PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 22 marksAP 风格选择题 + 安/卑省考短答 · 共 22 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show any working in the margin. For short-answer items, write concise answers using correct biological terminology. No calculator required for this section.本节包含选择题与短答题。选择题请圈出字母答案，可在空白处写出推理过程。短答题请使用准确的生物学术语简洁作答。本节不需要计算器。
A DNA template strand reads 3'-TAC GGA CCT-5'. What is the sequence of the mRNA produced during transcription?一段 DNA 模板链读序为 3'-TAC GGA CCT-5'。转录产生的 mRNA 序列是什么？
The Meselson-Stahl experiment used two isotopes of nitrogen ($^{14}$N and $^{15}$N) to track newly synthesised DNA strands. Bacteria were grown in $^{15}$N medium for many generations, then switched to $^{14}$N medium and allowed to divide.梅塞尔森-斯塔尔实验使用两种氮同位素（$^{14}$N 和 $^{15}$N）追踪新合成的 DNA 链。细菌在 $^{15}$N 培养基中生长多代，然后转移至 $^{14}$N 培养基中继续分裂。
(a)Describe the density of the DNA molecules expected after ONE generation in $^{14}$N medium, and explain why.描述在 $^{14}$N 培养基中繁殖一代后，DNA 分子的预期密度，并说明原因。[3]
(b)Describe the density pattern expected after TWO generations in $^{14}$N medium.描述在 $^{14}$N 培养基中繁殖两代后，DNA 分子的预期密度模式。[3]
(c)State the model of replication that these results support, and explain how they rule out the conservative model.陈述这些结果所支持的复制模型，并解释它们如何排除了保留性复制模型。[4]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分
Section B · Extended ResponseB 部分 · 简答题
Show each step of reasoning. Use correct biological terminology throughout. Where asked to write sequences, use standard 5' to 3' convention and specify direction. Where asked to explain, give the mechanism not just the outcome.每一步推理都要写出。全程使用准确的生物学术语。书写序列时使用标准 5' 至 3' 约定并标注方向。当被要求解释时，给出机制而不仅仅是结果。
An mRNA has the following sequence: 5'-AUG-UGC-GAA-CCG-UAG-3'. Use the standard genetic code: UGC = Cys, GAA = Glu, CCG = Pro.一条 mRNA 的序列为：5'-AUG-UGC-GAA-CCG-UAG-3'。使用标准遗传密码：UGC = Cys（半胱氨酸）、GAA = Glu（谷氨酸）、CCG = Pro（脯氨酸）。
(a)Identify the start codon and the stop codon in this sequence. State the role of each.识别该序列中的起始密码子和终止密码子。说明各自的作用。[3]
(b)Write the amino acid sequence of the resulting polypeptide chain, in order from N-terminus to C-terminus.按从 N 端到 C 端的顺序写出所得多肽链的氨基酸序列。[2]
(c)Write the anticodon sequence on the tRNA that delivers the amino acid Glu (GAA codon). Explain codon-anticodon base pairing.写出携带氨基酸 Glu（对应密码子 GAA）的 tRNA 上的反密码子序列。解释密码子-反密码子碱基配对。[3]
Consider the original mRNA sequence: 5'-AUG-GCU-AAG-UAA-3', which codes for Met-Ala-Lys (stop). Using the standard genetic code: GCU = Ala, AAG = Lys, GCC = Ala, AAA = Lys, GAU = Asp, UAA = stop.考虑原始 mRNA 序列：5'-AUG-GCU-AAG-UAA-3'，编码 Met-Ala-Lys（终止）。使用标准遗传密码：GCU = Ala、AAG = Lys、GCC = Ala、AAA = Lys、GAU = Asp、UAA = 终止。
(a)A point substitution changes the third codon from AAG to AAA. Classify this mutation and predict its effect on the protein. Justify your answer.一个点替换将第三个密码子从 AAG 改变为 AAA。对该突变进行分类，并预测其对蛋白质的影响。给出论证。[3]
(b)A different substitution changes the second codon from GCU to GAU. Classify this mutation and predict its effect on the protein.另一个替换将第二个密码子从 GCU 改变为 GAU。对该突变进行分类，并预测其对蛋白质的影响。[3]
(c)A single cytosine base is inserted between the first and second codons. Classify this type of mutation and state its typical consequence for a protein.在第一个和第二个密码子之间插入一个胞嘧啶碱基。对该突变类型进行分类，并陈述其对蛋白质的典型后果。[2]
The lac operon in E. coli controls the expression of genes for lactose metabolism. In the absence of lactose, a repressor protein binds to the operator region.大肠杆菌的 lac 操纵子控制乳糖代谢基因的表达。在没有乳糖的情况下，阻遏蛋白结合到操作子区域。
(a)Describe what happens to gene expression when lactose IS present in the cell. Include the role of the inducer molecule.描述当乳糖存在于细胞中时基因表达发生的变化。包括诱导分子的作用。[3]
(b)Explain why this operon model is an example of differential gene expression, and state one advantage of this regulatory mechanism for the bacterium.解释为什么该操纵子模型是差异基因表达的一个例子，并陈述这种调控机制对细菌的一个优点。[3]
