Full practice exams

Answer: B.

1. A

   This treats the larvae as held up by buoyancy. They hang from the surface film, and a drop of detergent does not make pond water noticeably denser; denser water would hold them up better, not worse.

2. B

   Correct. Cohesion through hydrogen bonds between neighboring water molecules creates surface tension, the film the larvae hang from. Detergent molecules come between the surface water molecules, so fewer hydrogen bonds hold the film together and it can no longer hold the larvae at the surface.

3. C

   This treats adhesion as a downward pull. Adhesion is attraction between water and another surface; it does not drag an object down, and the larvae sink because the cohesive film weakens.

4. D

   This confuses the covalent bonds within a water molecule with the hydrogen bonds between molecules. Detergent does not split water into hydrogen and oxygen; it interferes with the attractions between molecules.

Answer: C.

1. A

   Carbon is in every kind of biological macromolecule, so a carbon label would show up in the new DNA as well as in the proteins.

2. B

   This reverses the two molecules. Phosphorus is used to build nucleic acids and phospholipids, so it would label the new DNA, which is exactly what the design must avoid.

3. C

   Correct. Sulfur is used in building proteins, and nucleic acids contain none, so a sulfur label marks the new proteins while the new DNA stays unlabeled.

4. D

   Nitrogen is used to build nucleic acids, so a nitrogen label would appear in the new DNA. It cannot separate proteins from DNA.

Answer: D.

1. A

   This shows water being released as the bond breaks. Water is released when monomers are joined, not when they are separated; hydrolysis uses water.

2. B

   This pictures the whole water molecule becoming a bridge between the monomers. In hydrolysis, water is split, and its parts end up on two monomers that are no longer joined.

3. C

   This is dehydration synthesis, the reverse reaction. It joins two monomers and releases water rather than breaking the bond.

4. D

   Correct. In hydrolysis, water is added across the bond: the hydroxyl group goes to one monomer and the hydrogen to the other, and the two monomers separate.

Answer: A.

1. A

   Correct. Moving down the leg, the temperature falls from 37 to 30 to 20 to 10 °C while the unsaturated share rises from 48 to 55 to 68 to 79 percent, so the two change in opposite directions at every step.

2. B

   This reads the trend backward. The coldest sample, just above the hoof, has the highest unsaturated share (79 percent), not the lowest.

3. C

   This assumes a straight-line pattern the table does not show. Cooling 7 degrees (37 to 30 °C) adds 7 points, but cooling 10 degrees (30 to 20 °C) adds 13 points, about 1.0 versus 1.3 points per degree.

4. D

   This mixes up saturated and unsaturated. The hoof sample is 79 percent unsaturated, so only about 21 percent saturated, the lowest saturated share of the four.

Answer: C.

1. A

   These are the correct partner bases, but written in the wrong direction. Paired stretches run antiparallel, so the partner of 5'-GCAU-3', read from its own 5' end, starts with the base that pairs with the U.

2. B

   This has the right order but uses thymine. Thymine is a DNA base; in RNA, adenine pairs with uracil.

3. C

   Correct. Pairing G-C, C-G, A-U and U-A gives the partner bases C, G, U, A lined up against 5'-GCAU-3'. Because the folded strand runs the opposite way, the partner reads 3'-CGUA-5', which is 5'-AUGC-3'.

4. D

   This assumes a stretch pairs with an identical copy of itself. Bases pair with complementary bases (G with C, A with U), not with the same base.

Answer: A.

1. A

   Correct. Quaternary structure comes from interactions between polypeptides. Replacing the hydrophobic contact surface with positive charges on both identical chains removes the hydrophobic interactions and makes the surfaces repel, so the active pair does not form.

2. B

   This assumes charged R groups always attract. The two chains are identical, so both surfaces now carry positive charges, and like charges repel.

3. C

   This treats tertiary structure as the whole story. The enzyme works only as a pair, and all four levels of structure, including the quaternary level, determine function.

4. D

   This contradicts the stem, which says each chain still folds normally. Secondary structure comes from hydrogen bonds in the backbone, and a change on the contact surface does not remove it.

Answer: B.

1. A

   Lysosomes are separate sacs that digest material taken into the cell, not a connected system of tubules, and detoxifying drugs is not their role.

2. B

   Correct. Smooth ER is a network of tubules without bound ribosomes, and one of its functions is detoxification. Cells processing a large daily load of the drug would be expected to build more of it.

3. C

   Rough ER is the membrane studded with ribosomes, and the micrographs show those sacs did not change in amount, so this does not match.

4. D

   The Golgi complex is a stack of flattened sacs that modifies and packages products for trafficking; it is not the tubule system that detoxifies chemicals.

Answer: D.

1. A

   The stem gives both cells the same volume. A flat shape changes the surface, not the amount of cytoplasm to supply.

2. B

   This reverses the geometry. For a given volume a sphere has the least possible surface area, so the flattened cell has more membrane, and more membrane means more exchange, not less.

3. C

   Changing a cell's shape does not change the thickness of its membrane; the gain comes from having more membrane area.

4. D

   Correct. Flattening spreads the cytoplasm over more membrane, so the surface area-to-volume ratio is higher and more oxygen can cross the membrane at the same time.

Answer: A.

1. A

   Correct. The model shows that polar heads face water and nonpolar tails avoid it. At an oil-water boundary, a single layer lets every head sit in the water and every tail sit in the nonpolar oil.

2. B

   This copies the membrane picture without asking why it forms. A bilayer forms when there is water on both sides; at an oil-water boundary, the tails can sit in the oil directly, so one layer is enough.

3. C

   This treats the whole phospholipid as nonpolar. Its polar head does not mix with oil, so the molecules line up at the boundary instead of dissolving.

4. D

   This reverses the two regions. The phosphate head is polar and hydrophilic, so it is drawn to the water, while the nonpolar tails avoid it.

Answer: D.

1. A

   This imposes a pattern the data do not follow. From 25 to 50 mM the rate fell from 11.4 to 7.2, about 63 percent of its value, and from 50 to 100 mM it fell to 1.8, a quarter of its value; neither step is a halving.

2. B

   This reads the table backward. The highest rate, 15.0 per minute, is in plain water, and the lowest is at 100 mM.

3. C

   This misreads the middle rows. The rate had already fallen from 15.0 to 7.2 per minute by 50 mM.

4. D

   Correct. The mean rate drops at every step, from 15.0 to 11.4 to 7.2 to 1.8 emptyings per minute, as sucrose rises from 0 to 100 mM.

Answer: C.

1. A

   This is the difference in emptyings per minute (15.0 - 7.2 = 7.8), not a percentage. It has to be divided by the starting rate.

2. B

   This is 7.2 divided by 15.0, which is the share of the original rate that remains (48 percent), not the share that was lost.

3. C

   Correct. Percent decrease = (15.0 - 7.2) / 15.0 x 100 = 7.8 / 15.0 x 100 = 52%.

4. D

   This divides the change by the final rate instead of the starting one: 7.8 / 7.2 x 100 = 108%. A decrease cannot be more than 100 percent of where it started.

Answer: B.

1. A

   Osmosis is passive: water moves toward the cytoplasm's lower water potential with no direct input of energy, so it keeps entering.

2. B

   Correct. Water enters passively by osmosis whether or not ATP is made, but filling a vacuole depends on pumping ions with ATP, which is active transport. With entry continuing and removal stopped, a cell with no wall swells and can burst.

3. C

   This reverses the direction of osmosis. The cytoplasm has more solute than the surrounding water, so water moves into the cell, not out of it.

4. D

   Half of this is right: osmosis needs no ATP. But removing the water depends on active ion pumping, and without it the incoming water builds up.

Answer: A.

1. A

   Correct. A cell that cannot assemble a chloroplast from its own membranes, and gets new ones only by division of existing chloroplasts or of the plastids they develop from, is behaving as if the chloroplast were once an independent cell that reproduced on its own by dividing.

2. B

   Which organisms have chloroplasts tells us where they are found, not where they came from. Either origin, engulfed cell or folded-in membrane, could fit that distribution.

3. C

   This links the organelle to its job, which shows how chloroplasts are used, not how they originated. An organelle of any origin could be more common where it is needed.

4. D

   Every organelle sits in cytoplasm, so this feature does not distinguish an engulfed prokaryote from a structure the cell built itself.

Answer: D.

1. A

   This reads the activation energy as the height of the peak above zero. Activation energy is measured from the reactants up to the peak, so it is 75 - 50 = 25 units.

2. B

   This takes the difference between the two peaks (120 - 75 = 45), which is how much the enzyme lowered the activation energy, not the activation energy itself, which is 75 - 50 = 25 units.

3. C

   The activation energy is right, but 55 units is the drop from the peak to the products (75 - 20). The energy released is the difference between reactants and products, 50 - 20 = 30 units.

4. D

   Activation energy runs from the reactant level to the peak: 75 - 50 = 25 units. The energy released runs from the reactants down to the products: 50 - 20 = 30 units, the same with or without the enzyme, because the enzyme changes only the peak.

Answer: C.

1. A

   Cellobiose has the higher mean, but its range of 5.6 to 6.8 overlaps the maltose range of 4.9 to 6.3. Overlapping error bars of 2SE mean the data do not show a clear difference, so "faster on cellobiose" goes further than the data allow.

2. B

   Different means alone do not show different rates: every mean carries sampling error. The cellobiose and maltose error bars overlap, so those two rates cannot be told apart from these tubes.

