Biology by Bradford
Unit 1: Chemistry of Life. Practice questions with scoring guidelines.
196 original AP-style questions on Unit 1, written to the CED learning objectives: 138 multiple-choice, 58 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 31 questions
- 1.2 Elements of Life 31 questions
- 1.3 Introduction to Macromolecules 31 questions
- 1.4 Carbohydrates 26 questions
- 1.5 Lipids 28 questions
- 1.6 Nucleic Acids 31 questions
- 1.7 Proteins 39 questions
In the bank for Unit 1
- 138 multiple-choice
- 58 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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Four purified polymers are each completely broken down by adding water in the presence of the appropriate enzymes (hydrolysis). The type of monomer released from each polymer is recorded in the table.
Which polymer is a protein, and what evidence from the table supports this?
| Polymer | Monomer(s) released |
|---|---|
| J | Amino acids |
| K | Glucose molecules |
| L | Nucleotides |
| M | Glycerol and fatty acids |
- Polymer L, because proteins are made of nucleotides
- Polymer M, because proteins are made of glycerol and fatty acids
- Polymer J, because proteins are polymers of amino acid monomers
- Polymer K, because proteins are made of glucose
Two double-stranded DNA samples, P and Q, were heated slowly and the percentage of DNA that had denatured (strands separated) was recorded and plotted, with percent denatured on the y-axis and temperature on the x-axis. The tabulated values from the graph are shown below.
Two double-stranded DNA samples, P and Q, were slowly heated and the percentage of the DNA that had 'melted' (separated into single strands) was recorded and plotted against temperature. Both curves are S-shaped, rising from near 0% at low temperature towards 100% at high temperature. According to the graph, which sample has the higher melting temperature (the temperature at which 50% of the DNA has separated)?
| Temperature (°C) | Sample P (% denatured) | Sample Q (% denatured) |
|---|---|---|
| 60 | 2 | 0 |
| 70 | 18 | 4 |
| 80 | 62 | 15 |
| 90 | 97 | 45 |
| 95 | 100 | 68 |
| 100 | 100 | 92 |
- Sample P, reaching 50% at a higher temperature
- Sample P, because it reaches 100% melted first
- Both samples melt at exactly the same temperature
- Sample Q, reaching 50% melted at about 91 °C
A polypeptide is assembled from 150 amino acid monomers joined in a single unbranched chain, with each linkage formed by dehydration synthesis. How many peptide bonds are formed, and how many molecules of water are released, when this polypeptide is built from free amino acids?
- 149 peptide bonds and 149 water molecules
- 150 peptide bonds and 149 water molecules
- 149 peptide bonds and 150 water molecules
- 150 peptide bonds and 150 water molecules
A plant breeder compares the seeds of two varieties of an oilseed crop, K and M. Ten samples of dried seeds from each variety were analyzed, and the mass of each of three storage molecules was measured as a percentage of the dry seed mass. The results are shown in Table 1 as mean ± 2SE.
A student concludes: "Variety M seeds contain significantly more of the polymer built from amino acids than variety K seeds, because that mean is 2.2 percentage points higher. The 1.8-point difference in the polymer built from glucose is too small to be significant." Which evaluation of the student's conclusion is best supported by the data?
| Storage molecule | Variety K | Variety M |
|---|---|---|
| Oil (triglyceride) | 44.0 ± 1.6 | 49.5 ± 1.8 |
| Protein | 24.6 ± 2.4 | 26.8 ± 2.2 |
| Starch | 11.4 ± 0.8 | 9.6 ± 0.9 |
- It is correct, because a larger difference between two means shows a significant difference whatever the size of the error bars on each mean.
- It is incorrect for both: neither is significant, since the protein intervals (19.8 to 29.4 and 22.4 to 31.2) and the starch intervals (9.8 to 13.0 and 7.8 to 11.4) overlap.
- It is incorrect for both: both are significant, since the oil intervals (42.4 to 45.6 and 47.7 to 51.3) do not overlap, so the varieties differ in every store.
- It is the wrong way round: the protein intervals (22.2 to 27.0 and 24.6 to 29.0) overlap, but the starch intervals (10.6 to 12.2 and 8.7 to 10.5) do not.
