Biology by Bradford
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.
- 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.
- 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
- A four-function, scientific or graphing calculator is permitted.
- Section II: write your responses in the spaces provided.
- Points are not printed per part on the real paper; they live in the scoring guidelines.
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.
The code on the paper rebuilds this exact mock and its markscheme. Build a fresh mock draws a new one from the same rules, so a second sitting is never the first one memorised.
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The mock
Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.
- A four-function, scientific or graphing calculator is permitted.
- Section I and Section II each count for 50% of the exam score.
Answer all questions. For each question, choose the one best answer.
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. DNA is negatively charged and migrates toward the positive electrode, and the gel sieves fragments by size so that smaller fragments travel farther. The sizes of the fragments detected in each lane are listed in the table.
A circular plasmid is cut with restriction enzymes and the fragments are separated by gel electrophoresis. The gel includes a size-marker lane (a DNA ladder) and three sample lanes; smaller fragments migrate farther from the wells. The table lists the fragment sizes (in base pairs) seen in each lane. Based on the number of fragments produced by enzyme B alone, how many recognition sites for enzyme B are present in this circular plasmid?
| Lane | Fragment sizes detected (bp) |
|---|---|
| DNA ladder | 4000, 3000, 2000, 1000, 500 |
| Enzyme A only | 6000 |
| Enzyme B only | 3500, 2500 |
| Enzyme A + Enzyme B | 3500, 1500, 1000 |
- 0 sites
- 1 site
- 2 sites
- 4 sites
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?
- Enzyme A cuts at two sites, giving equal fragments
- Enzyme A does not cut the plasmid at all
- Enzyme A cuts at exactly one site, linearizing it
- Enzyme A degrades the plasmid into nucleotides
In the lane containing plasmid cut with both enzyme A and enzyme B, three fragments are present: 3500 bp, 1500 bp, and 1000 bp. Which of these three fragments will have migrated the farthest from the well, and why?
- All three fragments migrate the same distance because they came from one plasmid
- The 3500-bp fragment, because larger fragments carry more negative charge and move faster
- The 1500-bp fragment, because intermediate-sized fragments always travel farthest
- The 1000-bp fragment, because smaller fragments move more easily through the gel matrix
At room temperature, water (molar mass 18 g/mol) is a liquid, whereas carbon dioxide (44 g/mol) and methane (16 g/mol) are gases, even though these molecules have comparable or greater molar masses. Which property of water best explains why it is a liquid at room temperature while these molecules are gases?
- Water is nonpolar with covalent bonds between molecules
- Water's lower molar mass keeps its molecules moving
- Polar water molecules hydrogen-bond with one another
- CO₂ and methane hydrogen-bond and evaporate readily
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?
- The variation is neutral because enzyme structure does not affect survival.
- Individuals with the heat-stable enzyme survive and reproduce more in the spring.
- The heat-stable enzyme lowers survival because all proteins denature above 50 °C.
- The variation spreads only if the bacteria choose to express the new enzyme.
A biotechnology team wants to make a precise change at one specific location in a cell's genome. They use a system in which a short guide RNA directs a Cas9 protein to a matching DNA sequence, where the protein cuts the DNA so that a new sequence can be introduced at that exact site. Which technology are they using?
- CRISPR-Cas9, guided to a target by a guide RNA
- Bacterial transformation, moving whole plasmids
- The polymerase chain reaction, copying a region
- Gel electrophoresis, sorting fragments by size
Human skin pigmentation varies continuously from very light to very dark. This trait is influenced by several different genes, each contributing a small additive effect to the amount of pigment produced. Which term best describes this pattern of inheritance?
- Codominance
- Epistasis
- Polygenic inheritance
- Incomplete dominance
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.
Which inhibitor is competitive, and what is the evidence?
- P, because at high substrate concentration the rate approaches the uninhibited maximum
- Q, because it reduces the rate of reaction at every substrate concentration that was tested
- P, because it lowers the rate more than Q does at the lowest substrate concentrations tested
- Q, because its curve levels off at exactly half of the uninhibited maximum rate of reaction
Which statement best describes how inhibitor Q reduces the rate of reaction?
- It binds to the active site and is displaced when substrate concentration rises
- It binds to a site other than the active site and changes the shape of the active site
- It denatures the enzyme permanently by breaking its peptide bonds
- It lowers the activation energy of a competing reaction that uses up substrate
The concentration of inhibitor P is doubled and the experiment is repeated. Which change in the curve for P is predicted?
- The maximum rate falls to about 25 µmol min⁻¹
- The curve rises more slowly but still reaches the same maximum rate
- The curve becomes identical to the curve with no inhibitor
- The rate at every substrate concentration falls to zero
A large forest is split by roads and fields into several small, isolated patches. Interior-dwelling songbirds decline sharply even though the total forested area lost is modest. Which mechanism best explains this decline?
- Fragmentation raises edge effects and isolates birds.
