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ACT Science Review Questions and Answers 2022

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The correct answer is Uracil. Uracil is one of the four nitrogenous bases found in RNA. The other three nitrogenous bases in RNA are also found in DNA, but uracil is not found in DNA. Nitrogenous base that occurs in RNA but not in DNA (A) Deoxyribose (B) Ribose (C) Uracil (D) Cytosine (E) Thymine The correct answer is cytosine. The only base that is always found in equal quantities with guanine is its complement, cytosine (recall that the nitrogenous bases in DNA pair, A with T and C with G.) A nitrogenous base that occurs in DNA in equal quantities with guanine (A) Deoxyribose (B) Ribose (C) Uracil (D) Cytosine (E) Thymine 00:48 01:13 The correct answer is succession. Within a given area, the vegetation changes over time, tending towards the climax type of vegetation for that particular area. (Repeated disturbances, of course, interrupt this process, and may in effect reset the clock, so that succession begins anew.) The change in plant types inhabiting an area over time, resulting in a climax community (A) Succession (B) Dispersion (C) Fertilization (D) Speciation (E) Mutation The correct answer is speciation. Recall that the functional definition of a species is a group of organisms that can interbreed with each other but do not interbreed with others. The inability to interbreed after long separation is a sign that speciation has occurred; the two populations now represent two different species. The inability of two populations to interbreed after being separated by a barrier for a long period of time (A) Succession (B) Dispersion (C) Fertilization (D) Speciation (E) Mutation The correct answer is succession. This is a specific example of succession, the gradual filling in of a wetland to become a meadow or a forest. Plants growing in and around a pond eventually filling in the pond and changing it to a terrestrial habitat (A) Succession (B) Dispersion (C) Fertilization (D) Speciation (E) Mutation The correct answer is Producers (e.g., grasses). Organisms that comprise the greatest biomass in a terrestrial food chain are the producers. Net productivity (photosynthesis minus respiration) creates biomass. Organisms that comprise the greatest mass of living substance (biomass) in a terrestrial food chain (A) Decomposers (e.g., bacteria) (B) Producers (e.g., grasses) (C) Primary consumers (e.g., mice) (D) Secondary consumers (e.g., snakes) (E) Tertiary consumers (e.g., hawks) The correct answer is Decomposers (e.g., bacteria). Of the list of organisms given, decomposer bacteria are the only ones that can covert nitrogen-containing organic molecules into ammonia. Ammonia, which is a form of nitrogen-containing molecule, is subsequently converted to nitrate that plants can use to synthesize new organic compounds with nitrifying bacteria. Organisms that convert nitrogen-containing organic molecules into nitrates (A) Decomposers (e.g., bacteria) (B) Producers (e.g., grasses) (C) Primary consumers (e.g., mice) (D) Secondary consumers (e.g., snakes) (E) Tertiary consumers (e.g., hawks) The correct answer is 7.4. The question gives you the information that blood is slightly basic. 10.6 is very basic (recall that pH is a logarithmic scale), 7.0 is neutral, 6.4 is slightly acid, and of course 4.6 is the most acid of the five choices. The pH of human blood is slightly basic. Which of the following is most likely to be the pH of human blood? (A) (B) (C) (D) (E) The correct answer is a stable dominance hierarchy, (A). Ritualized combat, such as that found in many wild animal populations, leads to a stable hierarchy; generally one individual is acknowledged to be dominant, and true combat is avoided. This benefits both the winner and the loser, as neither faces death or serious injury in a mock battle. In animals, ritualized contests with little risk of serious injury or death to participants within the species lead to (A) a stable dominance hierarchy (B) biological altruism (C) adaptive radiation (D) instinctive behavior (E) a broader habitat The correct answer is natural selection, (B). Favorable traits increase in a population over time because the organisms bearing those favorable traits have some advantage over other members of the population, and thus leave more surviving offspring than do individuals without the trait. This is sometimes called "survival of the fittest" but it is important to realize that survival alone is only part of the equation; the gene frequency is most influenced by the differential reproductive success of these survivors (that is, by the fact that individuals with the trait leave more offspring than individuals without the favorable trait). Lamarck, (while a very important man in his time, and one who struggled to come up with an explanation that made sense) has been shown to have been wrong about the inheritance of acquired characteristics, so choice (A) is clearly wrong. Adaptive radiation involves the spread of organisms into open niches and subsequent speciation, so (C) is not a correct choice. Genetic recombination is part of sexual reproduction, whether or not gene frequencies are changing, so (D) is not a correct choice. And finally, segregation of alleles does not explain a change in gene frequency over time. Choice (E) is not correct. Which of the following correctly explains how a favorable genetic trait can increase in frequency in a population? (A) Lamarck's principle (B) Natural selection (C) Adaptive radiation (D) Genetic recombination (E) Segregation of alleles The correct answer is decomposition, (E). Decomposition breaks down organic material such as sewage. Succession of vegetation types (A) does not explain this process. Eutrophication (B) might well result from the dumping of sewage into a body of water, but does not explain the observation that the stream is free from pollutants a few miles downstream. Evaporation (C ) would result in the loss of water from the stream but not the loss of pollutants. Photosynthesis (D) takes place in aquatic plants and algae, sometimes using the products of decomposition, but it does not result in the breakdown of organic material. A stream is free of pollutants within a few miles downstream of a point at which a small amount of sewage is being dumped into it. This is most likely the result of (A) succession (B) eutrophication (C) evaporation (D) photosynthesis (E) decomposition The correct answer is that it lowers the pH in ponds and limits the survival of many organisms (A). Choice (B) correctly identifies the lower pH, but incorrectly couples it with water temperature. Choices (C) and (D) incorrectly state that acid rain will raise the pH. Choice (E) suggests that somehow carbon dioxide will removed from the air when acid rain is formed. Carbon dioxide within bodies of water is at equilibrium with carbon dioxide in the air. As rain mixes with chemicals such as sulfur dioxide in the air, acid rain is produced. This may result in (A) lowering the pH in ponds, thus limiting the survival of many organisms (B) lowering the pH in ponds, thus affecting water temperature (C) raising the pH in ponds, thus encouraging the growth of organisms (D) raising the pH in ponds, thus limiting animal development (E) depleting atmospheric carbon dioxide available for photosynthesis

