2023

AP Biology
®

Sample Student Responses

and Scoring Commentary

Inside:

Free-Response Question 1
R Scoring Guidelines
R Student Samples
R Scoring Commentary

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AP® Biology 2023 Scoring Guidelines

Question 1: Interpreting and Evaluating Experimental Results

with Experimental Design 9 points

In eukaryotic microorganisms, the PHO signaling pathway regulates the expression of certain genes. These
genes, Pho target genes, encode proteins involved in regulating phosphate homeostasis. When the level of
extracellular inorganic phosphate (Pi) is high, a transcriptional activator Pho4 is phosphorylated by a complex
of two proteins, Pho80–Pho85. As a result, the Pho target genes are not expressed. When the level of
extracellular Pi is low, the activity of the Pho80–Pho85 complex is inhibited by another protein, Pho81,
enabling Pho4 to induce the expression of these target genes. A simplified model of this pathway is shown in
Figure 1.

Figure 1. A simplified model of the regulation of expression of Pho target genes in (A) a high-phosphate
(high-Pi) environment and (B) a low-phosphate (low-Pi) environment

To study the role of the different proteins in the PHO pathway, researchers used a wild-type strain of yeast to
create a strain with a mutant form of Pho81 (pho81mt) and a strain with a mutated form of Pho4 (pho4mt).
In each of these mutant strains, researchers measured the activity of a particular enzyme, APase, which
removes phosphates from its substrates and is encoded by PHO1, a Pho target gene (Table 1). They then
determined the level of PHO1 mRNA relative to that of the wild-type yeast strain, which was set to 10.

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AP® Biology 2023 Scoring Guidelines
TABLE 1. APase ACTIVITY AND RELATIVE AMOUNTS OF PHO1 mRNA IN WILD-TYPE AND MUTANT STRAINS OF
YEAST IN HIGH- AND LOW-PHOSPHATE ENVIRONMENTS
APase Activity in APase Activity in Relative Amounts Relative Amounts
High- Pi Low- Pi of PHO1 mRNA in of PHO1 mRNA in
Yeast Environment Environment High- Pi Low- Pi
Mutation
Strain (mU/mL/OD600) (mU/mL/OD600) Environment Environment
±2SE x ±2SE x ±2SE x ±2SE x
Wild-type None 0.5 ± 0.1 17.3 ± 0.9 0.1 ± 0.0 10 ± 2.0
Nonfunctional
pho81mt 0.4 ± 0.1 0.6 ± 0.1 0.7 ± 0.2 0.9 ± 0.8
Pho81
Nonfunctional
pho4mt 0.5 ± 0.0 0.8 ± 0.2 0.6 ± 0.4 0.3 ± 0.1
Pho4

(a) Describe the effect the addition of a charged phosphate group can have on a protein that 1 point
would cause the protein to become inactive.
• It changes the structure/shape of the protein.
Explain how a signal can be amplified during signal transduction in a pathway such as the 1 point
PHO signaling pathway.
• Each enzyme (in a signal transduction pathway) can act on many copies of a protein.
Total for part (a) 2 points

(b) Based on Table 1, identify a dependent variable in the researchers’ experiment. 1 point
Accept one of the following:
• APase activity
• (Relative) amount of PHO1 (mRNA)
Justify the researchers’ using the wild-type strain for the creation of the mutant strains. 1 point
Accept one of the following:
• It ensures that any observed differences (in experimental results) between the strains
are due to the introduced mutations (and not to other genetic differences between
the yeast strains).
• It ensures that the strains are genetically identical except for the introduced
mutations.
Justify the researchers’ using mutant strains in which only a single component of the 1 point
pathway was mutated in each strain.
Accept one of the following:
• It allows them to test the effect of each mutation separately.
• It allows them to (better) determine which component is responsible for any observed
differences.
Total for part (b) 3 points

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AP® Biology 2023 Scoring Guidelines

(c) Based on the data in Table 1, identify the yeast strain and growth conditions that lead to 1 point
the highest relative amount of PHO1 mRNA.
• Wild-type yeast in a low-Pi environment
Calculate the percent change in APase activity in wild-type yeast cells in a high-Pi 1 point
environment compared with that of wild-type cells in a low-Pi environment.
Accept one of the following:
• 3,360% [(17.3–0.5)/0.5 × 100%]
• -97% [(0.5–17.3)/17.3 × 100%]
Total for part (c) 2 points

(d) In a follow-up experiment, researchers created a strain of yeast with a mutation that 1 point
resulted in a nonfunctional Pho85 protein. Based on Figure 1, predict the effects of this
mutation on PHO1 expression in the mutant strain in a high-Pi environment.
• It/PHO1/Target genes will be expressed.
Provide reasoning to justify your prediction. 1 point
• (In a high-Pi environment) a nonfunctional Pho85 will be unable to
phosphorylate/inhibit Pho4.
Total for part (d) 2 points

Total for question 1 9 points

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Q1 Sample A 1 of 2
Q1 Sample A 2 of 2
Q1 Sample B 1 of 1
Q1 Sample C 1 of 1
AP® Biology 2023 Scoring Commentary

Question 1

Note: Student samples are quoted verbatim and may contain spelling and grammatical errors.