- Fragmentation has no effect if some forest remains.
- Fragmentation increases the energy available.
- Splitting the forest raises interior humidity.
A sample of double-stranded DNA is analyzed and found to contain 22% adenine. Based on complementary base pairing, what are the expected percentages of thymine, guanine, and cytosine in this sample?
- Thymine 22%, guanine 22%, cytosine 22%
- Thymine 78%, guanine 11%, cytosine 11%
- Thymine 28%, guanine 22%, cytosine 28%
- Thymine 22%, guanine 28%, cytosine 28%
An electron micrograph of a heart muscle cell shows numerous organelles, each bounded by two membranes, the inner of which is extensively folded into cristae. Which statement best links this structure to the cell's high demand for ATP?
- The folds increase the volume of cytoplasm surrounding the organelle.
- The folded inner membrane increases the surface area for ATP synthesis.
- The outer membrane folds in order to package proteins for secretion.
- The cristae store the cell's genetic information for use during cell division.
A student wants to distinguish a sample of DNA from a sample of pure starch using only elemental analysis. Which pair of elements, if detected in a sample, would confirm that the sample is DNA rather than starch?
- Carbon and oxygen
- Carbon and hydrogen
- Sulfur and phosphorus
- Nitrogen and phosphorus
A storm blows a small number of lizards from a large mainland population onto a distant island, where they establish a new population. The island population has a very different frequency of a neutral color allele than the mainland, even though color has no effect on survival. Which process best accounts for this difference?
- Genetic drift resulting from a founder effect
- Natural selection strongly favoring the island color pattern
- Directional selection acting against the mainland color pattern
- An increased mutation rate caused by stress from the storm
Scientists compare the amino acid sequence of the respiratory protein cytochrome c across several animal species. Species A and B differ at 2 positions, while Species A and C differ at 14 positions. Assuming a roughly constant rate of change, what is the most reasonable inference?
- Species A and B share a more recent common ancestor than Species A and C do
- Cytochrome c sequences cannot be used to infer evolutionary relationships
- Species A and C are the most closely related pair in the comparison
- Species B must have evolved directly from Species C
A herbicide blocks the flow of electrons between photosystem II and photosystem I in the thylakoid membrane of a plant. Which immediate consequence is most likely in the light-dependent reactions?
- Production of NADPH and ATP declines because electrons cannot reach photosystem I.
- Oxygen consumption in the stroma rises sharply.
- The Calvin cycle immediately speeds up because of the excess NADPH available.
- Water splitting at photosystem II increases to compensate for the block.
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 ______.
- nondisjunction; an abnormal number of chromosomes (for example n + 1)
- independent assortment; a normal haploid set in a new combination
- synapsis; a tetrad of four paired chromatids
- crossing over; a recombinant chromosome carrying a new mix of alleles
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?
- The G₂ checkpoint, just before mitosis
- The G₀ phase, a resting state
- The S phase, when DNA is replicated
- The M phase, when the cell divides
When the human gene for insulin is inserted into bacterial cells, the bacteria transcribe and translate the gene and produce functional human insulin protein. Which conclusion about the hereditary machinery of these two very different organisms is best supported?
- Bacteria evolved directly from human cells.
- Humans and bacteria must have exchanged the insulin gene recently by horizontal transfer.
- The genetic code and the machinery for expressing DNA are largely shared.
- The insulin gene arose independently in bacteria by convergent evolution.
Elemental analysis of an unknown biological molecule shows that it contains only carbon, hydrogen, and oxygen, with hydrogen and oxygen present in an approximate 2:1 ratio, matching the formula (CH₂O)ₙ. To which class of macromolecule does this molecule most likely belong?
- Protein
- Nucleic acid
- Phospholipid
- Carbohydrate
A paleontologist plots the average shell height of a mollusc lineage against time across 8 million years of fossil layers. Shell height stays essentially unchanged for long spans, then shifts rapidly to a new stable value at two points that each coincide with the appearance of a new species. Which model of the pace of evolution do these data best illustrate?
- Artificial selection carried out by humans
- Punctuated equilibrium: stasis then rapid change
- Gradualism: slow steady change throughout
- Convergent evolution of unrelated lineages
The pesticide DDT is absorbed in small amounts by algae, and its concentration in body tissues rises at each higher trophic level, reaching its highest levels in fish-eating birds. This human-caused pattern is best described as which of the following?
- A density-independent factor acting on birds
- The 10% rule of energy transfer between levels
- Primary succession triggered by the pesticide
- Biomagnification up the trophic levels
A graph shows a person's blood glucose concentration over 12 hours. After each of three meals the glucose rises sharply, then returns to about the same baseline value within two hours, producing a series of peaks that each settle back toward one level. Which feature of the graph is the best evidence that blood glucose is regulated by negative feedback?
