AP BIOLOGY
Section II
8 Free-Response Questions
Time—80 minutes
Directions: Questions 1 and 2 are long free-response questions that should require about 20 minutes each to answer.
Questions 3 through 8 are short free-response questions that should require about 6 minutes each to answer. Read each question carefully and write your response in the space provided following each question. Only material written in this space will be scored. Answers must be written out. Outline form is not acceptable. It is important that you read each question completely before you begin to write.
1) In 1952, Alfred Hershey and Martha Chase conducted a series of experiments using a T2 bacteriophage. At that time it was generally thought that proteins were the molecules responsible for genetic instructions, but other felt that DNA was responsible. Hershey and Chase conducted a set of experiments to determine which molecule was responsible.
The structure of the T2 bacteriophage was already known, its host was easy to culture, and the method of infection had been visualized using transmission electron microscopy. The T2 bacteriophage infects Escherichia coli. This simple phage consists of a protein outer coat and a double strand of DNA.
These experimental procedures used radioisotopes of phosphorus and sulfur. Two cultures were set up. In one, the phage and initial host E. coli were cultured with radioactive sulfur. In the other, the phage and initial host E. coli were cultured with radioactive phosphorus.
a) Explain, in detail, why radioisotopes of carbon or nitrogen would not have provided the same conclusion and justify your explanation.
In the 1952 experiment, the phage was allowed to infect, replicate, and lyse the initial host E. coli within a culture media that contained radioactive sulfur. Then the culture was separated to harvest the radioactive phage from the remnants of the host cells. Next, the radioactive phages were introduced into a new E. coli culture that did not contain radioactive isotopes. The phages were allowed to infect, but not rupture, the host cells. The culture wasa agitated to dislodge the now empty protein outer coat from the infected bacteria. Then the sample was centrifuged and separated into supertnatant and cells. The supernatant contained the less dense material including the culture media and the empty viral coat.
b) Explain, in detail, why the culture wasn’t allowed to progress beyond infection? Justify your explanation.
c) In the second half of the experiment, the same growth protocol was followed, but instead of radioactive sulfur, the culture media contained radioactive phosphorus. Instead of simply separating the sample into supernatant and cells, the cells were ruptured and further separated into organelles, proteins, and nucleic acids. Predict the location of the largest amount of the radioactive phosphorus at the end of the experiment. Justify your answer.
2) In an investigation of fruit-fly behavior, a covered choice chamber is used to test whether the spatial distribution of flies is affected by the presence of a substance placed at one end of the chamber. To test the flies’ preference for glucose, 60 flies are introduced into the middle of the choice chamber at the insertion point indicated by the arrow in the figure above. A cotton ball soaked with a 10 percent glucose solution is placed at one end of the chamber, and a dry cotton ball with no solution is placed at the other end. The positions of flies are observed and recorded every minute for 10 minutes.
(a) Predict the distribution of flies in the chamber after 10 minutes and justify your prediction.
(b) Propose ONE specific improvement to each of the following parts of the experimental design and explain how the modifications will affect the experiment.
- Experimental control
- Environmental factors
(c) The experiment described above is repeated with ripe bananas at one end and unripe bananas at the other end. Once again the positions of the flies are observed and recorded every minute for 10 minutes. The positions of flies after 1 minute and after 10 minutes are shown in the table below.
DISTRIBUTION OF FLIES IN CHOICE CHAMBERTime (minutes) / Position in Choice Chamber
End with Ripe Banana / Middle / End with Unripe Banana
1 / 21 / 18 / 21
10 / 45 / 3 / 12
Perform a chi-square test on the data for the 10-minute time point in the banana experiment. Specify the null hypothesis that you are testing and enter the values from your calculations in the table below.
(d) Explain whether your hypothesis is supported by the chi-square test and justify your explanation.
(e) Briefly propose a model that describes how environmental cues affect the behavior of the flies in the choice chamber.
3) Survival of organisms depends on adaptive behavior and species interactions. Briefly explain why taxis/kinesis and agnostic behavior increase the survivorship of individuals of a species.
4) In science, it is not sufficient to say data is “close enough” to what was expected; rather, we impose rigorous tests to support the validity of results. One of these is the Chi-Square Goodness of Fit test. Briefly explain how probability is associated with Chi-squared critical values.
5) The processes of DNA replication, transcription, and translation are all enzyme-regulated. Summarize the role of at least four of these enzymes in terms of their biochemical reactions (i.e., “unzip” is not acceptable; explain EXACTLY what is happening at a biochemical level).