Overview

Question 1 described the PHO signaling pathway, which regulates phosphate homeostasis in yeast. The question
stimulus presented a simplified model of the signal transduction pathway and a data table from an experiment
designed to study the roles of Pho81 and Pho4, two proteins in the PHO pathway.

In part (a) students were expected to describe the effect of adding a charged phosphate group to a protein (Skill
1.A; Learning Objective [LO] SYI-1.C from the AP Biology Course and Exam Description [CED]). Students
were also expected to explain how signals can be amplified in a signal transduction pathway (Skill 1.C;
LO IST-3.D).

In part (b) students were expected to demonstrate understanding of experimental design by identifying a
dependent variable, justifying the researchers’ using a wild-type strain of yeast as the background for creating
mutant strains, and justifying the use of mutant strains that each contained a mutation to a single component of the
PHO pathway (Skill 3.C).

In part (c) students were expected to describe data from the table by identifying the yeast strain and growth
conditions that led to the highest relative amount of PHO1 mRNA (Skill 4.B). Students were then asked to
calculate the percent change in APase activity in wild-type cells exposed to a high extracellular inorganic
phosphate (high- Pi ) environment compared with those exposed to a low- Pi environment (Skill 5.A).

In part (d) students were expected to predict the results of a follow-up experiment that tests the effects of a loss-
of-function mutation to Pho85, another protein in the PHO pathway (Skill 3.B; LO IST-3.G). Students were then
expected to justify their predictions (Skill 6.C).

Sample: 1A
Score: 8

The response earned 1 point in part (a) for describing that “The addition of a charged phosphate group likely
changes the tertiary folding structure of a protein.” The response did not earn a point in part (a) because it does not
explain that each enzyme can act on many copies of a protein. The response earned 1 point in part (b) for identifying
a dependent variable as APase activity. The response earned 1 point in part (b) for justification by stating that the
only genetic difference between the two strains is the induced mutation. The response earned 1 point in part (b) for
justifying that “by using mutant strains that had only a single component mutated, they could know that any
significant differences in enzyme activity were for sure caused by that one mutation.” The response earned 1 point in
part (c) for identifying that the wild-type yeast strain in a low- Pi environment leads to the highest relative amount of
PHO1 mRNA. The response earned 1 point in part (c) for calculating the percent change as 3,360%. The response
earned 1 point in part (d) for predicting a nonfunctional Pho85 would allow for the transcription of “the Pho target
gene.” The response earned 1 point in part (d) for justifying that the mutant Pho85 would not phosphorylate Pho4,
thus allowing the latter to promote transcription of PHO1.

Sample: 1B
Score: 5

The response did not earn a point in part (a) because it does not correctly describe how the addition of a phosphate
group affects protein structure. The response did not earn a point in part (a) because it does not correctly explain how
a signal can be amplified during signal transduction. The response earned 1 point in part (b) for identifying APase
activity. The response earned 1 point in part (b) for justifying the use of the wild-type strain to create mutant strains
to “ensure no other mutations were present in the strain.” The response earned 1 point in part (b) for justifying the

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AP® Biology 2023 Scoring Commentary

Question 1 (continued)

use of mutant strains in which only a single component of the pathway was mutated “so that they can accurately
determine whether or not it is causing differences in APase activity ... If multiple components were mutated, the
researchers would not know which component is causing the difference.” The response earned 1 point in part (c)
for identifying “wild-type, low-pi environment.” The response earned 1 point in part (c) for correctly calculating
the percent change as 97.11%. The response did not earn a point in part (d) because it does not correctly predict
the effects of a mutation of Pho85 on PHO1 expression in a high- Pi environment. The response did not earn a
point in part (d) because it does not justify that Pho4 will not be inhibited.

Sample: 1C
Score: 2

The response did not earn a point in part (a) because it does not correctly describe how the addition of a phosphate
group affects protein structure. The response did not earn a point in part (a) because it does not correctly explain how
a signal can be amplified during signal transduction. The response earned 1 point in part (b) for identifying a
dependent variable as APase activity. The response did not earn a point in part (b) because it does not correctly
justify the use of the wild-type strain to create mutant strains. The response did not earn a point in part (b) because it
does not correctly justify the use of mutant strains in which only a single component of the pathway was mutated.
The response earned 1 point in part (c) for identifying that the wild-type yeast strain in a low- Pi environment leads
to the highest relative amount of PHO1 mRNA. The response did not earn a point in part (c) because it incorrectly
calculates the percent change in APase activity in wild-type yeast cells in a high- Pi environment compared with
wild-type yeast cells in a low- Pi environment. The response did not earn a point in part (d) because it does not
correctly predict the effects of a mutation of Pho85 on PHO1 expression in a high- Pi environment. The response
did not earn a point in part (d) because it does not justify that nonfunctional Pho85 will be unable to inhibit Pho4.

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