AP Biology 细胞周期

AP 生物学· AP 生物学 CED — 细胞通讯和细胞周期· 14 分钟阅读

1. 细胞周期的核心定义和阶段★★☆☆☆⏱ 4 min

细胞周期是真核体细胞经历的有序、受调控的生长、DNA复制和分裂序列，用于产生基因相同的子细胞。它约占AP生物学第4单元内容的三分之一，占AP考试总分的10-15%，经常出现在选择题和自由作答题部分。

📘 定义

细胞周期

真核体细胞中产生两个基因相同子细胞的有序生长、DNA复制和核/质分裂序列。

例:

人类皮肤细胞通过细胞周期定期分裂，替换受损或死亡组织。

G1（第一间隙期）：细胞生长，合成复制机制，通过G1检查点；若未接收到分裂信号则退出进入G0期。G0可以是暂时的（肝细胞），也可以是永久的（神经元、骨骼肌细胞）。
S（合成期）：所有核DNA完成复制；每条染色体现在有两个连接在同一个着丝粒上的相同姐妹染色单体。
G2（第二间隙期）：细胞继续生长，合成有丝分裂所需机制，通过G2检查点，确认DNA复制无错误。
有丝分裂（M）期：包括有丝分裂（核分离）和胞质分裂（细胞质分裂），产生两个子细胞。

📐 例题

一位研究人员测量了二倍体山羊体细胞在G1早期的总DNA含量，记录值为6.8皮克（pg）。G2晚期山羊细胞的预期DNA含量是多少，有丝分裂胞质分裂刚结束的山羊子细胞的预期DNA含量是多少？

1
G1早期发生在DNA复制之前，因此6.8 pg是基线二倍体DNA含量。
2
所有DNA都在S期复制，因此总DNA加倍，并在分裂前的G2期一直保持加倍状态。
3
$6.8 \text{ pg} \times 2 = 13.6 \text{ pg}$
4
After mitosis and cytokinesis, each daughter cell receives half the G2 DNA content, equal to the parent's original G1 content:
5
$13.6 \text{ pg} \div 2 = 6.8 \text{ pg}$
6
最终答案：13.6 pg（G2晚期），6.8 pg（胞质分裂后的子细胞）

Exam tip:

如果考试问题要求染色体数而非DNA含量，请记住着丝粒的数量（不是染色单体）决定染色体数，因此S期后染色体数不会改变。

2. 细胞周期检查点和调控★★★☆☆⏱ 5 min

细胞周期检查点是调控控制点，细胞在此处暂停进程，直到条件有利于继续进行。三个主要检查点控制细胞周期进程：

G1（限制点）检查点：G1末期，在进入S期前检查细胞大小足够、营养充足、有生长信号、DNA未受损。
G2检查点：G2末期，在进入M期前确认所有DNA已完全准确复制。
纺锤体（M）检查点：中期末期，在进入后期前检查所有染色体动粒都正确连接到纺锤体微管。

Regulation of checkpoints relies on two key protein groups: cyclins (regulatory proteins whose concentration oscillates throughout the cell cycle) and cyclin-dependent kinases (CDKs) (protein kinases that are always present in the cell in an inactive form). CDKs only become active when bound to a specific cyclin, and active cyclin-CDK complexes phosphorylate target proteins to push the cell through the checkpoint.

📘 定义

M期促进因子（MPF）

MPF

驱动细胞从G2进入M期的关键活性细胞周期蛋白-CDK复合物，由结合CDK的细胞周期蛋白B组成。

例:

MPF活性在中期达到峰值，M期结束后细胞周期蛋白B降解，活性降至低水平。

📐 例题

Researchers measure cyclin B concentration and MPF activity in asynchronous (unsynchronized) dividing hamster cells, with the following data:
| Phase | Cyclin B Concentration (μM) | MPF Activity (units) | |-------|------------------------------|------------------------| | G1 | 0.2 | 0.1 | | S | 0.7 | 0.6 | | G2 | 1.3 | 1.2 | | M | 0.3 | 0.2 |
解释细胞周期蛋白B浓度和MPF活性之间的关系，以及该关系存在的原因。

1
比较趋势：从G1到G2，细胞周期蛋白B浓度升高，MPF活性成比例升高；M期细胞周期蛋白B浓度下降，MPF活性也随之下降。
2
MPF是由细胞周期蛋白B（调节亚基）和CDK（催化亚基）组成的复合物，CDK在细胞周期所有阶段浓度保持恒定。
3
CDK仅在结合细胞周期蛋白B后才有活性，因此更高的细胞周期蛋白B浓度意味着更多活性MPF复合物，更低的浓度意味着更少活性复合物。
4
At the end of M phase, cyclin B is targeted for degradation, which removes the cyclin, inactivates MPF, and allows the cell to exit M phase and return to G1 for the next cycle.

