Chapter 17 -- Part 4: Review of Laboratory Investigations
The rate of a reaction is measured in this investigation.
The breakdown of hydrogen peroxide into H2O and O2 is investigated.
The H2O2 is broken down by two kinds of enzymes.
The peroxidase oxidizes H2O2 using an organic reducing agent.
The first two equations compare these reactions.
The organic substance is represented by the letter A.
One option is to use peroxidase with guaiacol.
H2O2 is broken down to H2O as four guaiacol are converted to a different substance.
The amount of product produced can be measured by the color intensity of the product.
The rate of a reaction can be determined by measuring the amount of one product or the amount of the other.
The rate of a reaction is the slope of the linear part of the graph that shows how much product is accumulated as time goes on.
Reaction rate can be affected by a number of factors.
A baseline can serve as a comparison to the reactions you investigate in parts II and III.
The H2O2 reaction can be measured by the disappearance of H2O2 or theAccumulation of H2O or O2.
If a reducing agent is used, the product can be measured.
In this investigation of H2O2 decomposition, you document the progress of the reaction by measuring the amount of O2 or tetraguaiacol.
The intensity of a color is compared to a color palette.
A series of test tubes with increasing color intensity are correlated with the concentration of the product.
To make a series of product dilutions, run the H2O2 decomposition reaction with peroxidase and guaiacol.
The color of the product at a concentration proportional to the reactant can be used to evaluate the progress of the reaction.
Measure absorbance to determine color intensity.
The amount of product produced can be determined by measuring the absorbance of the solution.
The more product produced, the darker the solution.
The oxygen product can be measured directly as it is being produced.
An organic reducing agent is not used.
To determine the rate of a reaction in the presence of an enzyme, follow these steps.
Take pictures for 1 minute at a time.
The amount of product formed at the end of each time interval can be determined by comparing pictures with the color palette.
Line up the best fit through the plotted points.
The rate of the reaction is determined by the slope of the line.
You can investigate the effect of different treatments on the rate of H2O2 decomposition if you have a baseline for comparison.
The effect of pH is investigated in this part of the investigation.
You prepare a series of test tubes, substituting various pH solutions for the H2O component in the baseline solution.
For each pH investigated, record the amount of accumulated product and graph that data to determine the rate of each reaction.
The rates are plotted on a graph as a function of pH.
The rate of peroxidase activity is influenced by a variable of your choice in this part of the investigation.
Consider the concentration of the two substances.
You should only test one variable at a time.
To maintain a constant volume, be sure to reduce the amount of H2O that you add by the same amount.
A review of the material presented in this chapter is provided by the questions that follow.
They can be used to evaluate how well you understand the concepts.
AP multiple-choice questions are often more general, covering a broad range of concepts.
The two practice exams in this book are for these types of questions.
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.
A bag is filled with a solution.
The bag is immersed in water with a 1% IKI solution.
IKI is a yellow-brown color but becomes blue in the presence of starch.
The bag is impermeable to starch.
The water potential inside the bag is negative when the bag is placed in the beaker.
The solution in the beaker is yellow-brown when the bag is first placed in it.
The solution in the bag becomes blue after 15 minutes.
The mass of the bag decreased after 15 minutes.
There are five potato cores that are placed in five beakers.
The water potential of the cells in the potato core is more negative than that of the solution in the beaker.
The pressure potential of cells in the potato core has increased after 24 hours in the beaker.
The solutions have a negative water potential.
The water potential of the potato core is zero if the net movement of water into it is zero.
Use the following graph to answer questions 3-6.
The rate will increase.
At a slower rate, the reaction will occur.
The reaction will not occur at a rate that is significantly different from the reaction rate with no enzyme at all.
The rate will not change.
There are four differentarrangements of ascospores, all resulting from a cross between a strainhomozygous for the wild type of spores and a strainhomozygous for the Mutant color.
An experiment that measures the rate of photosynthesis is referred to in question 8.
The lower energy state of DPIP is blue and the higher energy state is clear.
Light is used to illuminate samples that contain chloroplasts, DPIP, and anappropriate buffer.
The results of the experiment can be seen here.