3. C

   Mean plus or minus 2SE gives cellobiose 5.6 to 6.8, maltose 4.9 to 6.3 and sucrose 0.1 to 0.7 micromoles per minute. The first two ranges overlap, so they are not clearly different; neither overlaps sucrose, so both are clearly higher. The enzyme’s active site fits the shapes of cellobiose and maltose far better than the shape of sucrose.

4. D

   Error bars that do not overlap are evidence of a real difference. The sucrose range, 0.1 to 0.7, lies far below the other two, so at least those comparisons are clear.

Answer: A.

1. A

   At 0, 10 and 20 °C the Antarctic enzyme is faster (4.0, 7.5 and 9.0 against 1.0, 2.5 and 5.0). At 30, 40 and 50 °C it is slower (3.0, 0.5 and 0.0 against 9.5, 12.0 and 6.0).

2. B

   This assumes an enzyme works fastest at its body temperature. The Antarctic enzyme runs at 4.0 at 0 °C but 9.0 at 20 °C, so 0 °C is not its highest rate.

3. C

   The pupfish enzyme rises from 1.0 to 12.0 up to 40 °C, but drops to 6.0 at 50 °C, so it does not rise at every step.

4. D

   Only the Antarctic enzyme peaks at 20 °C (9.0). The pupfish enzyme is highest at 40 °C (12.0), so the two optimum temperatures differ.

Answer: B.

1. A

   This subtracts 9.0 - 3.0 = 6.0 and reports the difference as a percentage. A percent change divides the change by the starting value.

2. B

   Percent change = (new - old) / old x 100 = (3.0 - 9.0) / 9.0 x 100 = -66.7%, so the rate fell by about 67%.

3. C

   This is 3.0 / 9.0 = 33%, the share of the rate that remains, not how much it fell.

4. D

   This divides the change by the new value instead of the starting value: 6.0 / 3.0 x 100 = 200%. A rate cannot fall by more than 100%.

Answer: D.

1. A

   This changes temperature along with pH. A difference in rate could come from either one, so the effect of pH cannot be separated out.

2. B

   This changes the amount of enzyme along with pH. More enzyme alone raises the rate, so any effect of pH is confounded with enzyme amount.

3. C

   This repeats the temperature series at a single pH. Taking the fish from alkaline pools does not change the pH of the reaction, so it cannot test pH.

4. D

   Only pH differs among the tubes, while temperature, enzyme amount and substrate are held the same as before. Any difference in rate can then be attributed to pH.

Answer: B.

1. A

   Heating makes molecules move faster, not slower, and collisions become more frequent. The same substrate also sped up the pupfish enzyme over the same range.

2. B

   Temperatures above an enzyme's optimum disrupt the hydrogen bonds that hold its shape, changing the active site so substrate no longer fits. The Antarctic enzyme's structure is disrupted above 20 °C, while the pupfish enzyme keeps its shape up to about 40 °C, so its rate keeps rising with more frequent collisions.

3. C

   Enzymes are not consumed by the reactions they catalyze, and each temperature was a separate tube with the same amount of enzyme.

4. D

   This treats an enzyme as something that needs added energy. Heat speeds collisions for both enzymes; the difference is whether the enzyme keeps its shape at that temperature.

Answer: C.

1. A

   This confuses lowering activation energy with supplying energy. Lowering activation energy is the enzyme's job, and the stem says the reaction needs an energy input that a catalyst cannot provide.

2. B

   This treats an enzyme as an energy source. Enzymes lower activation energy but supply none, and the glow ending when ATP is gone shows ATP is what is being spent.

3. C

   An energy-requiring reaction can run when it is coupled to an energy-releasing one. The light stops when ATP runs out, so ATP breakdown is the energy source, and luciferase only catalyzes the reaction.

4. D

   This reverses the direction of energy flow. ATP is used up, not made, as the tube glows, so energy moves from ATP into light.

Answer: A.

1. A

   Pyruvate cannot reach the matrix, so the Krebs cycle releases less CO₂ from glucose. Glycolysis still makes NADH, and much of the pyruvate that builds up in the cytosol is converted to lactate, which also regenerates NAD⁺.

2. B

   With no pyruvate entering the mitochondria, the Krebs cycle delivers less NADH and FADH₂ to the electron transport chain, so oxygen use falls rather than rises, and the pyruvate trapped in the cytosol is turned into lactate.

3. C

   The Krebs cycle takes place in the mitochondrial matrix and cannot move to the cytosol, so the CO₂ it releases from glucose falls, and lactate rises rather than falls.

4. D

   Glycolysis makes pyruvate in the cytosol; it does not need the transporter. Blocking transport changes what happens to pyruvate afterward, not whether glycolysis runs, and the pyruvate that piles up is converted to lactate.

Answer: D.

1. A

   This treats a different target cell type as the mark of a hormone. Local regulators also act on other cell types; what matters is that the signal only reaches nearby cells.

2. B

   Platelets travel in blood, but the protein is released at the wound and is gone within a fraction of a millimeter. A signal carried in the blood would reach fibroblasts far away.

3. C

   The protein is released into the tissue around the wound, not passed between joined cells. Platelets and fibroblasts are not linked by channels.

4. D

   A local regulator acts on target cells in the vicinity of the cells that release it. The protein is broken down within a fraction of a millimeter, so only nearby fibroblasts respond.

Answer: C.

1. A

   The stem says both cell types take up C, so C crosses both membranes. Being nonpolar, it can cross any plasma membrane.

2. B

   This assumes different cell types carry different genes. Body cells of one organism hold the same genes; they differ in which proteins they make.

3. C

   A cell responds to a signal only if it has a receptor that recognizes it. In the model, C acts only after binding R, so a cell that takes up C but makes no R cannot form the complex and its genes do not change.

4. D

   In this model the receptor sits in the cytoplasm, not on the surface. A nonpolar hormone that crosses the membrane needs no surface receptor.

Answer: B.

1. A

   This treats any chemical that binds the receptor as enough. Mutant 2's receptor can be bound, but its intracellular domain cannot change shape, so the signal is never passed on.

2. B

   Mutant 1 lacks only the signal; its receptor and pathway are normal, so A substitutes for G and the stems grow long. In mutant 2 A can bind, but the intracellular domain cannot change shape, so transduction never starts and the stems stay short.

3. C

   This reverses the two mutants. Mutant 2 cannot respond to any ligand, because its defect lies after binding, in the step that starts transduction.

4. D

   This ignores that mutant 1 has a normal receptor. Chemical A binds it and changes its shape as G would, so the pathway runs in mutant 1.

Answer: D.

1. A

   This reverses the feedback. T inhibits gland A, so less T means less inhibition and more S, not less.

2. B

   This treats negative feedback as acting in one direction only. The loop responds to T falling below its set point as well as rising above it.

3. C

   Stimulation cannot replace a missing raw material. However much S reaches gland B, it cannot make T without the nutrient.

4. D

   Gland B cannot make enough T, so T falls. T normally inhibits gland A, so with little T, gland A is released from inhibition and S rises above normal, a negative feedback loop responding to the drop.

Answer: A.

1. A

   In positive feedback the response intensifies the stimulus that caused it. Each opened channel lets in sodium that changes the voltage further and opens more channels, moving the system ever further from its starting point.

2. B

   Sodium inflow increases the voltage change that opened the channels; it does not reduce it, so this is not negative feedback.

3. C

   The label is right but the reason is wrong. Positive feedback amplifies change rather than holding conditions steady; stability is what negative feedback does.

4. D

   The voltage moves further from where it started at each step; nothing here returns it to a set point.

Answer: C.

1. A

   This adds 500 cells per cycle (500 + 4 x 500). Each division doubles every cell present, so the increase grows with each cycle.

2. B

   This doubles only three times (500 x 2 x 2 x 2). 72 hours holds 72 / 18 = 4 full cycles.

3. C

   There are 72 / 18 = 4 cycles, and mitosis doubles the cell number each cycle: 500 x 2 x 2 x 2 x 2 = 500 x 16 = 8,000 cells.

4. D

   This doubles five times, counting the starting culture as a division. Four cycles give four doublings.

Answer: B.

1. A

   A cleavage furrow pinches the cell inward from the outside, and animal cells have no cell wall. This structure grows from the center outward.

2. B

   Two finished nuclei mean telophase is done, and a structure built from fusing vesicles that grows from the center out to the cell wall is a cell plate, how plant cells divide their cytoplasm.

3. C

   Metaphase comes before the new nuclei form. The drawing shows a complete nucleus at each end, so the cell is past mitosis, and the line in the middle is vesicles, not chromosomes.

4. D

   In prophase there is one nucleus, not two. The drawing shows two complete nuclei, so the cell is at the end of division, not the start.

Answer: A.

1. A

   A control should differ from the treated group in only the variable being tested. Same radiation, same solvent and no Z leaves Z as the only difference, so any change in the treated group can be credited to Z.

2. B

   This group differs from the treated group in three ways at once, so it cannot separate the effect of Z from the effect of the solvent or the radiation.

3. C

   This tests whether a larger dose has a larger effect. Both groups still get the solvent, so neither shows what the solvent alone does.

4. D

   This changes the radiation and keeps Z and the solvent, so it tests what Z does to undamaged cells, not whether the solvent caused the change.

Answer: B.

