Chapter 18: Practice Exam

Four possible answers or sentence completions are followed by each of the following questions or statements.

The one best answer or sentence is what you choose.
Continue with Part B when you are done with Part A.

There is a variety of cells with the tropomyosin.

The primary transcript that codes for tropomyosin is changed in a way specific to the type of cell.
Smooth muscle cells contain exons 1, 2, 4, 5, and 6.
In striated muscle cells, exons 1, 3, 4, 5, and 6 are missing.

The figure shows the results of the process.

The same sequence is needed for exons 2 and 3.

Exons 2 and 3 have the same number of nucleotides.

The number of nucleotides in exons 2 and 3 needs to be rearranged.

The number of nucleotides in exons 2 and 3 must be equal to 3.

There is a build up of mucus in the lungs of people with cystic fibrosis.

In addition, mucus build up from the cells of the sphinx blocks the production of the sphinx.
Individuals who inherit two copies of the same trait have a life expectancy of 30 to 40 years.

The disorder is caused by a defect in theCFTR.

The opening of a channel that allows Cl- to flow is triggered by the binding of ATP to CFTR.
In epithelial lung cells, the channel allows the movement of Cl- out of the cell, but in other cell types, the channel can be structured so as to promote Cl- into the cell.

In normal individuals, epinephrine binding to a GPCR on the lung cells starts a signal transduction pathway.

The GTP is exchanged for a GDP on a nearby GProtein.
The GTP has activated the GProtein that is bound to the GPCR.
The release of a G subunit allows it to bind to adenylyl cyclase.
The conversion of ATP to cAMP and cAMP is done by a drug called adenylyl cyclase.
The opening of a gated channel in the CFTR and the passive passage of Cl- out of the cell can be achieved with the help of two ATP molecules.

Water is forced out of the cell by the movement of Cl- into the fluid.

The Na+/K+ pump helps regulate the electrochemical gradient to maintain an appropriate surface liquid on the cells.

In cells with the cystic fibrosis genetic abnormality, the CFTR is not functional.

muscle cells contract and cough due to a build up of a drug.

There is a build up of water in the cells of the Airways.

Cells unable to move Cl- across the die produce thick mucus.

The mucus develops in the lungs because the water doesn't cross themembrane.

Sweat is produced by cells of the sweat glands and transported to the skin surface through ducts.

The sweat of people with cystic fibrosis is very high in salt, so that it can be analyzed to see if they have the disease.

There are four chromosomes, each consisting of two chromatids, arranged midway between the opposite poles of a cell.

A1, B1, C1, and D1 move toward one pole of the cell while chromatids A2, B2, C2, and D2 move toward the opposite pole.

There are two poles of the cell, one for chromosomes A and A2 and the other for chromosomes B and B2 and D and D.

The lizards and spiders eat the same things.

The lizards and spiders were surveyed on Caribbean islands.
On islands with lizards, they found more spider and spider species than on islands with no lizards.
They introduced lizards to four of the eight lizard-free islands and returned every 2 years to survey.
They found lizards and spiders in the same area.
The results are summarized in the figure.

Spiders can't escape from their island habitat.

Spiders are less likely to emerge from hiding in the dark.

Spider populations are reduced where lizards are present.

Spiders don't like areas where lizards are present.

The researchers found evidence of insects in some lizards, but no spiders.

The lizards don't eat spiders.

Spiders were eating lizards.

The lizards carried parasites.

Both lizards and spiders have the same resources.

The numbers of spiders are stable.

The survival of spiders on islands is increasing.

There is a factor other than lizards that contributes to the number of spiders reported.

The amount of food available to spiders is being reduced.

A biologist uses paper chromatography.

The leaf extracts are streaked near the bottom edge of the paper.

The paper is made of a polar molecule.

The bottom edge of the paper is mounted to make contact with the solvent.

The solvent and the leaf extract move up the chromatography paper after the paper touches it.

The leading edges of four pigments are shown in the figure.

The ratio of the distance traveled by a pigment to the distance traveled by the solvent is called the Rf value.

Long periods of studying or other types of mental activity can cause brain tissue to build up.

The binding of adenosine to a G protein-coupledreceptor on brain cells can cause it to replace GDP with a GTP.
The GProtein is bound to and activated by the adenylyl cyclase.

A second messenger, cAMP, is converted by the edylyl cyclase.

Different types of cells have different responses to cAMP.
In brain cells and other cells of the central nervous system, cAMP slows brain activity and causes drowsiness.
Normally, cAMP concentrations in the cell are kept low by the PDE, which converts cAMP to regularAMP.
During periods of mental fatigue or other types of stress, high levels of cAMP can be attained.

The structure of caffeine is similar to that of adenosine, so it can bind to it, but it can't be activated.

The binding of adenosine to the adenosine receptor is blocked by caffeine.

Overstimulating the adenosine signal transduction pathway is what causes it to fail.

The effect of adenosine is restored by the substitution for it as a signaling molecule.

There is no effect on the activity of adenosine.

