Chapter 19: 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.

• The graph plots the size of the stomatal aperture as a function of time of day for a typical plant.
• Normal conditions, very dry soil, and a low CO2 environment are investigated.

• During normal times, stomata are open during the daytime and closed at night.

• When it reacts with sugar, the solution turns from brown to blue, but not when it reacts with starch.
• A bag made from a permeable material is placed into a beaker with a solution of sugar and starch.

• The beaker had Glucose in it.

• The bag is not made of starch.

• The bag is not made of starch.

• The bag is impermeable to IKI.

• The following figure shows aMAP signaling cascade.
• Substances that cause celldivision are called mitogens.

• The figure shows two signaling molecule binding to a transmembrane receptor.

• They provide energy to thereceptors that will be used to turn on the relays and cascades.

• The signal transduction pathway is disabled by them.

• They cause them to start a signal transduction pathway.

• They cause them to open a gated ion channel, which allows the passage of specific molecules.

• The binding of a relay protein is allowed by the activated receptor proteins.
• The relayProtein causes the activation of the GProtein called Ras.

• The molecule binding to Ras is from the relay molecule.

• A GDP molecule is replaced with a GTP molecule.

• Two active proteins are formed by the split of ras.

• The active form of the protein is formed when ras bind to a second relay protein.

• There is a signaling cascade illustrated.

• The signal is more specific because of the cascade.

• The signal is amplified by the cascade.

• The overall reaction is reduced by the cascade.

• The necessary energy is generated by the cascade.

• The final step in the pathway is the transcription of an mRNA.

• The signaling cascade produces something.

• The signaling cascade is needed for transcription.

• The signaling cascade causes the mRNA to be activated.

• There is a product of the signaling cascade.

• Humans have some of the same defects in the Ras protein.
• A form of Ras that does not respond to the relay protein is permanently activated by one of thesemutations.

• The following figure shows the replicative cycle of a retroviruses such as HIV.

• The nucleus and the cytoplasm are where the replication of genes takes place.

• The genes of the viruses are much different than the genes of a human.

• uracil is used by the RNA viruses.

• The proofreading functions of DNA replication are missing.

• There is a table that shows the presence orabsence of X, Y, and Z in species I, II, and III.
• The number 1 indicates that a character is present, and 0 indicates that it is not.

• The species II is related to the species I.

• The species II is related to the species III.

• The outgroup is related to Species II.

• Both species I and III are related by species II.

• Most animal cells are small and the same size.

• Smaller cells avoid dehydration.

• Smaller cells have a lower surface-to-volume ratio.

• Smaller cells have a bigger surface-to-volume ratio.

• Smaller cells fit together better.

• The figure illustrates the evolution ofnine species from a single ancestral species.
• There are different areas of the figure.

• An animal chamber was made of clear plastic.
• The chamber had an opening at one end.
• The graduated burrette passed through the stopper.
• The movement of a solution inside the burrette could be used to measure gas production or consumption.

• Twenty hamsters were put into a chamber.
• Half of the chambers were maintained at 10 and 25 degrees.

• Oxygen consumption was recorded for each hamster every 30 seconds.
• The data is shown in a graph.

• The hamsters' respiration rate increases as time goes on.

• The weight of hamsters increases the temperature.

• The lowest trophic level is used to represent people.

• The lowest trophic level is for plants.

• A single tree in a forest is the lowest trophic level.

• The ecological state of an early successional stage is represented by the pyramid.

• Mitochondria and chloroplasts divide independently of the host cell.

• Mitochondria and chloroplasts have ribosomes that are very similar to those ofbacteria.

• Mitochondria and chloroplasts are not the same as the host cell.

• Mitochondria, chloroplasts, andbacteria have a single, circular chromosome.

• Mitochondria isolated from cells can be used to carry out respiration if an appropriate substrate is provided and a low temperature is maintained.
• This experiment uses isolated mitochondria to investigate respiration.

• As electrons are transferred to FAD for its reduction to FADH2, succinate is oxidation to fumarate.
• In this experiment, a substitute electron acceptor is provided.

• The blue DPIP accepts electrons from the air.