1. A

   This treats the separation of homologs as if it needed the centromere proteins to break down. In anaphase I the sister chromatids are supposed to stay joined; homologs part without that breakdown, so meiosis I can finish.

2. B

   In anaphase I homologous chromosomes separate while sister chromatids remain attached, so meiosis I is unaffected. In anaphase II the proteins at the centromeres must break down for sister chromatids to be pulled apart, and the chemical prevents that.

3. C

   This reverses the two divisions. Keeping the centromere proteins intact makes sister chromatids harder to separate, not easier, and sisters do not separate in anaphase I in normal meiosis either.

4. D

   Crossing over is an exchange between non-sister chromatids of homologs during prophase I. It does not depend on the centromere proteins breaking down, which happens later, in anaphase II.

Answer: C.

1. A

   Turning both pairs still puts both red chromosomes in one cell and both blue in the other, the same two combinations as before. A new combination needs the pairs to line up independently of each other.

2. B

   Lining chromosomes up singly is how they sit in mitosis, not in metaphase I, where homologs line up as pairs. This would not separate homologs at all.

3. C

   Each pair lines up independently of the other at metaphase I. Turning one pair gives gametes with long red plus short blue and long blue plus short red, two combinations the first arrangement could not produce.

4. D

   Trading segments models crossing over, which makes recombinant chromatids within a chromosome. It does not change which whole chromosomes go to each pole, which is what random assortment is.

Answer: D.

1. A

   This is true of every offspring of any two diploid parents, so it cannot tell heterozygous parents from homozygous ones. A true statement is not evidence unless it bears on the claim.

2. B

   This confuses dominance with how common an allele is. Which phenotype is more common among these offspring says nothing about allele frequency, and it does not show that either parent carries s.

3. C

   A spotted phenotype only shows each parent has at least one S. Two copies of S would contradict the claim, and the unspotted offspring rule it out.

4. D

   An unspotted offspring is ss, so it got an s from the mother and an s from the father. Each spotted parent must therefore carry one s along with its S, which makes both Ss.

Answer: D.

1. A

   A 1:1:1:1 ratio is what unlinked genes would give in this cross, but here two classes have about 400 offspring each and two have about 45, roughly 9 to 1.

2. B

   This reads the 9 to 1 contrast between classes as if it were a contrast between colors. Adding the classes gives 458 black (412 + 46) and 442 brown (398 + 44), about 1:1.

3. C

   A 3:1 ratio is expected when two heterozygotes are crossed, not in a cross with a homozygous recessive parent. The table gives 456 long (412 + 44) and 444 short (398 + 46), about 1:1.

4. D

   Black is 412 + 46 = 458 and brown is 398 + 44 = 442; long is 412 + 44 = 456 and short is 398 + 46 = 444. Each gene alone gives about 1:1, as expected when a heterozygote is crossed with a homozygous recessive.

Answer: B.

1. A

   This counts only one recombinant class, 46 / 900. Both black-short and brown-long offspring are recombinant, and both belong in the count.

2. B

   The parent inherited B with L and b with l, so black-short and brown-long are recombinant: 46 + 44 = 90 of 412 + 398 + 46 + 44 = 900 offspring. 90 / 900 = 0.10, or 10 percent, which is 10 map units.

3. C

   This divides the recombinants by the parental offspring, 90 / 810. Map distance is the percent of all offspring that are recombinant, so the denominator is all 900.

4. D

   This is the percent of offspring with parental combinations, 810 / 900. Map distance comes from the recombinant share, not the parental share.

Answer: C.

1. A

   Codominance describes two alleles of one gene both showing in a heterozygote. These are two different genes, each with a dominant allele, so codominance does not apply.

2. B

   Every gamete of the bbll parent carries b and l, so it cannot supply a recombinant combination. The offspring classes reflect the gametes of the BbLl parent.

3. C

   A crossover between B and L swaps segments between two non-sister chromatids, turning one into B-l and the other into b-L. Every such event makes one of each, so the two recombinant gamete types, and the offspring classes they give, occur about equally.

4. D

   Nothing in the cross suggests recombinant gametes are fertilized less often. Even if they were, that would lower both recombinant classes and would not explain why they equal each other.

Answer: A.

1. A

   Adenine pairs with thymine and guanine with cytosine, so every rung is one double ring and one single ring. Each pair is the same width, which keeps the width of the double helix constant.

2. B

   Adenine and guanine are purines, with a double-ring structure. Only the pyrimidines (cytosine, thymine and uracil) have a single ring.

3. C

   A-T and G-C pairs do not form the same number of hydrogen bonds, and the number of bonds is not what sets the width of a pair. The ring sizes of the paired bases set it.

4. D

   One strand can hold the four bases in any amounts. Base pairing makes A equal T and G equal C across the two strands together, and the model's problem is the pairing, not the amounts.

Answer: C.

1. A

   This is the effect of losing topoisomerase, which relaxes supercoiling ahead of the fork. Without helicase the strands are not unwound, so supercoiling ahead of the fork would not grow.

2. B

   Joining lagging-strand fragments is the job of ligase, not helicase. Without helicase, no new fragments would be made to join.

3. C

   Helicase unwinds the two strands at each replication fork. When it stops, no new single-stranded template is exposed, so DNA polymerase soon runs out of template on both the leading and lagging strands.

4. D

   Both strands need single-stranded template, and helicase supplies it for both. The leading strand is made continuously, but only as fast as helicase opens the DNA ahead of it.

Answer: D.

1. A

   This assumes only coding sequence matters. The tail codes for no amino acids, but it affects how long the mRNA survives, and so how much protein is made from it.

2. B

   This gives the tail the job of the cap. The GTP cap helps the ribosome recognize the mRNA, and the cap is added normally here, so some protein Q is still made.

3. C

   The stem says splicing is normal, so the introns are removed. Losing the tail does not put introns back into the mature mRNA.

4. D

   The poly-A tail makes mRNA more stable. Without it, each mRNA for protein Q lasts a shorter time, so fewer rounds of translation happen and less protein Q is made.

Answer: B.

1. A

   These are the right bases written in the wrong direction. The anticodon pairs antiparallel with the codon 5'-GCA-3', so its 3' end lines up with the codon's 5' end: 3'-CGU-5'.

2. B

   The mRNA is complementary to the template: 5'-AUG GCA UUC-3'. The second codon is 5'-GCA-3', and the anticodon pairs with it antiparallel, A with U and G with C, giving 3'-CGU-5'.

3. C

   This copies the codon instead of pairing with it. An anticodon is complementary to the codon, so G must face C and A must face U.

4. D

   This uses thymine, a DNA base. tRNA is RNA, so it carries uracil where the DNA template has thymine.

Answer: A.

1. A

   Methylation fell every week, from 82% to 15%, while mRNA rose every week, from 3 to 120 units. The largest weekly rise, 74 units, came between weeks 2 and 3.

2. B

   Methylation fell by 21, 23 and 23 points (82 to 61 to 38 to 15), about the same amount each week. Halving would give about 82, 41, 20 and 10.

3. C

   The weekly increases are 8, 35 and 74 units (11 - 3, 46 - 11, 120 - 46), so the rise grows larger each week rather than staying the same.

4. D

   This reverses the relationship. Methylation was highest, 82%, at week 0, when mRNA was lowest, 3 units.

Answer: A.

1. A

   The independent variable is whether the enhancer is present. Liver cells from the same culture with the enhancer intact differ only in that, so any drop in gene K mRNA in the deletion group can be traced to the missing enhancer.

2. B

   This adds temperature as a second variable while keeping the deletion. It gives no comparison with cells that still have the enhancer.

3. C

   Measuring the same treated cells later is a repeat measurement, not a control. Every group in it still lacks the enhancer.

4. D

   This changes the cell type, not the enhancer, so it cannot show what deleting the enhancer did in liver cells. A control should differ from the treated group only in the independent variable.

Answer: C.

1. A

   Viruses also pass through the filter, so virus-carried transfer would have worked across it. It did not.

2. B

   Free DNA passes through the filter, so transformation would have given resistance in the filter setup too. It did not.

3. C

   The filter lets free DNA and viruses pass, yet no resistance appears until the cells can touch. That points to conjugation, in which DNA such as a plasmid passes directly from one cell to another.

4. D

   Transposition moves DNA segments within or between DNA molecules of a cell, but it does not bring a gene from another cell. Strain 2 had no resistance gene to move until it received one.

Answer: B.

1. A

   This has the charge backward. A longer fragment has more phosphate groups and so more total negative charge, yet it moves more slowly. The difference comes from the gel, not the charge.

2. B

   Every DNA fragment is negatively charged along its sugar-phosphate backbone, so all move toward the positive electrode. The gel is a mesh, and shorter fragments pass through it faster, so they travel farthest in the same time.

3. C

   DNA fragments of every size are negatively charged. None is held back by attraction to the negative electrode; all move away from it.

4. D

   Fragments move through the gel toward the positive electrode, not upward. How far a fragment travels depends on its size moving through the mesh, not on sticking or floating.

Answer: D.

1. A

   Egg count is a proxy, not fitness. Fitness is offspring that survive to reproduce, and in fish years the small form wins on that measure, so the large form cannot steadily take over.

2. B

   Doing well in some years does not freeze the shares. Each year one form leaves more surviving offspring than the other, so the shares keep moving with the conditions.

3. C

   Body form here is inherited. Selection changes which forms are common among the offspring; it does not make each individual adjust its own size to the year.