Increased blood pressure, breathing rate, and the rate of cell metabolism are all associated with the fight-or-flight response.

A G protein-coupled receptor is a ligand for epinephrine.
The epinephrine signal transduction pathway is similar to that of adenosine and produces cAMP.

In muscle cells, the release of sugar is triggered.

In muscle cells, there is a production of fatigue.

In brain cells, it causes drowsiness.

Vasoconstriction is caused by the narrowing of the blood vessels.

In response to temperature changes, the fluidity of the plasma membrane can vary.

The concentration of solutes should be increased.

Change the ratio of phospholipids with double covalent bonds in their tails to those without double covalent bonds in their tails.

The following graph plots the rate of a reaction controlled by a humanidase as a function of temperature.

With increasing temperature, more molecule collide with each other.

As their energy increases, they become larger targets, increasing their chances of colliding.

There are more available to carry out the reactions at higher temperatures.

As the temperature increases, fewer inhibitory molecules are available to block the active site.

The rapid reaction rate quickly consumes all of the reactants at high temperatures.

The binding of additional reactants is prevented at high temperatures when the substrate is locked into the active site.

The reactants can't settle into the active site at high temperatures because they are moving so fast.

The secondary and tertiary structures of the enzyme are disrupted at high temperatures.

Questions 20-21 refer to a figure that shows the movement of materials with two embedded proteins.

The active and inactive states are shown.

The structure of the cell's main signaling molecule is nearly identical to that inbacteria.

Mitochondria are derived from ancient aerobicbacteria.

convergent evolution is an example of the similarity of the organisms.

Mitochondria and chloroplasts were able to escape from cells.

The R group is indicated by an R, while the side chains are shaded.

The cytosol has a normal pH of 7.2.

The majority of the lysosome are only active at a pH of 5.

The following figure shows parts of amitochondrion and a chloroplast.

There is a chemiosmotic gradient in both processes.

The electron transport chain in the mitochondria and chloroplasts causes the gradient.

For respiration and photosynthesis, electrons come from H2O and end in NADPH.

For respiration, electrons come from H2O and end in NADH; for photosynthesis, they come from NADPH and end in H2O.

They end in H2O.

For respiration, the electrons come from sugar, and for photosynthesis, they come from CO.

The leaves of these plants were exposed to various intensities ofPAR and CO2 was measured to determine rates of photosynthesis.

The leaves from the plant that was grown in the sun reached a higher maximal photosynthetic rate.

The shade leaves were exposed to less light than the sun leaves.

The sun leaves had more surface area than shade leaves.

The leaves grown in the sun were thicker than the shade leaves.

The leaves grown in the shade were more efficient than those grown in the sun.

Shade leaves have more surface area than sun leaves.

High rates of respiration in leaves grown in the shade result in higher rates of CO2 assimilation than leaves grown in the sun.

At low light intensities, leaves grown in the shade have lower rates of respiration than leaves grown in the sun.

The leaves grown in the sun are more efficient than those grown in the shade.

In response to low light intensities, leaf stomata close.

The rates of respiration are higher than the rates of photosynthesis.

The Calvin cycle releases CO2 at low light intensities.

There is no CO2 assimilated at light intensities below 50 MMol.

The bonds between the mino acids are called peptides.

Questions 31-32 refer to the figure of the glycophorin Aprotein, which extends from the cytosol to the extracellular fluid.

The circles and squares represent the side groups.

It causes the solution to be basic.

It increases the concentration of hydrogen in the solution.

It becomes green.

It has no effect on the solution.

Light is making the BTB more acidic.

The table shows the distribution of blood types among Native Americans.

The Blackfeet of Montana were an isolated population.

Various groups of Native Americans gathered in Montana to create a genetic mix.

The Blackfoot population began as a group of individuals whose genetic makeup did not correspond to the population from which they originated.

Due to chance, the Blackfoot population's genetic makeup began to resemble that of other Native American populations.

Red blood cells carry oxygen throughout the body.

The blood types A, B, and O identify red blood cells that have specific sugars attached to them.
Four phenotypes, A, B,AB, and O, are produced by the combination of any two of the three alleles.

There are some possible explanations for the evolution of the ABO allele group.

A nonfunctional gene is most likely the result of multiple nucleotide substitutions during evolution.

The graph below shows the temperature of water as a function of heat energy absorbed.

More hydrogen bonds are breaking.

The water doesn't absorb heat energy.

The water is getting hotter and hotter.

The water is losing energy at a rapid rate.

High fructose corn syrup is often added to soft drinks to make them taste better.

Microtubules are made from tubulin.

The assembly of microtubules is prevented by binding to tubulin.

The data in the table compares the sequence of a short segment of DNA from five different species.

The total number of nucleotides in the segment was divided by 100 to determine the number of differences between the two species.

There are two lines attached to a circle that represent alipid.

The circle and the two wavy lines represent the groups.

The end product of the pathway is alcohol.

A lot of energy remains in the product, but it's a waste product for the yeast cells that produce it.

CO2 is used for photosynthesis.

CO2 is provided for cellular respiration.