A student performs elemental analysis on four purified biological samples. Each sample is a different class of macromolecule: a protein, a carbohydrate, a phospholipid, and a nucleic acid. A plus sign (+) indicates that the element is present as a significant component of the macromolecule, and a minus sign (−) indicates it is not. The results are shown in the table.
Samples 2 and 4 can be distinguished from each other using the data. Which identification is correct?
| Sample | Carbon | Hydrogen | Oxygen | Nitrogen | Phosphorus |
|---|---|---|---|---|---|
| 1 | + | + | + | + | + |
| 2 | + | + | + | + | − |
| 3 | + | + | + | − | − |
| 4 | + | + | + | − | + |
- Both are proteins because both contain nitrogen
- Neither can be a macromolecule at all
- Sample 2 is the protein and Sample 4 the phospholipid
- Sample 2 is the phospholipid and Sample 4 the protein
Protein T is a single polypeptide. In its folded shape, the amino acids at positions 23 and 71 both lie in the interior, away from water: position 23 carries a large nonpolar R group and position 71 a small nonpolar R group. Researchers made three variants of protein T and measured the melting temperature (Tm) of each, the temperature at which half of the protein molecules have unfolded; a higher Tm means a more stable folded shape.
The researchers propose that, in the folded protein, the R groups at positions 23 and 71 are packed directly against each other, even though the two positions are far apart in the sequence. Which finding from the table most strongly supports this proposal?
| Protein | R group at position 23 | R group at position 71 | Tm (°C), mean ± SE |
|---|---|---|---|
| Wild type | Large nonpolar | Small nonpolar | 62.0 ± 0.4 |
| Variant 1 | Small nonpolar | Small nonpolar | 51.2 ± 0.5 |
| Variant 2 | Large nonpolar | Large nonpolar | 48.9 ± 0.6 |
| Variant 3 | Small nonpolar | Large nonpolar | 61.3 ± 0.5 |
- Variant 1 is less stable than the wild type, so the large R group at position 23 helps hold the folded shape
- Variant 3 is nearly as stable as the wild type, although each of its two changes on its own lowered the Tm
- Variant 2 is the least stable protein, so a large R group at position 71 does not fit in the protein interior
- Variants 1 and 2 are both less stable than the wild type, so the R groups at both positions affect stability
When table salt (NaCl) dissolves in water, each sodium ion (Na⁺) and each chloride ion (Cl⁻) becomes surrounded by a shell of water molecules. Which statement correctly describes how the water molecules next to each ion are oriented, and why?
- Hydrogen faces both ions, because hydrogen bonds form with any ion
- Oxygen faces Na⁺ and hydrogen faces Cl⁻, as opposite partial charges attract
- Hydrogen faces Na⁺ and oxygen faces Cl⁻, as opposite partial charges attract
- Oxygen faces both ions, because oxygen is the largest atom in water
A student analyzes an unknown biological macromolecule and finds that it is built from subunits, each of which contains an amino group and a carboxyl group, and that the sample contains a substantial proportion of nitrogen. To which class of macromolecule does the sample most likely belong?
- Protein
- Lipid (triglyceride)
- Carbohydrate
- Simple sugar
Water tends to move into a cell more strongly the larger the total number of dissolved solute particles in the cell. Each dissolved molecule or ion counts as one particle, whatever its size, and a molecule packed into a solid granule does not count as a dissolved particle at all. Many plant storage cells hold glucose as starch, packed into solid granules. Each starch molecule contains many thousands of glucose units.