• DPIP turns clear after accepting the electrons.
• A spectrophotometer can be used to determine the degree of color change.
• A higher transmittance percentage shows a solution with less DPIP.

• Threevettes are prepared with a buffer.
• The contents of a fourth cuvette are the same except that the suspension is returned to a low temperature.
• Distilled water was added to bring the contents to the same volume.

• Every 10 minutes, transmittance is measured.
• The following graph shows the results of the experiment.

• The reduction of electron acceptors are only part of the Krebs cycle.

• The substrates for the other steps are consumed before they are added.

• FAD is the only electron acceptor.

• There is only one electron acceptor for the Krebs cycle.

• H2O and CO2 can be exchanged between the inside of a plant leaf and the environment.

• Between 18 and 30 months, the population size changed due to disease.

• Between 18 and 30 months, the population size changed due to competition.

• 1,200 people make up the carrying capacity of the environment.

• The growth rate is zero for 18 and 30 months.

• The figure that follows shows the differences between humans, Chimpanzees, gorillas, and orangutans.
• The banding patterns in boxes 1 and 4 are nearly identical for all four species, but there are differences in boxes 2 and 3.

• The mitochondrion and the vesicle are related.

• The lysosome and the vesicle are related.

• The contents of the vesicle are shuttled to the nucleus.

• The cell's contents are released to be eaten.

• There are more than a dozen species of Darwin's finches in the Galapagos Islands.
• Each species has an adapted way of getting food.

• There are different foods on the mainland than there are on the Galapagos Islands.

• South American predator limited the evolution of the mainland finch.

• There is no reproductive isolation on the mainland.

• The available niches that the Darwin's finches exploited on the Galapagos Islands were already occupied by other species of birds on the mainland.

• There are circles and squares in the picture.
• The male and female are connected by a horizontal line.
• There is a descending vertical line that branches to the offspring.
• A circle or square is indicative of the individual's trait.

• The water temperature is the same.

• The water's pH goes down.

• Oligomycin is an antibiotic that blocks the cristae of mitochondria.

• Water production would increase.

• H+ would increase inside the intermembrane space.

• The matrix would increase in H+.

• Outside mitochondria, H+ would increase.

• During interphase, DNA replication occurs.

• Interphase takes longer than all phases combined.

• Prophase takes 10% of the time it takes to complete a cell cycle.

• During the cell cycle, Metaphase consumes the least amount of time.

• In Pompe disease, there is a normal breakdown in the lysosomes, but there is also a build up of glycogen in the lysosomes.

• As the blood pH increases, it decreases.

• Oxygen saturation of hemoglobin is unaffected by blood pH.

• Hemoglobin can't release oxygen.

• Oxygen is released to the tissues by hemoglobin.

• The saturation of hemoglobin is higher.

• The saturation of hemoglobin is lower.

• In tissues with high respiration, more oxygen is released by hemoglobin.

• In tissues where respiration is high, less oxygen is released by hemoglobin.

• Respiration is higher when the pH is higher than normal.

• Respiration is lower in tissues with a lower pH.

• A piece of potato is dropped into water.

• The pure water has a negative water potential.

• The pure water has a positive water potential.

• The potato has a positive water potential.

• The potato has a negative water potential.

• The water moves from potato to potato.

• The water moves into the potato.

• The absorption and action spectrum for individual pigments are shown in the graph.

• The graph shows changes in species abundance with time from early to late successional stages of a forest community in the eastern United States.

• Fruit flies have X-linked genes for eye color.
• A female with red eyes is crossed with a white-eyed male.

• The results are shown in the table.

• The null hypothesis should be rejected.

• The inheritance of this gene is not linked to X.

• The mixing of genes was not random.

• A researcher is looking into the success of a procedure that transfers plasmids.
• She uses a plasmid that has a resistance to the antibiotic tetracycline.

• The success of the procedure is determined by the preparation of two batches ofbacteria, one with and one without the gene.
• There are two kinds of agar plates, one with and one without.
• The two kinds ofbacteria are transferred to the two kinds of agar plates.

• The number of colonies is shown in the results of the experiment.

• The following answer choices are used for the questions.

• The treatedbacteria are resistant to the antibiotic.

• Somebacteria are resistant to a drug.