The defense against invadingbacteria is provided by yeast cells.

The NAD+ is needed for glycolysis to continue.

A student puts plant leaves in a beaker of water in a laboratory exercise.

A drop of soap is added to the water.

The soap molecule enters through the leaf's stomata.

The soap bonds to air bubbles on the leaf's surface.

Excess H+ is allowed to bond to the leaf surface by decreasing the water's pH.

The soap molecule's hydrogen bonds with the water molecule can be seen on the waxy leaf surface.

The reaction J - K + L is regulated by the Enzyme E.

The maximum rate is maintained from 6 to 9 minutes.

There is a cleft formed by the active site of the enzyme.

The maximum concentration of reactant J is.

All of the available molecule are bound to something.

The temperature of the reaction is very high.

Theidase is not working.

Products K and L are converted to reactant J.

All chemical activity has ended after the reaction is complete.

A molecule of K and a molecule of L combine to form a molecule of J.

K and L will be converted to other substances with the addition of enzymes.

J will be converted to a different molecule with the addition of anidase.

Raising the temperature of the solutions will add energy to the reaction.

In the process of generating ATP, hydrogen ion are transported across the inner mitochondrial and the thylakoid membranes by a protons pump.

The following table shows the various supplements for each of the four Mutants.

Each of the Mutants has a single mutations that inactivates a single conversion.
Growth occurs when any one of the supplements indicated by a plus sign is added to the minimum culture medium.

If citrulline, argininosuccinate, or arginine are added to the minimal medium, Mutant II will grow.

The fruits of a plant are either red or white and have either a smooth or spiny surface.

The table shows the results of crossing true-breeding parents with the F1 offspring.

Three groups of students used computer simulations to predict the numbers of each phenotype when F1 offspring were tested to see if they were compatible with each other.

Table II shows the results of the computer simulations.

No F1 individuals express the white or spiny trait according to the results of the cross of true-breeding parents.

White individuals and spiny individuals are always carriers.

The allele for red-colored fruits is more dominant than the allele for white-colored fruits, and the allele for smooth fruits is more dominant than the allele for spiny fruits.

Only one of the two genes can be passed on at the same time.

There are people with smooth fruits.

The results shown in Table II for each of the computer simulations are different because the students made different assumptions about how the traits of color and texture are inherited.

The students assumed that zygotes can't survive or develop normally.

The students assumed that the red and white alleles were codominant and the smooth and spiny alleles were codominant.

During cell division, the genes for color and texture were assumed to be independent.

Only the results of simulation 1 can be considered correct because it is the only simulation that generated enough offspring to establish reliability.

Red, spiny, and white fruits were not generated in simulation 2.

Simulation 2's total number of offspring is not large enough to produce reliable results.

The students assumed that the genes for color and texture are the same.

Some individuals can be affected by the results of a recombination of two chromosomes.

Recombination of all genes was blocked.

The numbers are different between simulation 1 and simulation 3.

The students assumed that the alleles for red and white are the same as those for smooth and spiny.

The students wanted the number of offspring generated by simulation 3 to be half that of simulation 1.

Students assumed that crossing over genes occurred during cell division.

The students assumed that the genes for color and texture were independent.

The correct answer is calculated for each of the six questions.

You can use a four-function calculator with square root capability and reference the provided equations and formulas pages.

A dominant allele that codes for the expression of carotenoids determines the yellow color of the corn kernels.

The color is white if no Carotenoids are produced.
The number of each type of plant is determined by the farmer after harvesting.
The answer should be between 0 and 1 to the nearest hundredth.

Your answer should be sent to the nearest tenth.

Light intensities and leaf temperature can affect photosynthetic activity.

The rate of respiration is plotted on the dashed line.
When the rates of respiration and photosynthesis are the same, the compensation point occurs.
Your answer should be sent to the nearest tenth.

The transfer of energy is shown in the figure.

The answer should be rounded to the nearest hundredth.

The equation describes the process of hydrolysis.

Humans have the same amount of ATP as their body weight.

Your answer should be a positive whole number.

Simpson's index is a method for determining the species diversity of a community.

Express your answer to the nearest tenth using the relative abundances given in the following table for a community of six mammal species.

The first two questions are worth 10 points each.

The first three questions are worth 4 points each and the second three are worth 3 points.

For all your answers, use complete sentences.

When questions have more than one part, you should separate your answers to each part and use the letter for that part.

diagrams can be used to supplement your answers, but a diagram alone is not adequate.

Smooth muscle surrounds the cells that form the walls of blood vessels.

Contraction of the smooth muscle reduces blood flow.
The smooth muscle relaxes and increases blood flow.
Nerve cells release the neurotransmitter acetylcholine at their terminal ends to control smooth muscle activity.

Acetylcholine opens a signaling pathway that leads to the release of Ca2+ into the cytoplasm from the endoplasmic reticulum.

The release of NO from the arginine is stimulated by the Ca2+.

The NO molecule acts as a signaling molecule for the smooth muscle cell.

