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Pre AP Biology -CELLS 2.1 Cell Structure and Function- MCQ Exam Style Questions -New Syllabus

Pre AP Biology -CELLS 2.1 Cell Structure and Function- MCQ Exam Style Questions – New Syllabus 2025-2026

Pre AP Biology -CELLS 2.1 Cell Structure and Function- MCQ Exam Style Questions – Pre AP Biology – per latest Pre AP Biology Syllabus.

Pre AP Biology – MCQ Exam Style Questions- All Topics

Question

Which of these terms best describes a seed?
a. an endosperm
b. a mature ovary
c. a mature ovule
d. a mature megaspore
▶️ Answer/Explanation
Detailed solution

The correct option is c. a mature ovule.
In seed plants, the ovule is the structure that contains the female gametophyte.
After fertilization occurs, the zygote develops into an embryo.
The integuments of the ovule thicken and harden to become the seed coat.
Therefore, a seed is biologically defined as a ripened or mature ovule.
By contrast, a mature ovary typically develops into the fruit.
The endosperm is merely a nutrient tissue found inside many seeds.
A megaspore is a haploid cell that precedes the formation of the ovule’s contents.

Question

Which of the following occurs during translocation of sucrose-rich phloem sap?
a. Companion cells pump sucrose into sieve tube members.
b. Sap flows toward a source as pressure builds up at a sink.
c. Sucrose diffuses into companion cells, and \(H^+\) simultaneously leaves the cells by a different route.
d. Companion cells use energy to load solutes at a source, and the sucrose then follows the concentration gradients to sinks.
▶️ Answer/Explanation
Correct Option: a
Detailed solution

Translocation begins with phloem loading, where sucrose is moved into the phloem.
Companion cells use ATP to actively pump \(H^+\) ions out, creating a gradient.
Sucrose is then moved into companion cells via cotransport with these \(H^+\) ions.
From companion cells, sucrose is pumped into sieve tube members against its concentration gradient.
This high solute concentration causes water to enter from the xylem, increasing turgor pressure.
The resulting pressure gradient drives the bulk flow of sap from source to sink.

Question

A plant cell with a solute potential of $-0.65\text{ MPa}$ maintains a constant volume when bathed in a solution in an open container that has a solute potential of $-0.3\text{ MPa}$. Which of these conclusions can you form from this information?
a. This cell has a pressure potential of $+0.65\text{ MPa}$.
b. This cell has a pressure potential of $+0.35\text{ MPa}$.
c. This cell has a water potential of $-0.65\text{ MPa}$.
d. This cell has a water potential of $0\text{ MPa}$.
▶️ Answer/Explanation
Detailed solution

The correct answer is b.
In an open container, the pressure potential of the solution ($\Psi_{p,\text{sol}}$) is $0\text{ MPa}$.
Therefore, the water potential of the solution is $\Psi_{w,\text{sol}} = \Psi_{s,\text{sol}} = -0.3\text{ MPa}$.
Since the cell maintains a constant volume, it is in dynamic equilibrium with the solution.
This means the cell’s water potential ($\Psi_{w,\text{cell}}$) must also be $-0.3\text{ MPa}$.
Using the formula $\Psi_w = \Psi_s + \Psi_p$, we set $-0.3 = -0.65 + \Psi_{p,\text{cell}}$.
Solving for pressure potential gives $\Psi_{p,\text{cell}} = -0.3 – (-0.65) = +0.35\text{ MPa}$.

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