• In the absence of tetracycline, both normal and treatedbacteria are able to grow.

• Some treatedbacteria did not absorb the plasmid.

• 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.

• The composition of the DNA was analyzed by a biochemist.
• The sample contained 28% adenine.

• Attach your answer to the nearest whole number.

• A variation of the X-linked allele is lethal.
• The male and female fetuses that inherit the all genes are aborted before birth.
• The answer should be written as a decimal to the nearest hundredth.

• The rate of the reaction is shown in the above graph.
• The answer should be written as a decimal to the nearest tenth.

• The flow of energy is shown in the figure.

• Your answer should be written as a percent to the nearest tenth.

• There are 300 sheep in the population in the equilibrium.

• The carrying capacity of the population is 2,000.

• The first two questions are worth 10 points each.
• The first two questions are worth 4 points each and the remaining four 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.

• Dissolved oxygen is measured from water samples collected at various depths in a freshwater lake.

• The temperatures of samples are the same.

• The graph shows the early growth of a population, labeled I, with two possible outcomes, labeled II and III.

• The structure of a molecule is described by how it interacts with each other and with the environment.

• Ceviche is not cooked with heat.
• It is prepared by soaking raw fish in a strong acid solution made from limes or other juices.

• A town in southern England had a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness that was a form of deafness Several families from the town migrated to North America in the 1600s and settled on Martha's Vineyard.

• The deafness of the descendants of the original families was high compared to other populations in the United States.

• The population on Martha's Vineyard did not interact with the mainland during this time.
• The families lived on the island.
• Many new families arrived on the island in the early 1900s, while some families moved away.
• By 1950, few people with deafness remained.

• Plants and animals can respond to changes in the environment by using a variety of mechanisms.

• There are many groups of organisms.

• A portion of a DNA molecule is shown next to its normal type.
• For reference, the genetic code is shown below.

• The stomata are open during the day and closed at night.
• It makes sense that the stomata must be open to allow CO2 to diffuse into the leaf.
• The graph shows that the stomata are open for a longer period of time when CO2 concentrations are low.
• CO2 isn't needed since photosynthesis doesn't happen at night.

• The solution in the beaker turned blue because the IKI diffused through the bag and into the solution in the beaker.
• The contents of the bag did not turn blue, so you know that the beaker did not get into the bag.
• The bag is not impermeable to starch.

• No conclusion can be drawn about the bag's permeability toglucose since no test was reported.

• The signaling molecule is a small molecule thatbinds to a larger molecule.
• There is a change in the shape of the receptor protein when the signaling molecule is binding to it.
• You don't need to know that the signaling molecule causes twoRTKs to form a pair.
• The next step is autophosphorylation, which allows the subsequentphosphorylation of relay proteins.

• In turn, the phosphorylates relay proteins can be achieved by the activation of this receptor protein.
• The GDP is exchanged for a GTP on a nearby GProtein, which is the RasProtein.

• In a signaling cascade, multiple copies of the second protein kinase can be activated.
• Multiple copies of the third are able to be activated by each copy of the second.
• Each step of the cascade magnifies the response because each function is repeated until it is dephosphorylated.

• A signaling cascade is intended to elicit a cellular response.
• The activation of a second messenger or the inhibition of transcription factors are some of the cellular responses.
• The signaling cascade that is shown in the figure is activated by the nucleus.
• The cellular response is stimulated by the figure since it designates a signaling cascade.
• Cell division can be initiated with the help of transcription factors.
• The products of transcription are the messenger RNAs that are translated.

• The figure shows a signaling cascade of products that stimulates cell division.
• There would be a constant supply of mitogens, which would result in growth and cell division.
• The defining characteristics of cancer are uncontrolled cell growth and division.

• The reverse transcriptase is found in the retroviruses'RNA genome.
• In the figure, reverse transcriptase makes a DNA complement of the viral RNA molecule.
• In the figure, reverse transcriptase makes a second DNA molecule to complete a viral double-stranded DNA molecule.

• The genes of the host are transcribed by the same way that the viral genes are.

• The assembly of an HIV virus requires two copies of the viral RNA with the reverse transcriptase enzyme.
• Some of the genes that are translated to produce the proteins end up in the reticulum, where they are modified into glycoproteins.
• The virus leaves the cell and the glycoproteins are merged with it.