The production of cGMP is made from the GTP.
The relaxation of smooth muscle is caused by cyclicGMP.
The signal transduction pathways are summarized in the figure.

Acetylcholine doesn't initiate the pathway itself.

The signal of acetylcholine can be transmitted by the G proteincoupled receptor.

There are two ways in which NO and acetylcholine differ.

Patients suffering from heart conditions where blood vessels that supply blood to heart muscle are blocked are more likely to be treated with nitrate.

The tablets are placed on the tongue and absorbed quickly.
The mitochondrial aldehyde dehydrogenase breaks down nitragine in the cells.

High altitude sickness can be a problem for mountain climbers.

There is a form of high-altitude sickness.
The lungs are where the condition develops.
One of the drugs is expected to help this condition.

Drug 1 competes with acetylcholine for binding to the receptors.

Drug 2 prevents its attachment to the endoplasmic reticulum.

Drug 3 blocks the degrader.

Drug 4 releases citrulline.

Wolves were eliminated from the park in 1925 due to hunting, poisoning, and bounty programs.

With the elimination of this major natural predator, park management was concerned that the elk population would grow and exceed the carrying capacity of the park environment.
Park management culled the herds in response to this concern.
Food shortages, disease, weather, and predation by bears and coyotes were some of the pressures remaining on elk survival.

In 1968, park management decided to allow nature to proceed with minimal human influence.

The introduction of 11 wolves from Canada in 1995 was in response to the 1973.

The population of animals went up from 1960 to 1995.

The data shows that the population of animals decreased from 1995 to 2010.

There is an explanation for the decrease.

The data showed a decline in the wolf population.

The cell is large and consists of a cap, a slender stalk, and a base.

Animals use prostaglandins for communication.

The attractor that signals to males that a female is receptive to mate is a chemical form of japonilure.
A second form is an enantiomer, or structural mirror image, of the first form and is used to signal that the female is no longer receptive.

The use of a volatile chemical to attract mates may be a disadvantage for a species.

Plants use colors, shapes, scents, and sweet rewards to attract insects.

The dates that the eggs hatched were recorded by researchers.

The hatchlings identified the sex of the fish when they reached sexual maturity.
The hatch-time data for males and females is shown in the graph.
Two standard errors of the mean are represented by the horizontal error bars.

After the egg is fertilized, it grows from embryo to adult hermaphrodite with a total of 959 cells.

Growth of the oocyte from its initial formation in the female uterus to the adult stage occurs by the expansion of cells.
The growth of the Nematode is shown in the graph by cell number and cell volume.
At each tick mark, the vertical axis increases by a factor of four.

The graph shows that the size of the oocyte increases by more than 1000 times when it becomes fertilized.

Some large cells are able to replicate their DNA many times without being affected by cytokinesis.

There are a variety of cell functions.

They originate from centrosomes and grow outward by adding individual tubulin proteins to a growing microtubule.

Adding or removing tubulin monomers can make microtubules unstable.

Extending from their origin in a more or less random pattern, they are stable when a growing microtubule contacts a target.

The main components of the cytoskeleton are microtubules.

Taxol, a substance obtained from the bark of yew trees, binding to microtubules, stabilizing them and preventing tubulin monomers from being added or removed.

Taxol can be used to stop the growth of cancer cells.

A, B, C, D, and E are the five species that have the same short segment of DNA.

The total number of nucleotides in the segment was divided by 100 to determine the number of differences between the two species.

The exon that follows the alternative exons is the same.

Exon 1 may not end with a complete codon of three nucleotides and exon 4 may not begin at the start of a new codon.

Both exons 2 and 3 must add where exon 1 ends to where exon 4 begins.
This requires that exons 2 and 3 have sufficient nucleotides to code for complete codons and that extra nucleotides are needed at the end of exon 1 and at the beginning.
Only answer choice D allows for the completion of codons at the end of exon 1 or the beginning of exon 4.
If the codons 1 and 4 are not completed by exon 2 or 3, the resulting processed mRNAs will have a frame shift, which will result in a nonfunctional, protein.
On the other hand, either exons 2 or 3 can have additional codons that would result in the addition of one or more additional amino acids into the tropomyosin protein, and that might influence its function.
The answer choices B and C don't address the possibility that the codon that terminates exon 1 and the codon that begins exon 4 might not be complete.

The signaling molecule is called a ligand.

The signal transduction pathway begins when it is initiated by it.
At the end of the pathway, thePKA is activated which opens the Cl- channel.

The function of the CFTR is to generate an electrochemical gradient that allows the flow of water.

In response, water moves across the membrane.
Thick mucus develops when water does not flow, because the surface liquid on the cells is deficient.

The sweat of people with cystic fibrosis is salty.

Normal individuals have sweat- producing cells.
The movement of the Cl- by the CFTRProtein must be in the opposite direction as it is in the lung cells.
In sweat glands that express a mutated form of the CFTR protein, Cl- does not return to cells and remains in the sweat.
Answer choices A and C would result in decreasing the amount of sweat in individuals with cystic fibrosis.
The answer choice D is incorrect because the causes of cystic fibrosis are not the same.