- Glucose returns to the same set point
- The glucose concentration never changes
- Glucose increases more with each meal
- Each peak is higher than the one before
The following table gives the amino acid or signal specified by several mRNA codons. Use it to answer the questions in this set.
A cell-free system is given the mature mRNA 5'-AUG UUU GAA AAG UGC UAA-3' along with ribosomes, tRNAs, and amino acids. Using the codon assignments in the table, translation begins at the start codon and proceeds until a stop codon is reached. How many amino acids will the completed polypeptide contain?
| mRNA codon | Amino acid / signal |
|---|---|
| AUG | Met (start) |
| UUU | Phe |
| GAA | Glu |
| GAG | Glu |
| AAG | Lys |
| UGC | Cys |
| UAG | Stop |
| UAA | Stop |
- 4
- 5
- 6
- 7
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?
- UGC changed to UGA (Cys to Stop)
- AAG changed to AUG (Lys to Met)
- UUU changed to CUU (Phe to Leu)
- GAA changed to GAG (Glu to Glu)
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?
- The polypeptide is one amino acid longer
- One amino acid is substituted, length unchanged
- Translation stops early, truncating the protein
- No effect, because the code is redundant
Two phospholipids are identical except that one has fully saturated fatty acid tails (no C=C double bonds) and the other has tails containing several C=C double bonds. Membranes rich in the second phospholipid remain more fluid at low temperatures. Which statement best explains this difference in function?
- Kinks stop tight packing, keeping the membrane fluid
- The double bonds let tails pack more tightly
- The number of carbons, not double bonds, matters
- The two tail types are built from different monomers
A peptide hormone normally binds a cell-surface receptor to trigger a response. A point mutation changes the shape of the region of the hormone that fits into the receptor's binding site. Which outcome is most likely for cells exposed to the mutated hormone?
- The hormone will bind its receptor more tightly and increase the response, because the new shape fits better.
- The mutation affects only transcription of the hormone gene, not signaling.
- The hormone cannot bind its receptor, so the downstream response is not triggered by it.
- The receptor will activate even in the complete absence of the hormone, because its binding site is now empty.
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?
- No feedback, because oxytocin is released only a single time.
- Positive feedback, because the response amplifies the original stimulus.
- Negative feedback, because oxytocin levels steadily decline over time.
- Negative feedback, because the response reduces the original stimulus.
A large population of 2,000 grove snails is in Hardy-Weinberg equilibrium for a shell-color gene. The yellow allele (y) is recessive to the brown allele (Y); 320 snails are yellow and the rest are brown.
In a large population of 2,000 grove snails, shell color is controlled by a single gene with two alleles. The yellow allele (y) is recessive to the brown allele (Y). Biologists count 320 yellow snails; the remainder are brown. The population is assumed to be in Hardy-Weinberg equilibrium. What is the frequency of the recessive yellow allele (y)?
- 0.16
- 0.40
- 0.60
- 0.84
Using the same grove snail population, approximately how many of the brown snails are expected to be heterozygous (Yy)?
- 320
- 720
- 960
- 1,280
A bird species arrives that preferentially eats brown grove snails against the dark soil but rarely finds the yellow snails. Which Hardy-Weinberg assumption is violated, and what is the predicted effect on the population?
- No natural selection is violated; the yellow allele frequency will increase
- No gene flow is violated; the yellow allele frequency will decrease
- Random mating is violated; genotype frequencies will remain constant
- No mutation is violated; the brown allele frequency will increase
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?
- CO₂, which is split during fixation in the Calvin cycle.
- H₂O, which is split during the light-dependent reactions.
- The phosphate groups of ATP.
- Glucose synthesized in the Calvin cycle.
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?
- RNA has deoxyribose and thymine; DNA ribose, uracil
- RNA has ribose and uracil and is single-stranded
- RNA and DNA differ only in their bound proteins
- RNA is double-stranded and DNA single-stranded
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?
- Folding is set only by hydrogen bonds with water
- The number of amino acids determines the shape
- Same amino acids must give the same function
- The order of amino acids determines the folding
Why are ecosystems with high species diversity generally more resistant to disturbance than those with low diversity?
- High-diversity ecosystems contain fewer predators, so prey populations remain stable when conditions change
- Every species in a diverse ecosystem is equally abundant, so no single species can be lost during a disturbance
- If one species declines, others with similar roles can maintain ecosystem functions such as decomposition and pollination
- Diverse ecosystems have simpler food webs with fewer links, so a disturbance to one species affects fewer others
Ground squirrels increase the frequency of upright vigilance postures when the scent of a coyote is present in the air. This behavioral change is best described as a response to which type of cue?
- A permanent developmental change fixed at birth
- An external cue signaling increased predation risk
- An internal metabolic signal unrelated to surroundings
- A random behavior with no environmental trigger
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?
- 400 cells per hour
- 50 cells per hour
- 100 cells per hour
- 0.0025 cells per hour
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?
