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The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
What is the frequency of the recessive homozygote?
A. 15 percent
B. 19 percent
C. 25 percent
D. 40 percent
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B: 100 - 45 - 36 = 19 percent.
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Biology/88
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Biology
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The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
What would be the approximate frequency of the heterozygote condition if this population were in Hardy-Weinberg equilibrium?
A. 20 percent
B. 45 percent
C. 48 percent
D. 72 percent
|
C: 36 percent of the population is AA. Taking the square root of 0.36, we find the frequency of the A allele to be 0.6. This means that the a allele's frequency must be 1 - 0.6, or 0.4. From these numbers, we can calculate the expected Hardy-Weinberg heterozygous frequency is 2pq = 2(A)(a) = 2(0.6)(0.4) = 48.0 or 48 percent.
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Biology/89
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Biology
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The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
Is this population in Hardy-Weinberg equilibrium?
A. Yes
B. No
C. Cannot tell from the information given
D. Maybe, if individuals are migrating
|
B: The expected heterozygous probability does not match up with the actual. This population is not in Hardy-Weinberg equilibrium.
|
Biology/90
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Biology
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The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
Which of the following processes may be occurring in this population, given the allele frequencies?
A. Directional selection
B. Homozygous advantage
C. Hybrid vigor
D. Allopatric speciation
|
B: The homozygous frequency is higher than expected; one explanation for this is that the homozygotes are being selected for.
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Biology/91
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Biology
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If the pedigree is studying an autosomal recessive condition for which the alleles are A and a, what was the probability that a child produced by parents A and B would be heterozygous?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.5000
|
D: The mother (person B) must be heterozygous Aa because she and her husband (aa) have produced children that have the double recessive condition. This means that person B (the mother) must have contributed an a and that the cross is Aa ×aa-and the probability is 1/2.
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Biology/92
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Biology
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Imagine that a couple (C and D) go to a genetic counselor because they are interested in having children. They tell the counselor that they have a family history of a certain disorder and they want to know the probability of their firstborn having this condition. What is the probability of the child having the autosomal recessive condition?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.3333
|
D: To answer this question, we must first determine the probability that person D is heterozygous. We know she is not aa because she does not have the condition. Since we know that the father has the condition, we know for certain that his genotype is aa. Both of mother D's parents must be heterozygous since neither of them have the condition, but they have produced a child with the condition. The probability that mother D is heterozygous Aa is 2/3. The probability that a couple with the genotypes Aa ×aa have a double recessive child is 1/2. The probability that these two will have a child with the condition is 1/2 ×2/3 = 1/3 = 0.333
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Biology/93
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Biology
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Imagine that a couple (C and D) have a child (E) that has the autosomal recessive condition being traced by the pedigree. What is the probability that their second child (F) will have the autosomal recessive condition?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.5000
|
D: If the couple has a child (person E) with the recessive condition, then we know for certain that mother D must be heterozygous. It is definitely an aa ×Aa cross, leaving a 50 percent chance that their child will be aa.
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Biology/94
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Biology
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The bold line that point C intersects is known as the
A. biotic potential.
B. carrying capacity.
C. limiting factor.
D. maximum attainable population
|
B
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Biology/95
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Biology
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On the basis of what happens at the end of this chart, what is the most likely explanation for the population decline after point E?
A. The population became too dense and it had to decline.
B. There was a major environmental shift that made survival impossible for many.
C. Food became scarce, leading to a major famine.
D. The population had become too large.
|
B
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Biology/96
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Biology
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The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and sodium chloride, but not to sucrose. Side A is filled with a solution of 0.6 M sucrose and 0.2 M sodium chloride (NaCl), and side B is filled with a solution of 0.2 M sucrose and 0.3 M NaCl. Initially, the volume on both sides is the same.
At the beginning of the experiment,
A. Side A is hypertonic to side B.
B. Side A is hypotonic to side B.
C. Side A is isotonic to side B.
D. Side A is hypotonic to side B with respect to sucrose
|
A: The total solute potential for side A is 1.0 MPa (remember that for NaCl, i = 2), and the total solute potential for side B is 0.8 MPa. Therefore, side A has a higher concentration of solute (hypertonic).
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Biology/97
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Biology
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The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and sodium chloride, but not to sucrose. Side A is filled with a solution of 0.6 M sucrose and 0.2 M sodium chloride (NaCl), and side B is filled with a solution of 0.2 M sucrose and 0.3 M NaCl. Initially, the volume on both sides is the same.
If you examine side A after a couple of days, you will see
A. an increase in the concentration of NaCl and sucrose and an increase in water level.
B. a decrease in the concentration of NaCl, an increase in water level, and no change in the concentration of sucrose.
C. no net change.
D. an increase in the concentration of NaCl and an increase in the water level.
|
D: Water will move from a hypotonic solution (side B) toward a hypertonic solution (side A). Sodium will diffuse from a region of more sodium (side B) to a region of less sodium (side A).
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Biology/98
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Biology
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If inhibitor 1 is able to bind to the active site and block the attachment of the substrate to the enzyme, this is an example of
A. noncompetitive inhibition.
B. competitive inhibition.
C. a cofactor.
D. a coenzyme.
|
B: In competitive inhibition, an inhibitor molecule resembling the substrate binds to the active site and physically blocks the substrate from attaching.
|
Biology/99
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Biology
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Which of the following is not a change that would affect the efficiency of the enzyme shown above?
A. Change in temperature
B. Change in pH
C. Change in salinity
D. Increase in the concentration of the enzyme
|
C: The other four are the four main factors that can affect enzyme efficiency.
|
Biology/100
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Biology
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Which of the following points on the preceding energy chart represents the activation energy of the reaction involving the enzyme?
A. A
B. B
C. C
D. D
|
B: The activation energy of a reaction is the amount of energy needed for the reaction to occur. Notice that the activation energy for the enzymatic reaction is much lower than the nonenzymatic reaction.
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Biology/101
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Biology
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What kind of inheritable condition does this pedigree appear to show?
A. Autosomal dominant
B. Autosomal recessive
C. Sex-linked dominant
D. Sex-linked recessive
|
B: It is not autosomal dominant because in order for the second generation on the left to have those two individuals with the condition, one parent would need to display the condition as well. It is probably not sex-linked because it seems to appear as often in females as in males. Autosomal recessive seems to be the best fit for this disease
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Biology/102
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Biology
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What is the probability that couple C and D will produce a child that has the condition?
A. 0
B. 0.125
C. 0.250
D. 0.333
|
D: One first needs to determine the probability that person C is heterozygous (Bb). We know that person D is double recessive because she has the condition. We know that the parents for person C must be Bb and Bb because neither of them has the condition, but they produced children with the condition. The probability of person C being heterozygous is 2/3, because a monohybrid cross of his parents (Bb × Bb) gives the following Punnett square:
https://img.crackap.com/ap/biology/a5/Image00211.jpg
Since you know that he doesn't have the condition, he cannot be bb. This leaves just three possible outcomes, two of which are Bb. A cross must then be done between the father (person C) Bb and the mother (person D) bb. The chance of their child being bb is 50 percent or 1/2. This means that the chance of these two having a child with the condition is 2/3 × 1/2 or 1/3.
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Biology/103
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Biology
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Which of the following conditions could show the same kind of pedigree results?
A. Cri-du-chat syndrome
B. Turner syndrome
C. Albinism
D. Hemophilia
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C: Albinism is the only autosomal recessive condition on this list.
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Biology/104
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Biology
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If child E does in fact have the condition, what is the probability that child F will also have it?
A. 0
B. .250
C. .500
D. .750
|
C: It is 1/2, because finding out that one of their children has the condition lets us know that the father (person C) is definitely Bb. This changes the probability of 2/3 to 1, meaning that the probability of the two having another child with this condition is simply the result of the Punnett square of Bb ×bb, or 1/2.
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Biology/105
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Biology
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An experiment involving fruit flies produced the following results:
Vestigial wings are wild type, crumpled wings are mutant.
Gray body is dominant, black body is mutant.
From the data presented above, one can conclude that these genes are
A. sex-linked.
B. epistatic.
C. holandric.
D. linked.
|
D: When you see a ratio like the one in this problem-7:7:1:1 (approximately)-the genes are probably linked. The reason the crumpled, gray, and vestigial black flies exist at all is because crossover must have occurred.
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Biology/106
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Biology
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An experiment involving fruit flies produced the following results:
Vestigial wings are wild type, crumpled wings are mutant.
Gray body is dominant, black body is mutant.
What is the crossover frequency of these genes?
A. 10 percent
B. 20 percent
C. 30 percent
D. 35 percent
|
A: To determine the crossover frequency in a problem like this, simply add up the total number of crossovers (75 + 45 = 120) and divide that sum by the total number of offspring (120 + 555 + 525 = 1200). This results in 120/1200 or 10 percent.
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Biology/107
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Biology
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An experiment involving fruit flies produced the following results:
Vestigial wings are wild type, crumpled wings are mutant.