The phosphate-containing head of a phospholipid is polar, and in some phospholipids the head also carries a net electric charge. In phospholipid N, the head carries one negative charge and one positive charge, so it has no net charge. In phospholipid S, the head carries a net negative charge. The two phospholipids have identical fatty acid tails. Researchers made vesicles, tiny hollow spheres each bounded by a phospholipid bilayer, from mixtures of N and S. They added either protein K, a small water-soluble protein whose surface carries many positively charged groups, or protein K0, which is identical except that each of those positively charged groups is replaced by a polar group with no charge. After 10 minutes they measured the percentage of each protein that was bound to the vesicles. The results are shown in Table 1.
| Phospholipid S in the vesicles (% of phospholipid) | Protein K bound (%) | Protein K0 bound (%) |
|---|---|---|
| 0 | 4 ± 2 | 3 ± 2 |
| 10 | 22 ± 4 | 4 ± 2 |
| 25 | 51 ± 6 | 5 ± 2 |
| 50 | 86 ± 5 | 6 ± 3 |
An insect's body is covered by a cuticle whose outer surface carries a thin layer of cuticular hydrocarbons, waxy molecules made only of carbon and hydrogen atoms. Researchers collected worker ants of one species from two sites: a dry grassland and a humid forest. For 12 ants from each site, they measured the amount of cuticular hydrocarbon, the water content of the body, and the rate at which the ants lost water when kept in dry air. The results are shown in Table 1 as mean ± 2SE.
| Measurement | Dry grassland | Humid forest |
|---|---|---|
| Cuticular hydrocarbon (µg per mg of dry body mass) | 3.1 ± 0.3 | 1.9 ± 0.3 |
| Body water content (% of body mass) | 66 ± 3 | 69 ± 3 |
| Rate of water loss in dry air (% of body water lost per hour) | 1.2 ± 0.2 | 2.4 ± 0.3 |
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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.
- C: Proteins are polymers of amino acids, so the polymer releasing amino acids (J) is the protein.
- D: Sample Q reaches 50% melting at about 91 °C, higher than sample P at about 78 °C, so Q has the higher melting temperature.
- A: Joining n monomers in a chain requires n − 1 bonds, and each dehydration reaction releases one water molecule, so 150 amino acids form 149 peptide bonds and release 149 water molecules.
- D: the polymer built from amino acids is protein and the polymer built from glucose is starch (the triglyceride in oil is not a polymer). The ±2SE intervals for protein are 22.2 to 27.0 (K) and 24.6 to 29.0 (M), which overlap from 24.6 to 27.0, so the larger 2.2-point difference is not significant. The starch intervals are 10.6 to 12.2 (K) and 8.7 to 10.5 (M), which do not overlap (a gap of 0.1), so the smaller 1.8-point difference is significant. Significance depends on the spread of each group as well as on the size of the difference between the means. A ignores the error bars altogether. B doubles values that are already 2SE, so its intervals are twice as wide as they should be. D uses the oil row, whose intervals do separate, as evidence about protein and starch; each molecule has to be judged from its own intervals.
- C: The protein contains nitrogen (from amino acids) but no phosphorus, whereas the phospholipid contains phosphorus, so Sample 2 is the protein and Sample 4 is the phospholipid.
- B: Making position 23 smaller (Variant 1) or position 71 larger (Variant 2) each lowered the Tm by more than 10 °C, but making both changes together (Variant 3) restored a Tm close to wild type: the larger R group at 71 fills the space left by the smaller one at 23. The effect of each change depends on the other only if the two R groups touch in the folded (tertiary) structure. B, C and D are true of the data but each would hold for any two interior positions, touching or not.
- B: The partially negative oxygen of a water molecule is attracted to the positive Na⁺ ion, and the partially positive hydrogens are attracted to the negative Cl⁻ ion, so polar water molecules form an oriented shell around each ion.
- A: Monomers bearing both an amino group and a carboxyl group are amino acids, so the nitrogen-rich polymer is a protein.
- Each glycosidic linkage that forms joins two particles into one, so many thousands of glucose molecules become a single starch molecule, and once it is packed into a solid granule it does not count as a dissolved particle at all; the cell holds the glucose while adding almost no dissolved particles.
- Glucose stored as starch is in a few huge, undissolved molecules instead of very many dissolved monomers, so the number of dissolved particles, and so the tendency of water to move in, stays low.
Total for part (A): 1 point
- It will increase, because each linkage that is hydrolyzed releases another glucose molecule that dissolves, so the number of dissolved particles in the cells rises greatly.
- Water will tend to move in more strongly, since breaking starch down into its glucose monomers turns one undissolved molecule into thousands of dissolved particles.