- Direct transfer of a phosphate group from glucose to ADP.
- The splitting of water molecules in the mitochondrial matrix.
- The fermentation of pyruvate to lactate.
- A proton gradient across the inner mitochondrial membrane.
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
- increased allele diversity, because the population grew rapidly and new mutations accumulated
- reduced allele diversity, because the survivors carried only a fraction of the original alleles
- no change in allele frequencies, because population size returned to its original value of 500
- a higher mutation rate, because the population was under stress and needed new alleles
A signaling pathway involves the following molecules: an extracellular protein that binds the cell, a membrane protein that changes shape when the extracellular protein binds, and a small molecule called cAMP that increases inside the cytoplasm and activates a kinase. Which molecule in this pathway functions as a second messenger?
- The kinase that becomes activated
- The extracellular protein that binds the cell
- The small molecule cAMP in the cytoplasm
- The membrane protein that changes shape
Root hair cells in plants each have a single long, thin extension that projects into the soil. This shape most directly benefits the cell in which way?
- It provides internal space for chloroplasts.
- It increases the membrane surface area for absorption.
- It reduces surface area so less water is lost.
- It increases cytoplasm volume for storing starch.
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?
- It raises the number of trophic levels supported
- It outcompetes natives, reducing their diversity
- It increases diversity by adding one species
- It has no effect on a stable wetland community
Himalayan rabbits carry an allele for an enzyme that produces dark fur pigment, but the enzyme is active only at cooler temperatures. These rabbits have dark fur on their cooler extremities (ears, nose, feet) and light fur on their warmer body core. Which statement best explains this coat pattern?
- A frameshift mutation changes fur color only in the warm regions
- The genotype is different in the warm and cool regions of the body
- The phenotype results from genotype interacting with temperature
- The rabbit carries two different genes whose effects blend together
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?
- Cytoplasm is continuously pumped into the lysosome to raise its internal pH.
- Proton pumps in the lysosomal membrane keep the enclosed interior acidic.
- The lysosome has no membrane, so its contents mix freely with the cytoplasm.
- The enzymes inside generate a new membrane around themselves as they work.
Two populations of fish in the same lake spawn at different depths and rarely interbreed, although no physical barrier separates them. This situation is best described as
- postzygotic isolation caused by hybrid inviability at the two depths
- allopatric speciation driven by a geographic barrier between the depths
- gene flow maintaining a single population across the whole lake
- prezygotic isolation that could lead to sympatric speciation
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?
- 0
- 1/4
- 1/2
- 3/4
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?
- The binding of epinephrine to its receptor on the liver cell membrane.
- The breakdown of glycogen into glucose within the liver cell.
- The relay of the signal through the G protein, cAMP, and protein kinase A.
- The release of glucose from the liver cell into the bloodstream.
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?
- 1/4
- 1/2
- 3/4
- 2/3
What is the probability that III-1 is a carrier of the recessive allele?
- 1/3
- 1/6
- 1/4
- 1/2
Red blood cells were placed in a series of sodium chloride (NaCl) solutions of different concentrations for 10 minutes. The relative volume of the cells, compared with their normal volume in blood plasma (taken as 100%), was then measured. Red blood cells lack a cell wall. Table 1 shows the results.
Which NaCl concentration is closest to being isotonic to the cytoplasm of the red blood cells?
| NaCl concentration (%) | Relative cell volume (%) |
|---|---|
| 0.00 | lysed |
| 0.45 | 132 |
| 0.90 | 100 |
| 1.50 | 79 |
| 3.00 | 58 |
- 0.45%
- 0.90%
- 1.50%
- 3.00%
The cells placed in 0.00% NaCl (distilled water) lysed (burst). Which statement best explains this outcome?
- Sodium chloride rushed into the cells, causing them to swell and burst.
- The cells actively pumped water inward until the internal pressure burst them.
- Water entered the cells by osmosis because the solution was hypotonic.
- Water left the cells because distilled water is hypertonic to the cytoplasm.
In the 3.00% NaCl solution the cells shrank to 58% of their normal volume. Which statement correctly describes the water movement responsible for this change?
- No water moved; the cells shrank because salt crystals formed inside them.
- Sodium chloride was actively pumped out, carrying the water with it.
- Water moved into the cells because the solution was hypotonic to the cytoplasm.
- Water moved out of the cells down its water potential gradient.
Motile soil bacteria swim toward regions of higher nutrient concentration. Surface receptors detect the concentration of the nutrient, and when the bacterium senses that the concentration is increasing, a signaling pathway adjusts the rotation of its flagella so that it continues moving in that direction. Which statement best describes the role of the environment in this behavior?
- A nutrient gradient is detected and alters movement
- The bacteria move randomly, ignoring nutrients
- The nutrient permanently disables the flagella
- The behavior needs direct cell-to-cell contact
Which of the following is the best measure of the evolutionary fitness of an individual?
