Biology  by Bradford
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AP Biology · full-length mock · 3 hours

AP Biology full-length mock. The whole syllabus, at the real length.

A complete Full-length practice exam built to the Course and Exam Description: 94 points across 66 questions, in 3 hours. It follows the real paper's structure and rubric below, and every question is original and tagged to the syllabus. Sit it on screen against the clock and mark yourself, print it for a class, or build a fresh one with nothing repeated.

What the real paper requires

3 hours · Section I and Section II each 50% of the score.

  1. Section I90 minutes 60 multiple-choice questions. Four options, no penalty for guessing. Standalone items and stimulus sets of two to four questions sharing one figure or scenario.
  2. Section II90 minutes 6 free-response questions. Fixed order: Q1 and Q2 long (8 to 10 points, Q2 usually with graphing), Q3 to Q6 short (4 points each): scientific investigation, conceptual analysis, analyse a model, analyse data.

On the front of the paper

How this mock is built

Section I is weighted to the CED unit ranges (Unit 7 the largest, Units 1 and 5 the smallest) rather than spread evenly, with stimulus sets kept intact and the running order shuffled as on the real exam. Section II is the six named types in the fixed College Board order, scored on screen point by point with the guidelines.

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AP Biology · practice exam
full length · Section I multiple choice and Section II free response
180 minutes94 marks

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.

Covers All eight units, CED weighted
  • A four-function, scientific or graphing calculator is permitted.
  • Section I and Section II each count for 50% of the exam score.
Name:
Section I · Multiple choice · 90 minutes

Answer all questions. For each question, choose the one best answer.

1.

The figure shows part of a leaf cell. In the chloroplast, O₂ is released into the space inside the thylakoids. Some of this O₂ diffuses to the matrix of the mitochondrion in the same cell. How many membranes, each a single phospholipid bilayer, must an O₂ molecule cross on this path, and how does it cross them?

[1]
part of a leaf cell in section (not to scale) air space in the leaf cell wall plasma membrane cytosol chloroplast thylakoid stroma mitochondrion matrix central vacuole
  1. Four, the two membranes that surround each of the organelles, crossing each by simple diffusion
  2. Six, counting the plasma membrane as well, crossing each one by simple diffusion between phospholipids
  3. Five, counting the thylakoid membrane as well, crossing each by simple diffusion between phospholipids
  4. Five, crossing each one through channel proteins, since O₂ cannot pass through a bilayer unaided
2.

A researcher digests a circular plasmid with two restriction enzymes, A and B, either alone or together, and separates the resulting fragments by gel electrophoresis alongside a DNA size ladder. The sizes of the fragments detected in each lane are listed in the table.

Based on the number of fragments produced by enzyme B alone, how many recognition sites for enzyme B are present in this circular plasmid?

[1]
LaneFragment sizes detected (bp)
DNA ladder4000, 3000, 2000, 1000, 500
Enzyme A only6000
Enzyme B only3500, 2500
Enzyme A + Enzyme B3500, 1500, 1000
gel electrophoresis of the digested plasmid − + 4000 3000 2000 1000 500 ladder enzyme A enzyme B A + B bp
  1. 2 sites
  2. 0 sites
  3. 4 sites
  4. 1 site
3.

Enzyme A alone produces a single 6000-bp band from the circular plasmid, and no smaller fragments appear in that lane. What does this single-band result indicate about enzyme A's recognition sites in the plasmid?

[1]
  1. Enzyme A cuts at exactly one site, linearizing it
  2. Enzyme A degrades the plasmid into nucleotides
  3. Enzyme A does not cut the plasmid at all
  4. Enzyme A cuts at two sites, giving equal fragments
4.

The enzyme B digest is repeated, but the reaction is stopped early, so some plasmid molecules have been cut only once by enzyme B while the rest have been cut completely. Which band would appear in this lane in addition to the 3500-bp and 2500-bp bands?

[1]
  1. A band at 1000 bp
  2. No additional band
  3. A band at 6000 bp
  4. A band at 3000 bp
5.

A population of bacteria lives in a hot spring. A mutation produces a version of a key metabolic enzyme whose active site keeps its shape at 65 °C, whereas the ancestral enzyme denatures near 50 °C. How does this molecular variation most likely affect the population over time?

[1]
  1. The variation is neutral because enzyme structure does not affect survival.
  2. The variation spreads only if the bacteria choose to express the new enzyme.
  3. Individuals with the heat-stable enzyme survive and reproduce more in the spring.
  4. The heat-stable enzyme lowers survival because all proteins denature above 50 °C.
6.

Once peas are harvested, an enzyme in them slowly breaks down the compounds that give them their green color and fresh flavor. Food scientists divided one batch of freshly harvested peas into three lots. Lot 1 was frozen at −18 °C. Lot 2 was blanched (dipped in water at 100 °C for 2 minutes), cooled, and then frozen at −18 °C. Lot 3 was blanched in the same way but not frozen. An extract was made from lots 1 and 2 after 6 months of frozen storage, and from lot 3 on the day it was blanched, and the activity of the enzyme in each extract was measured at 25 °C with an excess of its substrate. The results are shown in Table 1.

During the 6 months in the freezer, lot 1 changed only slowly in color and flavor, and lot 2 did not change. After thawing, the peas of lot 1 lost much of their green color and fresh flavor within a day at room temperature, but the peas of lot 2 did not. Which explanation is best supported by this information?

[1]
Table 1. Activity of a Pea Enzyme in Extracts Assayed at 25 °C with Excess Substrate (percent of the activity in freshly harvested peas)
PeasEnzyme activity (%)
Freshly harvested100
Lot 1: frozen at −18 °C for 6 months, not blanched92
Lot 2: blanched, then frozen at −18 °C for 6 months4
Lot 3: blanched, not frozen (tested on the day it was blanched)4
  1. Blanching washed the enzyme's substrate out of the peas, so in lot 2 the enzyme had nothing left to act on after the peas were thawed.
  2. Frozen storage slowly denatured the enzyme in both lots and blanching sped this up, so too little enzyme was left in lot 2 to act after thawing.
  3. Freezing denatured the enzyme in lot 1, so the color and flavor that lot 1 lost after thawing were not lost through the action of the enzyme.
  4. At −18 °C the enzyme kept its shape but rarely met its substrate, so it acted little; it worked again on thawing, but blanching denatured it.
7.

The graph shows the rate of an enzyme-catalyzed reaction at increasing substrate concentrations with no inhibitor and in the presence of two inhibitors, P and Q, each at a fixed concentration. All other conditions were held constant.

At a substrate concentration of 2 mmol dm⁻³, inhibitors P and Q each reduce the rate to about half of the uninhibited rate. Which statement best describes how their effects compare at 16 mmol dm⁻³?

[1]
Substrate concentration / mmol dm⁻³Rate of reaction / µmol min⁻¹024681012141601020304050Vmax½ Vmaxno inhibitorinhibitor Pinhibitor Q
  1. Q now lowers the rate by only about one fifth, while P still lowers it by about half.
  2. P and Q each still lower the rate to about half of the rate with no inhibitor present.
  3. Neither P nor Q has any measurable effect on the rate at this substrate concentration.
  4. P now lowers the rate by only about one fifth, while Q still lowers it by about half.
8.

Which statement best describes how inhibitor Q reduces the rate of reaction?

[1]
  1. It binds to a site other than the active site and changes the shape of the active site
  2. It binds to the active site and is displaced when substrate concentration rises
  3. It denatures the enzyme permanently by breaking its peptide bonds
  4. It lowers the activation energy of a competing reaction that uses up substrate
9.

The concentration of inhibitor P is doubled and the experiment is repeated. Which change in the curve for P is predicted?

[1]
  1. The curve rises more slowly but still reaches the same maximum rate
  2. The rate at every substrate concentration falls to zero
  3. The maximum rate falls to about 25 µmol min⁻¹
  4. The curve becomes identical to the curve with no inhibitor
10.

The figure shows the free-energy change (ΔG) of each step in a four-step pathway that breaks down compound A to compound E in a cell; each step is catalyzed by a different enzyme. Under the conditions in the cell, making ATP from ADP and inorganic phosphate (Pᵢ) has ΔG = +30 kJ/mol. The cell can make ATP only by coupling its synthesis to an individual step of the pathway. What is the greatest number of ATP molecules the cell could make for each molecule of A broken down to E?

[1]
Free-energy change of each step in the breakdown of A free energy A B C D E step 1 ΔG = −12 kJ/mol step 2 ΔG = −34 kJ/mol step 3 ΔG = −20 kJ/mol step 4 ΔG = −36 kJ/mol progress of the pathway
  1. Four, because every step is exergonic, so each step can be coupled to making one molecule of ATP
  2. Two, because only steps 2 and 4 each release more free energy than the 30 kJ/mol needed for one ATP
  3. Three, because the pathway releases 102 kJ/mol in all, which is enough free energy for three ATP
  4. Three, because steps 2 and 4 can each make one ATP and steps 1 and 3 together release enough for one
11.

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?

[1]
  1. Thymine 22%, guanine 28%, cytosine 28%
  2. Thymine 28%, guanine 22%, cytosine 28%
  3. Thymine 78%, guanine 11%, cytosine 11%
  4. Thymine 22%, guanine 22%, cytosine 22%
12.

In water, almost every glucose molecule is in one of two ring forms, α and β, which differ only in whether the –OH group on carbon 1 points below or above the ring. The ring can open, at the bond between carbon 1 and the oxygen atom in the ring, and then close again, and each time it closes it can form either ring. Pure crystals of each form were dissolved in water at 25 °C, and the percentage of the dissolved glucose in the α form was measured for 6 hours. The results are shown in the figure.

Once carbon 1 of a glucose unit is joined to another unit by a glycosidic linkage, the ring of that unit can no longer open. Disaccharide G is made of two glucose units: carbon 1 of unit 1 is joined by an α linkage to carbon 4 of unit 2, and carbon 1 of unit 2 is not joined to anything. A solution of G, freshly made with unit 2 in the β form, is kept at 25 °C for 24 hours with no enzyme present. Which prediction is best supported?

[1]
Glucose Molecules in the α Ring Form After Crystals Are Dissolved (water at 25 °C; each point is the mean of 3 solutions) 0 1 2 3 4 5 6 0 20 40 60 80 100 Time after dissolving (hours) Molecules in α ring form (%) Started from α crystals Started from β crystals
  1. Both units will change until each carbon 1 is in a mix of α and β forms, as in free glucose, since both units of G are glucose.
  2. Neither unit will change at all, since glucose units that are joined in a disaccharide can no longer hold their ring structures.
  3. The linkage will stay α, while unit 2, whose carbon 1 is free, will shift until it is in a mix of α and β forms, as free glucose is.
  4. The linkage will change to a mix of α and β orientations, as free glucose does, while unit 2 will stay in its β form throughout.
13.

The ribosomes of all cells contain RNA. When the sequence of one ribosomal RNA is compared across thousands of species of bacteria, archaea, and eukaryotes, most regions of the molecule differ widely among species. One stretch of about 20 nucleotides, at the site where the ribosome joins amino acids into a chain, is almost identical in every species. Which explanation accounts for both observations?

[1]
  1. The stretch arose separately in each domain, because only one sequence can join amino acids, so the lineages converged on the same sequence
  2. All of the species live in similar conditions, so the same environment produced the same sequence in each one of them, but not in other regions
  3. Mutations cannot occur in that stretch, because the ribosome shields that part of its RNA from damage, while the rest of the molecule is exposed
  4. The stretch was inherited from a common ancestor of bacteria, archaea, and eukaryotes, and cells with changes to it rarely left offspring
14.

A child is born with three copies of chromosome 21 (trisomy 21). Chromosome analysis shows the extra chromosome 21 was contributed by the egg. This condition most commonly arises when a chromosome pair fails to separate properly during gamete formation. The failure of homologous chromosomes (or sister chromatids) to separate during meiosis is called ______, and by itself it produces a gamete that has ______.