The chromosomes in the figure are arranged in pairs.

A1 and A2 are the same as B1 and B2 and represent a pair of chromosomes.
Another pair of chromosomes is represented by chromosome C and D.

There is an arrangement of pairs between the poles of a cell.

The next step in the process is for one member of each pair to migrate to a different pole.

A pair of chromosomes are derived from each parent and have the same genes.

Assuming that no copying errors occurred when the second chromatid was copied from the first, each chromosome consists of two chromatids that are identical.
The chromatids A1, A2, B1, and B2 all have the same genes.
C1, C2, D1, and D2 are the same genes and are not listed among the answer choices.

The figure only shows the Homologous chromosomes pair.

All of the chromosomes align along a plane between the poles.
During cytokinesis the chromosomes have moved toward the poles and are no longer positioned between the poles.
During interphase, the chromosomes are not visible, but they are long and thin.

The introduction of lizards is believed to have caused a decline in the size of spider populations on islands where lizards were introduced.

The answer choice C provides a simple, straightforward explanation that the lizards were eating the spiders.
There was no data presented for spider populations on habitats other than islands.
Answer choices B and D suggest that spiders hide to avoid being eaten by lizards, but no observations are presented to support that conclusion and the data shows that they were being eaten.

Although it is clear that the island lizards were not eating spiders, that explanation does not explain why the spider populations were smaller on the islands with the introduced lizards.

The lizards and spiders were competing for the same resources.
The amount of food available for the spiders decreased when the lizards were introduced to the islands.
According to answer choice C, the size of the spider populations declined as a result of an introduced parasites.

There is no data to support this conclusion.

The introductory paragraph makes it clear that spiders and lizards eat the same food.

The number of reported spiders on the islands with and without lizards seem to be increasing and decreasing at the same time.

It is possible that some other factor is affecting the number of surviving spiders from year to year.
None of these factors can be identified with the data provided in the figure.
Answer choice D cannot be accepted.

The ratio of the distance traveled by a pigment to the distance traveled by a solvent is called the Rf value.

The Rf value of Carotene is.

The reason that one pigment moves farther than the other is because it has a lower affinity for the chromatography paper.

The shorter the distance between the origin and the band of pigments, the stronger the affinity between them.
The remaining answer choices are not responsible for the primary separation of pigments as they are for increasing or decreasing the separation between bands.
There are some interactions between the solvent and the pigments, but none of the answer choices addresses this.

The shorter the distance between the origin and the band, the greater the attraction.

The GDP attached to the inactive G protein is replaced by a GTP when adenosine is binding to it.

The GTP for a GDP is activated.
The GTP and Ga bind to the adenylyl cyclase and cause it to be activated.

The effect of adenosine is blocked by caffeine.

When caffeine binding to the adenosine receptor, it blocks the G protein-coupled receptor from starting the adenosine signaling pathway.
drowsiness does not occur as a result of this.

Different cell types can have different effects on a single signaling molecule.

There are only three types of membrane receptors, but there are more than 700 known signaling molecules for each of them.
Each signaling molecule causes a signal pathway that is specific to the signaling molecule and cell type.
The PKA produced by the adenosine pathway is going to be different than the one produced by the epinephrine signal transduction pathway.
The fight-or-flight response is associated with the release of glucose from glycogen.
Fightor-flight activities use the release of glucose as a source of energy.
Reducing blood flow to the heart does not promote the fight-orfight response.

A sphinx with an sphinx with one or more double covalent bonds creates a kink in the tail that spreads adjacent sphinxes apart.

The conjugates with saturated fatty acids pack more closely together.
Answer choice A is incorrect because an increase in cytosol solute concentration won't keep the fluid.
The answer choice C is incorrect because the sugar is not found in the blood.
Replacing phospholipids with sugar would destroy the integrity of the membranes.
The answer choice D is incorrect because the two layers of the plasma are asymmetric.
Changing thelipids from one side to the other would damage the integrity.

The higher the temperature, the greater the average kinetic energy of the molecule.

The greater the molecule's speed, the greater the kinetic energy.
Molecules move faster when the temperature increases because they collide more frequently with the necessary activation energy.

As the temperature rises, the bonds that hold the enzyme together break down.

The three-dimensional shape can no longer function and is said to be denatured.

Facilitated transport is an example of the activity of protein II.

In contrast, the transport of Na+ and K+ is carried out by a single cotransporter, and each of them requires a certain amount ofATP.
Steroids don't need the help of protein carriers because they are nonpolar and don't present a barrier.

The substance to be transported is specific to the specificprotein that is used to transport it.

The substances to be transported are controlled by the plasma membrane.
Different kinds of cells have different types of proteins.
Only two genes require a downhill gradient.
The transport of large, un charged polar substances requires the use of proteins.
O2 and CO2 can diffuse without the help of a transport protein.

Aerobicbacteria have the same energy generating metabolic pathway as mitochondria.