- The number of its offspring that survive to reproduce, relative to other individuals
- Its body size and strength compared with other individuals of the same age in the population
- The length of time it survives compared with the average lifespan for its species
- The number of different environments it can tolerate compared with other individuals
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?
- Only the bacterium possesses ribosomes, which are needed for protein synthesis.
- The bacterium contains a membrane-bound nucleus, while the plant cell does not.
- The plant cell lacks a plasma membrane, relying on its wall to enclose the cytoplasm.
- Only the plant cell contains membrane-bound organelles such as a nucleus.
In the Miller–Urey experiment, a mixture of gases was subjected to electrical discharge and amino acids were recovered. Which conclusion does this result support?
- Organic monomers can form from inorganic precursors under abiotic conditions
- Living cells arose spontaneously within the apparatus from the gases supplied to it
- Amino acids can only be produced by living organisms, which must have contaminated it
- Proteins assembled themselves in the apparatus without any input of energy
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?
- Competitive exclusion for a shared resource
- A predator–prey cycle between the species
- Resource partitioning allowing coexistence
- Mutualism, with both species benefiting
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?
- Energy is created by the tube worms and passed downward to the bacteria
- The animals photosynthesize using faint light emitted by the vents
- Chemoautotrophs convert chemical energy into biomass that heterotrophs consume
- The community needs no energy input because it is completely isolated
Answer all questions. Write your responses in the spaces provided.
Ecologists measured the net primary productivity (NPP) of grassland plots under three grazing regimes maintained for several years. NPP is the energy that producers store as new biomass and make available to the rest of the community. The table reports the mean NPP of the plots in each regime, with standard error (SE).
The following information applies to parts B, C, and D.
The following information applies to parts B, C, and D. In a plot of these data, the error bars represent ±2 SE of the mean.
| Grazing regime | Net primary productivity (g·m⁻²·yr⁻¹), Mean ± SE |
|---|---|
| Ungrazed control | 820 ± 25 |
| Moderate grazing | 910 ± 30 |
| Heavy grazing | 540 ± 28 |
A student compared bacterial transformation efficiency under four treatment combinations using plasmid DNA carrying an antibiotic resistance gene. Five replicate plates were used for each treatment. The table shows the mean number of resistant colonies per plate and standard deviation (SD).
| Treatment | Mean colonies per plate (± SD) |
|---|---|
| No CaCl₂, no heat shock | 1 (±1) |
| CaCl₂ only, no heat shock | 4 (±2) |
| CaCl₂ + heat shock | 85 (±12) |
| CaCl₂ + heat shock + recovery | 140 (±15) |
In humans and many other mammals, the frequency of offspring born with trisomy (three copies of one chromosome) increases with the age of the mother. Trisomy can result when nondisjunction during meiosis produces an egg that carries an extra chromosome. Researchers propose that eggs from older females undergo nondisjunction more often than eggs from younger females. The researchers work with a laboratory strain of mouse in which 2n = 40, so a normal egg contains 20 chromosomes. They can collect mature eggs from females of any age, stain the chromosomes, and count the number of chromosomes in each egg.
Starch and cellulose are two polysaccharides found in plants. Both are polymers built entirely from the monosaccharide glucose, but the glucose monomers are joined by different types of glycosidic linkages. Starch stores energy inside plant cells and is readily digested by many animals, whereas cellulose forms strong fibers in plant cell walls and cannot be digested by most animals.
A model of the control of one bacterial gene states that transcription occurs only when an activator protein is bound to the DNA and a repressor protein is not. The activator binds only when the sugar arabinose is present; the repressor binds only when glucose is present. The table gives the amount of mRNA produced under four combinations of the two sugars.
| Arabinose | Glucose | Relative amount of mRNA |
|---|---|---|
| absent | absent | 0.0 |
| absent | present | 0.0 |
| present | absent | 1.0 |
| present | present | 0.1 |
To test whether a man (F1 or F2) is the father of a child (C), short tandem repeat (STR) regions were amplified by PCR from the DNA of the mother (M), the child and both men, and the products were separated by gel electrophoresis. The figure shows the gel. A ladder of fragments of known length was run in the first lane.
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- C — 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.
- C — 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.
- D — The gel separates DNA by size, and the smallest fragment (1000 bp) passes through the matrix most easily, so it migrates farthest from the well.
- C — Water's polarity allows extensive hydrogen bonding between molecules, holding them together as a liquid, whereas nonpolar CO₂ and CH₄ lack such attractions and are gases.
- B — 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.
- A — CRISPR-Cas9 uses a guide RNA to direct the Cas9 nuclease to a matching genomic sequence and cut it, enabling targeted editing at a chosen site.
- C — When several genes each add a small effect to a single trait, the result is continuous (quantitative) variation, the hallmark of polygenic inheritance.
