ACT Science : How to find synthesis of data in biology

Study concepts, example questions & explanations for ACT Science

varsity tutors app store varsity tutors android store varsity tutors ibooks store

Example Questions

Example Question #81 : How To Find Synthesis Of Data In Biology

Predator prey relationships can often influence the survivorship of species. Models are created to better visualize these relationships and their effects on population growths and declines. One way to display this information is to plot the density of the prey population against the number of prey consumed by a certain predator. As seen in Figure 1, three characteristic curves have been observed using this method. Curve Type I curve is the most unrealistic and exists when the number of prey consumed increases in direct proportion to the number of prey, with no limit on consumption. Curve Type II curve is characterized by a trend that shows the number of prey consumed per predator increasing quickly, but as the prey density increases the predators become satiated and the number of prey consumed stabilizes. Curve Type III resembles Type II in that it has an upper limit, but predators consume relatively few prey at lower densities due to various reasons.

Consumption_number

Figure 1

When prey exist in low numbers their evasiveness rises. With fewer number of prey in one area they have increased opportunities to hide from predators; furthermore, predators have a decreased ability to create a search image of their prey. A search image is a mental development that dictates a predator's main source of prey through multiple interactions. Which curve in Figure 1 exhibits these characteristics?

Possible Answers:

Type I

None of the choices exhibit these characteristics.

Type II

Type III

Correct answer:

Type III

Explanation:

A Type III curve exhibits these characteristics. The proportion of prey consumed at low densities is lower than that consumed at medium densities. This indicates that when there is a low number of prey, the have an increased ability to evade predators. This is supported by the passage.

Example Question #82 : How To Find Synthesis Of Data In Biology

     Osmoregulation is a necessary function to maintain homeostasis in organisms. Fish gills form the boundary between fish body fluid and environmental conditions. The regulation of water transport across barrier epithelial membranes in gills is assisted by membrane composition and affected by solute concentration on either side of the membrane. Fish body fluids are isotonic and have compositions of about 300 milliosmoles of solutes per liter. Freshwater environments are hypotonic and typically have compositions of about 1 milliosmole of solutes per liter. Saltwater environments are hypertonic and typically have a composition of about 1,000 milliosmoles of solutes per liter. Based on observed osmoregulation across varying concentration gradients, it is expected that gills exposed to isotonic solutions maintain steady water weight, while those exposed to hypotonic solutions increase in water weight, and those exposed to hypertonic solutions decreased in water weight.

Study 1

     Gills dissected from a fish were exposed to different ambient environments of salinity including an isotonic control solution (301.5 milliosmoles), a hypotonic solution (1.5 milliosmoles), and a hypertonic solution (601.5 milliosmoles). Percent water weight change between treatments was observed and calculated.

According to the passage, what will happen to the gills exposed to the isotonic solution?

Possible Answers:

None of the choices are expectations supported by the passage.

Their water weight will remain constant.

Their water weight will decrease.

Their water weight will increase.

Correct answer:

Their water weight will remain constant.

Explanation:

The solute composition of the gills and their outside environment are nearly identical; therefore, the gills are at equilibrium with their environment. When the gill cells reach equilibrium with the environment, the water entering the cells will equal the water leaving the cells and water weight will remain constant.

Example Question #83 : How To Find Synthesis Of Data In Biology

     Osmoregulation is a necessary function to maintain homeostasis in organisms. Fish gills form the boundary between fish body fluid and environmental conditions. The regulation of water transport across barrier epithelial membranes in gills is assisted by membrane composition and affected by solute concentration on either side of the membrane. Fish body fluids are isotonic and have compositions of about 300 milliosmoles of solutes per liter. Freshwater environments are hypotonic and typically have compositions of about 1 milliosmole of solutes per liter. Saltwater environments are hypertonic and typically have a composition of about 1,000 milliosmoles of solutes per liter. Based on observed osmoregulation across varying concentration gradients, it is expected that gills exposed to isotonic solutions maintain steady water weight, while those exposed to hypotonic solutions increase in water weight, and those exposed to hypertonic solutions decreased in water weight.

Study 1

     Gills dissected from a fish were exposed to different ambient environments of salinity including an isotonic control solution (301.5 milliosmoles), a hypotonic solution (1.5 milliosmoles), and a hypertonic solution (601.5 milliosmoles). Percent water weight change between treatments was observed and calculated.

According to the passage, what will happen to the gills exposed to the hypertonic solution?

Possible Answers:

Their water weight will increase.