Gray body is dominant, black body is mutant.
How many map units apart would these genes be on a linkage map?
A. 5 map units
B. 10 map units
C. 20 map units
D. 30 map units
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B: One map unit is equal to a 1 percent recombination frequency
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Biology/108
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Biology
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A population of rodents is studied over the course of 100 generations to examine changes in dental enamel thickness. Species that are adapted to eat food resources that require high levels of processing have thicker enamel than do those that eat softer, more easily processed foods. Answer the following questions using this information and the curves that follow
How is average enamel thickness changing in this population?
A. There is no real change.
B. The color and size are changing.
C. It is increasing.
D. It is decreasing.
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C: The average enamel thickness started at 10, increased to 12, and then increased to 15. It is therefore increasing overall.
|
Biology/109
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Biology
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A population of rodents is studied over the course of 100 generations to examine changes in dental enamel thickness. Species that are adapted to eat food resources that require high levels of processing have thicker enamel than do those that eat softer, more easily processed foods. Answer the following questions using this information and the curves that follow
You randomly pick one data point from all three sets of data (all three generations), and the individual’s enamel thickness score is 15. Which of the following can be inferred?
A. The individual comes from generation 1.
B. The individual comes from generation 50.
C. The individual comes from generation 100.
D. The individual could be from any of these generations.
|
D: The average enamel thickness does not describe the range of possible values; an individual with a thickness of 15 could reasonably come from any of the three generations (if we took into account probability, we could say that the individual most likely came from the 100th generation because this population has the highest frequency of individuals with this thickness; however, the question does not ask for probabilities).
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Biology/110
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Biology
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A population of rodents is studied over the course of 100 generations to examine changes in dental enamel thickness. Species that are adapted to eat food resources that require high levels of processing have thicker enamel than do those that eat softer, more easily processed foods. Answer the following questions using this information and the curves that follow.
What inference can you make about this species’ diet?
A. Its food resources are getting softer and easier to process.
B. Its food resources are getting harder and more difficult to process.
C. The population is growing.
D. The population is shrinking.
|
B: Because thicker enamel in this species indicates foods that are more difficult to process, the answer is B. Answer E is incorrect because our model has no predictive power; if the food resources change, the enamel thickness may as well, to either a thicker or thinner average (enamel thickness could also stay the same).
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Biology/111
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Biology
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A student sets up a lab experiment to study the behavior of slugs. She sets up a large tray filled with soil that measures 1 square meter and has four sets of conditions, one in each quadrant:
She places 20 slugs in the tray, 5 in each quadrant. Use this information to answer the following questions:
What is this lab setup called?.
A. A gel sheet
B. A choice chamber
C. A potometer
D. An incubation chambe
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B: Experimental setups where individuals are given a choice as to where to move are called "choice chambers."
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Biology/112
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Biology
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A student sets up a lab experiment to study the behavior of slugs. She sets up a large tray filled with soil that measures 1 square meter and has four sets of conditions, one in each quadrant:
She places 20 slugs in the tray, 5 in each quadrant. Use this information to answer the following questions:
After 5 minutes, there are 5 slugs in each quadrant. Which of the following is not a viable explanation for this finding?
A. The slugs haven’t had time to move yet.
B. The slugs have no preference for temperature or salinity conditions.
C. The slugs can’t move from one area of the tray to another.
D. The slugs do not like to live in high-temperature areas.
|
D: All the answers except D are possible, and are important things to consider when setting up an experiment. For example, it is important to allow your study animals enough time to move and/or get used to their new surroundings and conditions before drawing conclusions about their behavior. D is not a good answer because half of the slugs started in a high-temperature area and haven't moved.
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Biology/113
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Biology
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A student sets up a lab experiment to study the behavior of slugs. She sets up a large tray filled with soil that measures 1 square meter and has four sets of conditions, one in each quadrant:
She places 20 slugs in the tray, 5 in each quadrant. Use this information to answer the following questions:
After 20 minutes, 20 slugs are in the high-temperature, low-salinity quadrant. What kind of animal behavior has this experiment displayed?
A. Kinesis
B. Taxis
C. Survival
D. Feeding
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A: Kinesis is the movement of animals in response to current conditions; animals tend to move until they find a favorable environment, at which point their movement slows.
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Biology/114
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Biology
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In the following cladogram, a common ancestor (*) and species derived from it are illustrated. How many species have four or more common ancestors with Iguanodon?
A. 1
B. 2
C. 4
D. 6
|
D Iguanodon had one common ancestor with Hypsilophodon before Hypsilophodon diverged. Iguanodon had two with Tenontosaurus because one came from where Tenontosaurus diverged and they both share the common ancestor with Hypsilophodon. By that logic, Iguanodon has three common ancestors with Zalmoxes, four with Dryosaurus, and five with Camptosaurus. It will also share those five common ancestors with the other species on the tree, because everything above Camptosaurus still shares those original five. The higher levels also share the ancestor that existed when Iguanodon diverged. Either way, six species share at least four common ancestors with Iguanodon. Remember, donât count Iguanodon itself.
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Biology/115
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Biology
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Which of the following are true statements according to the figure?
I. r-strategists are unlikely to die young.
II. The death rate for a c-strategist is constant.
The following graph demonstrates 3 different strategies for survival.III. Humans are an example of a k-strategist.
A. I only
B. II only
C. II and III
D. I, II, and III
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C Most r-strategists die early in their lifetime, (A), but most k-strategists live long lives and then die when old. Humans are an example of k-strategists. The c-strategist survivorship line declines evenly, meaning that they are just as likely to die when old as they are when young, (C).
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Biology/116
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Biology
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Questions below refer to the following synthetic pathway of nRNA pyrimidine, cytidine 5' triphosphate, CTP. This pathway begins with the condensation of two small molecules by the enzyme aspartate transcarbamylase (ATCase).This enzymatic phenomenon is an example of
A. transcription
B. feedback inhibition
C. dehydration synthesis
D. photosynthesis
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B This enzymatic phenomenon is an example of feedback inhibition. Feedback inhibition is the metabolic regulation in which high levels of an enzymatic pathwayâs final product inhibit the activity of its rate-limiting enzyme. Transcription, (A), is the production of RNA from DNA. Dehydration synthesis, (C), is the formation of a covalent bond by the removal of water. In photosynthesis, (D), radiant energy is converted to chemical energy.
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Biology/117
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Biology
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During labor, pressure on the cervix and oxytocin form a positive feedback loop as shown below.Which of the following other pathways also demonstrate positive feedback?
A. Glycolysis leads to the production of ATP. ATP, in turn, turns off the enzyme phosphofructokinase, which catalyzes a key phosphorylation step in glycolysis.
B. The anterior pituitary gland in the brain releases adrenocorticotropic hormone (ACTH). ACTH then causes the adrenal cortex to release glucocorticoids. Glucocorticoids then prevent the pituitary from releasing more ACTH.
C. Luteinizing hormone triggers ovulation and the formation of the corpus luteum, which is a hormone-producing structure formed during ovulation. The corpus luteum secretes progesterone, which inhibits LH. The drop in LH causes the degradation of the corpus luteum.
D. When a tissue is injured, it releases chemicals that activate platelets. Activated platelets themselves then release chemicals that activate more platelets. These activated platelets then release chemicals to activate more platelets.
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D Positive feedback occurs when a process creates an end product whose production stimulates the process to create even more of the end product. In (D), the platelets lead to more platelet creation. Choices (A), (B), and (C) illustrate negative feedback.
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Biology/118
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Biology
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The following diagram demonstrates the ecological succession that occurs in an environment over time as it is colonized by different species.Why does it take 75 years for a beech-maple to occur in the figure above?
A. Beech seeds have a very long period of dormancy prior to germination.
B. It takes an average of 75 years for conifer trees to become extinct.
C. Agriculture was the predominant industry, and hardwood trees were removed.
D. Maple trees grow better in a pine forest than they do in a grassland.
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D Few species can grow in barren areas, but once they are colonized with plants, then more species can grow and live there. It takes a long time for the beech and maple trees to grow because first species need to change the environment, allowing other species to colonize. The environment was not hospitable until it was a pine forest, which took a long time to reach as well.
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Biology/119
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Biology
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Embryogenesis is a carefully timed and well-organized process. As a single-celled zygote divides and grows into hundreds and thousands of cells, a process called differentiation occurs wherein certain areas of the embryo become specialized to become different types of tissue. As differentiation continues, the level of specificity increases, and the cell potency decreases until highly specialized unique tissues and organs develop. The figure below shows 12 stages of development of human embryos.A totipotent embryonic cell has the most cell potency. Which of the following is most likely to be totipotent?
A. 8-cell zygote
B. Inner cell mass
C. Mesoderm
D. Digestive tract
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A Potency is the potential to become many things. The most potent cells are the least differentiated or the least specialized. The more specialized they become, the more they lose their potency. The 8-cell zygote is the least specialized and has the most potency.