[1]
  1. crossing over; a recombinant chromosome carrying a new mix of alleles
  2. independent assortment; a normal haploid set in a new combination
  3. nondisjunction; an abnormal number of chromosomes (for example n + 1)
  4. synapsis; a tetrad of four paired chromatids
15.

Mature nerve cells and many other specialized cells in the human body stop dividing and remain in a nondividing state, carrying out their normal functions but not progressing through the cell cycle. This nondividing state is best described as which of the following?

[1]
  1. The G₂ checkpoint, just before mitosis
  2. The M phase, when the cell divides
  3. The G₀ phase, a resting state
  4. The S phase, when DNA is replicated
16.

A suspension of the single-celled green alga Chlamydomonas reinhardtii was sealed in a chamber at 25 °C in a medium holding a large reserve of dissolved CO₂, and the concentration of dissolved O₂ was recorded while the light was changed, as shown in the figure. At 55 minutes, with the dim light still on, compound C was added. Compound C stops the mitochondrial electron transport chain and has no effect on the chloroplasts. Assume that the algae respire at the same rate in light and in darkness. Which prediction for the next 10 minutes is best supported?

[1]
O₂ in the sealed chamber of algae dark dim light bright light dim light compound C added 0 10 20 30 40 50 60 200 220 240 260 280 300 time / min dissolved O₂ / μmol per L
  1. O₂ rises at about 6 μmol per L per minute, the rate of photosynthesis in bright light, because none of the O₂ released is now used.
  2. O₂ stays near 280 μmol per L, because in dim light the algae carry out no photosynthesis and their respiration is now blocked.
  3. O₂ falls at about 2 μmol per L per minute, as in the dark, because the chloroplasts cannot make sugar without ATP from mitochondria.
  4. O₂ rises at about 2 μmol per L per minute, because in dim light photosynthesis releases O₂ as fast as respiration was using it.
17.

When a DNA copy of the protein-coding sequence of the human insulin gene is inserted into bacterial cells, the bacteria transcribe and translate it and make a polypeptide with exactly the amino acid sequence that human cells make from that gene. Which conclusion about the hereditary machinery of these two very different organisms is best supported?

[1]
  1. Bacteria evolved directly from human cells.
  2. The genetic code and the machinery for expressing DNA are largely shared.
  3. The insulin gene arose independently in bacteria by convergent evolution.
  4. Humans and bacteria must have exchanged the insulin gene recently by horizontal transfer.
18.

Eukaryotic cells contain ribosomes free in the cytosol and ribosomes bound to the endoplasmic reticulum (ER). Researchers used electron micrographs to count the ribosomes in two types of cell, A and B, from the same animal, and recorded what percent of each cell's ribosomes were bound to the ER. Table 1 shows the mean values. A student studying the table writes: "Cell type A must be the one secreting large quantities of protein, because it has twice as many ribosomes as cell type B, and ribosomes are where proteins are made."

Which statement best identifies the flaw in the student's reasoning?

[1]
Table 1. Ribosomes in Two Types of Cell from the Same Animal (mean per cell)
Cell typeRibosomes per cell (millions)Ribosomes bound to the ER (% of the cell's ribosomes)
A1020
B580
  1. It counts every ribosome, but ribosomes bound to the ER make proteins for secretion, and B has four times as many as A, 80 against 20 percent.
  2. Its conclusion is right for another reason: free ribosomes make proteins for secretion, and A has 8 million free ribosomes, B only 1 million.
  3. There is no flaw: free and bound ribosomes both release the proteins they make into the ER, so the cell with more ribosomes secretes more.
  4. It counts every ribosome, but ribosomes bound to the ER make proteins for secretion, and B has 4 million of these, A only 2 million.
19.

The graph shows a person's blood glucose concentration over 12 hours, with the times of three meals marked by arrows. Which feature of the graph is the best evidence that blood glucose is regulated by negative feedback?

[1]
0246810126080100120140Time / hoursBlood glucose / mg dL⁻¹mealmealmeal
  1. Glucose increases more with each meal
  2. Each peak is higher than the one before
  3. Glucose returns to the same set point
  4. The glucose concentration never changes
20.

The following table gives the amino acid or signal specified by several mRNA codons. Use it to answer the questions in this set.

The mRNA 5'-AUG UUU GAA AAG UGC UAA-3' is added to a cell-free translation system that contains ribosomes, all of the amino acids, and tRNAs for every codon in the table except GAA: no tRNA in the system can pair with GAA. Which outcome is expected?

[1]
mRNA codonAmino acid / signal
AUGMet (start)
UUUPhe
GAAGlu
GAGGlu
AAGLys
UGCCys
UAGStop
UAAStop
  1. No amino acids are joined because GAA is unreadable
  2. The ribosome stalls at GAA after joining Met and Phe
  3. Any tRNA can pair at GAA, so a full chain is made
  4. The ribosome skips GAA and makes Met-Phe-Lys-Cys
21.

Four different single-base substitutions each change one codon of the mRNA 5'-AUG UUU GAA AAG UGC UAA-3' as shown below. Which substitution is a silent (synonymous) mutation that does not change the polypeptide?

[1]
  1. UUU changed to CUU (Phe to Leu)
  2. AAG changed to AUG (Lys to Met)
  3. GAA changed to GAG (Glu to Glu)
  4. UGC changed to UGA (Cys to Stop)
22.

A mutation changes the fourth codon (AAG) of the mRNA 5'-AUG UUU GAA AAG UGC UAA-3' into UAG. Which of the following best describes the effect on the polypeptide and on the resulting phenotype?

[1]
  1. No effect, because the code is redundant
  2. One amino acid is substituted, length unchanged
  3. The polypeptide is one amino acid longer
  4. Translation stops early, truncating the protein
23.

Cultured cells respond to hormone G, which binds a receptor on the cell surface and starts a signaling cascade inside the cell. Researchers exposed cultures to six concentrations of G and measured the percentage of the cells' receptors that had G bound and the size of the cells' response, expressed as a percentage of the largest response the cells can give. The results are shown in Table 1.

The researchers then treated cells with a drug that binds permanently to 80 percent of the receptors, so that G cannot bind them; the other receptors are unaffected. Which prediction about the response of the treated cells to 100 nM G is best supported by the data?

[1]
Table 1. Receptor Binding and Response of Cultured Cells Exposed to Hormone G (mean ± 2SE, n = 5 cultures)
Hormone G (nM)Receptors with G bound (%)Response (% of maximum)
001 ± 1
0.010.5 ± 0.29 ± 3
0.15 ± 156 ± 6
133 ± 493 ± 3
1083 ± 599 ± 3
10098 ± 2100 ± 3
  1. Most of the maximum, because about 20 percent of all receptors would bind G, and 5 percent bound gave 56 percent
  2. No response, because the receptors bound by the drug block the signal sent on by the receptors that still bind G
  3. The full maximum, because at 100 nM G binds the receptors held by the drug as well as the free receptors
  4. About 20 percent of the maximum, because only one fifth of the receptors on each cell are still able to bind G
24.

Bacterium S has channel proteins in its plasma membrane that open only when the membrane is stretched; when open, each channel lets ions and other small solutes through. Cells of the normal strain and of a mutant strain that lacks these channels, all with intact cell walls, were grown in a medium with a high solute concentration and then moved suddenly into distilled water or into fresh growth medium. In some cells, ATP synthesis was blocked for 10 minutes before the move. The table shows the percentage of cells surviving 10 minutes after the move.

Which explanation of how the normal strain survives the move into distilled water is best supported by the data?

[1]
Survival of Bacterium S After a Sudden Move From a High-Solute Medium
StrainATP synthesisMoved intoCells surviving (%)
Normalnot blockedfresh growth medium99
Mutant (lacks the channels)not blockedfresh growth medium98
Normalnot blockeddistilled water92
Mutant (lacks the channels)not blockeddistilled water8
Normalblockeddistilled water90
Mutant (lacks the channels)blockeddistilled water7
  1. Stretching opens the channels, so solutes leave down their gradients without ATP and less water enters
  2. Open channels let water out as fast as it enters, so the cells take up no water and never swell at all
  3. The channels pump solutes out of the cells using energy from ATP, lowering the solute concentration inside
  4. The cell wall alone prevents bursting; the channels matter only for taking up solutes after the move
25.

During childbirth, stretching of the uterine wall stimulates the release of oxytocin, which increases the strength of uterine contractions. These stronger contractions stretch the uterus further, causing still more oxytocin to be released until the baby is delivered. Which term best describes this regulatory mechanism, and why?

[1]
  1. Negative feedback, because the response reduces the original stimulus.
  2. Positive feedback, because the response amplifies the original stimulus.
  3. No feedback, because oxytocin is released only a single time.
  4. Negative feedback, because oxytocin levels steadily decline over time.
26.

Two populations of a wild annual plant, a Coast population and an Inland population, have been separated by a mountain range for many thousands of years. Researchers grew plants from both populations in one greenhouse. They crossed Coast plants with Inland plants to produce F1 hybrids, then crossed F1 hybrids with each other to produce F2 hybrids. They measured the percent of pollen grains that were viable in 20 plants of each group. The figure shows the mean percent of viable pollen ± 2SE for each group.

The researchers' null hypothesis is that F1 hybrids and Coast plants have the same mean percent of viable pollen. Which statement about this null hypothesis is best supported by the data?

[1]
error bars: ± 2 SE (20 plants measured per group) 0 20 40 60 80 100 mean viable pollen (percent) 88 ± 4 Coast plants 86 ± 5 Inland plants 79 ± 6 F1 hybrids 58 ± 8 F2 hybrids group of plants
  1. Not rejected, because the F1 and Coast intervals overlap, so these data do not show that the two groups differ
  2. Rejected, because the F2 interval lies entirely below the Coast interval, so the hybrids are less fertile than Coast plants
  3. Shown to be true, because the F1 and Coast intervals overlap, so the two groups have equal mean pollen viability
  4. Rejected, because the mean viability of the F1 hybrids is 9 percentage points lower than that of Coast plants
27.

Mice of one inbred laboratory strain are genetically identical. Researchers took 20 male mice of this strain, all born in the same week, and divided them into two groups at weaning (3 weeks old). They reared one group at 25 °C and the other at 10 °C, with the same food, cages and light schedule. When the mice were 9 weeks old, the researchers measured three traits in every mouse. The results are shown in Table 1 as mean ± 2SE.

A student claims that rearing at 10 °C caused a statistically significant change in all three traits. Which evaluation of the claim is best supported by the data in Table 1?

[1]
Table 1. Three Traits of Inbred Mice Reared at 25 °C or at 10 °C (mean ± 2SE, n = 10 mice per group)
TraitReared at 25 °CReared at 10 °C
Tail length (mm)87 ± 374 ± 3
Ear length (mm)14.2 ± 0.513.0 ± 0.6
Body mass (g)24.8 ± 1.426.3 ± 1.6
  1. Supported, because the tail length intervals (84 to 90 mm and 71 to 77 mm) are far apart, and all three traits were measured on the same mice.
  2. Not supported: only tail length differs significantly, because doubling each ± value to give a 95 percent interval makes the other pairs overlap.
  3. Not supported: tail and ear length differ significantly, but the body mass intervals (23.4 to 26.2 g and 24.7 to 27.9 g) overlap.
  4. Supported, because every 10 °C mean differs from its 25 °C mean, and the 10 °C body mass mean (26.3 g) lies above the 25 °C interval.
28.

In a classic experiment, a plant is supplied with water containing a heavy isotope of oxygen, while its CO₂ contains only ordinary oxygen. The O₂ gas released during photosynthesis carries the heavy isotope. From which molecule does the released O₂ originate?

[1]
  1. The phosphate groups of ATP.
  2. Glucose synthesized in the Calvin cycle.
  3. CO₂, which is split during fixation in the Calvin cycle.
  4. H₂O, which is split during the light-dependent reactions.
29.