During photosynthesis, all of the light's wavelength is absorbed and utilized.
The most efficient wavelength is blue.
During photosynthesis, there are at least two wavelengths of light.
A blue solution is indicative of maximum activity.
An experiment designed to measure the respiratoryrate of crickets is the subject of questions 9-11.
Three respirometers are prepared.
The respirometers are immersed in the same water bath.
One end of the pipette is connected to the jar and the other end is open.
There is a semi-log graph of results from the standard restriction enzyme used in a gel electrophoresis procedure.
The endonuclease migrates 6 cm.
The following diagram shows the bandsproduced by an electrophoresis procedure using DNA from four human individuals.
The same restriction enzyme is used for each sample.
Individual 1 could be the offspring of 3 and 4.
Individual 1 is an offspring of 2 and 3.
The offspring of individuals 1 and 3 could be Individual 2.
Individual 3 is an offspring of 2 and 4.
The population consists of 20 individuals and all of them have at least one dominant trait.
There is a genetic drift.
The allele is deleterious.
All individuals with the same genes emigrate.
The rate of water loss for three plants is shown in the following graph.
One plant is exposed to normal conditions, another is exposed to high temperature and a third is exposed to high humidity.
The results of the experiment can be seen in the histogram.
The net primary productivity for a forest was measured.
The community's respiratory rate was determined to be 1,000 carbon fixed/L/day.
The questions that follow are typical of an entire AP exam question or just that part of a question that is related to this chapter.
There are two types of questions on the AP exam.
It takes about 20 minutes to answer a long free-response question.
Sometimes they offer you a choice of questions to answer.
6 minutes is the time it takes to answer a short free-response question.
diagrams can be used to supplement your answers, but a diagram alone is not adequate.
The transpiration rate is influenced by many factors.
Explain how the factors operate.
One of the two alleles in the model became zero after 10 generations of reproduction.
Provide an explanation in two or three sentences for why an allele would disappear after a short number of reproductive cycles.
DNA samples are placed on a gel and an electric voltage is applied.
The purpose of applying an electricvoltage is explained in two or three sentences.
An experiment was done to measure the effect of light.
Treatment I: healthy chloroplasts exposed to light Treatment II: healthy chloroplasts exposed to light Treatment III: healthy chloroplasts in darkness.
The DPIP turns from blue to clear when it is reduced.
The degree to which the DPIP was reduced in each treatment was determined using a spectrophotometer.
The amount of light that is transmitted through a sample is measured by a spectrophotometer.
The experiment was to measure light at a wavelength of 605 nm.
The data was collected for the healthy chloroplasts.
The healthy chloroplasts are exposed to light.
On the same set of axes, draw and label two additional lines that represent your prediction of the data obtained for treatments II and III.
Justify your predicted data for the second and third treatments.
The mass of the bag will increase after 15 minutes when the water moves from the beaker to the bag.
When the bag is placed in the beaker it has a negative water potential and a zero pressure potential.
When the net movement into and out of the potato core is zero, the water potentials inside and outside the potato core are the same, and there is no change in the mass of the potato core.
There is a change in mass on the graph.
When the net movement of water into a potato core is zero, the water potentials inside and outside the potato core are not the same.
The water potential of the potato cells must have been smaller than the water potential of the sucrose solution because the potato core gained weight.
After 24 hours, water that enters the potato core in the 0.2 M sucrose solution causes the potato cells to expand and gain weight.
Pressure potential increases when the cell walls exert restraining pressure on the cell contents.
The slope of the plotted curve at the beginning of the reaction is the initial rate of reaction.
The initial rate will be provided by the slope at any point along the curve from 0 to 2 seconds.
The slope is determined by the change in product and the change in time.
It will be damaged if it is heated to 100degC.
The reaction is not likely to happen now that the enzyme is no longer in use.
If activity continues, the reaction rate would be the same as if there was no activity.
Increasing the concentration of the product will slow the forward direction of the reaction and accelerate it.
The forward rate of the reaction will be increased by an increase in the substrate concentration.
The reaction rate will not be slowed if the concentration of the enzyme is increased.