4. D

   Selection follows the environment. In a fish year small copepods leave more surviving offspring, so the next generation holds a larger share of the inherited small form; in a fishless year the advantage flips to the large form. A fluctuating environment changes the direction of selection from one generation to the next.

Answer: A.

1. A

   Variation in a molecule changes how well an organism works in a particular environment. Where oxygen is scarce, mice whose hemoglobin picks up more of it can forage, stay warm and breed, so they leave more surviving offspring and the tight-binding variants become common at altitude.

2. B

   Nothing in the data ties the variants to body size, and size alone would not settle the question anyway. What counts is how many offspring survive to breed, and that is what the oxygen advantage changes.

3. C

   The data contradict this: the tight-binding variants are rare in lowland mice. A variation that helps in one environment can be neutral or harmful in another.

4. D

   This treats the environment as reshaping each mouse's protein. The variants are inherited differences that already exist; thin air decides which mice leave more offspring, not what their hemoglobin is.

Answer: D.

1. A

   Changes made to a plant during its life, such as trimming, are not inherited. Only the genetic variation passed in seed can change a line.

2. B

   Mutations are not produced to order. Growers chose among variation that was already there; wanting a trait does not cause the mutations that make it.

3. C

   This describes natural selection, where the environment decides who reproduces. In these crops the growers chose the seed plants on purpose.

4. D

   This is artificial selection: people, not the environment, chose which plants left offspring. Each group of growers kept a different trait, so inherited variation in each line moved in its own direction until the lines looked like different vegetables.

Answer: B.

1. A

   A shared ancestor does not hold frequencies still. The populations are alike now because pollen keeps mixing them; without that mixing they diverge.

2. B

   Gene flow keeps populations alike by mixing their alleles every generation. Once it stops, chance changes and selection by the different soils act on each population alone, so the allele frequencies drift and are pushed apart.

3. C

   Being apart is not the same as being reproductively isolated. Speciation takes many generations of divergence; pollen carried by hand between the hillsides right away could still fertilize the other population's flowers.

4. D

   Mutation arises within each population. Pollen moves alleles between populations, but it is not where new alleles come from.

Answer: D.

1. A

   This uses 0.16 as the frequency of g. The value 400/2,500 = 0.16 is q², the frequency of blue frogs, so q is its square root. Using q = 0.16 and p = 0.84 gives 2(0.84)(0.16)(2,500) = 672.

2. B

   This is the number of GG frogs, p²N = (0.6)²(2,500) = 900, not the heterozygotes.

3. C

   This takes half of the 2,100 green frogs as heterozygous. The green frogs split into p² and 2pq, which are not equal here.

4. D

   q² = 400/2,500 = 0.16, so q = 0.4 and p = 1 - 0.4 = 0.6. The heterozygote frequency is 2pq = 2(0.6)(0.4) = 0.48, and 0.48 × 2,500 = 1,200 frogs.

Answer: C.

1. A

   This would fit a fossil above the 41-million-year ash, as in layer 1. Layer 3 lies below that ash, so it was laid down before it.

2. B

   This reverses the order of the layers. Layer 3 lies above the 47-million-year ash, so it formed after that ash fell.

3. C

   In undisturbed layers, deeper rock is older. Layer 3 sits above the 47-million-year ash and below the 41-million-year ash, so it formed between those dates, and its fossils have the same age range.

4. D

   Fossils can be dated by the ages of the rocks around them, not only by dating their own layer. Dated layers above and below set the range.

Answer: B.

1. A

   This reads the lineage with the most derived traits as the oldest. S has the most changes from the out-group, so it branches last, not first.

2. B

   Shared derived characters group lineages. All four have fused wing covers, so they share an ancestor apart from O. Q, R and S also share 12-segment antennae, so P branches first; R and S also share lost eyes, so Q branches next; pale color is S's alone.

3. C

   This ignores a shared derived character. Q has 12-segment antennae, which it shares with R and S but not P, so Q groups with R and S and P is the one that branches off first.

4. D

   A trait that is lost during evolution is still a derived change from the ancestral condition. Losing eyes, shared by R and S, groups them like any gained trait.

Answer: C.

1. A

   Courtship and mating happen before a zygote forms, so this tests for a pre-zygotic barrier, not a post-zygotic one.

2. B

   Differences in appearance do not show whether two populations can produce viable, fertile offspring. Very different-looking populations can interbreed freely.

3. C

   A post-zygotic barrier acts after a zygote forms: hybrids fail to develop, die young or cannot reproduce. Raising hybrids and testing their fertility measures that directly, and crosses within each population are the control showing what normal survival and fertility look like.

4. D

   Genetic distance shows how far the populations have diverged, but not whether their hybrids survive and reproduce. Isolation has to be tested with offspring.

Answer: A.

1. A

   Genetic diversity raises the chance that some individuals already carry alleles that let them withstand a new pressure. Those colonies survived and kept reef X's cover up, while reef Y's single line responded to the heat almost as one.

2. B

   Organisms do not change their alleles in response to a pressure. The heat sorted among variation that existed before it came.

3. C

   Variation means the colonies differ from one another, not that each colony is equally tough. The heat killed some lines on reef X and spared others.

4. D

   Both percentages are shares of each reef's own starting cover, and the cover was about the same. The same death rate would leave the same share on both reefs.

Answer: A.

1. A

   The range is bounded by the earliest time life was possible and the earliest fossils. Moving the first bound to 3.7 billion years while the fossils stay at 3.5 billion years narrows the range to the time between them.

2. B

   Earth formed about 4.6 billion years ago, but life could not arise while conditions were too hostile. The new evidence moves the start of the range later, not earlier.

3. C

   Both lines of evidence set the range, not the fossils alone. A later end to the hostile period removes 3.9 to 3.7 billion years from it.

4. D

   The oldest fossil shows life was present by then, not that it began then. Life must have arisen some time before its earliest known fossils.

Answer: D.

1. A

   The data show the reverse: colony spiders leave more surviving offspring. Cooperative behavior spreads because it raises the fitness of the individuals that do it.

2. B

   Habits learned during a lifetime are not written into genes. The tendency spreads because it is already inherited and its carriers leave more offspring.

3. C

   Extra energy matters only if it turns into more young that live to breed. The researchers counted surviving offspring directly, and that count is what favors colony spiders.

4. D

   Behaviors that increase survival and reproduction are favored by selection. Here cooperators leave more offspring that survive, and the tendency is inherited, so each generation holds a larger share of spiders that join colonies.

Answer: C.

1. A

   This compares arrows that do not both touch the plant reservoir. Decomposition moves carbon between the soil and the atmosphere, so it does not decide whether the plants gained carbon.

2. B

   Respiration releases carbon from plants to the atmosphere. If it moved more carbon than photosynthesis, the plants would lose carbon, not store more.

3. C

   Matter is conserved, so a reservoir grows only if more carbon flows in than out. The only arrow into living plants is photosynthesis, and the two arrows out are respiration and the death of plant parts, so photosynthesis must exceed both together.

4. D

   This describes a gain in the soil reservoir, which could happen whether or not the plants gained carbon. It is not required by a gain in living plants.

Answer: B.

1. A

   The weekly increases are not equal: they rise from 20 to 180 fronds and then fall. A steady increase of the same number would plot as a straight line.

2. B

   From 20 to 40 to 80 to 156, each count is about twice the one before. After week 5 the weekly increases shrink (140, 58, 22) until the count stays near 690 from week 8 to week 10.

3. C

   This confuses doubling with the size of the increase. The first week added only 20 fronds; the largest increase, 180, came between weeks 4 and 5.

4. D

   Doubling stops after about week 3. Ten doublings from 20 would give more than 20,000 fronds, not about 700.

Answer: B.

1. A

   This divides the change by the week 1 count: 20/40 = 0.5. The equation uses the population size at the start of the interval, 20 fronds.

2. B

   dN/dt = 40 - 20 = 20 fronds per week and N = 20, so r_max = (dN/dt)/N = 20/20 = 1.0 per week. Each frond added about one new frond per week.

3. C

   This is the ratio of the two counts, 40/20 = 2.0. That is how many times larger the population became, not the per capita growth rate.

4. D

   This is dN/dt itself, 20 fronds per week. The per capita rate divides that change by N.

Answer: C.

1. A

   Repeating the experiment checks that the result is reliable, but space and nutrients run short together in both tanks, so it cannot tell the two explanations apart.

2. B

   Adding a competitor changes the population in a new way, but whatever happens to the count cannot tell shading apart from nutrient shortage, because the newcomer would both shade the duckweed and take up nutrients. The experiment does not isolate the variable the hypothesis names.

3. C

   Holding the volume of the same water constant keeps the total nutrients equal, while a wide, shallow tank gives the fronds more room to spread than a narrow, deep one before they shade one another. If shading limits growth, the wider tanks should level off at higher counts; if nutrients limit it, all tanks should level off near the same count.

4. D

   More frequent counts describe the growth curve in more detail but change nothing about space or nutrients, so they cannot test what causes the leveling off.

Answer: A.

1. A

   The carrying capacity depends on the resources of the tank, which are the same, so the count should again level off near 690. Starting closer to that number, it gets there in fewer weeks.

2. B

   Exponential growth cannot continue in a tank with limited resources. Ten doublings from 200 would far exceed what the first tank could hold.