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Biology/120
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Biology
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Embryogenesis is a carefully timed and well-organized process. As a single-celled zygote divides and grows into hundreds and thousands of cells, a process called differentiation occurs wherein certain areas of the embryo become specialized to become different types of tissue. As differentiation continues, the level of specificity increases, and the cell potency decreases until highly specialized unique tissues and organs develop. The figure below shows 12 stages of development of human embryos.The inner cell mass is what eventually forms the embryo. During development, the embryo differentiates into various types of cell layers. Which of the following is NOT one of them?
A. Mesoderm
B. Hypoblast
C. Blastomere
D. Endoderm
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C The blastomere is shown prior to the inner cell mass; therefore, it cannot be a differentiation of the inner cell mass. The hypoblast comes from the inner cell mass, and the endoderm and mesoderm also come from it.
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Biology/121
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Biology
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Diabetes mellitus is a disease characterized by an inability of the cells to properly produce (type I) or respond (type II) to insulin, a hormone produced by the pancreas in response to high levels of blood glucose. Without insulin, glucose accumulates in the blood. In situations of low blood glucose, another pancreatic enzyme, glucagon, is released, which triggers the process of gluconeogenesis, shown on the right side of the pathway below. The stimulators, activators, or inhibitors of each step are shown with + or - signs.Which of the following situations likely stimulates gluconeogenesis?
A. High levels of insulin
B. High levels of F-2,6-BP
C. High levels of glucagon
D. High levels of ADP
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C Gluconeogenesis is the formation of glucose. If the body had high insulin, then it would already have lots of glucose, so (A) is eliminated, but if the body had high glucagon, it would need glucose, (C). High ADP would inhibit gluconeogenesis as shown in the figure, (D). High F-2,6-BP would also inhibit gluconeogenesis, (B).
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Biology/122
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Biology
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Diabetes mellitus is a disease characterized by an inability of the cells to properly produce (type I) or respond (type II) to insulin, a hormone produced by the pancreas in response to high levels of blood glucose. Without insulin, glucose accumulates in the blood. In situations of low blood glucose, another pancreatic enzyme, glucagon, is released, which triggers the process of gluconeogenesis, shown on the right side of the pathway below. The stimulators, activators, or inhibitors of each step are shown with + or - signs.Patients with type I diabetes often require insulin injections. Which of the following situations would most require an insulin injection?
A. After eating a stalk of celery
B. After eating a cookie
C. After skipping breakfast
D. After drinking a lot of water
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B Insulin is needed when blood glucose gets too high. Eating a cookie would increase blood sugar more than eating celery, fasting, or drinking water.
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Biology/123
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Biology
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Diabetes mellitus is a disease characterized by an inability of the cells to properly produce (type I) or respond (type II) to insulin, a hormone produced by the pancreas in response to high levels of blood glucose. Without insulin, glucose accumulates in the blood. In situations of low blood glucose, another pancreatic enzyme, glucagon, is released, which triggers the process of gluconeogenesis, shown on the right side of the pathway below. The stimulators, activators, or inhibitors of each step are shown with + or - signs.Which of the following conditions would lead to increased production of fructose 1,6-bisphosphate?
A. High ATP and high citrate
B. High AMP and high citrate
C. High AMP and high F-2,6-BP
D. High ATP and high F-2,6-BP
|
C ATP inhibits its production since it inhibits phosphofructokinase. This eliminates (A) and (D). Citrate stimulates turning fructose 1,6-bisphosphate into fructose 6-phosphate, which eliminates (B). Choice (C) includes two things that stimulate phosphofructokinase and lead to production of fructose 1,6-bisphosphate.
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Biology/124
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Biology
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A population of goldfish in a large, isolated pond were studied in 1958 and again in 2008. The fishes' pigment level varied from pale white-orange fish to dark brown-orange fish. The color of each of the fish was recorded in the figure below.Which addition to the pond did NOT likely contribute to the change between 1958 and 2008?
A. A poisonous fish with a medium orange pigment
B. Runoff from fields that makes the water dark and murky
C. A light-orange water grass that grows in the pond
D. Predatory birds that can easily see medium-orange pigment
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A From 1958 to 2008, the graph shifted away from medium-orange and moved toward the extremes (light and dark). Choice (A) would make fewer orange fish be eaten, which would promote more medium-orange fish. Choices (B) and (C) would cause fewer light and dark fish to be eaten, respectively. This would promote the light and dark fish. Choice (D) would cause the medium-orange fish to be eaten more often and thus promote the light and dark extremes shown in 2008.
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Biology/125
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Biology
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A population of goldfish in a large, isolated pond were studied in 1958 and again in 2008. The fishes' pigment level varied from pale white-orange fish to dark brown-orange fish. The color of each of the fish was recorded in the figure below.Which of the following theories is supported by the evidence?
A. The pigment trait in fish demonstrates incomplete dominance.
B. The pigment trait in fish demonstrates classical dominance.
C. The pigment trait in fish demonstrates codominance.
D. None of the above
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D These graphs of phenotype tell us nothing about the mode of inheritance.
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Biology/126
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Biology
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A population of goldfish in a large, isolated pond were studied in 1958 and again in 2008. The fishes' pigment level varied from pale white-orange fish to dark brown-orange fish. The color of each of the fish was recorded in the figure below.Which of the following best describes the fish population in the year 1958?
A. All medium-orange in color
B. Mostly medium-orange fish with some nearly white and some nearly brown
C. Mostly white fish and brown fish with a few orange fish
D. Equal numbers of orange fish, white fish, and brown fish
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B The peak in 1958 is in the middle, corresponding to medium-orange fish. However, there are also some fish on each extreme (light or dark).
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Biology/127
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Biology
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A group of Daphnia, small crustaceans known as water fleas, was placed in one of three culture jars of different sizes to determine their reproductive rate. There were 100 females in the jar. The graph below shows the average number of offspring produced per female each day in each jar of pond water.The data in the figure above would best support which conclusion?
A. If you decreased the number of females, the container would have to remain constant.
B. The number of offspring produced scales proportionally with the container's size.
C. The number of offspring produced increases with time.
D. Daphnia prefer high-density conditions to have the most efficient reproductive rate.
|
B The passage does not say that the container would stay the same; in fact, that was the independent variable. Decreasing the number of females would not mean that the container could not be changed for a new experiment. The graph shows that both samples have a reproduction rate decrease after 10â20 days, so not (C). The highest density condition (same number of Daphnia but lower volume) shows no reproductive activity after trauma (so not (D)). Choice (B) is the correct answer because as the container size increased, the number of offspring increased.
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Biology/128
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Biology
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Scientists used embryology, morphology, paleontology, and molecular biology to create the phylogenetic tree below.Which of the sources of evolutionary evidence would be the most reliable of those listed?
A. Embryology
B. Morphology
C. Paleontology
D. Molecular biology
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D Because molecular biology demonstrates where the actual mutations occurred (rather than just what phenotype is observed), it is the most reliable; it allows you to detect changes that are not phenotypically observable.
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Biology/129
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Biology
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Scientists used embryology, morphology, paleontology, and molecular biology to create the phylogenetic tree below.Which of the following is the least closely related according to the phylogenetic tree?
A. B. physalus and B. brydei
B. B. boreales and B. brydei
C. B. musculus and B. brydei
D. B. musculus and Eschrichtius robustus
|
C Relatedness is shown on a phylogenetic tree by how far back you have to go to reach the common ancestor of the two species (shown by a branch node). Since the recent species are on the right, traveling to the left goes back in time. Because you have to go to the oldest common ancestor to connect B. musculus and B. brydei, they are the least closely related. Even though they are near each other on the tree, clades can rotate around a node so this does not matter.
|
Biology/130
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Biology
| null |
The rainfall and biomass of several trophic levels in an ecosystem were measured over several years. The results are shown in the graph below.If it rained 120 inches, what would you project the primary consumer biomass to be?
A. 150-200
B. 60
C. 45
D. 20
|
A The biomass seems to correlate fairly well with the rainfall. Therefore, a rainfall higher than any recorded would give a biomass higher than any recorded. The axis on the right shows the biomass. With a rainfall of 120, the biomass should be greater than 150.
|
Biology/131
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Biology
| null |
The rainfall and biomass of several trophic levels in an ecosystem were measured over several years. The results are shown in the graph below.Which of the following concepts is best demonstrated by this experiment?
A. Populations with higher genetic variation can withstand droughts better.
B. Meteorological impacts will affect the evolution of populations.
C. Environmental changes can affect all the levels of the ecosystem.
D. Unoccupied biological niches are dangerous because they attract invasive species.
|
C The graph does show a drought, but it says nothing about genetic variation. It is true that the rainfall could affect evolution of the population, but the graph doesnât address evolution. It shows only the decrease in biomass. Choice (C) is the best answer since it addresses how the environmental effects ripple through the levels of the ecosystem. Invasive species do not need an unoccupied niche, and that is not shown in the graph.
|
Biology/132
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Biology
| null |
An experiment was performed to assess the growth of two species of plants when they were grown in different pHs, given different volumes of water, and watered at different times of day over 6 weeks. Two plants were grown of each species and the average heights (in cm) are shown in the table.Which of the following would most improve the statistical significance of the results?