A student compares a molecule of DNA with a molecule of messenger RNA taken from the same cell. Which set of differences correctly distinguishes the RNA from the DNA?

[1]
  1. RNA has deoxyribose and thymine; DNA ribose, uracil
  2. RNA has ribose and uracil and is single-stranded
  3. RNA and DNA differ only in their bound proteins
  4. RNA is double-stranded and DNA single-stranded
30.

Two polypeptides are each 100 amino acids long and contain exactly the same set of amino acids, but arranged in a different order. When each folds, the two proteins adopt different three-dimensional shapes and carry out different functions. Which statement best explains this outcome?

[1]
  1. The order of amino acids determines the folding
  2. Same amino acids must give the same function
  3. The number of amino acids determines the shape
  4. Folding is set only by hydrogen bonds with water
31.

Why are ecosystems with high species diversity generally more resistant to disturbance than those with low diversity?

[1]
  1. Diverse ecosystems have simpler food webs with fewer links, so a disturbance to one species affects fewer others
  2. Every species in a diverse ecosystem is equally abundant, so no single species can be lost during a disturbance
  3. High-diversity ecosystems contain fewer predators, so prey populations remain stable when conditions change
  4. If one species declines, others with similar roles can maintain ecosystem functions such as decomposition and pollination
32.

A bacterial population of 200 cells is growing exponentially with a per-capita growth rate r = 0.5 per hour. Using the exponential model dN/dt = rN, what is the population's growth rate at this instant?

[1]
  1. 0.0025 cells per hour
  2. 100 cells per hour
  3. 400 cells per hour
  4. 50 cells per hour
33.

In a grassland, grasshoppers eat the plants, and a ground-nesting bird feeds its nestlings mainly on grasshoppers. Over 8 years, ecologists measured five variables at 30 grassland sites and used a statistical model to estimate the effect of each variable on the others. In the model, a solid arrow shows a statistically significant effect, a thicker solid arrow shows a larger effect, and + shows that an increase in the first variable increases the second. A dashed arrow shows an effect that was not statistically significant.

A rancher irrigates part of this grassland every spring, adding about as much water as an unusually wet spring would bring, while summer temperature is unchanged. Based on the model, which prediction for the irrigated part is best supported?

[1]
Model of Effects Among Five Grassland Variables over 8 Years spring rainfall summer temperature plant growth grasshopper density nestlings fledgedper nest ++++ solid arrow: statistically significant effect (thicker = larger effect) dashed arrow: effect not statistically significant + an increase in the first variable increases the second
  1. Grasshopper density will not change, because the arrow from spring rainfall to grasshopper density is dashed
  2. Grasshopper density and nestlings fledged will both rise, as more plant growth makes more energy available to consumers
  3. Grasshopper density will rise only in hot summers, because plant growth increases only when summer temperature also rises
  4. Grasshopper density will rise but nestlings fledged will not change, because the arrow from rainfall to nestlings is dashed
34.

A desert annual plant germinates after winter rain, flowers once, and dies. Some plants have hairy leaves and others have smooth leaves. The plant self-pollinates, and seeds from each form grow into plants of the same form. In one wet year and one later dry year, researchers marked 500 newly emerged seedlings of each form at the same site, recorded the percent that survived to flower, and counted the seeds produced by each flowering plant.

Which leaf form had the greater fitness in the wet year?

[1]
YearLeaf formSeedlings surviving to flower (%)Mean seeds per flowering plant
WetHairy8050
WetSmooth60100
DryHairy6040
DrySmooth2532
  1. Smooth, because it averaged 60 seeds per seedling, against 40 for hairy
  2. Neither, because each form did better than the other on one measure
  3. Hairy, because a larger percentage of its seedlings survived to flower
  4. Hairy, because its seedlings survived better in both the wet and dry year
35.

Suppose the next four years are wet, dry, wet, dry, and the plants in each year perform exactly as in the table for that type of year. Which pattern of change in the frequency of the smooth-leaf form is expected?

[1]
  1. It rises after each wet year and falls after each dry year as the favored form changes
  2. It falls every year, because a smaller share of smooth seedlings survive to flower each year
  3. It rises every year, because smooth plants set more seeds per flowering plant each year
  4. It stays the same, because each of the two forms is favored in one type of year only
36.

In which type of year did the two leaf forms differ more in fitness?

[1]
  1. Neither year, because the favored form led by the same margin in each year
  2. The wet year, when survival to flowering differed by 20 percentage points
  3. The wet year, when smooth plants set twice as many seeds per flowering plant
  4. The dry year, when hairy plants left three times as many seeds per seedling
37.

A student wrote two statements about bacteria.
Statement 1: "When a bacterium divides by binary fission, its chromosome is copied exactly, so all the cells descended from one bacterium have identical DNA."
Statement 2: "Because bacteria reproduce asexually, a bacterial cell can never gain genes from another bacterium."
Which of the following best evaluates the student's statements?

[1]
  1. Neither statement is wrong: a population that reproduces only by binary fission has no source of new DNA sequences, so every cell in it carries the same genes as its ancestor.
  2. Only statement 1 is wrong: uncorrected replication errors make some descendants differ, but bacteria cannot take in DNA from other cells because they reproduce only asexually, by binary fission.
  3. Both statements are wrong: uncorrected replication errors make some descendants differ, and cells can take in genes from other bacteria by conjugation, transformation, or transduction.
  4. Only statement 2 is wrong: proofreading by DNA polymerase removes every replication error, but cells can take in genes from other bacteria by conjugation, transformation, or transduction.
38.

Cells of a cultured line release a growth signal, G, and carry receptors for G; a cell divides only when G binds its receptors. Researchers also made two mutant lines: line M makes no G but has normal receptors, and line N makes normal G but has no receptors for G. Cells were spread so sparsely that no cell was near another, and in one dish line M cells were given medium taken from a crowded culture of line N. The percentage of cells that divided within 2 days is shown in the table. Which conclusion is best supported by the data?

[1]
Cells That Divided within 2 Days (mean of 4 dishes)
Cells in the dish (all spread sparsely)Cells that divided (%)
Normal cells only71
Line M only3
Line N only2
Normal cells and line M cells mixedNormal: 69; line M: 4
Line M only, in medium from a crowded culture of line N66
  1. Each cell responds to the G it releases itself, and G released by many crowded cells also acts on other cells
  2. G acts only through direct contact between cells, so a sparse normal cell divides without signals from other cells
  3. G released by sparse normal cells spreads through the medium and causes every cell in the same dish to divide
  4. Line M cells cannot respond to G, so they divide only when the medium carries another signal made by line N
39.

In aerobic respiration, most ATP is produced by oxidative phosphorylation rather than by substrate-level phosphorylation in glycolysis and the Krebs cycle. What directly powers ATP synthase during oxidative phosphorylation?

[1]
  1. A proton gradient across the inner mitochondrial membrane.
  2. Direct transfer of a phosphate group from glucose to ADP.
  3. The splitting of water molecules in the mitochondrial matrix.
  4. The fermentation of pyruvate to lactate.
40.

A population of 500 beetles is reduced to 12 individuals by a flood, then recovers to 500 over several generations. Compared with the original population, the recovered population is most likely to show

[1]
  1. increased allele diversity, because the population grew rapidly and new mutations accumulated
  2. a higher mutation rate, because the population was under stress and needed new alleles
  3. no change in allele frequencies, because population size returned to its original value of 500
  4. reduced allele diversity, because the survivors carried only a fraction of the original alleles
41.

Purple loosestrife, introduced to North American wetlands, grows rapidly and forms dense stands that crowd out native cattails and sedges, reducing food and nesting habitat for native wildlife. Which statement best describes its effect as an invasive species?

[1]
  1. It increases diversity by adding one species
  2. It has no effect on a stable wetland community
  3. It outcompetes natives, reducing their diversity
  4. It raises the number of trophic levels supported
42.

Cells of a cultured line respond to signal L, which binds a receptor on the cell surface. Researchers gave the cells three 1-minute pulses of fresh medium containing the same concentration of L, washing the cells after each pulse, and recorded the peak response to each pulse. When the pulses were 10 minutes apart, the second and third responses were 58% and 31% of the first. When the pulses were 60 minutes apart, they were 96% and 94% of the first. In cells treated with drug D, which stops a cell from taking proteins of its plasma membrane into the cell in vesicles, pulses 10 minutes apart gave second and third responses of 97% and 95% of the first. Which explanation is best supported by these results?

[1]
  1. Each response permanently harms part of the pathway inside the cell, so the cells respond more weakly every time they receive the signal.
  2. Each pulse uses up part of the signal L in the medium, so at short intervals the later pulses contain too little L to bind all of the receptors.
  3. Receptors that have bound L are taken into the cell, so for a time fewer receptors remain on the surface to bind L from the next pulse.
  4. Drug D causes the cells to make extra receptors for L, so the treated cells can replace the receptors that the earlier pulses used up.
43.

According to the endosymbiotic hypothesis, mitochondria and chloroplasts each descend from a free-living prokaryote that was taken in by a host cell. As each organelle evolved, many of its genes moved into the host cell's nucleus, where they stay even in a line of descendants that later loses the organelle. Researchers studied 400 species of single-celled eukaryotes drawn from all the major lines of eukaryotes. In every species, the mitochondrial DNA is most similar to the DNA of one group of aerobic bacteria; in every species that has chloroplasts, the chloroplast DNA is most similar to the DNA of cyanobacteria, which are photosynthetic bacteria. No species was found that has chloroplasts but no mitochondria. Table 1 summarizes the results.

Which sequence of events is best supported by the data?

[1]
Table 1. Organelles and Nuclear Genes in 400 Species of Single-Celled Eukaryotes
Number of speciesMitochondriaChloroplastsNuclear genes most similar to genes of aerobic bacteriaNuclear genes most similar to genes of cyanobacteria
138PresentPresentPresentPresent
262PresentAbsentPresentAbsent
  1. An aerobic prokaryote was taken in first, by an ancestor of all 400 species, and later the mitochondria in some of its descendants gained the ability to photosynthesize.
  2. A photosynthetic prokaryote was taken in first, by an ancestor of all 400 species, and an aerobic prokaryote was taken in later, by a cell that already had chloroplasts.
  3. An aerobic prokaryote and a photosynthetic prokaryote were taken in at the same time, by an ancestor of all 400 species, and 262 of the species later lost their chloroplasts.
  4. An aerobic prokaryote was taken in first, by an ancestor of all 400 species, and a photosynthetic prokaryote was taken in later, by a cell that already had mitochondria.
44.

Snowshoe hares grow a brown coat in summer and a white coat in winter. Researchers randomly assigned hares from one population to two rooms held at the same constant temperature (20 °C). In Room 1 the daily light period stayed at 16 hours; in Room 2 it was shortened gradually to 8 hours over several weeks. Only the hares in Room 2 molted into a white coat. Which conclusion is best supported by these results?

[1]
  1. Day length acts as an environmental cue that changes the coat color a genotype produces.
  2. Short days caused natural selection to favor white hares within a single generation.
  3. The hares in Room 2 carry an allele for white fur that the hares in Room 1 lack.
  4. A drop in air temperature is required to trigger the change to a white winter coat.
45.

A lysosome contains hydrolytic enzymes that work best at a pH of about 5, which is more acidic than the surrounding cytoplasm (pH ~7). Which statement best explains how the lysosome maintains this internal environment?

[1]
  1. The lysosome has no membrane, so its contents mix freely with the cytoplasm.
  2. Proton pumps in the lysosomal membrane keep the enclosed interior acidic.
  3. Cytoplasm is continuously pumped into the lysosome to raise its internal pH.
  4. The enzymes inside generate a new membrane around themselves as they work.
46.

Settlers brought rats and a small predatory mammal to an island on which no ground-hunting mammals had ever lived. Within a century, most of the island's ground-nesting bird species had become extinct, although the birds' forest habitat and food supply changed little. On a nearby continent, where such mammals are native, related ground-nesting birds have persisted alongside them for millions of years. Which explanation best accounts for the extinctions on the island?