Their water weight will remain constant.

Their water weight will decrease.

None of the choices are expectations supported by the passage.

Correct answer:

Their water weight will decrease.

Explanation:

When cells are exposed to a hypertonic solution, they are in an environment that contains more solute than that of the cell. Cells will lose water in order to make the outside salinity closer to that of the cells themselves. In doing so, cells will lose water weight and shrivel.

Example Question #84 : How To Find Synthesis Of Data In Biology

     Osmoregulation is a necessary function to maintain homeostasis in organisms. Fish gills form the boundary between fish body fluid and environmental conditions. The regulation of water transport across barrier epithelial membranes in gills is assisted by membrane composition and affected by solute concentration on either side of the membrane. Fish body fluids are isotonic and have compositions of about 300 milliosmoles of solutes per liter. Freshwater environments are hypotonic and typically have compositions of about 1 milliosmole of solutes per liter. Saltwater environments are hypertonic and typically have a composition of about 1,000 milliosmoles of solutes per liter. Based on observed osmoregulation across varying concentration gradients, it is expected that gills exposed to isotonic solutions maintain steady water weight, while those exposed to hypotonic solutions increase in water weight, and those exposed to hypertonic solutions decreased in water weight.

Study 1

     Gills dissected from a fish were exposed to different ambient environments of salinity including an isotonic control solution (301.5 milliosmoles), a hypotonic solution (1.5 milliosmoles), and a hypertonic solution (601.5 milliosmoles). Percent water weight change between treatments was observed and calculated.

According to the passage, what will happen to the gills exposed to the hypotonic soltion?

Possible Answers:

Their water weight will increase.

None of the choices are expectations supported by the passage.

Their water weight will remain constant.

Their water weight will decrease.

Correct answer:

Their water weight will increase.

Explanation:

When cells are exposed to a hypotonic solution, they are in an environment that contains less solute than that of the cell. Cells will take in water in order to maintain an equal amount of solute between the cells and the outside environment. Cells may even burst by taking in too much water.

Example Question #85 : How To Find Synthesis Of Data In Biology

    Species competition is driven by a variety of factors. Resources such as water, food, sunlight, and suitable habitat are among the top contributors that influence interspecific and intraspecific competition. Interspecific is competition between different species and intraspecific competition is between members of the same species.

     One interesting example of interspecific completion is that of two barnacle species that inhabit intertidal zones. Balanus balanoides inhabits the lower intertidal zone and Chthamalus stellatus inhabits the lower intertidal zone. A researcher attempts to study this phenomenon.

     The researcher removes the Balanus species from the lower intertidal zone and observes that theChthamalus species expands its range to inhabit the lower intertidal zone and the upper intertidal zone. The researcher then removes the Chthamalus species from the upper tidal zone of a different area and observes that the Balanus species does not extend its range. The researcher concludes that competition has allowed each species to exist simultaneously by forming specialized niches that promotes survivorship for each species.

An oil spill releases petroleum toxins to the entire intertidal zone under investigation. Cleaning procedures displace the barnacles. They are later reintroduced into the environment. Which of the following statements best describes the pattern of species dispersion?

Possible Answers:

Chthamalus will inhabit the upper zone and Balanus will inhabit the lower zone.

Chthamalus will inhabit the lower zone and Balanus will inhabit the upper zone.

Chthamalus will inhabit both the upper and lower zones.

Chthamalus and Balanus will be interdispersed between the two zones.

Correct answer:

Chthamalus will inhabit the upper zone and Balanus will inhabit the lower zone.

Explanation:

Chthamalus will inhabit the upper zone and Balanus will inhabit the lower zone.

This is the correct answer because it follows the pattern outlined in the passage. The species are reintroduced and there is no information that describes a difference in habitat or predation; therefore, it is safe to assume that the two species would compete as they did initially and separate themselves into their respective niches.

 

Example Question #86 : How To Find Synthesis Of Data In Biology

Carbonic anhydrase is an enzyme that is used by the human body to interconvert carbon dioxide (a gaseous compound) and bicarbonate (a compound composed of hydrogen, carbon, and oxygen), using water as a reactant. The human body needs to convert carbon dioxide to bicarbonate in the tissues to transport it in the blood to the lungs, as carbon dioxide is relatively insoluble. In the lungs, bicarbonate is converted back to carbon dioxide to be exhaled. In humans, there are two different carbonic anhydrase isomers, one that works in the lungs (isomer A) and one that works in the tissues (isomer B). Mutations in the chromosomes can prevent carbonic anhydrase from converting carbon dioxide to bicarbonate. In the following four experiments, the chromosomes of none, one, or both of the isomers were mutated per experiment.