• There are viral genes in the dsDNA.
• When the dsDNA enters the nucleus, it gets transcribed into the host's genes.

• The transcript of a viruses is copied.
• The ability to repair copying errors is not available.
• The genes of the viruses are different than the genes of the humans.

• The outgroup species is the most primitive of the taxa being studied.
• The outgroup doesn't have any of the recorded traits, so it branches from the tree first.
• Both species I and II have trait X and Y, while species III only has trait X.
• The species I and II are descendants of a common ancestor.
• There are two ways to represent this tree.

• The most closely related species are species I and II.

• The forward reaction will exceed the reverse reaction until equilibrium is reached again.
• If there was a lot of reactants, adding more enzyme would only help.

• Growing parts of an organisms is a good place to find it.
• Plants have cells that are 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- The early stages of embryo development are visible in animals.

• It is important that cells have a large surface area relative to their volume in order to maximize their ability to import and export.
• The greater the surface area, the greater the area for carrying out these processes.
• The surface area of a large cell may not be enough to fit the needs of the cell.

• The genome-to-volume ratio is a limitation to cell size.
• The size of the cell's genome and its ability to control the activity of the cell are fixed.
• The genome may not be able to accommodate the needs of a large cell.

• Horizontal changes show the divergence of species.
• The area indicated by period IV is where all of the horizontal changes occur.

• A period of time with little or no evolutionary change is represented by Area III.

• When a single species is introduced into an area with many niches, adaptive radiation can occur.
• As niches are exploited, rapid evolution of many species occurs.
• Evolutionary rates decline once all niches are filled.

• As time passes, the plots show the oxygen consumption.
• The respiratory rates are constant because the slopes are constant.

• Oxygen production is recorded in grams, which means the amount of oxygen produced is divided by the weight of the hamster.
• Oxygen production by hamsters with different sizes is averaged over their weights.
• The reported data does not show the differences in the hamsters' respiratory rates.

• Time and respiratory rate are dependent variables in this experiment.
• The temperature is an independent variable, but it was assigned two values for the purpose of examining how temperature affects respiratory rate over time.
• The temperature must be constant within each chamber.
• Time and respiratory rate are not allowed to change.
• The amount of oxygen in each chamber is allowed to change since the amount of respiratory rate is indirectly measured.
• KOH was added to the animal chambers to absorb any CO2 produced, thus maintaining a constant amount of CO2 in the chambers.

• A single tree can support large numbers of insects, which can in turn support many birds that eat the insects.
• A predator such as hawks may occupy the fourth tier.

• A frame shift is caused by a deletion or addition of a nucleotide in a DNA strand.
• The first, second, and third nucleotides in a codon will become the first, second, and third in the next codon.
• The final sequence of the polypeptide is likely to be affected by such an arrangement.
• A change in the third position of a codon will often code for the same amino acid, so answer choice D may not have any effect.
• Answer choice B will result in a missing acid, while answer choice C will change one acid to another.
• The changes may affect the effectiveness of the polypeptide, but not as much as changing many amino acids.
• The cause of the inherited disorder is the substitution of one of the two chains of hemoglobin with another, which reduces the effectiveness of hemoglobin in carrying oxygen.
• It would be useless if a frameshift in the coding of the genes for hemoglobin happened.

• The host cell does not have the ability to function alone.
• The host cell's genome contains some of the ancestral genes.
• The production of necessary enzymes by the host genome is necessary for respiration and photosynthesis to occur.

• The graph shows that the rate of respiration is higher when the concentration of substrate is higher.
• The percent transmittance of the solution is recorded on the vertical axis of the graph.

• Pyruvate is the beginning of the Krebs cycle.
• If NAD+ and pyruvate are continuously provided, all of the steps of the Krebs cycle can be reduced.
• For the reduction of FAD to FADH2, only succinate can contribute electrons.

• Water diffuses out of the leaf when the stomata are open.
• CO2 diffuses into the leaf.

• The low CO2 concentration in the atmosphere causes a small concentration gradient and a slow rate of CO2 into the leaf.
• CO2 is larger than H2O and diffuses more slowly.