[1]
  1. The island bird populations were large, so genetic drift removed their defensive alleles faster than elsewhere
  2. The birds had never been selected by such predators, so they had few defenses and could not adapt in time
  3. The predators caused new mutations for defense in the birds, but these arose too late to save most species
  4. The mammals competed with the birds for food, and being more efficient feeders, they starved the birds out
47.

A hillside meadow grows on soil poor in nitrogen, and the growth of its plants is limited by the nitrogen supply. Clover, a plant whose root partners add nitrogen to the soil, did not grow in the meadow. Ecologists sowed clover seed into 10 plots in the meadow and left 10 similar plots unsown. Table 1 shows measurements in the unsown plots and in the sown plots 2 and 15 years after sowing; the unsown plots changed little over the 15 years.

Which statement best describes how adding clover changed the structure of the meadow community?

[1]
TABLE 1. MEADOW PLOTS LEFT UNSOWN AND SOWN WITH CLOVER
Measurement (mean)Unsown plotsSown plots, Year 2Sown plots, Year 15
Nitrogen available to plants in the soil (mg per kg)122041
Plant biomass (g per m²)300420610
Plant species per m²282914
Ground covered by tall, fast-growing grasses (percent)101462
  1. The clover itself crowded out the other plant species, because by Year 15 it covered most of the ground and the number of species had fallen
  2. Adding one species raised productivity and so raised diversity, because a meadow with more plant biomass can support more plant species
  3. Adding one species had little lasting effect, because 2 years after sowing the sown plots had about as many plant species as unsown plots
  4. At first biomass rose with little other change, but over 15 years the extra nitrogen let tall grasses spread and halve the number of species
48.

In a certain plant, purple flower color (P) is dominant to white (p). Two plants heterozygous for flower color are crossed. What proportion of the offspring is expected to be white-flowered?

[1]
  1. 0
  2. 1/4
  3. 1/2
  4. 3/4
49.

When epinephrine binds a receptor on a liver cell, a G protein activates adenylyl cyclase, which produces cAMP; cAMP then activates protein kinase A, ultimately causing glycogen to be broken down into glucose that is released into the blood. Which of the following events represents the transduction stage of this signaling pathway?

[1]
epinephrine binds a receptor on the liver cell G protein activated adenylyl cyclase produces cAMP cAMP activates protein kinase A protein kinase A glycogen broken down glucose released into the blood
  1. The release of glucose from the liver cell into the bloodstream.
  2. The breakdown of glycogen into glucose within the liver cell.
  3. The binding of epinephrine to its receptor on the liver cell membrane.
  4. The relay of the signal through the G protein, cAMP, and protein kinase A.
50.

The pedigree shows the inheritance of a rare autosomal recessive disorder. Shaded symbols are affected individuals. II-4 married into the family and has no family history of the disorder; assume he does not carry the recessive allele. Use A and a for the alleles.

What is the probability that II-3 is a carrier of the recessive allele?

[1]
IIIIIII-1I-2II-1II-2II-3II-4III-1III-2unaffectedaffectedsquares male, circles female
  1. 1/2
  2. 3/4
  3. 1/4
  4. 2/3
51.

What is the probability that III-1 is a carrier of the recessive allele?

[1]
  1. 1/4
  2. 1/3
  3. 1/6
  4. 1/2
52.

Ecologists followed a population of a leaf-feeding moth in one forest for six years. Each spring they estimated the density of the caterpillars, and then they recorded the percentage of those caterpillars killed by parasitoid wasps, which lay their eggs inside caterpillars, and the percentage killed by heavy summer rainstorms, which wash caterpillars off the leaves.

Which mortality factor could keep this moth population near a stable density over many years, and why?

[1]
YearCaterpillars per m² in springKilled by parasitoid wasps (percent)Killed by rainstorms (percent)
160264
21405810
335156
4954131
51506218
6502125
  1. The rainstorms, because they kill caterpillars whatever the density, which keeps the population from ever growing too large
  2. The wasps, because across the six years they killed more caterpillars in total than the rainstorms killed in the same years
  3. The rainstorms, because in Year 4 they killed 31 percent of the caterpillars, the largest share that any storm season took
  4. The wasps, as the share of caterpillars they kill rises with density, so they slow growth when dense and ease off when sparse
53.

A green alga makes two membrane lipids, K and L. One of the two lipids contains phosphorus but no sulfur; the other contains sulfur but no phosphorus. Cells grown in a complete medium were transferred at day 0 either to a medium without phosphate, the only source of phosphorus, or to a medium without sulfate, the only source of sulfur; the two media were otherwise identical to the complete medium. The figure shows the amount of each lipid per cell over the next 8 days.

Cells from the same culture are transferred to a medium that lacks both phosphate and sulfate. Which prediction is best supported by the data?

[1]
0 2 4 6 8 0 10 20 30 40 50 time after transfer / days Medium without phosphate 0 2 4 6 8 0 10 20 30 40 50 time after transfer / days Medium without sulfate lipid per cell / relative units Lipid K Lipid L
  1. Lipid L will increase as it did in the medium without phosphate, because Lipid L contains no phosphorus
  2. Neither lipid will increase, because making more of either one needs an element that the medium lacks
  3. Lipid K will increase as it did in the medium without sulfate, because Lipid K contains no sulfur atoms
  4. Both lipids will increase, because in each medium the cells replaced the lipid that could not be made
54.

A soil bacterium grows well in a medium that contains only water and these substances: glucose (C₆H₁₂O₆), ammonium chloride (NH₄Cl), potassium phosphate (KH₂PO₄), magnesium sulfate (MgSO₄), and traces of a few metal ions. The bacterium can take up and break down glucose, glycerol (C₃H₈O₃), and the amino acid glycine (C₂H₅NO₂). Which single change to the medium would most reduce the number of new cells that the medium can support?

[1]
  1. Replacing the glucose with an equal mass of glycerol, a compound that this bacterium can also take up and break down
  2. Replacing the potassium phosphate with potassium chloride (KCl), so that the amount of potassium stays the same
  3. Replacing the ammonium chloride with enough glycine to supply the same mass of nitrogen as the ammonium chloride
  4. Doubling the amount of glucose in the medium while keeping the amounts of all of the other substances the same
55.

Strains of a soil bacterium differ in how many copies they carry of the gene for an enzyme that breaks down a toxic compound. Strains with four copies make about four times as much of the enzyme as strains with one copy. Which outcome is most likely when both strains grow in soil that contains the toxic compound?

[1]
  1. Both strains survive equally well, because every copy of the gene makes an identical enzyme.
  2. One-copy strains survive better, because a smaller amount of the enzyme is more specific.
  3. Four-copy strains break down the toxin faster, survive better and leave more descendants.
  4. Four-copy strains survive better only in soil where the toxic compound is completely absent.
56.

A biologist compares a bacterial cell with a plant cell using electron microscopy. Which structural difference best reflects the greater compartmentalization of the eukaryotic cell?

[1]
  1. Only the plant cell contains membrane-bound organelles such as a nucleus.
  2. The bacterium contains a membrane-bound nucleus, while the plant cell does not.
  3. The plant cell lacks a plasma membrane, relying on its wall to enclose the cytoplasm.
  4. Only the bacterium possesses ribosomes, which are needed for protein synthesis.
57.

When two protist species that require the same food resource are grown together in one culture, one species consistently drives the other to local extinction, even though each species thrives when grown alone. This outcome best illustrates which of the following?

[1]
  1. Mutualism, with both species benefiting
  2. A predator–prey cycle between the species
  3. Competitive exclusion for a shared resource
  4. Resource partitioning allowing coexistence
58.

A food scientist measured the melting points of six pure fatty acids that differ in the number of carbon atoms in the chain and in the number of carbon to carbon double bonds (C=C) it contains. A fatty acid is solid at 25 °C if its melting point is above 25 °C, and liquid if its melting point is below 25 °C. The results are shown in Table 1.

A student claims: “Whether a fatty acid is solid or liquid at 25 °C depends only on whether its chain has a C=C double bond. Fatty acids with no double bond are solid, and fatty acids with a double bond are liquid.” Which evaluation of the claim is best supported by the data?

[1]
Table 1. Melting Points of Six Fatty Acids
Carbon atoms in the chainC=C double bonds in the chainMelting point (°C)
8017
10032
14054
1610
20123
22134
  1. The data cannot test it, since no two of the fatty acids have the same chain length, so the effect of a double bond cannot be told apart from chain length.
  2. The data refute it, since an 8-carbon saturated fatty acid is liquid at 25 °C and a 22-carbon one with a double bond is solid, so chain length matters too.
  3. The data support it, since every fatty acid with a C=C double bond melts at a lower temperature than the saturated fatty acid with the closest chain length.
  4. The data refute it, since melting point rises with chain length among both kinds of fatty acid, which shows that a C=C double bond has no effect on it.
59.

In the 1800s, botanists knew that the living contents of a plant cell lose water and shrink in a concentrated sugar solution, but not which part of the cell controls what passes into it. Model 1 proposed that the cell wall is the selectively permeable boundary of the cell. Model 2 proposed that a thin boundary around the living contents, just inside the wall, is the selectively permeable boundary. A botanist placed living leaf cells in a concentrated sucrose solution colored with a blue dye whose molecules are about the size of sucrose molecules. Under the microscope, the living contents of each cell shrank away from the wall, the space between the wall and the shrunken contents filled with blue solution, and no blue color appeared in the contents themselves. How does this observation bear on the two models?

[1]
  1. It supports model 1, because the wall kept the dye away from the living contents of every cell.
  2. It supports model 2, because the dyed sucrose solution passed through the wall but not into the living contents.
  3. It supports both models equally, because water left the living contents whichever boundary controls entry.
  4. It refutes model 2, because a boundary that lets water out of the living contents cannot be selectively permeable.
60.

At deep-sea hydrothermal vents where no sunlight penetrates, bacteria oxidize hydrogen sulfide (H₂S) to fix carbon into organic molecules, and tube worms and other animals depend on these bacteria. Which statement best describes how energy enters this community?

[1]
  1. The community needs no energy input because it is completely isolated
  2. Energy is created by the tube worms and passed downward to the bacteria
  3. Chemoautotrophs convert chemical energy into biomass that heterotrophs consume
  4. The animals photosynthesize using faint light emitted by the vents
Section II · Free response · 90 minutes

Answer all questions. Write your responses in the spaces provided.

61.Interpreting and Evaluating Experimental Results

An annual wildflower of the mustard family grows on dry coastal hills. Each plant grows from a seed that germinates in the autumn rains, flowers, sets seed, and dies before the soil dries out in early summer, so each population produces one generation a year. In 2000, researchers collected seeds from many plants in two populations and stored them, cold and dry, in a seed bank. Population D grows on a hillside watered only by rain. Population M grows beside a spring, where the soil stays moist into the summer. From 2001 to 2006 the rains ended early every spring, and the soil on the hillside dried out about three weeks earlier than it had before; the soil beside the spring stayed moist. In 2007 the researchers collected seeds from the same two populations.

Experiment 1: The researchers first grew plants from all four seed collections together in one greenhouse, pollinated each plant by hand only with pollen from other plants of its own collection, and collected the seeds these plants made. From those seeds they grew 120 plants of each collection side by side in identical, well-watered conditions and recorded the number of days from germination to the opening of each plant's first flower (Table 1).

TABLE 1. DAYS FROM GERMINATION TO FIRST FLOWER OF GREENHOUSE-GROWN PLANTS FROM TWO POPULATIONS, COLLECTED IN 2000 AND 2007
PopulationSeeds collected in 2000 (days, mean ± 2SE)Seeds collected in 2007 (days, mean ± 2SE)
D (hillside watered only by rain)58.2 ± 1.451.6 ± 1.3
M (beside a spring)60.1 ± 1.659.4 ± 1.5

The following information applies to parts C and D.