Total for part (B): 1 point
- It increases 20-fold, because each molecule, one particle, becomes 20 separate dissolved chains, 20 particles, even though fewer than 1% of its linkages (19 of 3,999) were broken.
- A twentyfold increase, since the number of dissolved particles equals the number of separate chains, which rises from 1 to 20 for each original molecule; the length of each chain does not matter.
Total for part (C): 1 point
- The tendency of water to move in depends on the number of dissolved particles, not on their mass: the mass rises only slightly (by the mass of the 19 water molecules added in hydrolysis), but the number of separate molecules rises from 1 to 20, so water tends to move in much more strongly.
- Mass is the wrong measure: breaking the 19 linkages adds only 19 small water molecules' worth of mass, yet it creates 19 extra dissolved particles from each molecule, and each particle counts the same whatever its size.
Total for part (D): 1 point
Total for question 9: 4 points
- As the percentage of S increases from 0 to 50%, the percentage of protein K bound increases, from 4% to 86%.
- A positive relationship: the more phospholipid S the vesicles contain, the more protein K binds to them.
Total for part (A): 1 point
- Not significantly different: the intervals, 3 to 9% (50% S) and 1 to 5% (no S), overlap (from 3 to 5%).
- No significant difference, because the ±2SE intervals overlap, so the null hypothesis of no difference cannot be rejected.
Total for part (B): 1 point
- Binding of K rises with the percentage of S (from 4% with no S to 86% with 50% S), so it depends on the negatively charged heads; K0, which lacks only the positive charges of K, binds no more than about 6% at any percentage of S, so the positive charges on the protein are also needed.
- Adding the negatively charged phospholipid increases binding of K, but removing the positive charges from the protein (K0) abolishes that increase, so binding needs both the negative heads and the positive groups, as expected of an attraction between opposite charges.
Total for part (C): 1 point
- Very little, no more than to vesicles of N alone (about 4% or less), because the heads facing the water at the vesicle surface would carry positive charges, which repel the positively charged groups of K, and there would be no negatively charged heads to attract them.
- Low binding, at or below the 4% seen with no S, because like charges repel: the positively charged heads at the surface push the positively charged protein away instead of attracting it.
Total for part (D): 1 point
Total for question 10: 4 points
- Grassland ants have more cuticular hydrocarbon (3.1 vs 1.9 µg per mg) and lose water at about half the rate of forest ants (1.2 vs 2.4% per hour).
- The ants with more cuticular hydrocarbon (grassland) lose water more slowly, and the ants with less (forest) lose water faster.
- Hydrocarbon content is higher and the rate of water loss is lower in the grassland ants than in the forest ants.
Total for part (A): 1 point
- Not significant: the intervals 63 to 69% (grassland) and 66 to 72% (forest) overlap (from 66 to 69%).
- No significant difference, because the ±2SE intervals overlap, so the difference between the means could be due to chance (sampling variation) and the null hypothesis of no difference cannot be rejected.
Total for part (B): 1 point
- Hydrocarbons are nonpolar (electrons are shared about equally in C–H bonds), so they cannot form hydrogen bonds with polar water molecules; water does not dissolve in or pass easily through the hydrophobic layer, which slows evaporation from the body.
- The waxy layer is hydrophobic: polar water molecules are not attracted to the nonpolar hydrocarbons, so the layer acts as a barrier that keeps water in.
Total for part (C): 1 point
- The hexane-rinsed group, because nonpolar hexane dissolves the nonpolar hydrocarbons and removes the layer, whereas polar water cannot dissolve them, so the water-rinsed ants keep their barrier.
- The ants rinsed in hexane, because like dissolves like: hexane removes the waxy hydrocarbon layer and water does not.
Total for part (D): 1 point
Total for question 11: 4 points
Other AP Biology units
- Unit 2 Cell Structure and Function 260
- Unit 3 Cellular Energetics 195
- Unit 4 Cell Communication and Cell Cycle 187
- Unit 5 Heredity 203
- Unit 6 Gene Expression and Regulation 219
- Unit 7 Natural Selection 316
- Unit 8 Ecology 215
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