 

Experiment 1

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Bicarbonate was measured in the tissues but carbon dioxide was not measured in the lungs.

 

Experiment 2

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be abnormally high in the tissues and no bicarbonate was measured in the lungs.

 

Experiment 3

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be slightly low in the tissues and bicarbonate levels were abnormally high in the lungs.

 

Experiment 4

Radiation was not used, A probe was placed into the tissues and lungs of a model animal and found normal carbon dioxide and bicarbonate levels.

A slightly low carbon dioxide level in the tissues in Experiment 3 most likely indicates that __________.

Possible Answers:

isomer B is nonfunctional

isomer B is overactive

isomer A is nonfunctional

isomer A is overactive

Correct answer:

isomer B is overactive

Explanation:

Low carbon dioxide levels in the tissues paired with high bicarbonate is high in the lungs indicates that a greater-than-normal amount of carbon dioxide is being converted to bicarbonate in the tissues and transported to the lungs. Isomer B, the isomer that works in the tissues, must be overactive.

Example Question #87 : How To Find Synthesis Of Data In Biology

Carbonic anhydrase is an enzyme that is used by the human body to interconvert carbon dioxide (a gaseous compound) and bicarbonate (a compound composed of hydrogen, carbon, and oxygen), using water as a reactant. The human body needs to convert carbon dioxide to bicarbonate in the tissues to transport it in the blood to the lungs, as carbon dioxide is relatively insoluble. In the lungs, bicarbonate is converted back to carbon dioxide to be exhaled. In humans, there are two different carbonic anhydrase isomers, one that works in the lungs (isomer A) and one that works in the tissues (isomer B). Mutations in the chromosomes can prevent carbonic anhydrase from converting carbon dioxide to bicarbonate. In the following four experiments, the chromosomes of none, one, or both of the isomers were mutated per experiment.

 

Experiment 1

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Bicarbonate was measured in the tissues but carbon dioxide was not measured in the lungs.

 

Experiment 2

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be abnormally high in the tissues and no bicarbonate was measured in the lungs.

 

Experiment 3

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be slightly low in the tissues and bicarbonate levels were abnormally high in the lungs.

 

Experiment 4

Radiation was not used, A probe was placed into the tissues and lungs of a model animal and found normal carbon dioxide and bicarbonate levels.

A mutation that deactivates isomer B would most likely lead to the accumulation of what product and in what location?

Possible Answers:

Bicarbonate in the tissues

Bicarbonate in the lungs

Carbon dioxide in the tissues

Carbon dioxide in the lungs

Correct answer:

Carbon dioxide in the tissues

Explanation:

The passage tells us that isomer B is responsible for working in the tissues, meaning that it converts carbon dioxide to bicarbonate. Deactivating isomer B would lead to accumulation of carbon dioxide in the tissues, because it would not be converted to bicarbonate and carried to the lungs.

Example Question #81 : How To Find Synthesis Of Data In Biology

Carbonic anhydrase is an enzyme that is used by the human body to interconvert carbon dioxide (a gaseous compound) and bicarbonate (a compound composed of hydrogen, carbon, and oxygen), using water as a reactant. The human body needs to convert carbon dioxide to bicarbonate in the tissues to transport it in the blood to the lungs, as carbon dioxide is relatively insoluble. In the lungs, bicarbonate is converted back to carbon dioxide to be exhaled. In humans, there are two different carbonic anhydrase isomers, one that works in the lungs (isomer A) and one that works in the tissues (isomer B). Mutations in the chromosomes can prevent carbonic anhydrase from converting carbon dioxide to bicarbonate. In the following four experiments, the chromosomes of none, one, or both of the isomers were mutated per experiment.

 

Experiment 1

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Bicarbonate was measured in the tissues but carbon dioxide was not measured in the lungs.

 

Experiment 2

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be abnormally high in the tissues and no bicarbonate was measured in the lungs.

 

Experiment 3

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be slightly low in the tissues and bicarbonate levels were abnormally high in the lungs.

 

Experiment 4

Radiation was not used, A probe was placed into the tissues and lungs of a model animal and found normal carbon dioxide and bicarbonate levels.

If both isomers A and B were mutated, how would the levels of carbon dioxide in the tissues and bicarbonate in the lungs change over time?