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ACT Science Review Questions
A - Answer A certain type of bottom-dwelling microorganism thrives under the following
environmental conditions: low concentrations of Fe2+, high concentrations of O2, and a
neutral pH. Based on the table, at which of the following sediment depths would one
most likely find this microorganism?
A. 0 cm
B. 5 cm
C. 10 cm
D. 15 cm

B - Answer If the trends indicated in the table were to continue, one would predict the
pH of the sediments at a depth of 35 cm to be:
A. 1.5.
B. 3.5.
C. 4.5.
D.6.0.

C - Answer The graph and chart above best represents the relationship between
concentration and sediment depth for which of the following ions and dissolved gases?
A. Ferrous iron (Fe2+)
B. Oxygen (O2)
C. Carbon dioxide (CO2)
D. Sulfate (SO42-)

Correct answer is C. The other options describe processes that would not enable you to
determine whether or not the solution was supersaturated: you would observe the same
thing either way. Supersaturated solutions are not stable. If you add more solute (in this
case the crystal) to a supersaturated solution, the extra solute will precipitate out of the
solution, indicating that the solution was indeed supersaturated. - Answer To determine
whether a water solution of Na2S2O3 at room temperature is supersaturated, one can
(A) heat the solution to its boiling point
(B) add water to the solution
(C) add a crystal of Na2S2O3 to the solution
(D) acidify the solution
(E) cool the solution to its freezing point

D - Answer According to the information provided in the table, the concentration of
which of the following ions and dissolved gases is constant for sediment depths of 10
cm or more?

a. Sulfide (S2-)
b. Carbon dioxide (CO2)
c. Ferric iron (Fe3+)
d. Oxygen (O2)

,D - Answer According to the information provided in the table, the concentration of
which of the following ions and dissolved gases is constant for sediment depths of 10
cm or more?

