sample 1 contains the sugar deoxyribose and the base thymine; Sample 2 contains the sugar ribose and the base uracil. Which statement correctly compares the two nucleic acids?
- Sample 1 is DNA and Sample 2 is RNA; both are polymers of nucleotides
- Both samples are DNA because both are made of nucleotides
- Sample 1 is RNA and Sample 2 is DNA; only DNA contains phosphate groups
- Sample 2 is DNA because ribose is found only in DNA
Farmers add nitrogen-containing fertilizer to soil because plants need nitrogen to build certain macromolecules. Nitrogen taken up by a plant is used as a building-block element for which two classes of macromolecule?
- Lipids and proteins
- Proteins and nucleic acids
- Carbohydrates and lipids
- Carbohydrates and nucleic acids
The sugar in RNA is ribose, which carries a hydroxyl (–OH) group on its 2′ carbon, whereas the deoxyribose of DNA lacks this group. This extra –OH group makes RNA more chemically reactive and less stable than DNA. This structural difference helps explain why
- DNA cannot be copied without the 2′ hydroxyl
- RNA and DNA are chemically identical anyway
- RNA stores information more stably than DNA
- DNA suits long-term storage of information
A sample of double-stranded DNA is analyzed and found to contain 22% adenine. Based on complementary base pairing, what are the expected percentages of thymine, guanine, and cytosine in this sample?
- Thymine 22%, guanine 22%, cytosine 22%
- Thymine 78%, guanine 11%, cytosine 11%
- Thymine 28%, guanine 22%, cytosine 28%
- Thymine 22%, guanine 28%, cytosine 28%
Table salt (NaCl) and glucose both dissolve readily in water, whereas cooking oil does not. Which statement best explains why water dissolves salt and glucose but not oil?
- Water molecules are polar, so they surround ions and polar solutes but not nonpolar oil
- Water molecules are nonpolar and therefore attract other nonpolar substances such as oil
- Oil is denser than water, so it cannot form hydrogen bonds with sugars
- Water has a high specific heat, which is what allows it to dissolve ionic compounds
The model represents two amino acids being joined together. Each amino acid has an amino group (–NH₂), a carboxyl group (–COOH), and a variable side chain (R₁ or R₂). A small molecule labeled Y is released as the two amino acids are joined.
Starch and cellulose are two polysaccharides found in plants. Both are polymers built entirely from the monosaccharide glucose, but the glucose monomers are joined by different types of glycosidic linkages. Starch stores energy inside plant cells and is readily digested by many animals, whereas cellulose forms strong fibers in plant cell walls and cannot be digested by most animals.
The model represents a single nucleotide monomer, drawn as three joined shapes numbered 1, 2 and 3, and shows how nucleotides are joined to build a nucleic acid strand. As the strand is built, part 2 of one nucleotide bonds to part 1 of the next, forming the backbone of the strand. In this particular nucleotide the sugar is deoxyribose and the base is one of adenine, thymine, cytosine, or guanine.
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- A — Water's extensive hydrogen bonding gives it a high specific heat, so a water-rich cell warms more slowly and resists rapid temperature change.
- C — Triglycerides are more reduced, with proportionally less oxygen and many more C–H bonds than carbohydrates, so they yield more energy per gram when oxidized.
- D — Loss of activity despite normal folding shows that the specific side chain of one amino acid was essential to the enzyme's function.
- A — Deoxyribose with thymine identifies DNA and ribose with uracil identifies RNA, yet both are nucleotide polymers linked by phosphodiester bonds.
- B — Nitrogen is a component of the amino acids in proteins and the nitrogenous bases in nucleic acids, but not of typical carbohydrates or lipids.
- D — The 2′-OH of ribose makes RNA more reactive and less stable, so DNA's more stable deoxyribose backbone is better suited to long-term information storage.
- D — In double-stranded DNA adenine pairs with thymine (so T = 22%), leaving 56% split equally between guanine and cytosine (28% each).
- A — Water's polarity lets it form ion-dipole and hydrogen-bond interactions that surround polar and charged solutes, while nonpolar oil is excluded.
- Water
- H₂O
Total for part (A): 1 point
- Peptide bond
- Peptide bond/linkage (an amide bond)
Total for part (B): 1 point
- Yes, changing R₁ could alter the protein's folded shape and its function.
- Yes — the substitution could change how the protein folds and therefore what it does.
Total for part (C): 1 point
- A protein's shape (and thus its function) is determined by the sequence and chemical properties of its amino acids; changing one R group changes the interactions that fold the protein, so a single subunit change can alter the whole molecule's structure and function.
- Because the folded structure depends on each side chain, altering one R group can disrupt folding and change the protein's function, even though only one subunit changed.
Total for part (D): 1 point
Total for question 9: 4 points
- Glucose
- Glucose (a monosaccharide)
Total for part (A): 1 point
- Their glucose monomers are joined by different types of glycosidic linkages (α in starch, β in cellulose), giving the polymers different shapes and therefore different functions
- The way the identical monomers are linked/arranged differs, so the resulting polymers differ in structure and function
Total for part (B): 1 point
- No, it will not break down cellulose
- No / the enzyme cannot hydrolyze cellulose
Total for part (C): 1 point
- The enzyme's active site is specific to the shape of the α linkage in starch; cellulose's β linkages have a different structure that does not fit the active site, so the enzyme cannot hydrolyze it
- Because cellulose's β linkages differ in orientation from the α linkages the enzyme recognizes, the substrate does not fit the enzyme, so cellulose is not broken down
Total for part (D): 1 point
Total for question 10: 4 points
- A phosphate group, a (five-carbon/pentose) sugar, and a nitrogenous base
- Phosphate, sugar, and base
Total for part (A): 1 point
- Phosphodiester bond/linkage
- Phosphodiester bond
Total for part (B): 1 point
- The sugar would be ribose instead of deoxyribose, and the base thymine would be replaced by uracil
- Change deoxyribose to ribose and (if thymine) change thymine to uracil
Total for part (C): 1 point
- RNA is defined by containing the sugar ribose and using uracil in place of thymine, whereas DNA contains deoxyribose and thymine
- Ribose and uracil are characteristic of RNA; deoxyribose and thymine are characteristic of DNA, so those changes make it an RNA nucleotide
Total for part (D): 1 point
Total for question 11: 4 points
Other AP Biology units
- Unit 2 Cell Structure and Function 105
- Unit 3 Cellular Energetics 112
- Unit 4 Cell Communication and Cell Cycle 104
- Unit 5 Heredity 130
- Unit 6 Gene Expression and Regulation 116
- Unit 7 Natural Selection 149
- Unit 8 Ecology 131
All eight units → · IB Biology subtopics → · Open the exam maker →