• The introns found in the original DNA segment are missing from the cDNA.

• snRNPs are used to remove the intron nucleotides from the nucleus, which reduces the length of the finalRNA considerably.
• The original DNA segment will be the same regardless of the point deletion or addition of a single nucleotide.
• The primary transcript will have the same number of nucleotides as the original segment.

• There is no data to support the hypotheses that disease, competition, and carrying capacity are to blame for the decline between 18 and 30 months.
• The data shows that the rate of growth is zero at 18 months and 30 months.

• The bands and centromere in box 2 of the orangutan are in reverse order compared to the bands and centromere in box 2 of the human, Chimpanzees, and gorilla.
• The orangutan's chromosomes are inverted compared to the other species.

• Only the top surface of the leaf is exposed to air.
• All leaf cells interface directly or nearly directly with the surrounding water, so Xylem is not necessary.
• The mesophyll has a lot of air spaces that help with floating leaves.

• Vesicles that form by phagocytosis usually merge with a lysosome, and their contents are absorbed by the hydrolytic enzymes inside the lysosome.

• All available niches on the mainland used to be filled by species that were highly specialized for their niches.

• There are more specialized species of woodpeckers on the mainland for the same purpose as there are on the Galapagos Islands.
• evolution of the woodpecker finch would not have been possible if an actual woodpecker had arrived in the Galapagos Islands.

• All daughters and no sons will be X-linked if the father expresses the trait.

• His sons are not allowed to inherit the X-linked trait from him.
• All the remaining inheritance patterns are possible, as shown in the labeled pedigrees below.
• The sex and inheritance patterns of all possible genotypes are shown in each pedigree.
• Punnett squares should be used to confirm each pedigree.

• The water left behind cools because of the energy used to convert the water into gas.

• H+ can't move from the intermembrane space to the matrix if the channels associated with ATP synthase are blocked.
• The expectation is that H+ concentration will increase.

• The longer it takes for a phase to occur, the more often it appears.
• The data shows that metaphase took the least amount of time to occur than the other phases.
• According to the data presented, Interphase took the most time.

• The absence or malfunction of a lysosomalidase causes the abnormal accumulation of substances in lysosomes.
• lysosomal storage diseases include diseases where undigested materials accumulate due to the absence of a correctly functioning lysosomalidase.
• The source of the problems associated with the disease are not caused by the transport of cells.

• When a geographic barrier creates reproductive isolation between two parts of a single population, it's called allopatric speciation.
• Changes in the gene pools of each population can be caused by a variety of factors, such as genetic drift, natural selection, nonrandom mating, or gene flow from a third population.
• Changes in the genes of the populations may lead to behavioral, physiological, or structural differences that prevent individuals of the two populations from reproducing.
• The two populations are two different species.

• The graph doesn't tell you anything about cause and effect.
• It's possible that they increase together.

• Be careful with questions with acid or base information.
• You should note on the test which acid and basic it is.

• More CO2 is produced when respiration is high.

• H+ and HCO3 are formed by 2 with H2O.
• The increase in H+ lowers the pH in the blood.
• The lower plot, representing a more acidic blood, corresponds with tissues where respiration is higher.
• The lower plot shows a lower saturation.

• Pure water has a water potential of zero.
• The water potential is decreased by the presence of solutes in the water.
• The potato has a negative water potential.

• There is a region of higher water potential and a region of lower water potential.
• The water potential of the potato is less than that of the water potential of pure water.
• Water moving in the opposite direction out of the potato cells would cause plasmolysis.
• Water leaves the vacuoles, the vacuoles collapse, and the cell turgor drops when plasmolysis occurs.

• The plot of the action spectrum is the rate at which light is absorbed.
• The highest rate of activity occurs at 450 nm.

• Light that isn't absorbed is reflected.
• The wavelengths of light that are not absorbed in photosynthesis are found in the 525 to 575 nm region.
• This can be seen by looking at the plot for the photosynthetic rate.
• Green is the color we see when we look at a leaf.

• The energy from pyruvate is used to attach the electrons and protons to NAD+.
• If oxygen is present, the result is NADH, a molecule to be used in oxidation, where the reverse of this reaction occurs.
• The reverse of this reaction also occurs.
• The Krebs cycle can't occur in the absence of oxygen.
• When a net of 2 ATPs is generated, the NADH is converted back to its original state, allowing the process to continue.