a. Sulfide (S2-)
b. Carbon dioxide (CO2)
c. Ferric iron (Fe3+)
d. Oxygen (O2)

F is correct
By inspection, one can see that, at a depth of 0 cm, the concentration of Fe2+ is lowest
(0.5 ppm), the concentration of O2 is highest (2.0), and pH is neutral (7.0). - Answer
Passage I
The following table represents the concentration of ions and dissolved gases in the
sediment at the bottom of an ocean. A depth of 0 centimeters (cm) represents the top of
the sediment. The concentrations are expressed in parts per million (ppm). The acidity
of a solution is represented on a scale known as pH. A pH of 1 is very acidic, a pH of 7
is neutral, and a pH of 14 is very basic.A certain type of bottom-dwelling microorganism
thrives under the following environmental conditions: low concentrations of Fe2+, high
concentrations of O2, and a neutral pH. Based on the table, at which of the following
sediment depths would one most likely find this microorganism?

F. 0 cm
G. 5 cm
H. 10 cm
J. 15 cm

The best answer is H. Figure 3 indicates the number of trees that are present as a
function of the age of a stand for both pine trees and oak-hickory trees.

According to Figure 3, the oak-hickory trees reach a density of 13,000 trees per unit
area after approximately 150 years and a density of 15,000 trees per unit area after
approximately 160 years. Thus, H is the correct answer. - Answer Abandoned cornfields
have been the sites of investigations concerning ecological succession, the orderly
progression of changes in the plant and/or animal life of an area over time (see Figure
1).
(Note: The plants are ordered according to their appearance during ecological
succession.)

During the early stages of succession, the principal community (living unit) that
dominates is the pioneer community. Pioneer plants are depicted in Figure 2.

The final stage of ecological succession is characterized by the presence of the climax
community, the oak-hickory forest. Figure 3 depicts the gradual change from pine to
hardwoods.

, According to the information in Figure 3, a 150-year-old climax community would
contain oak and hickory trees with a density of approximately:

F. 3,000 trees per unit area.
G. 5,000 trees per unit area.
H. 15,000 trees per unit area.
J. 20,000 trees per unit area.

The change in the object's momentum is 2mV0 directed to the left. Momentum is a
vector quantity. Its magnitude for an object of mass m and speed V is mV. Taking the
positive direction to be to the right (the initial direction of the object), the object has
momentum +V0 before striking the wall and momentum -mV0 afterward. The change in
momentum is the final value minus the initial value: -mV0 - (+mV0) = -2mV0
This is a change of magnitude 2mV0 directed to the left. - Answer An object with mass
m and speed V0 directed to the right strikes a wall and rebounds with speed V0 directed
to the left.

The change in the object's momentum is
A.2mV0 Directed to the left
B. MV0 Directed to the left
C. Zero
D. MV0 Directed to the right
E. 2mV0 Directed to the right

The correct answer is (A). Without friction, there is still a constant force on the block.
This means a constant acceleration, so the speed curve will still be a straight line.
Removing friction will not cause the block to be in motion before it is pulled, so the initial
speed remains zero. Finally, since friction would act against the pulling force, the net
force on the block is greater without friction, so the acceleration and thus the slope of
the speed graph would be greater - Answer A block is pulled along a horizontal surface
with a constant horizontal force of magnitude F. The surface exerts a frictional force of
constant magnitude f on the block. The graph of speed V as a function of time t for the
block is shown above.

If, instead, the block slides without friction but is pulled with the same horizontal force of
magnitude F, which of the following would be a possible new graph of speed V as a
function of time t? (The dashed line represents the old graph.)

The correct answer is (B), a constant acceleration. At each instant the acceleration is
the slope of the velocity graph. Since the velocity curve is a straight line that has
constant slope, the acceleration is constant. Alternately, the description indicates that
both the force pulling the block and the frictional force are constant. Acceleration is
force divided by mass, so it also must be constant. - Answer A block is pulled along a
horizontal surface with a constant horizontal force of magnitude F. The surface exerts a

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