6) The table below shows the amino acid sequence of the carboxyl-terminal segment of a conserved polypeptide from four different, but related, species. Each amino acid is represented by a three-letter abbreviation, and the amino acid residues in the polypeptide chains are numbered from the amino end to the carboxyl end. Empty cells indicate no amino acid is present.
(a) Assuming that species I is the ancestral species of the group, explain the most likely genetic change that produced the polypeptide in species II and the most likely genetic change that produced the polypeptide in species III.
(b) Predict the effects of the mutation on the structure and function of the resulting protein in species IV. Justify your prediction.
7) The figure above represents a generalized hormone-signaling pathway. Briefly explain the role of each numbered step in regulating target gene expression.
8) For each of the following mutations, describe how the genetic information is altered and predict one specific effect that could occur with the error:
- nondisjunction
- base-pair deletion resulting in a frameshift within an intron
- base-pair substitution resulting in a different amino acid
AP Biology: Big Idea 3 Free Responses
Answer Section
ESSAY
1. a)
Both carbon and nitrogen are found in many cellular components.
The source of the radioactivity would have been inconclusive since they would have been found everywhere in the culture.
Sulfur is not found in DNA but is found in proteins.
Phosphorus is not found in proteins but is very prevalent in DNA.
b)
The components must be separated before the infected cells rupture.
Rupture would release newly formed phages into the supernatant and away from the enclosure that the infected cell provides.
Radioactivity would have been found in both the supernatant and the cell debris.
No conclusion could be drawn.
c)
Prediction: within the DNA portion
Justification: DNA contains a far greater quantity of phosphorus than proteins.
2. (a) Predict the distribution of flies in the chamber after 10 minutes and justify your prediction. (2 points maximum)
- 1 point for predicting the location of the flies in the choice chamber
- 1 point for justifying the prediction
(b) Propose ONE specific improvement to each of the following parts of the experimental design and explain how the modification will affect the experiment. (4 points maximum)
(c)
(d) Correct explanation with justification of why the stated null hypothesis is rejected or not rejected. Response must clarify each of the following:
degrees of freedom (df) = 2 and p = 0.05 (critical value = 5.99) OR
degrees of freedom (df) = 2 and p = 0.01 (critical value = 9.21
how the calculated test statistic compares to the selected critical value
whether the null hypothesis should be rejected
(e) Briefly propose a model that describes how environmental cues affect the behavior of the flies in the choice chamber. (1 point maximum)
Stimulus --> Response
Input --> (possible integration) --> Output
COMPLETION
3. Taxis/Kinesis
What: Identification of environmental stimulus/trigger (e.g. light, moisture, pH, nutrients, temperature, etc.) (2.5 points)
Why: Adaptiveness of response (e.g. escape from predators, find food, avoid dessication, etc.) (2.5 points)
Agnostic
What: Any behavior associated with competition (aggression, defense, fighting, submission, avoidance) (2.5 points)
Why: Can lead to the formation of social dominance hierarchies (wolves, dogs, etc.)
4. Critical values depend on degrees of freedom
Degrees of freedom = n -1
Critical values based on probability data was due to chance.
Critical values based on probability the difference between the observed and expected is statistically significant.
If X2 > critical value, reject H0 (statistically-significant difference / high probability data was due to chance)
If X2 < critical value, fail to reject H0 (statistically-insignificant difference / low probability data was due to chance)
5. DNA helicase: breaks hydrogen bonds between nitrogenous bases
topoisomerase: loosens the rotational tension in the DNA molecule
DNA polymerase: adds nucleotides to the new strand of DNA / hydrogen bonds new nucleotides to the template strand
DNA ligase: covalently bonds nucleotides together / fills in “nicks” in the new strand
RNA polymerase: adds nucleotides to the new strand of RNA / hydrogen bonds new nucleotides to the template strand
6.
7. Step 1 = hormone/ligand binding to receptor to initiate/trigger/induce signaling OR signal
reception
Step 2 = an intracellular cascade that transduces/amplifies/transfers the signal from plasma
membrane to nucleus (or other cellular effectors)
Step 3 = transcription/expression of target genes is stimulated/repressed
SHORT ANSWER
8. nondisjunction:
description: (1 pt)
- chromosomes fail to separate properly in meiosis
- one cell ends up with two copies of a chromosome, while the other gets none
- aneuploidy
prediction: (1 pt)
- down syndrome
deletion:
description: (1 pt)
- one nucleotide is removed
- reading frame is altered
prediction (1 pt)
- no phenotypic effect (intron)
- may shift past the intron into an exon
substitution:
description: (1 pt)
- one nucleotide is replaced with another
prediction (1 pt)
- could be neutral (wobble)
- could change amino acid and change primary structure of protein