Experiment 2: Using seeds from the greenhouse generation, the researchers grew 60 plants from each of the 2000 and 2007 collections of population D in each of two treatments. In the short-season treatment, watering stopped 65 days after germination; in the long-season treatment, watering continued for 90 days. They counted the seeds that each plant made (Table 2).

The following information applies to parts C and D.

TABLE 2. SEEDS MADE PER PLANT BY POPULATION D PLANTS GROWN IN TWO WATERING TREATMENTS
TreatmentSeeds collected in 2000 (seeds per plant, mean ± 2SE)Seeds collected in 2007 (seeds per plant, mean ± 2SE)
Short season (watering stopped at day 65)22 ± 541 ± 6
Long season (watering continued to day 90)96 ± 988 ± 10
(A) Identify the dependent variable in Experiment 1. Explain why the researchers grew plants from all four seed collections for one generation in the same greenhouse before measuring flowering time.
(B) Calculate the percent change in the mean number of days to first flower for population D from the 2000 collection to the 2007 collection. Determine whether the change in flowering time in population D between the 2000 and 2007 collections is statistically significant. Justify your answer. The researchers claim that the change in flowering time in population D was caused by the dry springs of 2001 to 2006 and not by the way the seeds were stored, grown, or measured. Support the claim using the data for population M.
(C) Using the data in Table 2, describe how the length of the season affects the difference in seed production between the 2000 and 2007 plants of population D. Explain how the dry springs of 2001 to 2006 led to the change in flowering time in population D.
(D) Suppose that from 2008 onward the rains end late every spring, and plants of population D make seeds as in the long-season treatment of Table 2. Predict whether mean flowering time in population D would return toward the 2000 value of 58 days. Justify your prediction. A student claims that the dry springs made each plant in population D flower earlier and that each plant then passed this earlier flowering on to its offspring. Evaluate the claim using the design and results of Experiment 1.
62.Interpreting and Evaluating Experimental Results with Graphing

A student grew Escherichia coli cultures in six concentrations of a lactose analog inducer and measured β-galactosidase activity after a fixed induction period. Five replicate cultures were used at each concentration. The table shows the mean enzyme activity and standard deviation (SD).

Inducer concentration / mmol dm⁻³Mean β-galactosidase activity / Miller units (± SD)
020 (±5)
0.1150 (±20)
0.5900 (±60)
11800 (±100)
52500 (±120)
102600 (±110)
(A) Using the grid provided, construct a scatter graph of mean β-galactosidase activity against inducer concentration. Label both axes with units, choose a scale that uses most of the grid, plot the means accurately and add error bars of ± 1 SD. Draw a curve through the plotted points that shows the trend in the data.
Mean β-galactosidase activity / Miller units
Inducer concentration / mmol dm⁻³
(B) Describe the relationship between inducer concentration and mean enzyme activity shown by the data. Identify the independent variable in this investigation and one variable that should be held constant. Using the error bars, determine whether the mean activity at 5 mmol dm⁻³ differs significantly from the mean activity at 10 mmol dm⁻³. Justify your answer.
(C) Predict the mean enzyme activity if cells were exposed to 20 mmol dm⁻³ of inducer. Justify your prediction, referring to the state of the operon.
(D) Explain why the mean enzyme activity with no inducer (20 Miller units) is low but not zero. Propose one change to the investigation that would confirm the plateau reflects a genuine steady state rather than cells reaching maximum growth.
63.Scientific Investigation

In a species of frog, tadpoles living in clear, shallow ponds have darker skin than tadpoles of the same species living in shaded ponds. A researcher hypothesizes that exposure to ultraviolet (UV) light during development causes tadpoles to make more of the dark pigment melanin. She has collected eggs from a shaded pond, in several clutches (egg masses), each laid by a different female. She can rear tadpoles in tanks under lamps that give off both visible and UV light, and she can cover each tank with either a clear filter that lets UV light through or a filter that blocks UV light but lets the same amount of visible light through. After 4 weeks, she will measure skin darkness as the percentage of light reflected from the skin of each tadpole; a lower percentage means darker skin.

(A) State the null hypothesis for this experiment.
(B) Identify the control group for this experiment, and explain why it is a better control than tadpoles reared in tanks kept in darkness.
(C) Explain why the researcher should divide the tadpoles from each clutch between the two filters, rather than rearing all the tadpoles of some clutches under one filter and all the tadpoles of the other clutches under the other filter.
(D) The researcher also rears tadpoles from eggs collected in a clear, shallow pond under the UV-blocking filter. If the darker skin of tadpoles in clear, shallow ponds is caused only by their exposure to UV light, and not by genetic differences between the two pond populations, predict how the skin darkness of these tadpoles will compare with that of the shaded-pond tadpoles reared under the UV-blocking filter.
64.Conceptual Analysis

In a species of flowering plant, genes N1, N2, and N3 encode proteins that root cells use to take up nitrogen-containing ions from the soil. The three genes lie on three different chromosomes, and each has its own promoter. The same short DNA sequence, element E, lies upstream of each of the three promoters. When the root cells run short of nitrogen, a transcription factor, protein T, binds element E and increases the transcription of the gene next to it; when nitrogen is plentiful, protein T does not bind element E. Students were asked to explain how the three genes are regulated. Parts A to D each describe a claim made by one student, and each claim contains an error.

(A) Student 1 claims that, because N1, N2, and N3 are switched on at the same time, they must be transcribed together into one mRNA, as the genes of a bacterial operon are. Explain the error in this claim, and explain how the three genes are switched on together.
(B) Student 2 claims that element E cannot affect how often a gene is transcribed, because element E is not the promoter and only the DNA sequence that RNA polymerase binds can control transcription. Explain the error in this claim.
(C) Student 3 predicts that in mutant plants that make no protein T, N1, N2, and N3 will be transcribed at a high rate at all times, because nothing will hold their transcription back. Predict the actual effect of this mutation on the transcription of the three genes when the root cells run short of nitrogen, and explain the error in Student 3's reasoning.
(D) Student 4 claims that protein T binding to element E is the only thing that can change how often gene N1 is transcribed. Explain the error in this claim by describing one other way in which a root cell could change how often gene N1 is transcribed.
65.Analyze a Model or Visual Representation

Geneticists crossed fruit flies heterozygous for four linked genes with flies homozygous recessive for all four, and recorded the percentage of offspring that were recombinant for each pair of genes. The model is the linkage map built from these data; one map unit corresponds to a recombination frequency of 1 %.

ABCD 082834 map position / map units (cM) A–B: 8 % recombinants B–C: 20 % recombinants Observed recombination frequency between A and D in a testcross: 31 %
(A) Using the model, determine the expected recombination frequency between genes A and C, and describe what the distance between two genes on the map represents.
(B) Explain why genes that are close together on a chromosome, such as A and B, are usually inherited together.
(C) A fly with genotype AB/ab (A and B on one homolog, a and b on the other) is testcrossed to an ab/ab fly. Predict the percentage of offspring expected to show each of the two recombinant phenotypes.
(D) The map places A and D 34 units apart, but the observed recombination frequency between A and D was only 31 %. Provide reasoning that accounts for this difference.
66.Analyze Data

Researchers soaked barley grains in water for 24 hours and placed equal masses of grains in insulated flasks, each plugged with cotton wool so that air could pass in and out, with a thermometer among the grains. Flask 1 held soaked grains. Flask 2 held soaked grains that had been boiled for 10 minutes, cooled, and rinsed in a disinfectant that kills bacteria and fungi. All of the flasks were kept in a room at 20 °C, with four flasks of each type. The graph shows the mean temperature inside each type of flask over 72 hours.

Mean temperature inside the flasks 0 12 24 36 48 60 72 19 20 21 22 23 24 25 26 27 28 29 time / hours temperature / °C Flask 1: soaked grains Flask 2: boiled, disinfected grains
(A) Identify the purpose of flask 2 in this investigation.
(B) Using the graph, calculate the mean rate of temperature increase in flask 1 between 24 hours and 72 hours. Include units.
(C) Explain, in terms of the energy stored in the grains, why the temperature in flask 1 rose.
(D) As the grains germinate, they build new cells, which are more highly ordered than the molecules the cells are made from. A student claims that this growth decreases disorder and so violates the second law of thermodynamics. Using the data, explain why the claim is incorrect.