Exam tip:

AP自由作答题几乎总是要求你解释细胞周期蛋白浓度与MPF活性的关系图——一定要明确说明CDK浓度是恒定的，只有细胞周期蛋白浓度周期性变化，因此活性依赖于细胞周期蛋白结合。

3. 细胞周期调控异常与癌症★★★☆☆⏱ 4 min

Cancer is defined as uncontrolled cell division caused by mutations that disrupt cell cycle regulation. For cancer to develop, mutations typically disrupt two classes of cell cycle regulatory genes:

Proto-oncogenes: Normal genes that code for proteins that stimulate cell division (e.g., growth factor receptors, cyclins). A gain-of-function mutation converts a proto-oncogene to an oncogene, causing constant, excessive stimulation of cell division even without growth signals. Only one mutated copy is needed for this effect.
Tumor suppressor genes: Normal genes that code for proteins that inhibit cell division, repair DNA errors, or trigger apoptosis (programmed cell death) for damaged cells (e.g., p53, Rb, BRCA1). Loss-of-function mutations that inactivate both copies of the gene remove the "brake" on cell division, allowing damaged cells to continue dividing and accumulate more mutations.

Normal cells follow density-dependent inhibition (stop dividing when they form a single layer) and anchorage dependence (must attach to a substrate to divide), while cancer cells ignore both rules.

📐 例题

The retinoblastoma (Rb) gene is a tumor suppressor gene that normally prevents excessive progression through the G1 checkpoint. A child inherits one mutated, non-functional copy of the Rb gene, and one normal functional copy. Explain why this child has a >90% lifetime risk of developing retinoblastoma (a retinal cancer), while a child born with two functional Rb copies has a <1% risk.

1
Tumor suppressor genes follow the "two-hit hypothesis": both copies of the gene must be inactivated to eliminate all functional protein.
2
The child already has one heritable non-functional copy (the first "hit") in all retinal cells.
3
Only one additional somatic mutation in the remaining functional copy (the second "hit") in a single retinal cell is needed to completely eliminate functional Rb protein.
4
Without functional Rb, the G1 checkpoint does not halt cell division in cells with damaged DNA, leading to uncontrolled growth and cancer. A child with two functional copies needs two independent mutations in the same cell to eliminate Rb function, which is extremely rare, hence the low risk.

Exam tip:

Always remember the difference between mutation types: gain-of-function for proto-oncogenes/oncogenes (one mutation is enough) vs loss-of-function for tumor suppressors (two hits required). AP exam writers love to test this distinction.

4. AP-Style Practice Problem: Flow Cytometry Calculation★★★★☆⏱ 5 min

📐 例题

Flow cytometry is a technique that measures the DNA content of individual cells in an asynchronous population of dividing cells. A researcher analyzes 10,000 rapidly dividing mouse fibroblast cells, with a total cell cycle length of 22 hours. They count 5,800 cells with 6 pg DNA, 2,700 cells with 12 pg DNA, and 1,500 cells with between 6 and 12 pg DNA. Calculate the length of S phase in these cells, and interpret your result.

1
First, assign DNA content to cell cycle phases: 6 pg = G1 (pre-replication), 12 pg = G2 and M phase (post-replication, pre-cytokinesis), 6-12 pg = actively replicating DNA in S phase.
2
Calculate the percentage of cells in S phase:
3
$\frac{1500}{10000} = 0.15$
4
For an asynchronous population, the percentage of cells in a phase equals the percentage of total cell cycle time spent in that phase.
5
Calculate S phase length:
6
$0.15 \times 22 \text{ hours} = 3.3 \text{ hours}$
7
Interpretation: In this population of mouse fibroblasts, S phase (the phase where DNA is replicated) lasts approximately 3.3 hours, which is consistent with observed S phase durations in mammalian somatic cells.