The only function of the chloroplasts is to carry out the energy generating metabolic pathway that occurs inbacteria.
The same machinery is used to accomplish these processes.
The observations support the idea that the cells are frombacteria.
The remaining answer choices don't support the observed similarities between the two types of cells, but there are no observations that show that the same ancestors used the same mechanisms.

There are nonpolar side chains on the shaded amino acids.

The side chains are facing away from the solution when they cluster together.

Water molecule sticks to other water molecule in order to form a vertical column in plant xylem vessels.

The downward pull of gravity is helped by this mechanism.

The lysosome has a pH of 5 and the enzymes are active.

The lysosome is maintained by the active transport of protons into it.

In cellular respiration, glucose is broken down and energy is captured and stored.

During the oxidation stage of respiration, electrons from FADH2 enter the electron transport chain.
The protons move from the matrix into the intermembrane space.
H2O is formed when the electrons combine with a protons and oxygen.
In contrast, the electrons in photosynthesis originate in H2O from inside the thylakoids of the chloroplasts.
The protons move into the stroma as the electrons pass by.
At the end of the ETC, the electrons combine with NADP+ and a protons.

The NADPH is used to generate sugar.

Both respiration and photosynthesis can be accomplished with electrons passing through an ETC.
The two processes are reversed for the origins and endpoints of the electrons involved.

The ability to read and understand data in graphical form is evaluated by this question.

Answer choices A, B, and D are not supported by the data presented in the figure.
Answer choice A is not supported by the data in the figure because the data compare shade and sun leaves exposed to the same amount of light at different intensities.
Answer choice B is not supported by the data in the figure because the data compare equal surface areas of plant leaves.
Answer choice D suggests that shade leaves have a higher maximal photosynthetic rate than sun leaves, but the data for light intensities above 200 MMol shows the opposite.
The hypothesis presented in answer choice C is the best of the four hypotheses presented.

The ability to read and understand data in graphical form is evaluated by this question.

The figure shows that shade leaves have more CO2 in them than sun leaves.
They are doing less cellular respiration than the sun leaves.
Under low light conditions, some cells may be doing photosynthesis while other cells may be doing cellular respiration.
The hypothesis in answer choice C describes a scenario where the shade leaves are releasing less CO2 because more cells are photosynthesizing than the sun leaves.
Shade leaves are more efficient than sun leaves under low light conditions.
Answer choice A is not supported by the data because the CO2 assimilation data in the figure is not related to the leaf surface area.
Answer choice B is incorrect because respiration releases CO2, the opposite of CO2 assimilation.
Answer choice D shows that shade leaves are more efficient than sun leaves in terms of light intensity.

One of the two plausible hypotheses that could explain the data in the figure is supported by answer choice C.

The rate of photosynthesis is so low that cells need more energy from respiration.

Other answer choices may explain why CO2 assimilation is low.

The structure of each amino acid is not expected to be recognized by the AP exam.

This question looks at your understanding of how the two acids bond.
The two amino acids form a bond.
A water molecule is formed and released as a covalent bond is created between two adjacent atoms.
The OH in the carboxyl group of one amino acid and the H of the NH2 of an adjacent amino acid combine to form a water molecule, as shown below.

HCO - 3 and H+ are formed by water and carbon dioxide.

The blue solution of BTB becomes acidic due to the increase in concentration of H+.

In the Calvin cycle phase of photosynthesis, the CO2 in the BTB solution is used to generate sugars.

When the CO2 is consumed, the solution becomes basic as H+ is converted to H2CO3.

Light must be present for the observed color change from yellow to blue in Reaction IV.

Light alone does not cause the color change observed in Reaction IV.

When light isn't available, cellular respiration becomes the dominant process.

The cellular respiration breaks down the sugar in the air.
As the solution becomes more acidic from the addition of CO2 the blue BTB becomes yellow.

Reaction II shows that darkness doesn't cause a blue BTB solution to yellow.

The plant is responsible for the change.

Gene flow is an example of the movement of alleles from one population to another.

The Blackfoot population has a high percentage of the A phenotype compared to other Native American populations.

The founder effect is when a group of genetically atypical individuals leave the larger population to create a new one.

Replacements for existing nucleotides do not increase or decrease the total number.

A deletion would result in a change in sequence.

The codons following the shift are different from those in the original sequence.
One of the codon changes creates a stop codon, which terminates the translation of the mRNA long before the end of it.
The frameshift errors introduced in other codons are responsible for the production of a nonfunctional protein.
There is no sugar attached to the protein.

Liquid water is represented by the line labeled X as it changes temperature.

The horizontal lines show that the energy is being absorbed but the temperature is not changing.

The absorbed energy is used to break the hydrogen bonds that hold the water together, changing the physical state of the water from solid to liquid and from liquid to gas.
The heat of fusion or the heat of vaporization is the energy absorbed during these periods.

Sweating cools bodies because the energy used to change the liquid state of the sweat to the gas state comes from our bodies.

We feel cooler when we lose body energy.
The stage that represents the change from liquid to gas is labeled Y.

High fructose corn syrup is made from corn.