• The primary structure of this polypeptide is a linear display of its amino acids.
• The three-dimensional shapes that are essential for their proper functioning are shown in the secondary, tertiary, and quaternary structures of polypeptides.

• There is a drawing of something.
• The left side of the head has a phosphate group on it.
• Two long hydrocarbon chains are extended to the right.
• A molecule with a strong affinity for water is called a hydrophilic molecule.
• The molecule resists mixing with water.

• The molecule is an a-glucose polysaccharide.
• There are two substances found in plant and animal cells.
• The b-glucose polysaccharide, shown, is a structural element in the cell walls of plants.

• Pioneer species that reproduce and mature quickly and produce many offspring are selected life history.

• When succession begins, these species are usually the first to invade a habitat, when a previously uninhabited region becomes available, or after a climax community has been destroyed.

• The forest community typical of the eastern United States has a lot of hardwood trees.

• The first requirement for the evolution of two species is reproductive isolation.

• An X-linked gene is only found on the X side of the Y chromosome.

• The expected number of each type of fly is 0.25 x 100, or 25.
• Others may have absorbed the plasmid, but they were not able to express the antibiotic-resistant gene.

• Somebacteria are resistant to a drug.
• Antibiotic resistance may have been acquired by way of a change in genetics.

• Another possibility is that the agar plate they were growing was contaminated with the treated-bacteria batches.

• Adenine and thymine are equal in their numbers.
• Since guanine is in their numbers, cytosine is in their numbers as well.
• If adenine and thymine are equal, then cytosine and guanine must be equal as well.
• There is no uracil in DNA.

• The slope of the plotted curve is the rate of reaction.
• Since the plot is a straight line, the rate of the reaction is constant and the slope at any point along the line will provide the rate.
• The slope is determined by the change in product and the change in time.

• The energy taken in by producers is 20,950 kcal and the energy taken in by carnivores is 328 kcal.
• The producers have obtained a fraction of the animals.
• 0.0157 x 100 is converted to a percent.

• The number of remaining black individuals must be 1 - 0.09.

• The logistic equation is used.

• To get points for your answers, use the standards given below.
• For each item listed below that matches the content and vocabulary of a statement or explanation in your essay, 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.

• Axes are labeled and scaled.

• Both plots are drawn correctly.

• Each plot has a label or legend.

• The title is given to the graph.

• For day and night values, 6.0 and 18.0 are used.

• Respiration occurs in both plants and animals.

• Less oxygen is available for animals at lower depths.

• Less food is available to animals at lower depths because most plants live closer to the surface.

• Resources are readily available in the beginning.

• Competition and disease are minimal in the beginning.

• Resources become more limited when the population approaches its peak size.

• Resources, living conditions, or competition may vary from season to season.

• Possible examples include snowshoe hares, elephants, and humans.

• Resources may have been exhausted, disease may have overwhelmed the population, or environmental conditions may have deteriorated.

• Any population can experience a population crash under certain conditions.

• The life-history strategy experience rapid growth and decline.
• Such species are pioneer or opportunist species that enter a habitat and reproduce quickly, producing many offspring that require little or no parental care, mature quickly, and have short life spans.

• Possible examples include insects, grasses, and weeds.

• Water is a polar molecule.

• The hydrogen bonds with water are formed by the amino acids.

The hydrogen bonds between water andProteins give their secondary structure

• The nonpolar regions of the amino acids will face the nonpolar molecule of the solvent.

• The structure of an enzyme will change when it is immersed in a nonpolar solvent.

• When immersed in a strong acid, the structure of aProtein will be disrupted, similar to the application of heat.

• Denatured fish proteins change from a translucent, gel like texture to a solid, opaque texture.

• If separated, indeterminate and determinate cleavages can produce daughter cells that can individually complete normal development.

• daughter cells that are limited to the development of definite cells that contribute to only a part of a complete embryo are produced by a determinate cleavage.

• The influence of the egg cytoplasm is related to the distribution of the cytoplasms of daughter cells.