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1. 2.9.A [1]
  • C: the O₂ crosses the thylakoid membrane, the inner and then the outer chloroplast membrane, and then the outer and inner mitochondrial membranes: five bilayers in all; it never leaves the cell, so it does not cross the plasma membrane, and as a small nonpolar molecule it passes through each bilayer by simple diffusion. A leaves out the thylakoid membrane; B adds the plasma membrane, which lies off the path; C wrongly says that O₂ needs channel proteins.
2. 6.8.A [1]
  • A: Cutting a circular molecule at each recognition site yields one fragment per cut, so two fragments (3500 bp + 2500 bp) means enzyme B cuts the plasmid at two sites.
3. 6.8.A [1]
  • A: A single cut in a circular molecule opens it into one linear fragment equal to the whole plasmid, so the lone 6000-bp band shows enzyme A has exactly one recognition site.
4. 6.8.A [1]
  • C: one cut in a circular plasmid opens it into a single linear molecule of full length, 6000 bp; molecules cut completely still give 3500 and 2500 bp. The two enzyme B sites divide the circle into 3500 and 2500 bp, so no 1000-bp or 3000-bp piece can form, and partly cut molecules add a new band rather than none.
5. 7.2.B [1]
  • C: Molecular variation that keeps an enzyme functional in the population's environment raises the fitness of individuals carrying it, so it tends to increase in frequency.
6. 3.2.A, 3.2.B [1]
  • D: after 6 months at −18 °C the extract of lot 1 still had 92 % of the fresh activity, so freezing left nearly all of the enzyme in its active shape. Lot 1 changed only slowly while frozen because at −18 °C molecules move so slowly that enzyme and substrate rarely collide; once the peas thawed, the enzyme acted at the much faster rate of room temperature, so lot 1 lost in a day more than it had lost in months in the freezer. Lots 2 and 3 had only 4 % of the activity whether or not they were frozen, so the 2 minutes at 100 °C, not the freezing, inactivated the enzyme: the heat disrupted the interactions that hold its shape, and the change was not reversed when the peas cooled. A is contradicted by lot 1's 92 %. C is contradicted by the assay: every extract was given an excess of substrate, so the low activity of lots 2 and 3 shows that the enzyme itself had been lost. D is contradicted by lot 1, which lost little activity in frozen storage, and by lot 3, which lost as much as lot 2 without being frozen at all.
7. 3.2.B [1]
  • D: at 16 mmol dm⁻³ the uninhibited rate is about 44 µmol min⁻¹; with P it is about 36 (a reduction of about 18 percent, roughly one fifth) and with Q about 22 (still about half), so raising substrate concentration overcomes much of the effect of P but none of the effect of Q.
8. 3.2.B [1]
  • A: Q lowers Vmax without being overcome by substrate, which is expected when an inhibitor binds an allosteric site and alters the active site so that bound substrate is not converted to product.
9. 3.2.B [1]
  • A: More competitive inhibitor means more substrate is needed to outcompete it, so the curve is shifted further to the right, but with enough substrate the same Vmax is still reached.
10. 3.3.A [1]
  • B: a coupled reaction proceeds only if its overall ΔG is negative. Coupling ATP synthesis (+30 kJ/mol) to step 2 gives −34 + 30 = −4 kJ/mol, and to step 4 gives −36 + 30 = −6 kJ/mol, so each of these steps can make one ATP (but not two, since −36 + 60 is positive). Steps 1 (−12) and 3 (−20) release less than 30 kJ/mol, so coupling either to ATP synthesis would give a positive ΔG; their free energy is released as heat. A ignores the size of each step. B and D treat the free energy of separate steps as if it could be pooled, but free energy released in one step is not stored for use in another.
11. 1.6.A [1]
  • A: In double-stranded DNA adenine pairs with thymine (so T = 22%), leaving 56% split equally between guanine and cytosine (28% each).
12. 1.4.A [1]
  • C: the figure shows that dissolved glucose, started from either pure form, ends up as the same mix (about 36 percent α and 64 percent β) after 4 to 6 hours, because its ring keeps opening and closing again in either form. In G, carbon 1 of unit 1 is held in the glycosidic linkage, so that ring cannot open and the α orientation of the linkage is fixed; carbon 1 of unit 2 is free, so its ring can open and close like that of free glucose, and after 24 hours molecules of G carry unit 2 in a mix of the α and β forms. A treats the linked carbon 1 as if it were free. B reverses which carbon 1 can change. C is wrong because joined units keep their rings; the linkage only stops the ring of unit 1 from opening. Because a linkage fixes the orientation at carbon 1, a polysaccharide keeps its α or β linkages, and so its shape, even though free glucose switches between forms.
13. 7.7.A [1]
  • D: a ribosomal RNA found in bacteria, archaea, and eukaryotes, with a shared sequence, points to inheritance from a common ancestor. Mutations arise throughout the molecule, but a change at the site that joins amino acids impairs the ribosome, so cells carrying one leave few descendants and the stretch stays the same, while changes elsewhere are tolerated and accumulate. A: independent origins would not explain why the whole molecule, not just this site, is shared by every species, and lineages arising independently would not be expected to match nucleotide for nucleotide. B: mutations occur at random with respect to their effects, so the stretch is not protected from changing; changes in it are removed by selection. C: the species live in very different conditions, and the environment does not produce particular DNA or RNA sequences.
14. 5.4.A [1]
  • C: Nondisjunction is the failure of chromosomes (or chromatids) to separate; it yields gametes with an extra (n + 1) or missing (n − 1) chromosome, and an n + 1 egg fertilized normally gives trisomy 21.
15. 4.5.A [1]
  • C: Cells that leave the cell cycle and stop dividing while remaining metabolically active are in the G₀ resting phase.
16. 3.4.B, 3.5.A [1]
  • D: in the dark, O₂ fell by 20 μmol per L in 10 minutes, so respiration uses about 2 μmol per L per minute. In dim light the concentration stayed constant, so photosynthesis was releasing O₂ at that same rate. Once compound C stops the mitochondrial electron transport chain, O₂ is no longer used as its final electron acceptor, and the O₂ still released by photosynthesis accumulates at about 2 μmol per L per minute. A mistakes a flat trace for no photosynthesis. B: compound C does not affect the chloroplasts, which make their own ATP in the light-dependent reactions. C uses the bright-light rate of photosynthesis (the net rise of 4 plus 2 used in respiration, 6 μmol per L per minute), but the light is still dim.
17. 7.7.A [1]
  • B: Bacteria read a human coding sequence and make a polypeptide with the same amino acid sequence that human cells make from it. This shows that the genetic code and the machinery for expressing DNA are largely conserved, which is evidence of a shared common ancestor.
18. 2.1.A [1]
  • D: ribosomes bound to the ER make proteins that enter the ER, the start of the route for proteins the cell secretes, while free ribosomes release theirs into the cytosol. A has 10 × 0.20 = 2 million bound and 8 million free ribosomes, and B has 5 × 0.80 = 4 million bound and 1 million free, so B has twice as many of the ribosomes that make proteins for secretion, although it has half as many ribosomes in all. A is wrong because free ribosomes release their proteins into the cytosol, not the ER; C compares the percentages bound instead of the numbers bound, which differ only twofold; D reverses the roles of free and bound ribosomes.
19. 4.4.A [1]
  • C: Negative feedback opposes the change and restores the variable toward a set point, seen here as glucose returning to the same baseline after each rise.
20. 6.4.A [1]
  • B: A codon is decoded only by a tRNA whose anticodon pairs with it, so after Met and Phe are joined the ribosome waits at GAA with no tRNA to deliver Glu; ribosomes do not skip codons or accept an unpaired tRNA, and the first two codons are still translated.
21. 6.7.A [1]
  • C: GAA and GAG both specify glutamic acid, so this substitution leaves the amino acid sequence unchanged; the code's redundancy makes it a silent mutation.
22. 6.7.B [1]
  • D: UAG is a stop codon, so this nonsense mutation truncates the protein after only three amino acids, and the loss of a functional protein can alter the phenotype.
23. 4.2.A [1]
  • A: G at 100 nM binds about 98 percent of the receptors left free, so about 20 percent of all the receptors would have G bound. The table shows that the response is far from proportional to binding: 5 percent bound gives 56 ± 6 percent of the maximum and 33 percent bound gives 93 ± 3 percent, because each bound receptor activates many molecules at each step of the cascade, amplifying the signal. About 20 percent bound therefore gives a response between these, most but not all of the maximum. A assumes the response is proportional to the receptors bound, which the data contradict. B: the blocked receptors are simply not activated; nothing in the data suggests that they stop the free receptors from signaling. C: the stem states that G cannot bind the receptors held by the drug.
24. 2.6.A, 2.7.A [1]
  • A: water entering the cells stretches the membrane and opens the channels; solutes then leave passively, down their concentration gradients, which lowers the solute concentration inside so that less water enters and the cells do not burst. Blocking ATP synthesis did not change survival (90 versus 92 percent), so no pumping is involved. A is ruled out by the ATP result, C by the channels opening only when the membrane is stretched, which needs the cells to swell, and D by the mutant, which has a wall yet mostly dies in water.
25. 4.4.A [1]
  • B: Each contraction intensifies the stimulus that triggered it, amplifying the response until delivery ends the cycle.
26. 7.10.C [1]
  • A: The F1 interval is 73 to 85 percent and the Coast interval 84 to 92 percent. They overlap, if only from 84 to 85, so the 9-point difference in means could be due to chance and the null hypothesis is not rejected. A compares the means without their intervals. B treats a null hypothesis that is not rejected as proved: a real difference too small for 20 plants per group to detect could still exist. D uses the F2 hybrids (50 to 66 percent), which do differ significantly from Coast plants, but the null hypothesis is about the F1 hybrids.
27. 5.5.A [1]
  • C: the ±2SE intervals are 84 to 90 mm (25 °C) and 71 to 77 mm (10 °C) for tail length, 13.7 to 14.7 mm and 12.4 to 13.6 mm for ear length, and 23.4 to 26.2 g and 24.7 to 27.9 g for body mass. The tail and ear intervals do not overlap (the ear intervals only just fail to, 13.6 against 13.7 mm), so those two differences are significant. The body mass intervals overlap from 24.7 to 26.2 g, so the null hypothesis of no difference in body mass is not rejected, even though the 10 °C mean is larger. A mean lying beyond the other group's interval is not the test (A); the ± values are already 2SE, so doubling them again is a misreading (C); and a significant difference in one trait says nothing about another (D). Because the mice are genetically identical, the significant differences in tail and ear length show the environment changing the phenotype a genotype produces.
28. 3.4.B [1]
  • D: Isotope labeling shows the O₂ released in photosynthesis comes from water molecules split during the light-dependent reactions.
29. 6.1.A [1]
  • B: RNA uses the sugar ribose, contains uracil instead of thymine, and is usually single-stranded, whereas DNA uses deoxyribose, contains thymine, and is double-stranded.
30. 1.7.A [1]
  • A: A protein's three-dimensional shape and function are determined by the specific sequence of its amino acid monomers, so reordering them changes the folded structure.
31. 8.5.A, 8.6.A [1]
  • D: Functional redundancy buffers ecosystem processes.
32. 8.3.A [1]
  • B: dN/dt = rN = 0.5 × 200 = 100 cells per hour.
33. 8.2.B [1]
  • B: the added water acts through the thick solid + arrow from spring rainfall to plant growth; plant growth has a thick solid + arrow to grasshopper density, which has a solid + arrow to nestlings fledged. More energy captured by the producers passes up the food chain, so each consumer level can increase. A and C read only the direct dashed arrows and ignore the effect that passes through plant growth and grasshoppers; D is wrong because the arrow from summer temperature to plant growth is dashed, so in the model plant growth does not depend on summer temperature.
34. 7.1.B [1]
  • A: fitness is reproductive success, which depends on both surviving and reproducing. Per seedling, smooth plants left 0.60 × 100 = 60 seeds and hairy plants 0.80 × 50 = 40, so the smooth form contributed more offspring in the wet year. A and C judge fitness by survival alone, and D ignores that the two measures combine into one outcome.
35. 7.1.B [1]
  • A: per seedling, smooth plants leave 60 seeds against 40 in a wet year (0.60 × 100 and 0.80 × 50) but 8 against 24 in a dry year (0.25 × 32 and 0.60 × 40), so selection favors smooth leaves in wet years and hairy leaves in dry years; a fluctuating environment changes the direction of selection from one generation to the next. A is wrong because hairy plants set more seeds per flowering plant in dry years, B judges fitness by survival alone, and C ignores that each year's selection changes the frequency.
36. 7.1.B [1]
  • D: seeds per seedling were 24 for hairy and 8 for smooth plants in the dry year (0.60 × 40 and 0.25 × 32), a threefold difference, but 60 for smooth and 40 for hairy plants in the wet year (0.60 × 100 and 0.80 × 50), only 1.5-fold. B and C each use one component of reproductive success, not the combined outcome, and C is wrong even on its own terms, since survival differed by 35 percentage points (60 against 25) in the dry year. D is wrong because the margins are 3 and 1.5.
37. 6.7.C [1]
  • C: Replication is very accurate but not perfect; an error that escapes proofreading and repair becomes a mutation that is passed to all of that cell's descendants, so cells descended from one bacterium do not all have identical DNA. Bacteria also gain genes from other cells without sexual reproduction: by conjugation (direct transfer between cells), transformation (uptake of free DNA), and transduction (transfer by viruses). A accepts both errors; C wrongly claims proofreading removes every error; D wrongly ties gene transfer to sexual reproduction.
38. 4.1.B [1]
  • A: Sparse normal cells divide (71%) even though no other cell is near them, while line M cells in the same dish do not (4%), so the G each normal cell releases acts on that cell itself and does not reach other cells at an effective level; medium from a crowded culture of line N, which releases G, makes line M divide (66%), so G from many cells together can also act on other cells. B is contradicted by the medium result, which involves no contact between cells. C is contradicted by the 4% of line M cells that divided in the mixed dish. D is contradicted by the stem, since line M has normal receptors, and by line M dividing in medium from line N, whose G it can receive.
39. 3.5.B [1]
  • A: In chemiosmosis, the electron transport chain builds a proton gradient across the inner membrane, and the return flow of protons through ATP synthase drives ATP synthesis.
40. 7.4.B [1]
  • D: A bottleneck samples the gene pool at random and the sample is small. Recovery restores numbers, not the alleles that were lost with the individuals who did not survive.
41. 8.7.B [1]
  • C: an aggressive invader lacking natural controls displaces native species and lowers diversity, disrupting the community's structure.
42. 4.2.A [1]
  • C: the response fell with pulses 10 minutes apart but not with pulses 60 minutes apart, so the loss is temporary, and it did not happen when drug D stopped the cells from taking membrane proteins inside. Receptors taken into the cell after binding L leave fewer on the surface to bind the next pulse, until receptors return to the surface or are replaced. A: each pulse was fresh medium with the same concentration of L. B: a permanent change could not recover when the pulses were 60 minutes apart. D: drug D acts by stopping membrane proteins from being taken in, and nothing in the data shows extra receptors being made.
43. 2.10.A [1]
  • D: every species has mitochondria and nuclear genes like those of aerobic bacteria, so an aerobic prokaryote was taken in by an ancestor of them all; chloroplasts occur only in cells that also have mitochondria, and the 262 species without chloroplasts carry no nuclear genes like those of cyanobacteria, so their lines never held chloroplasts, and the photosynthetic prokaryote was taken in later, by a cell in a line that already had mitochondria. A and C would make the 262 species descendants of cells with chloroplasts, so they should still carry cyanobacteria-like genes in their nuclei, and they do not; D is contradicted by the chloroplast DNA, which is most similar to that of cyanobacteria, not to that of aerobic bacteria.
44. 5.5.A [1]
  • A: Temperature was the same in both rooms and the hares were assigned at random from one population, so the only difference between the groups was day length; the same genotype produces a brown or a white coat depending on that environmental cue. A temperature trigger is ruled out by the design, random assignment makes a genetic difference between the rooms unlikely, and no selection took place because no hares were removed or bred.
45. 2.9.B [1]
  • B: the membrane creates an isolated compartment, and proton pumps maintain the acidic interior distinct from the cytoplasm.
46. 7.10.B [1]
  • B: the island birds evolved where no ground-hunting mammals lived, so there was no selection for traits such as wariness or safe nest sites, and few individuals carried such traits when the predators arrived; the environment changed far faster than selection could act, whereas the continental birds descend from ancestors that were selected by these predators for millions of years. B treats the predators as the cause of the mutations, but mutations arise at random, not in response to need. C is contradicted by the stem, which says the birds' food supply changed little. D is wrong on two counts: drift is weakest in large populations, and the birds had no history of selection for such defenses in the first place.
47. 8.6.B [1]
  • D: Two years after sowing, the clover had raised soil nitrogen and plant biomass, but the number of plant species was unchanged (29 against 28). By Year 15 soil nitrogen had more than tripled, tall, fast-growing grasses covered 62 percent of the ground instead of 10 percent, and species per m² had fallen from 28 to 14: adding one species changed an abiotic factor, nitrogen, and over the long term that let a few grasses that grow fast when nitrogen is plentiful crowd out many smaller species. A ignores the fall in species number by Year 15; B looks only at the short term; D has no support, as the table gives no clover cover, and the plant that spread was the tall grasses.
48. 5.3.A [1]
  • B: Pp × Pp gives 1 PP: 2 Pp: 1 pp; only pp (1/4) is white.
49. 4.2.A [1]
  • D: Transduction is the internal relay that converts the received signal into a cellular action via a molecular cascade.
50. 5.3.A [1]
  • D: Both parents are Aa. Among their unaffected children the genotype ratio is 1 AA: 2 Aa, so an unaffected child has a 2/3 probability of being a carrier.
51. 5.3.A [1]
  • B: III-1 is a carrier only if II-3 is a carrier (probability 2/3) and passes a to him (probability 1/2): 2/3 × 1/2 = 1/3.
52. 8.4.A [1]
  • D: Sorted by density, the wasps killed 15, 21, 26, 41, 58 and 62 percent at 35, 50, 60, 95, 140 and 150 caterpillars per m², so they are a density-dependent factor: they take a larger share of a dense population and a smaller share of a sparse one, pushing the population back toward a middle density. The share killed by storms (6, 25, 4, 31, 10 and 18 percent at the same densities) shows no relationship with density, so storms are density-independent: they can cut the population in any year but do not act harder when it is crowded, so they cannot regulate it. B is true (about 245 against about 87 caterpillars per m² over the six years) but total numbers killed do not show regulation; C picks one year; D describes a density-independent factor, which does not respond to density.
53. 1.2.A [1]
  • B: Without phosphate the cells lose Lipid K and build up Lipid L, and without sulfate they lose Lipid L and build up Lipid K, so K is the lipid that contains phosphorus and L is the lipid that contains sulfur. Each lipid could increase only when the medium supplied its element, so with neither phosphate nor sulfate supplied, neither lipid can be built up and the swap seen in each experiment is no longer possible. A ignores that L needs sulfur, B ignores that K needs phosphorus, and C assumes a replacement that needs an element the medium no longer supplies.
54. 1.2.A [1]
  • B: Potassium phosphate is the only source of phosphorus in the medium, and potassium chloride contains none. Every new cell needs phosphorus atoms from its surroundings to build its nucleic acids and the phospholipids of its membranes, so without a phosphorus source few new cells can be made. A swaps one carbon source for another with almost the same proportion of carbon. B swaps one nitrogen source for another that supplies the same nitrogen, along with some extra carbon. C adds more of the carbon source, which would not lower the number of cells the medium can support.
55. 7.2.B [1]
  • C: more gene copies give more enzyme molecules, so the toxin is broken down faster; variation in the number of a molecule in cells changes the ability to survive and reproduce in a particular environment, here soil that contains the toxin.
56. 2.10.A [1]
  • A: Membrane-bound organelles create separate internal compartments, a hallmark of eukaryotic cells absent in prokaryotes.
57. 8.5.B [1]
  • C: when two species compete for the same limiting resource, one may outcompete and exclude the other, illustrating competitive exclusion.
58. 1.5.A [1]
  • B: the claim says that saturation alone decides the state at 25 °C, so one fatty acid that breaks the rule is enough to refute it, and there are two. The saturated 8-carbon fatty acid melts at 17 °C, so it is liquid at 25 °C, and the 22-carbon fatty acid with one double bond melts at 34 °C, so it is solid. Among the saturated fatty acids, melting point rises with chain length (17, 32, 54 °C), and so it does among those with one double bond (0, 23, 34 °C), so chain length also affects melting point. The double bond still matters: the 16-carbon fatty acid with one double bond melts at 0 °C, far below the shorter, saturated 14-carbon one at 54 °C. A states a true pattern, but a lower melting point is not the same as being liquid at 25 °C, and the claim fails for the 8-carbon and 22-carbon fatty acids. C ignores the 16-carbon comparison, in which a longer chain with a double bond melts lower. D is wrong because a claim that one factor alone decides the outcome is refuted by a single counterexample; it does not need a controlled comparison.
59. 2.4.A, 2.4.B [1]
  • B: the blue solution filled the space inside the wall, so the wall let the dye and the sucrose through; the shrunken contents took up none of the blue dye while losing water, so the boundary around the contents, the plasma membrane, let water out but kept the solutes out, which is what selective permeability means. The wall is a structural boundary, and a barrier only to some much larger substances. A misreads where the dye stopped: it crossed the wall and stopped at the contents; B ignores the dye, which tells the two boundaries apart; C treats selective permeability as letting nothing through, when it means letting some substances, such as water, through and not others.
60. 8.2.C [1]
  • C: chemoautotrophs convert chemical energy in H₂S into biomass, and heterotrophs obtain that energy by consuming them, enabling energy flow with no light.
61. Interpreting and Evaluating Experimental Results 7.8.A, 7.1.B
(A)
3.C Identify the dependent variable in Experiment 1.
Accept one of the following:
  • The number of days from germination to the opening of the first flower
  • Flowering time (days to first flower)
3.C Explain why the researchers grew plants from all four seed collections for one generation in the same greenhouse before measuring flowering time.
Accept one of the following:
  • So that every plant measured grew from a seed of the same age made by parents in the same conditions; otherwise seven years of storage, or the dry conditions in which the 2007 seeds were made, could change flowering time, and a difference could not be attributed to heritable (genetic) differences
  • To remove effects of the parents' environment and of seed age: the 2000 seeds had been stored for years and the 2007 seeds came from mothers that grew in dry springs, so only after a generation in the same conditions do differences reflect inherited differences