- A — A competitive inhibitor binds the active site and is outcompeted when substrate is abundant, so Vmax is unchanged; Q lowers Vmax at all concentrations, the signature of a non-competitive (allosteric) inhibitor.
- B — 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.
- B — 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.
- A — dividing habitat raises the proportion of edge, shrinks interior habitat, and isolates small populations, all of which lower the abundance of interior specialists.
- D — In double-stranded DNA adenine pairs with thymine (so T = 22%), leaving 56% split equally between guanine and cytosine (28% each).
- B — cristae increase the inner-membrane surface area that holds the electron transport chain and ATP synthase, supporting high rates of ATP production.
- D — DNA contains nitrogen (in its nitrogenous bases) and phosphorus (in its phosphate groups), whereas starch, a carbohydrate, contains only carbon, hydrogen, and oxygen.
- A — a small, non-representative founding sample produces a chance shift in allele frequencies, the founder effect form of genetic drift.
- A — fewer sequence differences imply less time since divergence, so A and B share a more recent common ancestor.
- A — Blocking electron flow starves photosystem I of electrons, reducing NADPH production and the proton gradient that drives ATP synthesis.
- A — 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.
- B — Cells that leave the cell cycle and stop dividing while remaining metabolically active are in the G₀ resting phase.
- C — That a human gene can be read and expressed by bacteria shows the genetic code and expression machinery are conserved, which is evidence of a shared common ancestor.
- D — Containing only C, H, and O with H and O in a 2:1 ratio, matching (CH₂O)ₙ, is characteristic of carbohydrates.
- B — long stasis punctuated by rapid change associated with speciation events is the signature of punctuated equilibrium, in contrast to the slow steady change of gradualism.
- D — a persistent toxin that concentrates in tissues and increases up the food chain is undergoing biomagnification.
- A — 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.
- B — AUG-UUU-GAA-AAG-UGC code for Met-Phe-Glu-Lys-Cys and UAA is a stop codon, so five amino acids are joined before termination.
- D — 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.
- C — 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.
- A — C=C double bonds kink the fatty acid tails and prevent tight packing, so the membrane stays fluid at lower temperatures, altering its function.
- C — Altering the binding region's shape prevents the ligand–receptor fit, so reception and the resulting response fail to occur.
- B — Each contraction intensifies the stimulus that triggered it, amplifying the response until delivery ends the cycle.
- B — q² = 320/2,000 = 0.16, so q = √0.16 = 0.40.
- C — with p = 0.60 and q = 0.40, 2pq = 0.48; 0.48 × 2,000 = 960 heterozygotes.
- A — selective predation removes brown (Y) alleles, so the y allele frequency is predicted to rise, violating the no-selection condition.
- B — Isotope labeling shows the O₂ released in photosynthesis comes from water molecules split during the light-dependent reactions.
- 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.
- D — 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.
- C — Functional redundancy buffers ecosystem processes.
- B — the predator scent is an external environmental cue, and heightened vigilance is a behavioral response that adjusts to that change.
- C — dN/dt = rN = 0.5 × 200 = 100 cells per hour.
- D — 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.
- 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.
- C — A second messenger is a small, non-protein intracellular molecule, such as cAMP, that relays a signal from the receptor to downstream targets.
- B — the projection is a surface-area adaptation that increases the membrane available for absorbing water and minerals.
- B — an aggressive invader lacking natural controls displaces native species and lowers diversity, disrupting the community's structure.
- C — Every cell carries the same temperature-sensitive allele, but it yields different phenotypes in different body regions, showing that phenotype arises from the genotype interacting with the environment.
- B — the membrane creates an isolated compartment, and proton pumps maintain the acidic interior distinct from the cytoplasm.
- The barrier acts before fertilisation and there is no geographic separation, which is what makes it prezygotic and sympatric. Postzygotic isolation would require hybrids to form first.
- B — Pp × Pp gives 1 PP : 2 Pp : 1 pp; only pp (1/4) is white.
- C — Transduction is the internal relay that converts the received signal into a cellular action via a molecular cascade.
- 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.
- A — 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.
- B — at 0.90% the cells stay at their normal volume (100%), indicating no net water movement and therefore equal water potential inside and out.
- C — in hypotonic distilled water, water enters by osmosis; without a cell wall the animal cell cannot resist the pressure and bursts.
- D — the hypertonic 3.00% solution has a lower water potential, so water leaves the cells by osmosis and they shrink.
- A — Detection of the external nutrient gradient initiates a signaling pathway that alters flagellar rotation, changing the cell's behavior (chemotaxis).
- A — Fitness is relative reproductive success; survival matters only insofar as it leads to reproduction.
- D — Membrane-bound organelles create separate internal compartments, a hallmark of eukaryotic cells absent in prokaryotes.
- The experiment produced monomers, not cells and not polymers. It establishes that the building blocks are chemically reachable without life, which is a much narrower claim than the other options make.