Each glucose monosaccharide must be converted into a fructose monosaccharide if it is to be used in food.
This is a cheaper way of making a sweetener than by using sugar cane or sugar beets.

The components of the spindle apparatus are called microtubules.

The chromatids are separated from the chromosomes, and then the daughter cells are formed.

Microtubule assembly is prevented by colchicine.

This is a rough way to figure out the best tree.

Determine which two species are closest to each other.

cluster 1 is called because J and L have the most at 6.
The next closest pair will be K and M at 8.
Call it cluster 2.
Try to figure out if the two clusters can fit on one of the provided trees.
There is only one tree that provides a match for these two clusters.

There are other ways to make a tree, but they take too much time and require a lot of math.

Try to find pairs in the data that fit one of the provided trees.

A glycerol molecule is attached to a phosphate group.

The figure with a circle is a representation of the head of the phosphate group.
The "tails" are shown as wavy lines.
When immersed in water, the head of a phospholipid orients towards the water and the tails face away from the water.
The orientation of the bilayer is to have the heads facing the water and the tails facing the center of the bilayer.
When a flat bilayer closes to form a circle in three dimensions, it is the best orientation.

Once all of the NAD+ is converted to NADH, there is a finite amount of NAD+ in the cytoplasm.

In the presence of oxygen, there would always be enough NAD+ because NADH is converted back to NAD+ during oxidation.
Oxygen isn't available as the final acceptor of the electrons in the electron transport chain when it's absent.
Under conditions where oxygen is not present, there is a need for fermentation to replenish the NAD+.
The yeasts do not do photosynthesis.

Cells at the surface of a leaf have a waxy coating.

The leaves are not easy to submerge in water.
There are two regions in soap molecule.
When soap is added to water, the soap molecule's hydrophobic regions are over the waxy coating of the leaf, while the soap molecule's hydrophilic regions are away from the leaf surface.
In this way, the leaf is covered by soap with the water surrounding it.
The leaves can fall through the water.

There are no additional reactants that can be catalyzed once all of the available molecule are engaged with them.

During the interval between 6 and 9 minutes, the reaction rate remains constant.
The slope of the plotted curve is the amount of product formed per unit of time.

There are two reactions: forward and backward.

Concentrations of reactants and products determine which direction is dominant.
When the reaction reaches equilibrium, each reaction in the forward direction is matched by one in the reverse direction and the net activity is zero.

The reaction J - K + L is no longer in equilibrium when substances K and L are removed.

More of K and L will be produced by the forward reaction than the reverse reaction.

There are two reasons why you would expect the proton pumps inbacteria to be similar to the ones found in the chloroplasts and mitochondria in animals and plants.

The generating of ATP with pumps in the erythrocytes is only possible in the erythrocytes.
The water-soluble enzymes in the cytosol can be used to generate ATP.

Theplasmamembrane must be used bybacteria which do not have mitochondria or chloroplasts.

Mitochondria, chloroplasts andbacteria have a common evolutionary origin.
The theory explains that a non-photosynthetic bacterium merged with a second one to form a eukaryotic animal cell.
A non photosynthetic bacterium merged with a third bacterium to become a plant cell.
The ancestors of eukaryotes were the symbiotic assemblages.
There are similarities between their ribosomes, their DNA, and their membranes.

Mutant I cannot grow unless ornithine is added.

This suggests that the ornithine conversion is either nonfunctional or absent.
The DNA that codes for this enzyme is the cause of the induced mutation.
Adding any one of the supplements ornithine, citrulline, or argininosuccinate will result in the production of arginine.
Adding citrulline or argininosuccinate allows arginine to be produced because the enzymes that convert citrulline to argininosuccinate are not functional.
In a similar way, Mutant III and Mutant IV have nonfunctionalidases that convert citrulline to argininosuccinate.
Adding ornithine, citrulline, or argininosuccinate doesn't lead to the production of arginine because the required enzyme is not present.
Only answer choice A describes a pathway that is consistent with the data.

Theidase that converts ornithine to citrulline is missing in Mutant II.

Substances in the minimal medium will be converted to ornithine if citrulline is added to the medium.
The new ornithine can't be converted to citrulline because of a problem with theidase that converts it to citrulline.
Once citrulline is converted to argininosuccinate, growth stops and the newly generated ornithine is not converted to citrulline.

Mutant IV can't grow because it can't convert argininosuccinate to arginine.

People who are carriers for a trait do not express that trait.

The plants with red and smooth fruits are carriers for the white and spiny traits.
Answer choice C is incorrect because each parent contributes one of each chromosomes to every offspring and, as a result, every offspring receives two alleles for every gene.
Answer choice D is incorrect because the F1 plants must be Heterozygous in order to satisfy the results.

A Punnett square can be used to show all possible combinations of gametes if genes for different traits assort independently.

Answer choice D is incorrect because the question asks for the best explanation of the results, not if the results are accurate.

Simulations 2 and 3 have enough offspring to produce reliable results.

When genes are on the same chromosomes, they can't assort independently.