Total for part (A): 2 points

(B)
5.A Calculate the percent change in the mean number of days to first flower for population D from the 2000 collection to the 2007 collection.
Accept one of the following:
  • A decrease of about 11 percent ((51.6 − 58.2) ÷ 58.2 × 100 = −11.3 percent)
  • −11 percent (6.6 fewer days out of 58.2)
5.B Determine whether the change in flowering time in population D between the 2000 and 2007 collections is statistically significant. Justify your answer.
Accept one of the following:
  • It is significant: the intervals do not overlap (56.8 to 59.6 days for 2000 and 50.3 to 52.9 days for 2007)
  • Yes, because 58.2 − 1.4 = 56.8 is greater than 51.6 + 1.3 = 52.9
6.B The researchers claim that the change in flowering time in population D was caused by the dry springs of 2001 to 2006 and not by the way the seeds were stored, grown, or measured. Support the claim using the data for population M.
Accept one of the following:
  • Population M's seeds were stored, grown, and measured in the same way, but its soil stayed moist, and its flowering time did not change significantly (intervals 58.5 to 61.7 and 57.9 to 60.9 days overlap); only the population that experienced the dry springs changed
  • If storage or the greenhouse procedure caused the change, M would have changed too; M, which was not exposed to early drying, flowered at about the same time in both collections (60.1 and 59.4 days, overlapping intervals)

Total for part (B): 3 points

(C)
4.B Using the data in Table 2, describe how the length of the season affects the difference in seed production between the 2000 and 2007 plants of population D.
Accept one of the following:
  • In the short season the 2007 plants made significantly more seeds than the 2000 plants (35 to 47 versus 17 to 27 seeds per plant, no overlap), but in the long season the two collections did not differ significantly (87 to 105 versus 78 to 98, overlapping)
  • The 2007 plants have an advantage only when watering stops early (41 versus 22 seeds); with a long season both make similar numbers of seeds (96 and 88, overlapping intervals)
6.C Explain how the dry springs of 2001 to 2006 led to the change in flowering time in population D.
Accept one of the following:
  • Flowering time varied among plants and is heritable (the difference persisted in a shared greenhouse); in the dry springs, plants that flowered earlier set more seeds before the soil dried (as in the short season of Table 2), so alleles for early flowering were passed on more often and became more common over the six generations
  • Natural selection: the early end of the rains favored the plants that flowered earliest, which left the most offspring; because flowering time is inherited, the population's mean flowering time fell by about 6.6 days in six generations

Total for part (C): 2 points

(D)
6.E Suppose that from 2008 onward the rains end late every spring, and plants of population D make seeds as in the long-season treatment of Table 2. Predict whether mean flowering time in population D would return toward the 2000 value of 58 days. Justify your prediction.
Accept one of the following:
  • It would not return toward 58 days (or would change little), because in a long season early-flowering and late-flowering plants make about the same number of seeds (the intervals overlap), so there would be little or no selection favoring later flowering
  • Flowering time would stay near 52 days, since without a difference in reproductive success between early and late flowering plants, selection would not shift the mean back (any change would be slow or by chance)
6.D A student claims that the dry springs made each plant in population D flower earlier and that each plant then passed this earlier flowering on to its offspring. Evaluate the claim using the design and results of Experiment 1.
Accept one of the following:
  • The claim is not supported: the 2007 plants flowered earlier even after they and their parents grew in identical, well-watered greenhouse conditions, so the difference is inherited; it arose because plants that already flowered earlier left more offspring (selection on existing heritable variation), not because individual plants changed in response to drought and passed the change on
  • Individuals did not change; the population did. Experiment 1 removed effects of the environment on each plant, yet the difference remained, which shows a change in the frequencies of alleles in the population produced by natural selection rather than changes acquired by individual plants

Total for part (D): 2 points

Total for question 61: 9 points

62. Interpreting and Evaluating Experimental Results with Graphing 6.5.A, 6.5.B
(A)
4.A Using the grid provided, construct a scatter graph of mean β-galactosidase activity against inducer concentration. Label both axes with units, choose a scale that uses most of the grid, plot the means accurately and add error bars of ± 1 SD.
Accept one of the following:
  • Scatter graph with inducer concentration on the x-axis and mean β-galactosidase activity on the y-axis, both labelled with units, a scale filling most of the grid, all means plotted correctly and ± SD error bars shown
  • A correctly scaled and labelled scatter graph with the points plotted and ± 1 SD bars drawn
4.A Draw a curve through the plotted points that shows the trend in the data.
Accept one of the following:
  • A curve rising steeply between low and intermediate inducer concentrations and levelling off at the highest concentrations
  • A line of best fit that rises then flattens toward the highest inducer concentrations