✓ 快速检测

Test your understanding with this AP-style multiple choice question:

A researcher treats dividing mammalian cells with a chemical inhibitor that prevents cyclin B degradation at the end of metaphase. What effect will this treatment most likely have on the cell cycle?
- A. The cell will immediately exit the cell cycle into G0.
- B. The cell will be unable to enter anaphase and will arrest in metaphase.
- C. MPF will remain active, and the cell will be unable to exit M phase after mitosis.
- D. Cyclin B will bind to CDK to trigger entry into G2 phase from S phase.
显示答案
C —
Cyclin B is the regulatory subunit of MPF. Normally, cyclin B is degraded at the end of metaphase, which inactivates MPF and allows the cell to exit M phase. If cyclin B cannot be degraded, MPF remains active, so the cell cannot exit M phase. Anaphase is triggered by securin degradation, not cyclin B degradation, so B is incorrect.

5. 常见陷阱

错误做法:

Stating that chromosome number doubles after S phase because DNA replication produces two sister chromatids per chromosome.

原因:

Students confuse DNA content (mass) with chromosome number, which is defined by the number of centromeres.

正确做法:

Always count centromeres to get chromosome number – two sister chromatids connected at one centromere count as one chromosome, so chromosome number does not change after S phase.

错误做法:

Claiming CDKs are the proteins whose concentration oscillates during the cell cycle.

原因:

Students mix up the names and roles of the two regulatory complex components.

正确做法:

Remember "cyclIN cycles" – cyclin concentration cycles up and down, CDK concentration is always constant; only CDK activity changes based on cyclin binding.

错误做法:

Stating that a gain-of-function mutation in a tumor suppressor gene causes cancer.

原因:

Students mix up the roles of proto-oncogenes and tumor suppressor genes.

正确做法:

Always associate proto-oncogenes with gain-of-function mutations that produce oncogenes (excess cell division) and tumor suppressors with loss-of-function mutations that remove cell division brakes.

错误做法:

Claiming G0 is always a temporary phase that all cells eventually exit to re-enter the cell cycle.

原因:

Textbooks often describe G0 as a "resting phase", leading students to assume it is only a temporary pause.

正确做法:

Recognize that G0 can be permanent for fully differentiated cells like neurons or skeletal muscle cells, which never divide again in adulthood.

错误做法:

Stating that the spindle checkpoint occurs at the end of prophase, before metaphase.

原因:

Students misremember the order of M phase events and checkpoint function.

正确做法:

Recall the spindle checkpoint checks for attachment of all kinetochores to microtubules after chromosomes align at the metaphase plate, so it occurs at the end of metaphase, before anaphase begins.

6. 速查表

类别	规则/数值	注释
G1期DNA含量	$D N A_{G 1} = C$	复制前二倍体体细胞的基线值
G2期DNA含量	$D N A_{G 2} = 2 C$	DNA doubles in S phase, remains doubled through G2/M
染色体数目规则	Count by number of centromeres	Two sister chromatids = 1 chromosome; no change after S phase
G1检查点位置	End of G1	Checks growth signals, undamaged DNA before S entry
Spindle Checkpoint Location	End of metaphase	Checks kinetochore attachment before anaphase
MPF Composition	Cyclin B + CDK	Triggers entry into M phase from G2
Proto-oncogene Mutation	Gain-of-function (1 hit)	Creates oncogene, stimulates excess cell division
Tumor Suppressor Mutation	Loss-of-function (2 hits)	Removes cell division brake, allows uncontrolled growth

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · AP Biology
关于细胞周期蛋白调控和癌症的FRQ
2022 · AP Biology
关于细胞周期检查点的MCQ
2021 · AP Biology
关于DNA含量计算的FRQ

下一步

Understanding the cell cycle is foundational for many other core AP Biology concepts, including cell signaling, Mendelian genetics, and evolution. Cell cycle dysregulation is a common real-world context for exam questions, and mutations accumulated during cell division are the ultimate source of genetic variation that drives evolutionary change. This topic also connects directly to meiosis, the specialized cell division process that produces gametes for sexual reproduction, which shares core regulatory features with mitosis but has distinct outcomes for genetic diversity. Mastering cell cycle concepts will prepare you to tackle the challenging multi-concept FRQs that make up a large portion of your total AP Biology exam score.