A. Let the plants grow for a longer period of time.
B. Add more conditions to test, such as amount of light and amount of soil.
C. Test the same plants with more pHs and more volumes and times of day.
D. Increase the number of plants in each group.
|
D Choices (A), (B), and (C) would all give interesting information about the experiment, but the only one that would make the results more statistically significant is to increase the number of plants in each group. Two plants is not enough to be sure about the results of the experiment.
|
Biology/133
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Biology
| null |
An experiment was performed to assess the growth of two species of plants when they were grown in different pHs, given different volumes of water, and watered at different times of day over 6 weeks. Two plants were grown of each species and the average heights (in cm) are shown in the table.Which pH and volume were likely used for the watering time experiment?
A. pH 4 and 40 mL
B. pH 7 and 40 mL
C. pH 4 and 80 mL
D. pH 7 and 80 mL
|
D The plants in the watering time experiment seemed to be around 60 and 20 cm in height. This corresponds to a pH of 7 rather than 4 and a watering volume of 80 mL rather than 40 mL.
|
Biology/134
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Biology
| null |
An experiment was performed to assess the growth of two species of plants when they were grown in different pHs, given different volumes of water, and watered at different times of day over 6 weeks. Two plants were grown of each species and the average heights (in cm) are shown in the table.What are the preferred growth conditions for Species B?
A. pH 7, 40 mL, any time of day
B. pH 10, 40 mL, 7:00 A.M.
C. pH 7, 80 mL, any time of day
D. pH 10, 80 mL, 12:00 P.M.
|
A The top growth conditions for Species B are pH 7, 40 mL, and any time of day. In those conditions, the Species B plants grew the tallest.
|
Biology/135
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Biology
| null |
An experiment was performed to assess the growth of two species of plants when they were grown in different pHs, given different volumes of water, and watered at different times of day over 6 weeks. Two plants were grown of each species and the average heights (in cm) are shown in the table.For which conditions do the species have different preferences?
A. pH
B. Volume
C. Volume and watering time
D. pH and volume and watering time
|
B The two species differ in their preference for water volume. Species A prefers 80 mL, and Species B prefers 40 mL. The different watering times do not seem to play a role in growth as there is no obvious pattern and all of the times had similar growth. Both plants prefer pH 7.
|
Biology/136
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Biology
| null |
Pumping blood through the human heart must be carefully organized for maximal efficiency and to prevent backflow. In the figure below, the blood enters the heart through the vena cava (1), passes through the right atrium and right ventricle and then goes through the pulmonary artery toward the lungs. After the lungs, the blood returns through the pulmonary vein and then passes into the left atrium and the left ventricle before leaving the heart via the aorta.Blood is pumped via heart contractions triggered by action potentials spreading through the heart muscle. If there is a sudden increase in blood in chamber 3, which chamber of the heart received an increased number of action potentials?
A. Left atrium
B. Left ventricle
C. Right atrium
D. Right ventricle
|
C The action potentials cause the heart to contract. If chamber 3 is full of more blood, then that would indicate that the preceding chamber, chamber 2 (right atrium), had suddenly contracted a larger amount.
|
Biology/137
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Biology
| null |
Pumping blood through the human heart must be carefully organized for maximal efficiency and to prevent backflow. In the figure below, the blood enters the heart through the vena cava (1), passes through the right atrium and right ventricle and then goes through the pulmonary artery toward the lungs. After the lungs, the blood returns through the pulmonary vein and then passes into the left atrium and the left ventricle before leaving the heart via the aorta.Which of the following chambers or vessels carry deoxygenated blood in the human heart?
A. 1 only
B. 2 and 3
C. 1, 2, 3, 4
D. 4 and 5
|
C Deoxygenated blood flows through all chambers before going to the lungs. This means the correct answer is chambers 1, 2, 3, and 4.
|
Biology/138
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Biology
| null |
Note: + indicates a feature present in an organism.
Questions below refer to the data below concerning the general animal body plan of four organisms.The two most closely related organisms are
A. sea anemone and hagfish
B. eel and salamander
C. hagfish and eel
D. sea anemone and salamander
|
B The two most closely related organisms are the two with the most shared derived characteristics.
|
Biology/139
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Biology
| null |
Questions below refer to the following figures which show 5 species of insects that were discovered on a previously unknown island and were named as shown in the table. Proteomic analysis was performed on a highly conserved protein in the insects and the number of amino acid differences was calculated and included in the table below. The scientists used this data to create the phylogenetic tree shown below with positions labeled I, II, III, and IV as well as O, P, Q, R, and S.Which 2 species would you expect to have the most shared derived characters?
A. Snippeiq and Sqellert
B. Gerdellen and Sqellert
C. Snorflak and Gerdellen
D. Snorflak and Fixxels
|
D Snorflak and Fixxels, (D), would have the most shared derived characters because they have the fewest amino acid differences. Snippeiq and Sqellert would be expected to have the least since they are so different, which rules out (A). Sqellert and Gerdellen, (B), were close with only 4 differences but still not as low as Snorflak and Fixxels. Snorflak and Gerdellen, (C), must be an intermediate with 9 differences observed in the table.
|
Biology/140
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Biology
| null |
Questions below refer to the following figures which show 5 species of insects that were discovered on a previously unknown island and were named as shown in the table. Proteomic analysis was performed on a highly conserved protein in the insects and the number of amino acid differences was calculated and included in the table below. The scientists used this data to create the phylogenetic tree shown below with positions labeled I, II, III, and IV as well as O, P, Q, R, and S.Which species is the out-group?
A. Snippeiq
B. Fixxels
C. Snorflak
D. Sqellert
|
A Snippeiq, (A), is the out-group because it is the least related to the other species. The table shows that Snippeiq has the most differences from all of the other species.
|
Biology/141
|
Biology
| null |
Questions below refer to the following figures which show 5 species of insects that were discovered on a previously unknown island and were named as shown in the table. Proteomic analysis was performed on a highly conserved protein in the insects and the number of amino acid differences was calculated and included in the table below. The scientists used this data to create the phylogenetic tree shown below with positions labeled I, II, III, and IV as well as O, P, Q, R, and S.Which location contains the oldest common ancestor?
A. Position I
B. Position II
C. Position III
D. Position IV
|
C Position III, (C), contains the oldest common ancestor because it is the node for all the current species designated by the lettered positions.
|
Biology/142
|
Biology
| null |
Questions below refer to the following figures which show 5 species of insects that were discovered on a previously unknown island and were named as shown in the table. Proteomic analysis was performed on a highly conserved protein in the insects and the number of amino acid differences was calculated and included in the table below. The scientists used this data to create the phylogenetic tree shown below with positions labeled I, II, III, and IV as well as O, P, Q, R, and S.Based on the data, which is the LEAST possible location for Gerdellen to be placed?
A. Position O
B. Position P
C. Position R
D. Position S
|
D Since Snippeiq has the most differences with the other organisms it must be at position S. This means that Gerdellen cannot be there, which makes (D) the best answer. (Position Q or R is the best position for Gerdellen, though positions O and P could possibly be correct as well.)
|
Biology/143
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Biology
| null |
The cell cycle is a series of events in the life of a dividing eukaryotic cell. It consists of four stages: G1, S, G2, and M. The duration of the cell cycle varies from one species to another and from one cell type to another. The G1 phase varies the most. For example, embryonic cells can pass through the G1 phase so quickly that it hardly exists, whereas neurons are arrested in the cell cycle and do not divide.Since neurons are destined never to divide again, what conclusion can be made?
A. These cells will go through cell division.
B. These cells will be permanently arrested in the G0 phase.
C. These cells will be permanently arrested in the M phase.
D. These cells will quickly enter the S phase.
|
B Because neurons are not capable of dividing, it is reasonable to conclude that these cells will stay in the G1 phase (specifically in an extension of G1 called G0) phase. This is a reading comprehension question. The passage states that cells that do not divide are arrested at the G1 phase. Choice (A) is incorrect because these cells will not be committed to go through cell division. You can also eliminate (C) and (D), as the cells will not enter the M or S phase.
|
Biology/144
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Biology
| null |
The cell cycle is a series of events in the life of a dividing eukaryotic cell. It consists of four stages: G1, S, G2, and M. The duration of the cell cycle varies from one species to another and from one cell type to another. The G1 phase varies the most. For example, embryonic cells can pass through the G1 phase so quickly that it hardly exists, whereas neurons are arrested in the cell cycle and do not divide.If the cell cycle fails to progress, which of the following is NOT a possible explanation?