Total for part (A): 2 points

(B)
4.B Describe the relationship between inducer concentration and mean enzyme activity shown by the data.
Accept one of the following:
  • Enzyme activity rises sharply as inducer concentration increases and then levels off (plateaus) at the highest concentrations
  • Activity increases with inducer concentration, approaching a maximum around 2500–2600 Miller units
3.C Identify the independent variable in this investigation and one variable that should be held constant.
Accept one of the following:
  • Independent variable is inducer concentration; constants include the bacterial strain, incubation time and temperature, and initial cell density
  • IV: inducer concentration. Controlled: strain, incubation time/temperature, starting cell density
5.B Using the error bars, determine whether the mean activity at 5 mmol dm⁻³ differs significantly from the mean activity at 10 mmol dm⁻³. Justify your answer.
Accept one of the following:
  • The ± SD ranges (2380 to 2620, and 2490 to 2710) overlap, so the means are not significantly different
  • Not significantly different, because the error bars (± SD) overlap

Total for part (B): 3 points

(C)
6.E Predict the mean enzyme activity if cells were exposed to 20 mmol dm⁻³ of inducer.
Accept one of the following:
  • Similar to the activity at 10 mmol dm⁻³, e.g. roughly 2600–2650 Miller units
  • A value close to the current plateau, since activity appears to have reached a maximum
6.C Justify your prediction, referring to the state of the operon.
Accept one of the following:
  • Once enough inducer is present to remove essentially all repressor molecules from the operator, the operon is fully (maximally) induced, so adding more inducer cannot further increase the rate of transcription or enzyme activity
  • The lac operon is already maximally induced, so additional inducer has little further effect on activity

Total for part (C): 2 points

(D)
6.D Explain why the mean enzyme activity with no inducer (20 Miller units) is low but not zero.
Accept one of the following:
  • The repressor binds the operator reversibly, so it occasionally leaves it and RNA polymerase transcribes the operon at a low basal rate, making a small amount of β-galactosidase.
  • Repression is not complete: brief periods with the operator free allow some transcription even without inducer.
3.D Propose one change to the investigation that would confirm the plateau reflects a genuine steady state rather than cells reaching maximum growth.
Accept one of the following:
  • Measure enzyme activity at additional time points after induction at the highest inducer concentrations to confirm activity remains stable rather than still changing
  • Normalize enzyme activity to cell density (or measure at a fixed, controlled cell density) at each concentration

Total for part (D): 2 points

Total for question 62: 9 points

63. Scientific Investigation 5.5.A
(A)
3.B State the null hypothesis for this experiment.
Accept one of the following:
  • UV light has no effect on skin darkness: there is no difference in the mean percentage of light reflected from the skin between tadpoles reared under the UV-transmitting filter and tadpoles reared under the UV-blocking filter.
  • Exposure to UV light during development does not change the amount of melanin in the skin, so both groups of tadpoles will have the same mean skin darkness.

Total for part (A): 1 point

(B)
3.C Identify the control group for this experiment, and explain why it is a better control than tadpoles reared in tanks kept in darkness.
Accept one of the following:
  • Tadpoles reared under the UV-blocking filter (same lamps, same visible light); they differ from the treatment group only in UV exposure, whereas tanks kept in darkness would also lack visible light, so a difference in skin darkness could be caused by visible light rather than by UV.
  • The tanks covered with the filter that blocks UV: only UV differs between the two groups, while darkness would change two factors at once (UV and visible light).

Total for part (B): 1 point

(C)
3.C Explain why the researcher should divide the tadpoles from each clutch between the two filters, rather than rearing all the tadpoles of some clutches under one filter and all the tadpoles of the other clutches under the other filter.
Accept one of the following:
  • Tadpoles from different clutches have different parents and differ genetically, so they may differ in skin darkness for genetic reasons; splitting every clutch between the two filters gives both groups the same mix of genotypes, so a difference between the groups can be attributed to UV rather than to genetic differences.
  • Dividing each clutch controls for genotype: each treatment gets siblings from every female, so clutch (genetic) differences cannot cause the result.

Total for part (C): 1 point

(D)
6.E The researcher also rears tadpoles from eggs collected in a clear, shallow pond under the UV-blocking filter. If the darker skin of tadpoles in clear, shallow ponds is caused only by their exposure to UV light, and not by genetic differences between the two pond populations, predict how the skin darkness of these tadpoles will compare with that of the shaded-pond tadpoles reared under the UV-blocking filter.
Accept one of the following:
  • About the same (no significant difference in the percentage of light reflected); without UV exposure, tadpoles from the shallow pond will be as pale as those from the shaded pond, because the two populations do not differ genetically in this trait.
  • Their skin will be no darker than that of the shaded-pond tadpoles under the same filter, since in the same environment the same genotype gives the same phenotype.

Total for part (D): 1 point

Total for question 63: 4 points

64. Conceptual Analysis 6.5.A, 6.5.B
(A)
1.B Student 1 claims that, because N1, N2, and N3 are switched on at the same time, they must be transcribed together into one mRNA, as the genes of a bacterial operon are. Explain the error in this claim, and explain how the three genes are switched on together.
Accept one of the following:
  • The genes are on different chromosomes and each has its own promoter, so each is transcribed into its own mRNA; they are switched on together because each has element E next to its promoter, so the same transcription factor, protein T, binds near all three and increases their transcription at the same time when the root cells run short of nitrogen
  • Coordinated expression does not need an operon: separate eukaryotic genes that share a regulatory sequence (element E) are controlled by the same transcription factor (protein T), so each gene is transcribed separately but all three rise together

Total for part (A): 1 point

(B)
1.C Student 2 claims that element E cannot affect how often a gene is transcribed, because element E is not the promoter and only the DNA sequence that RNA polymerase binds can control transcription. Explain the error in this claim.
Accept one of the following:
  • Transcription is also controlled by regulatory sequences other than the promoter: element E is bound by a regulatory protein (protein T), and protein T bound there helps RNA polymerase bind the nearby promoter or start transcription, so element E changes how often the gene is transcribed although RNA polymerase does not bind it
  • RNA polymerase binds the promoter, but how often it binds and starts transcribing depends on transcription factors bound to regulatory DNA such as element E; when protein T binds element E, transcription of the gene next to it increases

Total for part (B): 1 point

(C)
6.E Student 3 predicts that in mutant plants that make no protein T, N1, N2, and N3 will be transcribed at a high rate at all times, because nothing will hold their transcription back. Predict the actual effect of this mutation on the transcription of the three genes when the root cells run short of nitrogen, and explain the error in Student 3's reasoning.
Accept one of the following:
  • The genes would stay at a low (basal) rate of transcription even when nitrogen runs short, because protein T is an activator that increases transcription, not a repressor that holds it back; without protein T nothing raises their transcription
  • Low transcription of all three genes, with no increase when nitrogen is short; Student 3 has treated protein T as a repressor, but it is a positive regulator, so removing it lowers rather than raises transcription

Total for part (C): 1 point

(D)
1.A Student 4 claims that protein T binding to element E is the only thing that can change how often gene N1 is transcribed. Explain the error in this claim by describing one other way in which a root cell could change how often gene N1 is transcribed.
Accept one of the following:
  • Chemical changes to the chromatin around N1: adding acetyl groups to the histones loosens the chromatin, so RNA polymerase and transcription factors reach the gene more easily and transcription increases (removing them has the opposite effect)
  • Adding methyl groups to the DNA of the N1 promoter, or removing acetyl groups from the nearby histones, packs the chromatin more tightly, so N1 is transcribed less
  • A different regulatory protein, such as a repressor bound to a silencer near N1, could lower its transcription, or another activator binding a different regulatory sequence near N1 could raise it

Total for part (D): 1 point

Total for question 64: 4 points

65. Analyze a Model or Visual Representation 5.4.A
(A)
2.A Using the model, determine the expected recombination frequency between genes A and C, and describe what the distance between two genes on the map represents.
Accept one of the following:
  • 28 % (0 to 28 map units). The distance represents how often crossing over separates the two alleles: the farther apart two genes are, the more likely a crossover occurs between them.
  • 28 %; map distance is proportional to the frequency of crossing over between the genes.

Total for part (A): 1 point

(B)
2.B Explain why genes that are close together on a chromosome, such as A and B, are usually inherited together.
Accept one of the following:
  • Linked genes on the same chromosome do not assort independently; a crossover must occur in the short region between them to separate the parental combinations, and that is rare, so the parental allele combinations are passed on together most of the time.
  • Crossing over between two nearby loci is infrequent, so their alleles remain in the parental combination in about 92 % of gametes.

Total for part (B): 1 point

(C)
6.E A fly with genotype AB/ab (A and B on one homolog, a and b on the other) is testcrossed to an ab/ab fly. Predict the percentage of offspring expected to show each of the two recombinant phenotypes.
Accept one of the following:
  • 4 % each (8 % recombinants in total, split equally between Ab/ab and aB/ab offspring)
  • Each recombinant class about 4 %

Total for part (C): 1 point

(D)
6.C The map places A and D 34 units apart, but the observed recombination frequency between A and D was only 31 %. Provide reasoning that accounts for this difference.
Accept one of the following:
  • Over long distances, double crossovers can occur between the two genes; a double crossover restores the parental combination of A and D, so those offspring are not counted as recombinants and the observed frequency underestimates the true map distance.
  • Some offspring experience two crossovers between A and D, which cancel out and appear non-recombinant, so recombination frequency is lower than the sum of the shorter intervals.

Total for part (D): 1 point

Total for question 65: 4 points

66. Analyze Data 3.3.A
(A)
3.C Identify the purpose of flask 2 in this investigation.
Accept one of the following:
  • It is a control: its grains are dead (boiling denatured their enzymes) and no microorganisms can grow on them, so it shows that the temperature rise in flask 1 is caused by the living grains, not by the flask, the soaking, or the room.
  • To show what happens to the temperature without living (metabolically active) grains, so that any rise in flask 1 can be put down to the activity of the living grains.
  • A negative control that rules out other sources of heat, such as the surroundings or microorganisms decomposing the grains.

Total for part (A): 1 point

(B)
5.A Using the graph, calculate the mean rate of temperature increase in flask 1 between 24 hours and 72 hours. Include units.
Accept one of the following:
  • About 0.15 °C per hour: (28.5 − 21.5) °C ÷ 48 h = 0.146 °C per hour (accept 0.14 to 0.15 °C per hour).
  • About 3.5 °C per day: a rise of 7 °C over 2 days (accept 3.4 to 3.6 °C per day).

Total for part (B): 1 point

(C)
6.D Explain, in terms of the energy stored in the grains, why the temperature in flask 1 rose.
Accept one of the following:
  • The germinating grains break down stored molecules such as starch (by cellular respiration) to release the energy they need to grow; only part of this energy is captured in ATP, and the rest, along with the energy of the ATP once it is used, is released as heat, which the insulated flask keeps in, so the temperature rises.
  • No energy transformation in the cells is 100 percent efficient: as the chemical energy of stored macromolecules is used to make ATP and to build new cells, some of it is released as heat.
  • Living grains respire their stored food continuously to power germination, and the energy that is not kept in new molecules leaves the cells as heat, warming the grains in the flask.

Total for part (C): 1 point

(D)
6.C As the grains germinate, they build new cells, which are more highly ordered than the molecules the cells are made from. A student claims that this growth decreases disorder and so violates the second law of thermodynamics. Using the data, explain why the claim is incorrect.
Accept one of the following:
  • The grains release heat to their surroundings (flask 1 warmed by about 8.5 °C, while flask 2 did not warm), and this heat, together with the breakdown of large stored molecules into smaller ones, increases the disorder of the surroundings by more than growth increases order inside the grains, so total disorder still increases.
  • The order in the new cells is built only with a constant input of energy from the stored molecules, and the rising temperature in flask 1 shows energy leaving the grains as heat; the second law applies to the grains together with their surroundings, whose disorder increases.
  • The second law is not violated because the grains are not a closed system: they use energy from stored food and give out heat (shown by the temperature rise), increasing the disorder of the surroundings more than the order they build.

Total for part (D): 1 point

Total for question 66: 4 points

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