If the nature of the enzyme is known, the rate of reaction can be increased or decreased.
There are two groups of the same color ascospores in the ascus.
If no crossing takes place, the order of ascospores is the same for each of the four chromatids.
If crossing over occurs, there will be a swap of ascospores between one pair and another pair that will look similar to the images in the other answer choices.
The graph shows that 450 and 600 nm each produce a high rate, and together produce the highest rate, which indicates that the process depends on these two wavelengths.
The light of 450 nm does induce some activity, but the light of 600 nm causes more activity.
A blue solution shows little or no photosynthesis because the DPIP has not been reduced.
An increase in CO2 gas can't be detected because it immediately reacts with KOH to produce solid K2CO3.
The amount of O2 gas detected by the respirometer will be decreased by a live cricket.
The water pressure on the pipette inlets is affected by temperature and atmospheric pressure, which causes the respirometer to register a change in volume.
Since there is no insect in respirometer 3, the purpose of this respirometer is to control all the variables that might influence the volume changes in respirometers 1 and 2, other than O2 consumption by insects.
The stage of rapid population growth referred to as competence is whenbacteria are most receptive to absorbing foreign DNA.
The standard curve has a vertical line at 6 cm and a horizontal line at 600 base pairs.
The increase of 100 base pairs is represented by the horizontal lines between 100 and 1,000.
This is an example of gel electrophoresis.
The horizontal bands represent fragments of DNA that have migrated across a gel with lighter fragments moving farthest.
The columns represent the fragments from a different person.
Fragments that have migrated the same distance are made up of the same DNA.
The offspring must match the fragments of one or the other parent in order to inherit their genes.
It is possible that individual 3 had his or her genes from 2 and 4.
Many different restriction enzymes are used so that many different DNA fragments can be compared.
25% of the offspring should be homozygous if only Heterozygotes mate.
This answer can't explain why there are no homozygous individuals in the population.
Genetics may be responsible for the small population.
Natural selection may explain why the population is not made up of individuals with the same genetic makeup.
The dominant allele is masked by the recessive one in the population.
If individuals leave the population for another location, there can be a explanation for the absence of homozygous individuals.
Water potential is highest in the soil and lowest in the air.
Water moves from the soil to the roots and through the plant to the lowest water potential area.
When water entering roots can't adequately replace the water loss by transpiration, the stomata is close to prevent it.
The tallest bar shows the preference of the greatest number of brine shrimp.
Brine shrimp in the wild are rarely found in water with low salt concentrations when laboratory experiments show that is where they prefer to be.
The answer is that under natural conditions, the shrimp are eaten or the shrimp avoid these waters because of the predator.
Brine shrimp can survive in water with higher salt concentrations, even if they prefer low-salt water.
The gross primary productivity is the sum of the net primary productivity and the respiratory rate.
The transpiration rate is influenced by temperature.
The hotter the environment, the faster the water molecule changes from a liquid state to a gas state.
The rate of movement of water from inside to the outside of the leaf is increased by the wind.
The surface area from which water flows is a factor.
The model's population size was probably less than 50 individuals.
Changes in allele frequencies can be influenced by genetic drift when populations are small.
There are negatively charged groups in the nucleotides.
The positively charged end of the voltage field attracts DNA fragments.
Because larger, heavier fragments move more slowly than smaller, lighter fragments, the various fragments in the sample are separated.
The chloroplasts were boiled.
The chloroplasts are damaged by boiling.
There are denatured catalysts in the membranes.
The secondary and tertiary structures of these molecules can no longer function.
In treatment II, photosynthesis is greatly reduced, occurring only in the few membranes that are still intact with functioning enzymes and pigments.
The transmittance of a treatment II sample is low, and only a small amount of DPIP is reduced.
In treatment III, there is no reduction in DPIP because photosynthesis can't happen without light.
If photosynthesis can't happen, the electrons in the pigment systems are not reduced.
The transmittance is low and the DPIP is low.
When healthy chloroplasts are exposed to light, the electrons in photosystem II are reduced.
Two electrons are passed to a primary electron acceptor.
The energy from the electrons is used to generate 1.5 ATP molecule on average.