A. There are inadequate phosphate groups available for the cyclin dependent kinase.
B. A tumor suppressor protein has signaled for apoptosis.
C. A cyclin is unable to release from its cyclin dependent kinase.
D. An inhibitor of a cyclin gene has been highly expressed.
|
C To induce cell cycle progression, an inactive CDK (cyclin-dependent kinase) binds a regulatory cyclin. Once together, the complex is activated, can affect many proteins in the cell, and causes the cell cycle to continue. To do this, the CDK transfers phosphate groups onto other molecules. If there are inadequate phosphate groups, CDK would not function properly and the cell cycle would not progress, which rules out (A). Tumor suppressor proteins inhibit cell cycle progression and can trigger apoptosis, which rules out (B). To inhibit cell cycle progression, CDKs and cyclins are kept separate. Together, they promote the cell cycle. If a cyclin is unable to release from its cyclin dependent kinase, the cell cycle would be promoted, not stopped, making (C) correct. Choice (D) couldâve been eliminated because an inhibitor of a cyclin gene would prevent expression of the cyclin gene. With lower levels of the cyclin protein, cell cycle progression would be inhibited.
|
Biology/145
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Biology
| null |
Questions below refer to the following bar graph, which shows the relative biomass of four different populations of a particular food pyramid.Which of the following would be the most likely result if there was an increase in the number of organisms in population C?
A. The biomass of population D will remain the same.
B. The biomass of population B will decrease.
C. The biomass of population A will steadily decrease.
D. The food source available to population C would increase.
|
B An increase in the number of organisms in population C would most likely lead to a decrease in the biomass of B because population B is the food source for population C. Make a pyramid based on the biomasses given. If population C increases, population B will decrease. Eliminate (A) and (C), as you cannot necessarily predict what will happen to the biomass of populations that are above population C. Choice (D) can also be eliminated because the food source available to population C would most likely decrease, not increase.
|
Biology/146
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Biology
| null |
It is difficult to determine exactly how life began. Answer the following questions as if the following data had been collected billions of years ago.Photosynthesis likely began ________ billion years ago when the first ________ appeared.
A. 4.5; autotrophs
B. 3.2; autotrophs
C. 3.5; heterotrophs
D. 3.2; heterotrophs
|
B The sign that photosynthesis began was when oxygen appeared. Oxygen was absent at 3.25 billion years and present at 3 billion years so photosynthesis must have begun between those times; eliminate (A) and (C). Autotrophs, not heterotrophs, perform photosynthesis. Heterotrophs rely on eating other things for their energy.
|
Biology/147
|
Biology
| null |
It is difficult to determine exactly how life began. Answer the following questions as if the following data had been collected billions of years ago.When is the earliest that functional ribosomes could have been found?
A. Between 4 billion and 3.5 billion years ago
B. Between 3.5 billion and 3.25 billion years ago
C. Between 3.25 billion and 3 billion years ago
D. Between 3 billion years ago and the present
|
B Functional ribosomes would not likely be found before the proteins they make. Therefore, this would be between 3.5 and 3.25 billion years ago.
|
Biology/148
|
Biology
| null |
It is difficult to determine exactly how life began. Answer the following questions as if the following data had been collected billions of years ago.Which best describes the origin of life according to the data?
A. Life required an environment with atmospheric oxygen and any type of nucleic acids.
B. Life required an environment with atmospheric carbon dioxide, but not atmospheric oxygen.
C. Life required an environment with self-replicating nucleic acids that can take on many shapes.
D. Life required an environment with nucleic acids and proteins, but not atmospheric oxygen.
|
C The origin of life refers to the period when the first life began. Life is shown to be present 3.5 billion years ago when no atmospheric oxygen was present yet, so (A) can be eliminated. Also, there was no life at 4 billion years despite having CO2 and no O2; therefore, something else must be required, and (B) can be eliminated. There were no proteins when the first life arose, so (D) is incorrect. Choice (C) states that self-replicating nucleic acids were necessary, which is why life could only occur when RNA appeared.
|
Biology/149
|
Biology
| null |
Which process is demonstrated in the figure below?
A. Krebs cycle
B. Photosynthesis
C. Glycolysis
D. Fermentation
|
B Photosynthesis requires light, carbon dioxide, and water and produces oxygen and glucose.
|
Biology/150
|
Biology
| null |
The following are important pieces of replication and transcription machinery:What might have occurred to produce the following situation?
A. Stalled DNA replication
B. Initiation of transcription
C. Repression of transcription
D. Crossing-over in meiosis
|
C The RNA polymerase is not bound, but there is a bound repressor. The transcription must be repressed.
|
Biology/151
|
Biology
| null |
The graph below shows two growth curves for bacterial cultures, A and B.What could explain the difference between culture A and culture B?
A. Culture B started with more bacteria than culture A.
B. Culture A was grown with a competitive inhibitor.
C. Culture B was not measured as often.
D. Culture A has not yet exhausted its space and resources.
|
D The growth of curve A is exponential, meaning that the bacteria are replicated as fast as possible without any restrictions. It looks like the populations started at the same level, and an inhibitor would not cause high growth as shown in culture A. Not being measured as often would not cause a graph like the one shown for culture B.
|
Biology/152
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Biology
| null |
The graph below shows two growth curves for bacterial cultures, A and B.Which of the following represents the carrying capacity for culture B?
A. 10
B. 50
C. 100,000
D. 1,000,000
|
C The carrying capacity is the maximum number of organisms of a given species that can be maintained in a given environment. Once a population reaches its carrying capacity, the number of organisms will fluctuate around it.
|
Biology/153
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Biology
| null |
The Loop of Henle is a structure within each of the million nephrons within a kidney. As shown in the figure, the two sides have different permeabilities, and there is differential movement across each membrane. The Loop acts as a counter-current multiplier that makes the medulla of the kidney very osmotic. The longer the loop, the higher and more powerful the osmolarity gradient that is created. The gradient is required for the reclamation of water from the urine collecting duct. On the right side of the figure is the urine collecting duct. If the body needs to retain water, anti-diuretic hormone makes this region permeable to water via the introduction of aquaporins, and the osmotic pull of the medulla reclaims the water, out of the collecting duct, which makes the urine more concentrated.What type of transport is occurring when water flows out of the descending tubule?
A. Simple diffusion
B. Facilitated diffusion
C. Active transport
D. Secondary active transport
|
B When water flows out of the tubule, it is moving by simple osmosis. However, because it is a polar molecule, it must pass through an aquaporin channel, which is facilitated diffusion.
|
Biology/154
|
Biology
| null |
The Loop of Henle is a structure within each of the million nephrons within a kidney. As shown in the figure, the two sides have different permeabilities, and there is differential movement across each membrane. The Loop acts as a counter-current multiplier that makes the medulla of the kidney very osmotic. The longer the loop, the higher and more powerful the osmolarity gradient that is created. The gradient is required for the reclamation of water from the urine collecting duct. On the right side of the figure is the urine collecting duct. If the body needs to retain water, anti-diuretic hormone makes this region permeable to water via the introduction of aquaporins, and the osmotic pull of the medulla reclaims the water, out of the collecting duct, which makes the urine more concentrated.Which of the following statements correctly describes the state of things near the top of the descending limb?
A. The fluid within the descending limb is hypotonic to fluid in the space surrounding the tubule.
B. The blood within the vasa recta is hypotonic to the filtrate within the descending limb.
C. The fluid in the area surrounding the tubule is hypertonic to the blood in the vasa recta.
D. The water in the area surrounding the tubule has a higher water potential than the water in the descending tubule.
|
A The figure shows that water leaves the descending limb and flows by osmotic pressure into the surrounding space and then into the vasa recta. This means that each of those places is more hypertonic than the previous. The most hypotonic is the descending tubule. (This is why the water leaves.)
|
Biology/155
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Biology
| null |
The affinity of hemoglobin for oxygen is reduced by many factors, including low pH and high CO2. The graph below shows the different dissociation curves that maternal (normal) hemoglobin and fetal hemoglobin have.How much pO2 would it take in an extremely CO2-rich environment to saturate hemoglobin 90 percent?
A. 15
B. 30
C. 45
D. 60
|
D The CO2-rich environment would decrease the affinity of hemoglobin for oxygen, so the curve should shift toward the right. The regular curve seems to hit 90% saturation around 45â50 pO2, so the CO2-rich curve would take more oxygen to get there.
|
Biology/156
|
Biology
| null |
The affinity of hemoglobin for oxygen is reduced by many factors, including low pH and high CO2. The graph below shows the different dissociation curves that maternal (normal) hemoglobin and fetal hemoglobin have.Hemoglobin's affinity for O2
A. decreases as blood pH decreases
B. increases as H+ concentration increases
C. increases as blood pH decreases
D. decreases as OH- concentration increases
|
A Hemoglobinâs affinity for O2 decreases as the concentration of H+ increases (or the pH decreases) and as the concentration of OHâ increases (or the pH increases). As the pH decreases, the affinity for oxygen will decrease, and as the pH increases, the affinity for oxygen will increase.
|
Biology/157
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Biology
| null |
The affinity of hemoglobin for oxygen is reduced by many factors, including low pH and high CO2. The graph below shows the different dissociation curves that maternal (normal) hemoglobin and fetal hemoglobin have.Which of the following processes would likely shift the normal dissociation curve to the right?