Possible Answers:

Carbon dioxide in the tissues remains the same and bicarbonate in the lungs decreases

Carbon dioxide in the tissues remains the same and bicarbonate in the lungs increases

Carbon dioxide in the tissues increase and bicarbonate in the lungs remains the same

Carbon dioxide in the tissues decreases and bicarbonate in the lungs remains the same

Correct answer:

Carbon dioxide in the tissues increase and bicarbonate in the lungs remains the same

Explanation:

If both isomers A and B were both mutated, no carbon dioxide would be converted to bicarbonate in the tissues, resulting in an accumulation of carbon dioxide in the tissues. Bicarbonate would also not be converted to carbon dioxide in the lungs, but since there is no bicarbonate produced by the tissues and transported to the lungs in the first place, the bicarbonate levels in the lungs would remain the same.

Example Question #89 : How To Find Synthesis Of Data In Biology

Carbonic anhydrase is an enzyme that is used by the human body to interconvert carbon dioxide (a gaseous compound) and bicarbonate (a compound composed of hydrogen, carbon, and oxygen), using water as a reactant. The human body needs to convert carbon dioxide to bicarbonate in the tissues to transport it in the blood to the lungs, as carbon dioxide is relatively insoluble. In the lungs, bicarbonate is converted back to carbon dioxide to be exhaled. In humans, there are two different carbonic anhydrase isomers, one that works in the lungs (isomer A) and one that works in the tissues (isomer B). Mutations in the chromosomes can prevent carbonic anhydrase from converting carbon dioxide to bicarbonate. In the following four experiments, the chromosomes of none, one, or both of the isomers were mutated per experiment.

 

Experiment 1

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Bicarbonate was measured in the tissues but carbon dioxide was not measured in the lungs.

 

Experiment 2

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be abnormally high in the tissues and no bicarbonate was measured in the lungs.

 

Experiment 3

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be slightly low in the tissues and bicarbonate levels were abnormally high in the lungs.

 

Experiment 4

Radiation was not used, A probe was placed into the tissues and lungs of a model animal and found normal carbon dioxide and bicarbonate levels.

According to the passage, the solubility of carbon dioxide in blood is likely __________.

Possible Answers:

low

high

moderate

There is not enough information given in the passage to determine the solubility of carbon dioxide in blood.

Correct answer:

low

Explanation:

The passage states that carbon dioxide must be converted to bicarbonate in order for it to be transported to the lungs, as carbon dioxide is relatively insoluble in blood.

Example Question #371 : Act Science

Carbonic anhydrase is an enzyme that is used by the human body to interconvert carbon dioxide (a gaseous compound) and bicarbonate (a compound composed of hydrogen, carbon, and oxygen), using water as a reactant. The human body needs to convert carbon dioxide to bicarbonate in the tissues to transport it in the blood to the lungs, as carbon dioxide is relatively insoluble. In the lungs, bicarbonate is converted back to carbon dioxide to be exhaled. In humans, there are two different carbonic anhydrase isomers, one that works in the lungs (isomer A) and one that works in the tissues (isomer B). Mutations in the chromosomes can prevent carbonic anhydrase from converting carbon dioxide to bicarbonate. In the following four experiments, the chromosomes of none, one, or both of the isomers were mutated per experiment.

 

Experiment 1

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Bicarbonate was measured in the tissues but carbon dioxide was not measured in the lungs.

 

Experiment 2

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be abnormally high in the tissues and no bicarbonate was measured in the lungs.

 

Experiment 3

Radiation was used to mutate one of the chromosomes of one or more isomers of carbonic anhydrase. A probe was placed into the tissues and lungs of a model animal. Carbon dioxide levels were found to be slightly low in the tissues and bicarbonate levels were abnormally high in the lungs.

 

Experiment 4

Radiation was not used, A probe was placed into the tissues and lungs of a model animal and found normal carbon dioxide and bicarbonate levels.

According to the passage, the solubility of bicarbonate in blood is likely __________ .

Possible Answers:

There is not enough information given in the passage to determine the solubility of bicarbonate in blood.

low than that of carbon dioxide

higher than that of carbon dioxide

the same as that of carbon dioxide

Correct answer:

higher than that of carbon dioxide

Explanation:

The passage states that carbon dioxide must be converted to bicarbonate in order for it to be transported to the lungs, as carbon dioxide is relatively insoluble in blood. Needing to convert carbon dioxide to bicarbonate to be transported in blood must mean that bicarbonate is more soluble in blood than carbon dioxide.

Learning Tools by Varsity Tutors