3. C

   The timing depends on how far the population starts below the carrying capacity. Starting with 200 fronds, it needs fewer weeks of growth than a tank starting with 20.

4. D

   Carrying capacity is set by the available resources, not by the starting population. An identical tank supports about the same number of fronds.

Answer: D.

1. A

   Sharing a food source shows the species could compete, but not that competition sets where C lives. Another factor, such as being eaten by predators below, could still explain it.

2. B

   This tests what limits species B in the upper zone, not what limits species C in the lower zone. It cannot show that B excludes C.

3. C

   Growth alone in a tank says nothing about what happens when the two species meet on the shore, so it does not show that B keeps C out.

4. D

   If species C can live in the lower zone once species B is gone, the lower zone itself is not too harsh for C. That points to the presence of B as what keeps C out, which is what the claim says competition does.

Answer: A.

1. A

   Correct. Each doubling of the diameter (0.25 to 0.5 to 1.0 to 2.0 mm) cut the height in half (120 to 60 to 30 to 15 mm), so height is inversely related to diameter.

2. B

   This reverses the trend. The narrowest tube, 0.25 mm, holds the tallest column, 120 mm.

3. C

   This assumes equal steps. The drops were 60, 30 and 15 mm, each half the one before, so the change shrinks with every doubling.

4. D

   This misreads the last two rows. The height still fell from 30 mm to 15 mm when the diameter went from 1.0 to 2.0 mm.

Answer: C.

1. A

   Removing water is dehydration synthesis, the reaction that builds a chain. Breaking the chain apart uses water instead.

2. B

   The amino acids in a chain are joined by covalent peptide bonds, not hydrogen bonds. Hydrolysis cleaves those covalent bonds.

3. C

   Correct. Digestion of a protein is hydrolysis: for each covalent bond broken, a water molecule is added, its hydrogen going to one amino acid and its hydroxyl group to the other.

4. D

   This treats water as a helper that is not used up. In hydrolysis, each water molecule is split and its atoms become part of the released monomers.

Answer: B.

1. A

   A linear polymer has no branch points, so an enzyme that cuts only at branch points would leave it intact rather than splitting it into many chains.

2. B

   Correct. Complete hydrolysis shows one kind of monomer, and cutting only at branch points breaks the molecule into many separate chains, so it must have had branches joining those chains.

3. C

   This ignores the hydrolysis result. Breaking every bond produced only one kind of monosaccharide, so the polymer is built from one kind.

4. D

   This mistakes one product of the enzyme for the whole molecule. The polymer fell apart into many 12-sugar chains, so it was much larger than any one of them.

Answer: D.

1. A

   Both parts are wrong. Nucleotides are added to the 3' end, and hydrogen bonds join bases on opposite strands, not neighbors on the same strand.

2. B

   This has the right kind of bond but the wrong end. Synthesis adds nucleotides only to the 3' end, where the sugar carries a free hydroxyl.

3. C

   This has the right end but the wrong bond. Hydrogen bonds pair bases across the two strands; neighbors within one strand are joined covalently through the sugar and phosphate.

4. D

   Correct. New nucleotides are added to the 3' end of the growing strand, and a covalent bond forms between the 3' hydroxyl of the last sugar and the phosphate of the incoming nucleotide.

Answer: C.

1. A

   Primary structure is the sequence of amino acids. Breaking a bond between two R groups leaves every peptide bond, and so the sequence, unchanged.

2. B

   The model shows the helices and sheet held by hydrogen bonds in the backbone, which the chemical does not break. The bridge joins R groups, not backbone atoms.

3. C

   Correct. Tertiary structure is the three-dimensional shape set by interactions among R groups, including disulfide bridges. Breaking the bridge frees the two regions of the chain it held together.

4. D

   Quaternary structure needs more than one polypeptide, and this protein has only one. The bridge joins two regions of the same chain.

Answer: B.

1. A

   The stem says ribosomes still make proteins in the cytosol, and many proteins are always made on free ribosomes, so the cytosolic enzymes are unaffected.

2. B

   Correct. Proteins made by ribosomes on the rough ER enter the endomembrane system, pass through the Golgi and leave in vesicles. With no ribosomes on the ER, the secreted enzyme stays in the cytosol, while cytosolic enzymes are made as before.

3. C

   This treats where a protein is made as unimportant. A secreted protein enters the endomembrane system as it is made on the rough ER; made in the cytosol, it never reaches the Golgi or a vesicle.

4. D

   Free ribosomes release proteins into the cytosol. Nothing in the mutation moves cytosolic enzymes into the endomembrane system.

Answer: A.

1. A

   Correct. P's rate is twice the concentration at every point (2.0 at 1 mM up to 40.0 at 20 mM), while G's rate climbs quickly at first and then barely changes, 15.0 to 16.0 from 10 to 20 mM.

2. B

   Only P's rate kept pace with concentration. G's rate rose by just 1.0 unit when concentration doubled from 10 to 20 mM.

3. C

   This swaps the two substances. G is the one that levels off; P doubles every time its concentration doubles.

4. D

   This holds only at low concentrations. At 10 and 20 mM, P's rate (20.0 and 40.0) is higher than G's (15.0 and 16.0).

Answer: D.

1. A

   Molecules in solution do not block one another from a membrane this way. Size alone does not explain a ceiling on the rate; a limited number of transport proteins does.

2. B

   Nothing in the data points to ATP, and the cells would not run short of ATP within one minute just because more G is outside. The leveling comes from a limited number of carriers.

3. C

   Polar molecules do not cross the hydrophobic interior on their own at any concentration, which is why G needs proteins. This reasoning also gives no reason for a limit.

4. D

   Correct. A large polar molecule like glucose needs transport proteins to cross, and a membrane has only so many. Once they are all in use, adding more G cannot speed uptake. Small nonpolar P passes through the lipid bilayer itself, so its rate keeps rising with its gradient.

Answer: A.

1. A

   Correct. Facilitated diffusion needs no energy input, while active transport requires it. If G's uptake is unchanged in cells that cannot make ATP, the hypothesis is supported; if uptake drops, it is not.

2. B

   A larger sample makes the measurements more precise but compares nothing; G's mechanism would still be unknown.

3. C

   Both facilitated diffusion and active transport use proteins that can become saturated, so confirming the plateau cannot tell the two apart.

4. D

   This studies the wrong substance. Whatever P does, it says nothing about how G crosses.

Answer: C.

1. A

   Pumps move ions and small molecules across a membrane; they do not remove whole patches of membrane. Moving lipids across the bilayer would not reduce its area anyway.

2. B

   Channel proteins let ions and polar molecules through the membrane; they are not a way for the membrane itself to shrink.

3. C

   Correct. The diagram shows that each fused vesicle adds its membrane to the cell surface. In endocytosis, the plasma membrane folds inward and pinches off as new vesicles, which takes membrane back in and balances what exocytosis adds.

4. D

   The diagram itself is exocytosis, and it adds membrane rather than removing it. More exocytosis would make the surface larger.

Answer: D.

1. A

   This would be true only if neither solute could cross. Equal starting concentrations stop the first net flow of water, but glucose then diffuses into arm 1 and the balance changes.

2. B

   Water moves toward the side with more dissolved particles, not toward smaller ones. Arm 2 loses glucose to arm 1, so it ends with fewer particles than arm 1.

3. C

   Solute molecules do not drag water with them through the membrane. Water moves by osmosis toward the lower water potential, which ends up in arm 1.

4. D

   Correct. Glucose diffuses down its gradient until its concentration is equal in the two arms. Arm 1 then holds that glucose plus all of the sucrose, so it has the higher solute concentration and lower water potential, and water moves into it by osmosis.

Answer: B.

1. A

   Making a structure larger lowers its surface area-to-volume ratio. The benefit here is internal membrane area for reactions, not the organelle's outer surface.

2. B

   Correct. Internal membranes increase the surface area where membrane-based reactions can occur. With little light available, more of that surface lets the chloroplast capture more of what light there is.

3. C

   The data concern membrane inside each chloroplast, not the number of chloroplasts, and internal membrane is not how chloroplasts divide.

4. D

   Plants in deep shade receive less light, so they do not photosynthesize more, and these membranes are sites of light-capturing reactions, not sugar storage.

Answer: C.

1. A

   The enzyme tube stops rising at 10.0 mM after 8 minutes, and the no-enzyme tube climbs from 0.3 to 4.6 mM, so neither half is right.

2. B

   The first half is right, but 4.6 mM is only the last reading. The no-enzyme values are still climbing (1.2, 2.4, 4.6), so the data show no plateau.

3. C

   With the enzyme, product reaches 10.0 mM at 8 minutes and stays there, which is all of the starting substrate. Without it, product climbs from 2.4 to 4.6 mM between 16 and 32 minutes and has not leveled off.

4. D

   The enzyme tube rises fastest in the first 2 minutes (6.0 mM) and does not rise at all after 8 minutes.

Answer: B.

1. A

   This describes a noncompetitive inhibitor. The stem says ethanol binds the same active site, so it competes with methanol there rather than acting at another site.

2. B

   Ethanol binds the same active site reversibly, so it competes with methanol. With a high concentration of ethanol, most enzyme molecules are occupied by ethanol at any moment, and methanol is turned into the toxic product much more slowly.

3. C

   The stem gives ethanol a role at the enzyme, not in the blood. Ethanol slows the formation of the toxic product rather than destroying it afterward.