A. Photosynthesis
B. Respiration
C. Fermentation
D. Mitosis
|
C The passage says that high CO2 and low pH lead to reduced affinity. A right shift curve would represent a reduced affinity. Photosynthesis consumes CO2, so this would not increase the CO2. Mitosis and respiration are not related. Fermentation is done in cells that are doing anaerobic cell respiration, so CO2 should be released, possibly along with lactic acid, which would lower pH.
|
Biology/158
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Biology
| null |
If a competitive inhibitor is bound to enzyme 1, which of the following would be decreased?
I. Protein A
II. Protein C
Consider the following pathway of reactions catalyzed by enzymes (shown in numbers):III. Protein X
A. I only
B. II only
C. II and III
D. I, II, and III
|
C If enzyme 1 were inhibited, then everything that comes after that in the pathway would be reduced. Substance A would not be reduced, but both C and X would. Choice (C) is correct.
|
Biology/159
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Biology
| null |
Consider the following pathway of reactions catalyzed by enzymes (shown in numbers):An increase in substance F leads to the inhibition of enzyme 3. All of the following are direct or indirect results of the process EXCEPT
A. an increase in substance X
B. increased activity of enzyme 6
C. decreased activity of enzyme 4
D. increased activity of enzyme 5
|
D If substance F leads to the inhibition of enzyme 3, then substances D and E and enzymes 3, 4, and 5 will be affected. The activity of enzyme 5 will be decreased, not increased.
|
Biology/160
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Biology
| null |
Consider the following pathway of reactions catalyzed by enzymes (shown in numbers):Which of the following situations represents feedback inhibition?
A. Protein D activating enzyme 4
B. Protein B stimulating enzyme 1
C. Protein 7 inhibiting enzyme C
D. Protein X inhibiting enzyme 2
|
D Feedback inhibition occurs when something created by a process then inhibits that process. Choices (A) and (B) involve stimulations, so they can be eliminated. Choice (C) doesnât make sense because it says that "C" is an enzyme and the passage says that the enzymes are the numbers shown between the steps.
|
Biology/161
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Biology
| null |
G-protein coupled receptors (GPCRs) are a common type of protein receptor containing seven transmembrane segments and bearing intracellular and extracellular portions. The structure of a typical GPCR and its associated partners is shown below. GPCRs are coupled to G-proteins consisting of three subunits (α, β, and γ) that bind the nucleotides GTP or GDP on the intracellular side of the GPCR. The presence or absence of a ligand on the extracellular side of the GPCR determines whether GDP or GTP will bind to the intracellular G-protein. Extracellular ligand binding initiates GDP being exchanged for GTP and the separation of the G-protein. One subunit will travel to join with a secondary partner and will initiate a cascade of signaling effects within the cell. In the example shown below, the secondary partner is adenylyl cyclase and the signaling is cAMP upregulation.What does a GPCR's capability of making a ligand-induced conformational change allow?
A. Binding of the ligand to the phospholipid head groups within the membrane
B. Communication between the outside of the cell and the inside of the cell
C. A solid attachment to the intermembrane region of the cell membrane
D. Intracellular binding of the GDP-associated G-protein complex
|
B The conformational change allows a message to be passed from the outside of the cell to the inside. The conformational change is a sign that the ligand is bound. The ligand does not bind to the membrane, so (A) is incorrect. The ligand triggers binding of the GTP-protein, so (D) is incorrect. A conformational change would not make an attachment more solid, which rules out (C).
|
Biology/162
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Biology
| null |
G-protein coupled receptors (GPCRs) are a common type of protein receptor containing seven transmembrane segments and bearing intracellular and extracellular portions. The structure of a typical GPCR and its associated partners is shown below. GPCRs are coupled to G-proteins consisting of three subunits (α, β, and γ) that bind the nucleotides GTP or GDP on the intracellular side of the GPCR. The presence or absence of a ligand on the extracellular side of the GPCR determines whether GDP or GTP will bind to the intracellular G-protein. Extracellular ligand binding initiates GDP being exchanged for GTP and the separation of the G-protein. One subunit will travel to join with a secondary partner and will initiate a cascade of signaling effects within the cell. In the example shown below, the secondary partner is adenylyl cyclase and the signaling is cAMP upregulation.Which of the following situations is most similar to extracellular binding to a GPCR?
A. A police officer pulling over a speeding car on the expressway
B. A celebrity chef visiting the kitchen of another restaurant
C. A driver giving their order to a person through a drive-up window
D. A taxi driver picking up a group of passengers
|
C The G-protein coupled receptor (GPCR) allows a message to be transmitted inside the cell without the ligand actually entering the cell. This is most like a drive-through window where the order is communicated to someone inside a restaurant, but the patron does not actually enter the building.
|
Biology/163
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Biology
| null |
G-protein coupled receptors (GPCRs) are a common type of protein receptor containing seven transmembrane segments and bearing intracellular and extracellular portions. The structure of a typical GPCR and its associated partners is shown below. GPCRs are coupled to G-proteins consisting of three subunits (α, β, and γ) that bind the nucleotides GTP or GDP on the intracellular side of the GPCR. The presence or absence of a ligand on the extracellular side of the GPCR determines whether GDP or GTP will bind to the intracellular G-protein. Extracellular ligand binding initiates GDP being exchanged for GTP and the separation of the G-protein. One subunit will travel to join with a secondary partner and will initiate a cascade of signaling effects within the cell. In the example shown below, the secondary partner is adenylyl cyclase and the signaling is cAMP upregulation.Based on Figure 1, which of the following is the first step in the activation of cAMP-dPK?
A. The exchange of GTP for GDP
B. The binding of epinephrine
C. The separation of α-GTP
D. Phosphorylation of intracellular enzymes
|
B According to the figure, the first step in the activation pathway was the binding of epinephrine, the external ligand.
|
Biology/164
|
Biology
| null |
G-protein coupled receptors (GPCRs) are a common type of protein receptor containing seven transmembrane segments and bearing intracellular and extracellular portions. The structure of a typical GPCR and its associated partners is shown below. GPCRs are coupled to G-proteins consisting of three subunits (α, β, and γ) that bind the nucleotides GTP or GDP on the intracellular side of the GPCR. The presence or absence of a ligand on the extracellular side of the GPCR determines whether GDP or GTP will bind to the intracellular G-protein. Extracellular ligand binding initiates GDP being exchanged for GTP and the separation of the G-protein. One subunit will travel to join with a secondary partner and will initiate a cascade of signaling effects within the cell. In the example shown below, the secondary partner is adenylyl cyclase and the signaling is cAMP upregulation.A setup like a GPCR is unnecessary in which of the following situations?
A. The signaling event is very specific.
B. The ligand is a hormone traveling in the bloodstream.
C. The effector molecules are located inside the cytoplasm.
D. The ligand is a small nonpolar molecule.
|
D G-protein coupled receptors (GPCRs) are necessary to bring a signal from outside the cell to inside the cell. If the ligand is a small nonpolar molecule, then it can go through the membrane without trouble and does not need an extracellular receptor.
|
Biology/165
|
Biology
| null |
Which of the following could NOT be represented by Figure 1?
A. A monomer used to build a protein
B. A nucleotide
C. A reactant used to synthesize nucleic acids
D. A chemical used to relay genetic information
|
A Figure 1 is adenine, which is a nucleotide. Nucleotides are the monomers of nucleic acids, so the image represents (B) and (C). According to the information in the passage, nucleic acids relay genetic information, so (A) is the only answer that incorrectly describes the figure.
|
Biology/166
|
Biology
| null |
According to Figure 1, within the protein ARKKKK60491, which of the following locations contain polypeptide N-termini?
I. Position A
II. Position B
The enzymatic catalysis of a reaction essential in the production of dog saliva is mediated by the protein ARKKKK60491. ARKKKK60491 is a homodimer, with each subunit being 337 amino acids. Figure 1, below, indicates a modular structure of the homodimer. Three positions are indicated on the model. Position A is the active site. Position B is a region known to have a large number of nonpolar residues. Position C is known to have a large number of charged residues.III. Position C
A. I only
B. I and II
C. II and III
D. I and III
|
C The protein is known to be a homodimer, so it must have two polypeptides in it. Each polypeptide has one N-terminus, so this dimer protein would have two. They would not occur at the active site, so Position A should not be in the answer.
|
Biology/167
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Biology
| null |
The enzymatic catalysis of a reaction essential in the production of dog saliva is mediated by the protein ARKKKK60491. ARKKKK60491 is a homodimer, with each subunit being 337 amino acids. Figure 1, below, indicates a modular structure of the homodimer. Three positions are indicated on the model. Position A is the active site. Position B is a region known to have a large number of nonpolar residues. Position C is known to have a large number of charged residues.Which of the following statements best predicts the effect on the protein's structure if the subunits of the homodimer fail to attach?