- A — when two species compete for the same limiting resource, one may outcompete and exclude the other, illustrating competitive exclusion.
- C — chemoautotrophs convert chemical energy in H₂S into biomass, and heterotrophs obtain that energy by consuming them, enabling energy flow with no light.
- NPP is the chemical energy stored as new producer biomass; it is the only energy that can be passed to consumers, and because energy is lost at each transfer, it caps the energy available to higher trophic levels
- Producers capture and store energy as NPP, and all consumers ultimately depend on that stored energy, so more NPP means more energy for the community
- Moderate grazing has the highest mean NPP (910), the ungrazed control is intermediate (820), and heavy grazing has the lowest (540)
- NPP is slightly higher under moderate grazing than the control and much lower under heavy grazing
Total for part (A): 2 points
- Significant, because the ±2 SE intervals (control ≈ 770–870; heavy ≈ 484–596) do not overlap
- Significant because the ±2 SE error bars for heavy grazing and the control do not overlap
- Not significant, because the ±2 SE intervals (control ≈ 770–870; moderate ≈ 850–970) overlap
- Not significant because the ±2 SE error bars overlap, so the means may not truly differ
Total for part (B): 2 points
- It would decrease
- The grassland would support a smaller herbivore population under heavy grazing
- Heavy grazing lowers NPP, so less energy is stored by producers and less is available to pass to consumers, supporting fewer/smaller herbivore populations
- Because consumer populations are limited by the energy producers make available, the reduced NPP under heavy grazing supports a smaller herbivore population
Total for part (C): 2 points
- Species diversity and resilience would decrease
- The community would become less diverse and less resilient
- Lower NPP supports fewer individuals and species, and a less diverse community has fewer species that can compensate after a disturbance, so resilience falls
- With less available energy the community sustains less complex structure and lower diversity, which reduces its ability to recover from disturbance
- Reduce grazing intensity (e.g., rotational grazing or lower stocking density) so producers can regrow and NPP recovers
- Rest or temporarily exclude grazers from degraded plots to let vegetation and NPP rebuild, restoring energy to the food web
Total for part (D): 3 points
Total for question 61: 9 points
- Bar graph with treatment on the x-axis and mean colonies per plate 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 bar graph with the points plotted and ± 1 SD bars drawn
- 'CaCl₂ + heat shock + recovery' labelled as greatest and 'No CaCl₂, no heat shock' labelled as least
- Highest bar identified as CaCl₂ + heat shock + recovery (140), lowest bar identified as the untreated control (1)
Total for part (A): 2 points
- Colony counts rise from the untreated control through CaCl₂ alone, then increase sharply once heat shock is added, and increase further when a recovery period follows heat shock
- Transformation efficiency is lowest without CaCl₂ or heat shock and highest with CaCl₂, heat shock and a recovery period
- 140 ÷ 85 ≈ 1.6 times greater
- About 1.6-fold greater (140/85 = 1.65)
- The ± SD ranges (0 to 2, and 2 to 6) overlap, so the means are not significantly different
- Not significantly different, because the error bars (± SD) overlap
Total for part (B): 3 points
- Still low, similar to the untreated control, e.g. roughly 1–2 colonies
- A value close to the no-treatment control, since CaCl₂ is needed for heat shock to be effective
- CaCl₂ neutralizes the negative charges on the bacterial cell membrane and on the DNA, allowing plasmid DNA to approach and enter the cell during heat shock; without CaCl₂, heat shock alone provides little increase in transformation
- Without CaCl₂ masking charge repulsion between DNA and the membrane, heat shock cannot substantially increase DNA uptake
Total for part (C): 2 points
- CaCl₂ neutralizes the negative charges on both the DNA and the bacterial cell membrane, reducing the electrostatic repulsion between them; the rapid temperature increase of heat shock then transiently destabilizes the membrane, creating pores through which the now-less-repelled plasmid DNA can enter the cell; the recovery period afterward allows cells to begin expressing the resistance gene before antibiotic selection, increasing the number of colonies detected
- CaCl₂ masks charge repulsion and heat shock disturbs the membrane, allowing more plasmid DNA to enter; recovery time lets resistance genes be expressed before selection, increasing colony counts
- Include a plasmid-free negative control, cells given CaCl₂ and heat shock but no DNA, plated on the same antibiotic medium, to confirm colonies do not arise from spontaneous resistance
- Add a no-plasmid control treated identically and check that it produces no colonies on the antibiotic plate
Total for part (D): 2 points
Total for question 62: 9 points
- If nondisjunction becomes more frequent as females age, then a higher percentage of the eggs collected from older females will contain an abnormal number of chromosomes (19 or 21 rather than 20) than the eggs collected from younger females.
- The percentage of eggs with a chromosome number other than 20 will increase with the age of the female from which the eggs were collected.