4. D

   Ethanol is a substrate of the enzyme, not a denaturing agent, and binding at the active site is reversible.

Answer: D.

1. A

   The data show all three are similar; they give no measure of which pair is more similar, so they cannot rank relatedness.

2. B

   Separate origins would not be expected to produce similar amino acid sequences for every enzyme. Shared sequences point to shared descent.

3. C

   The data show these species have glycolysis, not that it is their only source of ATP. All three can also make ATP in other ways.

4. D

   A pathway with similar enzymes in Bacteria, Archaea and Eukarya is best explained by inheritance from a common ancestor of all three domains, which is why core pathways such as glycolysis are evidence of common ancestry.

Answer: A.

1. A

   Tube 4 rose by 39 - 30 = 9 points, more than tube 2 (2 points) and far less than tube 1 (42 points).

2. B

   Tube 4 ended at 39 and tube 1 at 72. Tube 4 rose by 9 points, far less than tube 1's 42.

3. C

   Tube 4 rose from 30 to 39, while tube 2 rose only from 30 to 32, so tube 4 did not stay level with it.

4. D

   Tube 4 rose at every reading (30, 34, 37, 39); it never fell.

Answer: A.

1. A

   Average rate = change / time = (72 - 30) / 15 = 42 / 15 = 2.8% per minute.

2. B

   This is the rate over the first 5 minutes only, (47 - 30) / 5 = 3.4% per minute, when the change was fastest.

3. C

   This divides the final reading by the time (72 / 15 = 4.8). A rate of change uses the change, 72 - 30 = 42.

4. D

   This is the total change, 42 percentage points, without dividing by the 15 minutes.

Answer: C.

1. A

   That is the job of tube 5, which has light and dye but no chloroplasts. Tube 3 does contain chloroplasts.

2. B

   This confuses tube 3 with the dark tube. Boiling does not block light, and tube 3 sat in the light like tube 1. The light requirement is shown by tube 2, wrapped in foil.

3. C

   Tube 3 matches tube 1 except that its chloroplasts were boiled, which denatures their proteins. Its transmittance barely changed, so the dye change needs working proteins such as those of the electron transport chain, not just chloroplast material in the light.

4. D

   A tube set as the blank would hold no dye, since the dye is what is being measured. Tube 3 holds dye and is measured over time like the others.

Answer: B.

1. A

   DCPIP takes the place of NADP⁺ as the electron acceptor, so a shortage of NADP⁺ would not slow the dye change. The dye reports electron flow in the light reactions, not the Calvin cycle.

2. B

   The dye changes as it accepts electrons from the light reactions. Tube 4 changed by only 9 points against 42 in tube 1, under the same light, so H slowed electron flow; it still changed more than the boiled tube, so some flow remained.

3. C

   Tube 4 rose by 9 points while boiled tube 3 rose by only 1, so some activity remained with H. H did not destroy the chloroplasts as boiling did.

4. D

   This reads the rise in tube 4 as an increase, judging the tube against its own starting value instead of the control. Tube 4 rose far less than tube 1, so H slowed electron flow rather than speeding it.

Answer: D.

1. A

   Protons are pumped out of the cytoplasm, so it has the lower proton concentration and the higher pH. Protons also flow down their gradient, which here is inward.

2. B

   The pH is right, but protons through ATP synthase move down their gradient, from the high concentration outside back into the cytoplasm, not further out.

3. C

   The direction of flow is right, but the cytoplasm is the side protons were pumped out of, so it has the higher pH, not the lower.

4. D

   Pumping protons outward makes their concentration higher, and the pH lower, just outside the membrane. In chemiosmosis protons flow back down that gradient through ATP synthase into the cytoplasm, driving ATP synthesis, just as in mitochondria but across the plasma membrane.

Answer: C.

1. A

   The Krebs cycle oxidizes pyruvate, not lactate, and lactate formation is what happens instead of the Krebs cycle when oxygen is absent.

2. B

   This assumes the fermentation step makes ATP. The ATP comes from glycolysis; making lactate only regenerates NAD⁺ so glycolysis can continue.

3. C

   Glycolysis turns NAD⁺ into NADH. Without oxygen, fermentation is the only way this cell can turn NADH back into NAD⁺; with no lactate step, NAD⁺ runs out, glycolysis stops, and the cell makes no ATP.

4. D

   Glycolysis does not use oxygen; the normal strain grows without oxygen by glycolysis and fermentation.

Answer: D.

1. A

   This asks about purpose, which no experiment can measure, and it says nothing about how the cells signal one another.

2. B

   This can be tested, but it is about the amoebae's food, not about how starving cells signal one another.

3. C

   This can be tested, but crowding could change mound size whether the cells signal by a chemical or by touch, so it does not test the proposed chemical.

4. D

   If the cells release a chemical that spreads, agar that held starving amoebae should still hold it after they are gone, and other amoebae should crawl toward it with no cell there to touch. If they do not, the hypothesis is not supported.

Answer: B.

1. A

   This assumes a ligand must enter the cell to act. In the model, N binds sites on the outside, and what flows in is ions, not N.

2. B

   This is a ligand-gated channel: binding of N changes the receptor's shape so that its pore opens, and the ions that enter trigger the response. The receptor and the channel are the same protein.

3. C

   This describes a pathway with a second messenger. The model has no cAMP; the receptor itself is the channel.

4. D

   This describes a phosphorylation cascade. In the model the binding opens the channel directly, with no kinases.

Answer: A.

1. A

   In a phosphorylation cascade, adding a phosphate changes a protein's shape and so its activity. Activated kinase 1 activates kinase 2, which activates the transcription factor, relaying the signal from the receptor to the genes.

2. B

   This reverses the role of phosphorylation here. Each phosphorylation activates the next protein, carrying the signal forward rather than ending it.

3. C

   The pheromone stays outside, bound to the surface receptor. What moves inward is the signal, relayed from protein to protein.

4. D

   This treats phosphorylation as an energy delivery. The phosphate on the transcription factor changes its activity; it does not power transcription.

Answer: D.

1. A

   This assumes a cell still needs the signal to respond. Once Z is active on its own, the pathway runs in any cell that makes Z, whether or not the signal reaches it.

2. B

   This treats every mutation as a loss of function. The change keeps Z active, so it passes the signal on all the time instead of never.

3. C

   This assumes the pathway still stops when the signal leaves. A Z that cannot switch off keeps the pathway running after the signal is gone.

4. D

   A component downstream of the receptor that stays active runs the pathway with no signal at all. Every cell that makes Z, including the cells of the toes, is driven into apoptosis, so the toes do not form normally.

Answer: A.

1. A

   In positive feedback the response intensifies its own stimulus. If rising glucose caused more glucagon, the liver would release still more glucose, raising glucose further and calling for still more glucagon, moving the level away from its set point.

2. B

   This is the model as it stands: the response reduces its own stimulus, which is negative feedback.

3. C

   This adds a second way to raise glucose, but rising glucose would still reduce glucagon, so the loop stays negative.

4. D

   This removes feedback altogether. If glucagon release ignores glucose, nothing links the response back to the stimulus.

Answer: C.

1. A

   A cell that is not dividing waits outside the cycle in G0, not partway through mitosis. One finished mitosis would also give two cells, not thousands.

2. B

   Meiosis produces cells with half the chromosomes, each genetically different. Identical copies come from mitosis.

3. C

   A nondividing cell can sit in G0 and reenter the cycle in response to the right cue. Repeated rounds of the cycle, each with DNA replication and mitosis, give thousands of daughter cells with the same complete genome.

4. D

   Without S phase the DNA would not be copied, and the daughter cells could not each receive a complete genome.

Answer: B.

1. A

   Cyclin was not measured. The only difference reported between the cells is protein K.

2. B

   Cyclin-CDK complexes drive the cell through the cycle. Normal cells stop dividing when K, which keeps CDKs inactive, is high, tumor cells without K never stop, and removing K from normal cells makes them behave like the tumor cells. Taking K away causes the loss of the stop, so the evidence supports K acting as a brake on the cycle.

3. C

   The tumor cells keep dividing in the same kind of dish, so the stop is not explained by nutrients running out; it matches the rise in protein K.

4. D

   The stem says K keeps CDKs inactive, and the tumor cells without it divide more, not less.

Answer: C.

1. A

   In anaphase of mitosis each pole receives a full diploid set, six chromosomes with two of each kind. Here each pole gets only three, one of each kind.

2. B

   In anaphase I homologs separate, so each pole does get three different chromosomes, but each one still has two sister chromatids. These are single chromatids.

3. C

   Each pole receives one chromosome of each kind, a haploid set (n = 3), and each chromosome is a single chromatid. That is what results when sister chromatids are pulled apart in anaphase II.

4. D

   In metaphase II chromosomes are lined up at the plate with their sister chromatids still joined. The drawing shows single chromatids already moving toward the poles.

Answer: A.

1. A

   A cross of two double heterozygotes with independently assorting genes gives 9/16, 3/16, 3/16 and 1/16 of the four classes. The null hypothesis says any difference from that ratio is due to chance.

2. B

   A 1:1:1:1 ratio is expected from a test cross, a double heterozygote crossed with a homozygous recessive. Both parents here are RrTt.

3. C

   This is an alternative hypothesis, linkage, not the null. The null predicts no departure from the ratio for independent assortment.

4. D

   This states the opposite of a null hypothesis. The null says there is no significant difference between observed and expected counts.