A. The primary structure will change because the order of amino acids will differ.
B. The secondary structure will differ because hydrogen bonds will occur in different locations and the protein will fold into a different shape.
C. The tertiary structure will be destroyed because the side chain interactions will not occur and the protein will not fold into a 3-dimensional shape.
D. The quaternary structure will be destroyed because the different amino acid chains will not join together.
|
D A homodimer is formed when more than one polypeptide subunit come together. This is defined as a quaternary structure, (D).
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Biology/168
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Biology
| null |
The enzymatic catalysis of a reaction essential in the production of dog saliva is mediated by the protein ARKKKK60491. ARKKKK60491 is a homodimer, with each subunit being 337 amino acids. Figure 1, below, indicates a modular structure of the homodimer. Three positions are indicated on the model. Position A is the active site. Position B is a region known to have a large number of nonpolar residues. Position C is known to have a large number of charged residues.Which of the following is Position B most likely to assist the protein with?
A. Stabilization of the substrate into the transition state during catalysis
B. Transportation of catalyzed cargo from the nucleus to the mitochondria
C. Creation of a homodimer through catalysis of homodimerization
D. Attachment to the membrane by embedding into the phospholipid bilayer
|
D Position B has nonpolar residues and is not the active site, since Position A is the active site. Therefore, (A) is not correct since the substrate does not bind at Position B. It doesnât look like it is where the homodimer is, so (C) is incorrect. There is no information to indicate (B). Nonpolar residues are known to be helpful for inserting into the lipid bilayer, which makes (D) a strong answer.
|
Biology/169
|
Biology
| null |
The enzymatic catalysis of a reaction essential in the production of dog saliva is mediated by the protein ARKKKK60491. ARKKKK60491 is a homodimer, with each subunit being 337 amino acids. Figure 1, below, indicates a modular structure of the homodimer. Three positions are indicated on the model. Position A is the active site. Position B is a region known to have a large number of nonpolar residues. Position C is known to have a large number of charged residues.A mutation is discovered in the gene for ARKKKK60491 that converts positively charged lysine residue into negatively charged glutamic acid residue. This single change directly impacts the location where the substrate to ARKKKK60491 attaches during catalysis. Which position is likely affected by this change?
A. Position A
B. Position B
C. Position C
D. None of the positions will be affected.
|
A The question states that the change affects the place where the substrate binds, which is defined as the active site. Position A is identified as the active site.
|
Biology/170
|
Biology
| null |
A certain gene has been identified on chromosome 2 in a species of butterfly. The mRNA transcript has been found to be 1578 base pairs in length. A portion of the transcript is shown below. The area shown in grey is part of a known ribosome binding site.Where on the strand shown in Figure 1 is a transcription factor likely to bind?
A. Bases 0 to 15
B. Bases 20 to 25
C. Bases -20 to -15
D. None of the above
|
D A transcription factor binds to DNA before RNA is created during transcription. Since this is already a map of an RNA transcript, transcription has already happened and no transcription factors will bind at any location.
|
Biology/171
|
Biology
| null |
A certain gene has been identified on chromosome 2 in a species of butterfly. The mRNA transcript has been found to be 1578 base pairs in length. A portion of the transcript is shown below. The area shown in grey is part of a known ribosome binding site.RNA polymerase transcribes DNA into mRNA by matching base pairs of nucleotides. Which of the following would be the segment of DNA on the coding strand corresponding to bases 0-6 in Figure 1?
A. 5' AAUGCA 3'
B. 5' TTACGT 3'
C. 3' AATGCA 5'
D. 3' TTUGCT 5'
|
B The coding strand is the strand that is identical to the mRNA transcript, except that the mRNA has the nucleotide uracil in place of the nucleotide thymine. It is the partner strand to the coding strand and used as a template during transcription. Therefore, the answer should match the transcript shown in the image with "Uâs" in place of any "Tâs."
|
Biology/172
|
Biology
| null |
A certain gene has been identified on chromosome 2 in a species of butterfly. The mRNA transcript has been found to be 1578 base pairs in length. A portion of the transcript is shown below. The area shown in grey is part of a known ribosome binding site.A transcript was identified in the nucleus that is similar to this transcript. How is it likely different from the mRNA transcript?
A. The nuclear transcript likely contained introns.
B. The nuclear transcript likely contained exons.
C. The nuclear transcript likely contained a poly-A tail.
D. The nuclear transcript likely contained a 3' GTP-Cap.
|
A Before the transcripts are completed and leave the nucleus, they undergo certain modifications including the removal of introns and the addition of a 5â GTP cap and a 3â poly-A tail. The transcript in the nucleus would not have undergone these modifications yet, so it would contain introns.
|
Biology/173
|
Biology
| null |
A certain gene has been identified on chromosome 2 in a species of butterfly. The mRNA transcript has been found to be 1578 base pairs in length. A portion of the transcript is shown below. The area shown in grey is part of a known ribosome binding site.If bases 0-15 were deleted, what would be the likely consequence?
A. No transcription would be allowed to occur.
B. No translation would be allowed to occur.
C. Neither transcription nor translation would occur.
D. No consequence would occur as these bases are non-coding.
|
B The bases at the 5â side of the transcript are in the untranslated region because this is generally the site for ribosome binding. Without this region, the ribosome couldnât bind, and there would be no translation.
|
Biology/174
|
Biology
| null |
Which of the following populations does NOT demonstrate exponential growth?
I. Cougar
II. Chipmunk
In a heavily populated suburb, two cougars were once spotted roaming in a small field. Local wildlife experts, though not surprised, warned the public to be aware of their surroundings and to keep small pets protected. A group of local junior high school students were curious about the population of cougars in the area since no one they asked had ever seen one in the area. With the help of local wildlife enthusiasts and carefully placed motion-activated wildlife cameras, the group of students recorded sightings of animals in a local forest preserve for their entire four years of high school. The results are shown below.III. Deer
A. I only
B. I and II
C. I and III
D. I and II and III
|
D None of the populations demonstrate exponential growth because they each have a carrying capacity that sets a limit on their growth. Only populations growing as fast as they can reproduce at a maximal amount will grow at an exponential rate.
|
Biology/175
|
Biology
| null |
In a heavily populated suburb, two cougars were once spotted roaming in a small field. Local wildlife experts, though not surprised, warned the public to be aware of their surroundings and to keep small pets protected. A group of local junior high school students were curious about the population of cougars in the area since no one they asked had ever seen one in the area. With the help of local wildlife enthusiasts and carefully placed motion-activated wildlife cameras, the group of students recorded sightings of animals in a local forest preserve for their entire four years of high school. The results are shown below.It has been shown that the population size of chipmunks is directly tied to the number of acorns dropped. This can be scientifically summarized by which of the following statements?
A. Acorn number is a density-dependent population factor and affects the carrying capacity of the chipmunk population.
B. Acorn number is a density-dependent population factor and does not affect the carrying capacity of the chipmunk population.
C. Acorn number is a density-independent population factor and affects the carrying capacity of the chipmunk population.
D. Acorn number is a density-independent population factor and does not affect the carrying capacity of the chipmunk population.
|
A If the chipmunk is dependent on the acorns, then the acorns affect the carrying capacity. Since the quantity of acorns would affect a large population differently than a small population, this makes it a density-dependent factor.
|
Biology/176
|
Biology
| null |
In a heavily populated suburb, two cougars were once spotted roaming in a small field. Local wildlife experts, though not surprised, warned the public to be aware of their surroundings and to keep small pets protected. A group of local junior high school students were curious about the population of cougars in the area since no one they asked had ever seen one in the area. With the help of local wildlife enthusiasts and carefully placed motion-activated wildlife cameras, the group of students recorded sightings of animals in a local forest preserve for their entire four years of high school. The results are shown below.Which change will cause fewer skunks to be sighted in the future?
A. An increase in the number of cougars
B. An increase in the number of raccoons
C. An increase in the number of deer
D. It is impossible to make this determination.
|
D There is no information about what species is a predator of skunks, and the population seems to be stable over time, so this is impossible to determine. Even though the skunks and raccoons are both stable, there is no indication that they are related to each other.
|
Biology/177
|
Biology
| null |
In a heavily populated suburb, two cougars were once spotted roaming in a small field. Local wildlife experts, though not surprised, warned the public to be aware of their surroundings and to keep small pets protected. A group of local junior high school students were curious about the population of cougars in the area since no one they asked had ever seen one in the area. With the help of local wildlife enthusiasts and carefully placed motion-activated wildlife cameras, the group of students recorded sightings of animals in a local forest preserve for their entire four years of high school. The results are shown below.Although the data is limited, which of the following populations is the most likely to be a primary food source for the cougars?