Total for part (A): 1 point
- Independent variable = the age of the female mouse; dependent variable = the percentage of her eggs that contain an abnormal number of chromosomes (not 20); constants (any two) = the mouse strain (all females genetically similar), diet and housing conditions, the number of eggs scored per female, the staining and counting method, and having the same person count chromosomes without knowing the age of the female.
- IV is female age, DV is the proportion of eggs with 19 or 21 chromosomes, and the females in every age group must come from the same strain and be kept under the same conditions, with eggs collected, stained and counted in the same way.
Total for part (B): 1 point
- Use at least ten females in each of several age groups (for example 3, 6, 12 and 18 months); collect a fixed number of mature eggs (for example 50) from each female; stain the chromosomes and count the number in every egg; classify each egg as normal (20) or abnormal (19, 21 or other); calculate the percentage of abnormal eggs for each female and the mean (±SE) for each age group; compare the means across age groups.
- Collect equal numbers of eggs from young, middle-aged and old females of the same strain kept under identical conditions, count the chromosomes in each stained egg, and record the percentage of eggs per female with an abnormal chromosome number; compare the mean percentage among the age groups.
Total for part (C): 1 point
- Support: the mean percentage of eggs with an abnormal chromosome number rises with age, for example a low percentage in 3-month-old females and a several-fold higher percentage in 18-month-old females, with both 19- and 21-chromosome eggs found (no increase with age would refute the hypothesis). During nondisjunction, a pair of homologous chromosomes fails to separate at anaphase I (or sister chromatids fail to separate at anaphase II), so one egg receives both copies of that chromosome (21) while another receives none (19); when a 21-chromosome egg is fertilized by a normal sperm carrying 20, the zygote has three copies of that chromosome (2n + 1 = 41), which is trisomy.
- If the hypothesis is correct, older females produce a greater proportion of eggs with 19 or 21 chromosomes than younger females; an egg with 21 chromosomes arises when homologs (meiosis I) or sister chromatids (meiosis II) both move to the pole that becomes the egg, and fertilization of that egg by a normal haploid sperm gives an embryo with an extra copy of one chromosome.
Total for part (D): 1 point
Total for question 63: 4 points
- Glucose
- Glucose (a monosaccharide)
Total for part (A): 1 point
- Their glucose monomers are joined by different types of glycosidic linkages (α in starch, β in cellulose), giving the polymers different shapes and therefore different functions
- The way the identical monomers are linked/arranged differs, so the resulting polymers differ in structure and function
Total for part (B): 1 point
- No, it will not break down cellulose
- No / the enzyme cannot hydrolyze cellulose
Total for part (C): 1 point
- The enzyme's active site is specific to the shape of the α linkage in starch; cellulose's β linkages have a different structure that does not fit the active site, so the enzyme cannot hydrolyze it
- Because cellulose's β linkages differ in orientation from the α linkages the enzyme recognizes, the substrate does not fit the enzyme, so cellulose is not broken down
Total for part (D): 1 point
Total for question 64: 4 points
- Arabinose present and glucose absent
- Arabinose present, glucose absent
Total for part (A): 1 point
- The activator is bound because arabinose is present, but glucose being present also lets the repressor bind, and the model requires the repressor to be absent for transcription
- Both conditions must be met at once; the bound repressor overrides the bound activator, so almost no mRNA is made
Total for part (B): 1 point
- Without arabinose the activator cannot bind, and the model makes activator binding necessary for transcription, so the glucose condition cannot rescue it
- Activator binding is a requirement, so failing it gives no mRNA regardless of the repressor
Total for part (C): 1 point
- High, close to 1.0: the activator is bound and the only thing that suppressed transcription, the repressor, cannot be made
- Transcription would be high, since with no repressor the model's second condition is met automatically
Total for part (D): 1 point
Total for question 65: 4 points
- The child shares the 850 and 500 bp bands with the mother but has two additional bands, at 600 and 100 bp, that the mother lacks.
- Two of the child's four bands (850, 500) match the mother; the other two (600, 100) do not.
Total for part (A): 1 point
- F2: every band in the child that did not come from the mother (600 and 100 bp) is present in F2's profile, whereas F1 lacks the 600 bp band.
- F2, because the child's non-maternal bands at 600 and 100 both appear in F2 and only the 100 bp band appears in F1.
Total for part (B): 1 point
- DNA is negatively charged because of its phosphate groups, so it moves toward the positive electrode; shorter fragments pass through the pores of the gel more easily and so travel further in the same time.
- Fragments migrate toward the positive electrode at a rate that decreases with length, so smaller fragments are found closer to the bottom of the gel.
Total for part (C): 1 point
- Any single STR allele is shared by many unrelated people, so a match at a few loci could occur by chance; with more loci the probability that an unrelated man matches all of the child's non-maternal bands becomes vanishingly small.
- Each additional matching locus multiplies down the chance of a coincidental match, making the paternity conclusion far more reliable.
Total for part (D): 1 point
Total for question 66: 4 points