• If a frog cell is separated from a developing embryo, the gray crescent cytoplasmic substance must be present.

• The influence of one cell or group of cells over another is called embryonogenesis.

• The signaling molecule provides cell-to-cell communications that direct development and differentiation.

• Apoptosis is the programmed death of cells.

• The space between fingers and toes of humans undergoes programmed cell death to form separate, unattached digits.

• MicroRNAs are small segments ofRNA that bind to aProtein.

• 1 pt is the amount ofProtein complexes that block translation by binding to or breaking down the mRNA.

• The binding of groups to the histones is called methylation.

• It suppresses the unfolding of DNA and prevents its expression.

• The plant hormone auxin stimulates plant growth in stems.

• A plant's response to gravity is described.

• Abscission is the loss of leaves in winter.

• Abscission begins with the withdrawal of vitamins from leaves.

• Dormancy is a plant's response to unfavorable environmental conditions.

• A response to temperature, fire, scarification, or day length can break seed dormancy.

• Photoperiodism is the response to changes in the photoperiod.

• Photoperiodism requires the maintenance of a clock that measures the length of the day or night.

• melatonin is a hormone that is maintained in humans.

• Jet lag is the feeling of fatigue as a result of moving across time zones in humans.

• Jet lag is caused by the time required to realign the biological clock with the new time zone.

• Humans and other animals respond to cold temperatures by shivering.

• It warms the body by the heat produced by contracting muscles.

• Humans and other animals sweat.

• Sweating cools the body by removing heat from it.

• Hibernation is the behavior of entering a period of extended sleep or a reduction in activity.

• Hibernation is a response to cold temperatures.

• Estivation is the behavior of entering a period of extended sleep or a reduction in activity.

• Estivation is a response to hot summer temperatures.

• Estivation can be accomplished by burrowing into mud or tunnels.

• The characteristic describes each group of organisms.

G1/G2 checkpoint, G3P,GABA, gallbladder,gametes, 90, 91,gametic isolation, gap junctions, 36, 73,gas exchange, 182, gastric juice, 195, gastrin,gastrovascular cavity, gated channels

Document Outline

• Title Page

• Copyright Page

• Dedication

• Acknowledgments

• Contents

• Chapter 1: An Overview of the AP Biology Exam How You Should Use This Book What to Bring to the Exam Exam Format Exam Grading What's on the Exam Strategies for Multiple-Choice Questions Strategies for Grid-In Questions Strategies for Free-Response Questions Taking the Practice Exams

• Part I: Subject Area Reviews With Review Questions and Answers

• Chapter 2: Chemistry Review Review Questions Answers and Explanations

• Chapter 3: Cells Review Review Questions Answers and Explanations

• Chapter 4: Cellular Respiration Review Review Questions Answers and Explanations

• Chapter 5: Photosynthesis Review Review Questions Answers and Explanations

• Chapter 6: Cell Communication Review Review Questions Answers and Explanations

• Chapter 7: Cell Division Review Review Questions Answers and Explanations

• Chapter 8: Heredity Review Review Questions Answers and Explanations

• Chapter 9: Molecular Biology Review Review Questions Answers and Explanations

• Chapter 10: Evolution Review Review Questions Answers and Explanations

• Chapter 11: Biological Diversity Review Review Questions Answers and Explanations

• Chapter 12: Plants Review Review Questions Answers and Explanations

• Chapter 13: Animal Form and Function Review Review Questions Answers and Explanations

• Chapter 14: Animal Reproduction and Development Review Review Questions Answers and Explanations

• Chapter 15: Animal Behavior Review Review Questions Answers and Explanations

• Chapter 16: Ecology Review Review Questions Answers and Explanations

• Part II: Laboratory Review

• Chapter 17: Review of Laboratory Investigations Review Review Questions Answers and Explanations

• Part III: Ap Biology Practice Exams Equations and Formulas

• Chapter 18: Practice Exam 1 Section I Section II Answer Key for Practice Exam 1 Scoring Your Practice Exam Answers and Explanations for Practice Exam 1

• Chapter 19: Practice Exam 2 Section I Section II Answer Key for Practice Exam 2 Scoring Your Practice Exam Answers and Explanations for Practice Exam 2

• Index