A. Skunk
B. Deer
C. Raccoon
D. Chipmunk
|
B The cougar population is shown to be decreasing over time. One possible explanation for the decline may be due to a disappearing food source, so the food source should also show a decrease over time. The only animal that also shows a decreasing population is deer.
|
Biology/178
|
Biology
| null |
In a heavily populated suburb, two cougars were once spotted roaming in a small field. Local wildlife experts, though not surprised, warned the public to be aware of their surroundings and to keep small pets protected. A group of local junior high school students were curious about the population of cougars in the area since no one they asked had ever seen one in the area. With the help of local wildlife enthusiasts and carefully placed motion-activated wildlife cameras, the group of students recorded sightings of animals in a local forest preserve for their entire four years of high school. The results are shown below.Raccoons eat a diet rich in berries. Which statement is the most likely to describe the berry population?
A. A hard frost eliminated nearly all of the berries in 2007, but 2008 was a milder winter.
B. A virus was introduced in 2006 that eliminated a fly using the berry bushes as a niche habitat.
C. Cougars destroyed much of the berry habitat in 2005.
D. The berry population remained steady, with a slight decrease due to lower rainfall in 2008.
|
D The raccoon population seems steady, with a slight dip in 2008. If raccoons rely on berries as a primary food source, then it is likely that the berry population would have also remained stable.
|
Biology/179
|
Biology
| null |
A chain of three small islands was found to be the home to a small species of mouse. The angle of jaw opening was found to vary significantly. The average size angle of the maximal jaw opening found in mice at 10 locations on the three small islands is shown in Figure 1, below.Which of the following statements best supports that the jaw angle on Island B changed via punctuated equilibrium?
A. Fossil evidence has shown that over time the jaw angle on Island B slowly increased.
B. Fossil evidence has shown that the jaw angle of the mice on Island B increased very quickly.
C. Fossil evidence has shown that over time the jaw angle on Island B slowly decreased.
D. Fossil evidence has shown that the jaw angle of the mice on Island B fluctuates over time.
|
B Punctuated equilibrium is defined as an isolated episode of rapid development of a population. A slow change, described in (A) and (C), or a fluctuation, described in (D), do not suggest a quick change.
|
Biology/180
|
Biology
| null |
A chain of three small islands was found to be the home to a small species of mouse. The angle of jaw opening was found to vary significantly. The average size angle of the maximal jaw opening found in mice at 10 locations on the three small islands is shown in Figure 1, below.Skeletons have been found that indicate that mice with jaw angles less than 30 degrees were once found on Island B. Which explanation best accounts for this?
A. A small jaw angle hinders the survival of mice on Island B.
B. A large jaw angle hinders the survival of mice on Island A.
C. A small jaw angle encourages the survival of mice on Island C.
D. A large jaw angle encourages the survival of mice on Island A.
|
A Since the jaw angle on island B is now large, the smaller jaw angle must have disappeared. The small jaw angle is likely selected against, and only mice with large jaws survive on Island B. The survival on the other islands is not directly relevant.
|
Biology/181
|
Biology
| null |
A chain of three small islands was found to be the home to a small species of mouse. The angle of jaw opening was found to vary significantly. The average size angle of the maximal jaw opening found in mice at 10 locations on the three small islands is shown in Figure 1, below.If the mice from Island A and the mice from Island B were placed together, what would likely happen?
A. The mice would mate, but it is impossible to predict jaw angle.
B. The mice would mate, and the jaw angle would be approximately 27°.
C. The mice would mate, and the jaw angle would be either 21° or 32°.
D. The mice would not be capable of mating.
|
A It says in the passage that the mice are the same species. This means they should be able to mate with each other. As we do not know anything about the genetics and inheritance of jaw angle, it is not possible to predict how the jaw angle would be affected.
|
Biology/182
|
Biology
| null |
A chain of three small islands was found to be the home to a small species of mouse. The angle of jaw opening was found to vary significantly. The average size angle of the maximal jaw opening found in mice at 10 locations on the three small islands is shown in Figure 1, below.Which of the following might account for the variations in maximal jaw angle?
A. Different wind speeds on each island
B. Different heights of trees on each island
C. Different sizes of seeds on each island
D. Differences in altitude on each island
|
C The wind speed, tree height, and altitude do not tie directly into the function of the jaw and the size of the opening. Only the size of seeds that the mice eat would directly relate to the mouth size.
|
Biology/183
|
Biology
| null |
A gene responsible for production of hair pigment in dogs is called Fursilla (frsl). A map of the Fursilla locus is shown in Figure 1, below. When expressed, it results in darkly pigmented dog fur. When unexpressed, the hair is devoid of pigmentation and appears pure white. Expression of Fursilla (frsl) depends on the binding/lack of binding of several proteins: Nefur (NEFR), Lesfur (LSFR), and Dirkfur (DRKFR). Figure 2 shows the relative levels of frsl transcript as measured via RT-qPCR when each protein is overexpressed or unexpressed in the cell.How would the overexpression of BLD affect the transcript levels of Fursilla?
A. Fursilla transcript levels would increase.
B. Fursilla transcript levels would decrease.
C. Fursilla transcript levels would be unaffected.
D. Fursilla transcript would completely disappear.
|
A DRKFR is a transcriptional inhibitor, and it significantly decreases Fursilla transcript levels when overexpressed. If BLD binds to the transcriptional inhibitor, DRKFR, then it would reduce inhibition and increase the transcript levels of Fursilla.
|
Biology/184
|
Biology
| null |
A gene responsible for production of hair pigment in dogs is called Fursilla (frsl). A map of the Fursilla locus is shown in Figure 1, below. When expressed, it results in darkly pigmented dog fur. When unexpressed, the hair is devoid of pigmentation and appears pure white. Expression of Fursilla (frsl) depends on the binding/lack of binding of several proteins: Nefur (NEFR), Lesfur (LSFR), and Dirkfur (DRKFR). Figure 2 shows the relative levels of frsl transcript as measured via RT-qPCR when each protein is overexpressed or unexpressed in the cell.An additional protein, Baldidog (BLD), was identified and found to interact with the DRKFR protein. Based on this information, which of the following best describes why these proteins are able to bind together?
A. Complementary nucleotides sequences
B. Amino acid pockets with complementary conformations
C. Opposing regions rich in cytosines and thymines
D. Hydrophobic lipids and hydrophobic molecules
|
B Two proteins fitting together would not occur from similarities in their nucleotide sequences, which is referenced in (A) and (C). As proteins, they should not be formed from lipids, so (D) is incorrect. Proteins are formed from amino acids, and complementary conformations would allow them to bind together.
|
Biology/185
|
Biology
| null |
A gene responsible for production of hair pigment in dogs is called Fursilla (frsl). A map of the Fursilla locus is shown in Figure 1, below. When expressed, it results in darkly pigmented dog fur. When unexpressed, the hair is devoid of pigmentation and appears pure white. Expression of Fursilla (frsl) depends on the binding/lack of binding of several proteins: Nefur (NEFR), Lesfur (LSFR), and Dirkfur (DRKFR). Figure 2 shows the relative levels of frsl transcript as measured via RT-qPCR when each protein is overexpressed or unexpressed in the cell.Based on the data, which of the following most likely describes how the binding of proteins affects the production of Fursilla transcript?
A. Because Fursilla transcript levels change when certain proteins are overexpressed or unexpressed in the cell, Fursilla must interfere with the other proteins.
B. Because Fursilla transcript levels stay the same whether NEFR is overexpressed or unexpressed, NEFR must bind inhibit the production of Fursilla transcript.
C. Because Fursilla transcript levels decrease when LSFR is unexpressed in the cell, LSFR must compete for the binding site or RNA polymerase.
D. Because Fursilla transcript levels increase when DRKFR is unexpressed in the cell, DRKFR must compete for the binding site of RNA polymerase.
|
D If something competes with the polymerase, then it is likely an inhibitor and high amounts of it would decrease transcription. DRKFR is the only protein that significantly decreases Fursilla transcript levels when overexpressed.
|
Biology/186
|
Biology
| null |
A gene responsible for production of hair pigment in dogs is called Fursilla (frsl). A map of the Fursilla locus is shown in Figure 1, below. When expressed, it results in darkly pigmented dog fur. When unexpressed, the hair is devoid of pigmentation and appears pure white. Expression of Fursilla (frsl) depends on the binding/lack of binding of several proteins: Nefur (NEFR), Lesfur (LSFR), and Dirkfur (DRKFR). Figure 2 shows the relative levels of frsl transcript as measured via RT-qPCR when each protein is overexpressed or unexpressed in the cell.Which of the following would likely produce a dog with the darkest fur?
A. High levels of LSFR and high levels of DRKFR
B. High levels of LSFR and low levels of DRKFR
C. Low levels of LSFR and high levels of DRKFR
D. Low levels of LSFR and low levels of DRKFR
|
B Dark fur would be caused by high expression of fursilla. The triggers for that high expression would be an underexpression of DRKFR and high expression of LSFR.
|
Biology/187
|
Biology
| null |
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