Answer: B.

1. A

   This uses (o - e) / e without squaring the differences: 6/180 + 6/60 + 2/60 + 2/20 = 0.27. The formula squares each difference.

2. B

   Expected counts are 320 x 9/16 = 180, 320 x 3/16 = 60, 60 and 320 x 1/16 = 20. Then $\chi^2$ = 6²/180 + 6²/60 + 2²/60 + 2²/20 = 0.20 + 0.60 + 0.07 + 0.20 = 1.07.

3. C

   This divides each squared difference by the observed count: 36/186 + 36/54 + 4/62 + 4/18 = 1.15. The formula divides by the expected count.

4. D

   This adds the squared differences, 36 + 36 + 4 + 4 = 80, and never divides by the expected counts.

Answer: D.

1. A

   This uses 4 degrees of freedom, one per class. Degrees of freedom are the number of classes minus 1, so 4 - 1 = 3, and the critical value is 7.81.

2. B

   This reverses the rule. A chi-square value larger than the critical value means the difference is unlikely to be due to chance alone, so the null is rejected.

3. C

   A chi-square test never accepts or proves the null, and closeness to the critical value is not the test. Here 8.6 is above 7.81, so the null is rejected.

4. D

   With 4 classes there are 4 - 1 = 3 degrees of freedom, and the critical value at p = 0.05 is 7.81. Since 8.6 > 7.81, the departure from the 9:3:3:1 ratio is unlikely to be chance alone, so the null hypothesis is rejected.

Answer: B.

1. A

   Identical results from both directions, then a 3:1 ratio, are the pattern of a recessive nuclear allele following Mendel's laws. It argues against chloroplast inheritance.

2. B

   In plants, chloroplasts pass to offspring in the ovule and not in the pollen. If offspring always match the ovule parent, whichever way the cross is made, the trait follows the chloroplasts, as the claim says.

3. C

   Matching results in both directions mean it does not matter which parent gives the ovule. A trait carried in chloroplasts would depend on the ovule parent.

4. D

   Here the offspring match the pollen parent. Chloroplasts are not passed on in pollen, so this pattern would argue against the claim.

Answer: A.

1. A

   Clone 1 went from 42 to 118 micrometers, 118 / 42 = 2.8 times as tall, while clone 2 went only from 45 to 51. The same environmental change produced a large change in one genotype and little in the other.

2. B

   Clone 1 rose by 118 - 42 = 76 micrometers and clone 2 by 51 - 45 = 6, so the increase was not equal.

3. C

   Clone 2 was taller only in plain water, 45 versus 42 micrometers. With predator chemicals clone 1 was far taller, 118 versus 51.

4. D

   In plain water the clones were about the same, 42 and 45 micrometers. In predator water they differed widely, 118 and 51.

Answer: C.

1. A

   Histones do condense eukaryotic chromosomes, but the stem says all proteins were removed first, and the bacterial chromosome is not wound on histones in the first place.

2. B

   This gives eukaryotes the prokaryotic arrangement. Eukaryotic chromosomes are typically linear, with exposed ends.

3. C

   Prokaryotes typically have a circular chromosome, which has no free ends for the enzyme to start from. Eukaryotic chromosomes, such as yeast's, are linear, so their ends can be attacked.

4. D

   A linear strand has a 5' end and a 3' end, but in a circular molecule each strand is joined end to end, so there is no free end.

Answer: D.

1. A

   This reverses the direction of synthesis. A strand made on a template whose 3' end is on the left runs 5' to 3' from left to right, so it grows to the right, toward the right fork.

2. B

   A strand can grow in only one direction, 5' to 3'. On this template it grows rightward, the same way the right fork moves but away from the left fork, so it cannot be continuous at both.

3. C

   A template running 3' to 5' in the direction a fork moves supports continuous synthesis at that fork. Fragments form only at the fork moving the other way.

4. D

   The new strand is antiparallel to the template, so its 5' end is on the left and it grows toward the right. At the right fork it grows in the same direction the fork opens and is made continuously; at the left fork it grows away from the opening, so it is made in fragments that ligase joins.

Answer: A.

1. A

   RNA polymerase reads the template strand 3' to 5', so moving right means the template's 3' end is on the left. It builds RNA 5' to 3', so the first-made end, the 5' end, is the one trailing free to the left.

2. B

   RNA polymerase reads the template 3' to 5'. Moving to the right, it must start near the template's 3' end, so the left end is 3', not 5'.

3. C

   The template label is right, but RNA grows at its 3' end, which stays inside the polymerase. The end that trails away is the 5' end, made first.

4. D

   Both labels are reversed. Moving right along a template read 3' to 5' puts the template's 3' end on the left, and the free end of the RNA is its 5' end.

Answer: C.

1. A

   Bacteria have no nucleus, so their mRNA never has to pass through a nuclear pore.

2. B

   This reverses the two cell types. Bacteria have no nuclear pores for the drug to block, and in yeast translation happens on ribosomes in the cytoplasm and on the rough ER, outside the nucleus.

3. C

   Yeast mRNA is made in the nucleus and must move to ribosomes in the cytoplasm, so new mRNA is trapped. A bacterium has no nucleus, and its ribosomes begin translating each mRNA while it is still being transcribed.

4. D

   Ribosomes read mRNA, not DNA. Rough ER ribosomes in yeast are in the cytoplasm and still need mRNA that has left the nucleus.

Answer: D.

1. A

   Removing the inducer does not release a repressor in this model. Without inducer even a normal repressor stays on the operator.

2. B

   Constant expression is what a repressor that cannot bind the operator would cause. This repressor still binds the operator.

3. C

   Normal induction needs the inducer to bind the repressor. The mutant cannot bind it, so adding inducer has no effect.

4. D

   Release of the operator depends on the inducer binding the repressor. The mutant repressor binds the operator but cannot bind the inducer, so it is never released and RNA polymerase stays blocked.

Answer: B.

1. A

   The yeast genes are on different chromosomes and make separate mRNAs, so they cannot share one promoter. Only the bacterium's genes form an operon.

2. B

   The bacterium's genes form an operon: one promoter and one regulatory sequence control all three, so they are transcribed together. In eukaryotes, genes far apart can be regulated together when the same transcription factor acts on a regulatory sequence near each one.

3. C

   An inducer such as galactose acts through regulatory proteins, which then change how RNA polymerase transcribes the genes. It does not switch genes on by binding DNA or mRNA itself.

4. D

   This reverses the two species. The bacterium's genes make one mRNA from one promoter, and the yeast's genes each make their own mRNA, so nothing needs to be cut apart.

Answer: B.

1. A

   A silent mutation lies in a codon, and by definition it has no effect on the protein. It would not explain a drop in amount, and the promoter is not read as codons.

2. B

   The promoter is where RNA polymerase and transcription factors bind to start transcription. A change there can weaken that binding and lower how often the gene is transcribed, changing the amount of protein while leaving the coding sequence, and so the amino acids, unchanged.

3. C

   A frameshift in the coding region would change the amino acids after it. The sequence is unchanged, and a promoter is not translated.

4. D

   A premature stop codon would shorten the protein, changing its sequence. The promoter lies outside the part of the gene that is translated.

Answer: C.

1. A

   The mixed genes come from the two viral strains, not from the host. Host genes would not give particles carrying genes of both strains.

2. B

   Random mutation could not produce exact copies of many genes from another strain, and the mix appeared in a single round of infection.

3. C

   Related viruses that infect the same cell can recombine their genetic information. New particles assembled there can carry a mix of genes from both strains, which adds to genetic variation.

4. D

   The particles carry a mix of genes, not two complete genomes, and the mixing happens inside the host cell before release.

Answer: D.

1. A

   Heating separates the strands by breaking the hydrogen bonds between base pairs, with or without primers. The missing primers stop the extension step, not the heating step.

2. B

   Primers do mark the ends of the copied stretch, but they are also what lets DNA polymerase start. Without them it cannot copy any part of the template.

3. C

   DNA polymerase does not search for a sequence. The primers, by pairing with the template, set which stretch is copied.

4. D

   DNA polymerase can only add nucleotides to an existing strand paired with the template. Primers give it that starting point, so without them the extension step has nowhere to begin.

Answer: A.

1. A

   Selection acts on inherited variation. Long baskers do leave more surviving offspring, but their offspring are no more likely to bask long, so the next generation's average stays where burrow sites put it.

2. B

   Learned habits are not inherited, and the stem says offspring basking is unrelated to their parents'. Nothing passes the trait on.

3. C

   Differential reproduction alone is not enough. Natural selection changes a population only when the favored variation is inherited, and here offspring do not resemble their parents in basking time.

4. D

   This contradicts the stem: longer basking does go with more surviving offspring. The obstacle is that the variation is not inherited.

Answer: A.

1. A

   This is a bottleneck. The few survivors held only part of the population's alleles, chosen by chance, and the regrown population descends from them, so most of the lost variation is still missing.

2. B

   Regrowing in number does not bring back lost alleles. New variation comes only from mutation, which is far too slow to restore it in about a century.

3. C

   Mutation does not speed up because a population grows. A population that passed through 20 to 30 animals holds less variation, not more.

4. D

   A bottleneck is genetic drift: which seals survived was largely chance, not a sorting by alleles. The survivors' alleles are a random sample, not a selected set.

Answer: C.