If crossing over ensues, recombination can occur.
When the genes are very close together, it is not likely that they will cross over.
When the genes are linked and there is no crossing over, each parent can only provide two kinds of gametes.
One of the parents has only one kind of gamete because of their genetic makeup.
One parent is donating two kinds of gametes that combine (after fertilization) with only one kind of gamete from the other parent.
There are two types of offspring.
When the numbers of offspring produced are skewed to the opposite of their parents', it's a sign of crossing over.
Skewing is dramatic when there is no crossing over.

If genes are on the same chromosomes, they can cross over to each other.

The farther the genes are separated, the more likely they are to cross over.
When the ratio of phenotypes is about 7:0, crossing over causes offspring in ratios that may not look like typical crosses.
When thephenotypes of offspring are produced in numbers skewed toward thephenotypes of certain parents, there is a suspicion of crossing over.
In the previous question, suspect linkage with no crossing over was dramatic.
The question does not request it, but the crossing over frequencies in simulation 3 is 30%.

The total number of plants is 1232.

The total number of plants is 1232.

There is a difference between the two.

Even if the digit is 0, be sure to say your answer to the nearest tenth.
You can see the following graph.

The light intensity is 11.5 kJ/m2/min.

The increase is 40degC.

The producers occupy 2.5 x 104 of the energy entering the first trophic level.

The energy left at the trophic level and entering the second trophic level is 3 x 103.
As follows, the proportion is calculated.

The following calculation needs to be made first.

A straightforward way to do a problem like this is to set up a calculation with each factor arranged so that units in the numerators and dominators can be divided into one another.

The units of grams and moles can be eliminated by dividing them into one another in the first and second factors.
You will have set up the calculation correctly if you arrange the factors so that the units of interest remain after canceling other units.

To get points for your answer, use the standards given below.

For each item listed below that matches the content and vocabulary of a statement or explanation in your response, add the indicated number of points to your score.

Each question will have a score ranging from 0 to 10 points.

There are words in parentheses in the answers.

Acetylcholine is a polar molecule.

Acetylcholine can't pass across the membranes.

The message is sent to the reticulum.

The second messenger, IP3, signals the reticulum to open Ca+ channels.

Acetylcholine is a polar molecule that can't be passed across the membranes.

NO is a small molecule that is able to cross the plasma membrane.

Acetylcholine targets a receptor on the outside of a cell.

No targets an intracellular receptor.

Dilating blood vessels in the heart increases the flow of blood.

When broken down, nitroglycerin releases NO.

The guanylyl cyclase is not activated by 1 pt.

GTP is converted to cGMP with 1 pt.

The drug relaxes blood vessels.

Drug 3 2 points: By blocking the breakdown of cGMP, it is possible to maintain a relaxed state of smooth muscle.

The population of animals declined as a result of park management.

As a result of the absence of its major predator, the elk population increased.

This hypothesis would be tested if wolves were reintroduced.

Food shortages are caused by a decline in vegetation productivity due to changes in the weather.

An increase in disease is caused by an increase in herd densities.

The vegetation upon which the deer browse will decrease.

Increasing the number of animals requires more food.

There is a decline in vegetation availability as a result of excessive browsing.

The carrying capacity of the wolf population declined.

The wolf population declined because of disease.

The new cap type will be determined by the mix of the mRNA produced by both nuclei.

The different messages promote reproductive isolation.

There are different messages that keep the species from intermating.

A volatile chemical can attract males as well as females.

The same volatile chemical can be released by a predator to lure males into traps.

These orchids attract male bees by releasing a pheromone that mimics the female bees' pheromones.

Eggs from population 1 have the earliest hatch date.

Females hatch first in populations 1 and 2.

The hatch times for females and males are the same.

The earlier hatch time for females may be beneficial for some populations because it allows more time for females to feed and grow and thus provide more nutrition for producing eggs.

Depending on the location, the populations may experience different environmental factors such as water temperature, food, or shelter.

As a result of cell expansion, the growth occurs.

Material taken from its mother causes the growth.

It doesn't increase at all, it remains the same size.

Redundant copies of DNA give a larger-than-normal cell adequate amounts of genes.

Cells are in a constant state of change so they need flexibility in the processes that maintain their function.

Changes in the cell's internal environment need to be responded to quickly.

Microtubules that don't make contact with a structure must disassemble and try again.

Taxol causes the attachment of chromosomes to be hampered and prevents the completion of mitosis.

Cells that make up the bark are not affected by taxol.

Cells that make up the bark don't exchange substances with living cells.

Taxol is shed along with the bark.

There is a limit on the number of trees that can be created with rotation.

The outgroup for this tree, Species D, is placed furthest away because it is different from the other species.

The most representative evolutionary relationships are provided by the most minute and most subtle changes in the genome.

The most representative evolutionary relationships are provided by mino acid data.

The most representative evolutionary relationships can be found in the data from mino acid.

The most representative evolutionary relationships are provided by the data from the mino acid.

Morphological data provide the most representative evolutionary relationships because they express the final and most complete representation of all the interactions among genes.