1. A

   This uses genotype frequencies (0.29, 0.42, 0.29 against 0.25, 0.50, 0.25) in place of counts. Chi-square must be computed from counts of individuals.

2. B

   This leaves the differences unsquared: 40/250 + 80/500 + 40/250 = 0.48. Each (observed - expected) must be squared before dividing by the expected count.

3. C

   p = (2 × 290 + 420)/2,000 = 0.5, so the expected counts are 250 AA, 500 Aa and 250 aa. χ² = 40²/250 + 80²/500 + 40²/250 = 6.4 + 12.8 + 6.4 = 25.6, far above both critical values, so the null hypothesis is rejected: there are too few heterozygotes for equilibrium.

4. D

   The value is right, but the conclusion is reversed. A chi-square larger than the critical value means the observed counts differ from the expected ones by more than chance would explain.

Answer: D.

1. A

   Diet does not cause particular mutations. Mutations arise at random, so the environment could not write the same changes into three species.

2. B

   Similar selection can favor similar traits, but it cannot produce the identical set of mutations in the same positions. Those had to be inherited.

3. C

   Mutations arise at random positions, so the chance of the same set of disabling changes arising three separate times is very small.

4. D

   Sharing the same mutations at the same positions is best explained by inheriting them once from a common ancestor. Comparing DNA sequences this way is evidence of common ancestry.

Answer: B.

1. A

   Prokaryotes also store genes in DNA and build proteins on ribosomes. These features point to the common ancestry of all life, not to an ancestor that was a eukaryote.

2. B

   Linear chromosomes and genes containing introns are features typical of eukaryotic cells and rare or absent in prokaryotes. Finding the same features in three very different organisms points to an ancestor that already had them.

3. C

   Being made of many specialized cells arose more than once in different lineages, so it does not show these three shared one eukaryotic ancestor.

4. D

   Habitat and releasing energy from sugars are shared with many prokaryotes, so they cannot single out a eukaryotic ancestor.

Answer: A.

1. A

   Each node stands for the most recent common ancestor of the lineages that branch from it. X and Y meet at the 4-million-year node; the lineage leading to Z had split from theirs at the 9-million-year node.

2. B

   Lineages on neighboring branches are sister lineages that share an ancestor at their node. Neither evolved from the other.

3. C

   Every living lineage has been evolving for the same length of time since its shared ancestor. Branching early does not make a lineage less evolved.

4. D

   Z is a living lineage at a tip, not an ancestor. The node at 9 million years is an ancestor that X, Y and Z all share.

Answer: B.

1. A

   Sharing a field does not prevent speciation. Crosses produce offspring that are nearly sterile, so the two groups cannot exchange genes and are reproductively isolated.

2. B

   The cross forms zygotes, but the hybrids produce almost no viable seeds, so the barrier acts after fertilization. Isolation arose with no geographic separation, which is sympatric speciation.

3. C

   The crosses do produce offspring, so fertilization succeeds. The failure comes later, in the hybrids, which makes the barrier post-zygotic.

4. D

   Allopatric speciation needs geographic isolation. These plants grow among the diploids; coming from one plant is not the same as being separated by place.

Answer: D.

1. A

   This varies exposure to the fungus, not diversity, since every plot is equally diverse. It tests whether the disease kills barley, not whether diversity helps.

2. B

   This varies the number of plants per plot, not their genetic diversity, and breaks the plan of 50 seedlings in every plot.

3. C

   Diversity and exposure to the fungus both change, and together. Any difference in survival could come from either, so the variables are confounded.

4. D

   Genetic diversity is the only thing that differs: closely related seedlings of one self-fertilized parent against seedlings of many parents, at the same number of seedlings, under the same exposure to the fungus. Survival is the dependent variable.

Answer: C.

1. A

   The means do fall at each step, but the 10-snail range (2.7 to 3.5 mm) overlaps the 40-snail range (2.1 to 3.1 mm). The first rise in density is not shown to slow growth.

2. B

   One overlapping pair does not erase the other comparisons. The 160-snail range, 0.8 to 1.4 mm, lies well below both other ranges, so density clearly matters at the highest level.

3. C

   Mean plus or minus 2SE gives 2.7 to 3.5 mm at 10 snails, 2.1 to 3.1 mm at 40 and 0.8 to 1.4 mm at 160. Only the first two overlap. At the highest density snails compete for the same algae, so food per snail, and with it growth, drops: a density-dependent effect.

4. D

   A difference of 0.5 mm between means is smaller than the spread the error bars show. The two ranges overlap, so these data do not show the middle density slowed growth.

Answer: A.

1. A

   Fixation by root bacteria moves nitrogen from the largest reservoir, the atmosphere, into ammonia the beans assimilate. When the plowed plants decompose, ammonification releases ammonium, which nitrification can turn to nitrate for the corn: matter is cycled and conserved.

2. B

   Decomposing tissue releases nitrogen as ammonium through ammonification. Most plants cannot use nitrogen gas, so the corn could not take it up from the air.

3. C

   This ignores fixation, the reason for planting beans: the root bacteria bring nitrogen in from the atmosphere, so the soil gains nitrogen it did not hold before.

4. D

   Photosynthesis fixes carbon, not nitrogen, and no process makes new atoms. Matter is conserved; the beans move nitrogen into the field from the air.

Answer: D.

1. A

   Winter survival does change, from 71% to 30%; only its first step, from 71% to 68%, is small.

2. B

   The drops grow with density instead of staying equal: survival falls by 3 points at the first doubling and by 25 at the last.

3. C

   Both columns decrease down the table. Crowding increases competition for food and space rather than helping the voles.

4. D

   Young per female falls by 1.2, then 2.1, then 2.6, and winter survival falls by 3, then 13, then 25 percentage points. Both decline as density rises, and the largest drops come between 200 and 400 voles per hectare, a sign of density-dependent limits.

Answer: B.

1. A

   This is Σ(n/N)² alone, 0.35. The index subtracts that sum from 1.

2. B

   The proportions are 12/24 = 0.5, 6/24 = 0.25, 4/24 ≈ 0.167 and 2/24 ≈ 0.083. Their squares sum to 0.25 + 0.0625 + 0.0278 + 0.0069 ≈ 0.347, and 1 - 0.347 ≈ 0.65.

3. C

   This is 1 - 1/4, the value four equally common species would give. The species here are not equally common, which lowers the index.

4. D

   This squares N but not n: 1 - (12 + 6 + 4 + 2)/24² ≈ 0.96. Each proportion n/N must be squared as a whole.

Answer: C.

1. A

   The model says figs are under 2% of the trees, so they are not the main producers by mass. A keystone's importance is out of proportion to its abundance.

2. B

   Rarity alone does not make a species a keystone. Most rare species have small effects; what matters is the role figs play for many others.

3. C

   A keystone species has an effect far larger than its abundance. Figs carry many fruit-eaters through the lean months, and those animals spread the seeds of other trees, so losing the figs would ripple through much of the forest.

4. D

   The model says the reverse: in lean months many birds, monkeys and bats depend on figs.

Answer: B.

1. A

   Two very high plates (107 and 64) stand out, but most separate cultures had none, and the mean, 17.5, is close to the samples' 17.0.

2. B

   The separate cultures total 175 colonies on 10 plates, a mean of 17.5; the samples total 170, a mean of 17.0. The means are nearly equal even though the separate counts are spread very unevenly.

3. C

   Six separate cultures had no colonies, but every sample of the large culture had at least 13.

4. D

   The highest count, 107, came from separate culture 5. The highest sample count was 26.

Answer: C.

1. A

   Under this hypothesis the drug causes mutations on the plate, so some colonies would still grow. A lack of earlier contact with the drug does not mean zero resistance.

2. B

   This is the result the researchers got, and it fits mutations arising at random times during growth: a culture where one arose early holds many resistant descendants.

3. C

   If mutations arose only on the plates, every plate would start with similar cells exposed in the same way, and each would get a similar, small number of resistant colonies. Separate cultures would then vary no more than samples of one culture.

4. D

   Samples of one well-mixed culture each hold a similar share of its cells, so nothing in either hypothesis makes them vary more than separate cultures.

Answer: A.

1. A

   An adaptation is favored only in a particular environment. On drug R resistant cells survive while the rest die, but without the drug they were outreproduced by susceptible cells in all 5 cultures, so the same allele is helpful in one environment and costly in another.

2. B

   Mutations are not undone because the environment changes. The resistant share fell because susceptible cells outreproduced them, not because resistant cells turned back into susceptible ones.

3. C

   Fitness depends on the environment. On drug R plates only resistant cells formed colonies at all, so resistance is a large advantage there, even though it is a cost in broth without the drug.

4. D

   Selection acts whenever kinds of cells differ in how many offspring they leave. A fall from 50% to about 20% in all 5 cultures is consistent, not random, so it points to lower fitness of resistant cells without the drug.

Answer: D.

1. A

   The cultures had never met drug R either, yet most held resistant cells. Resistance arises before exposure, so a lack of contact does not mean none is present.

2. B

   Individual cells do not acquire resistance by adjusting. The drug selects cells that already carry resistance mutations.

3. C

   Rare at the start does not mean rare for long. Once the drug kills the susceptible cells, the resistant ones face little competition and can take over.

4. D

   About 100 million cells grown without the drug held about 17 resistant cells on average, so an infection this size likely holds some. The drug kills the rest, and the resistant cells multiply: the population evolves resistance.