A forensic scientist amplifies a DNA sample using PCR and then separates the products using gel electrophoresis.法医科学家使用 PCR 扩增 DNA 样品，然后用凝胶电泳分离产物。
(a)Name the three steps of one PCR cycle in order, and state the purpose of each step.按顺序命名一个 PCR 循环的三个步骤，并说明每个步骤的目的。[3]
(b)On a gel, smaller DNA fragments travel farther from the wells than larger fragments. Identify which band (closest to wells vs. farthest from wells) represents the largest fragment, and explain why.在凝胶上，较小的 DNA 片段比较大的片段迁移距离更远。确定哪条条带（最靠近上样孔还是最远离上样孔）代表最大片段，并解释原因。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
These questions require synthesis across multiple sections. Where asked to draw or describe a diagram, label all key components. Conclude each answer with a sentence connecting your result to its biological significance.这些题目需要综合多个知识点。当被要求绘制或描述图示时，标注所有关键组件。每个答案以一句话总结结果的生物学意义。
The following table summarises the key enzymes in DNA replication. Use the Honors SBI4U level of detail to answer the questions.下表总结了 DNA 复制中的关键酶。使用荣誉 SBI4U 级别的细节回答问题。
(a)Name the enzyme that unwinds the double helix at the replication fork, and describe how it does so.命名在复制叉处解开双螺旋的酶，并描述其作用方式。[2]
(b)Explain why a short RNA primer is required before DNA polymerase III can begin synthesis. Name the enzyme that lays down this primer.解释为什么 DNA 聚合酶 III 开始合成前需要一小段 RNA 引物。命名铺设该引物的酶。[3]
(c)Explain what happens to the RNA primers after DNA synthesis is complete. Name the two enzymes involved in this process.解释 DNA 合成完成后 RNA 引物的命运。命名参与此过程的两种酶。[3]
A researcher discovers a mutation in a eukaryotic gene that causes premature termination of translation. The original mRNA sequence contains the codon GGA (Gly) at position 12. The mutation changes this to UGA. Use the standard genetic code: GGA = Gly, UGA = stop.研究人员发现真核基因中的一个突变导致翻译提前终止。原始 mRNA 序列在第 12 位含有密码子 GGA（Gly，甘氨酸）。突变将其改变为 UGA。使用标准遗传密码：GGA = Gly、UGA = 终止。
(a)Classify the type of point mutation that caused this change (substitution, insertion, or deletion), and identify what type of effect it has on protein function.对导致此变化的点突变类型进行分类（替换、插入或缺失），并确定其对蛋白质功能的影响类型。[2]
(b)The original protein contains 18 amino acids. Predict the length of the mutant protein and explain the consequence for protein function.原始蛋白质含 18 个氨基酸。预测突变蛋白质的长度，并解释对蛋白质功能的后果。[3]
(c)If this mutation occurs in a somatic cell of an adult organism, will it be inherited by the organism's offspring? Explain why or why not.如果这个突变发生在成年个体的体细胞中，它会被该个体的后代遗传吗？解释原因。[3]
A forensic biologist starts with a single double-stranded DNA molecule containing a target sequence of interest. She performs PCR using specific primers that flank the target region.一位法医生物学家从一个含有目标序列的单个双链 DNA 分子开始。她使用侧翼靶区的特异引物进行 PCR。
(a)Calculate the number of double-stranded DNA molecules present after 5 cycles of PCR. Show your reasoning using the formula $2^n$.计算经过 5 个 PCR 循环后存在的双链 DNA 分子数量。用公式 $2^n$ 展示你的推理。[2]
(b)Explain why Taq polymerase (from the bacterium Thermus aquaticus) is used in PCR instead of human DNA polymerase. What property of Taq polymerase is essential for the PCR denaturation step?解释为什么 PCR 中使用 Taq 聚合酶（来自嗜热菌 Thermus aquaticus）而不是人类 DNA 聚合酶。Taq 聚合酶的哪种特性对 PCR 变性步骤至关重要？[3]
(c)A germline mutation changes one base in the target sequence. After PCR amplification and gel electrophoresis, explain whether the band position on the gel would change. Justify your answer.一个生殖细胞突变改变了目标序列中的一个碱基。经过 PCR 扩增和凝胶电泳后，解释凝胶上的条带位置是否会改变。给出论证。[2]
(d)Explain the difference between a germline mutation and a somatic mutation in terms of heritability. In the context of this forensic case, which type of mutation is more significant for identifying a suspect, and why?从可遗传性方面解释生殖细胞突变与体细胞突变的区别。在这个法医案例的背景下，哪种突变对识别嫌疑人更重要，为什么？[3]
🇺🇸 US NGSS美国 NGSSHS-LS1-1 · HS-LS3-1
🇨🇦 Ontario安大略SBI4U Strand D · D3.1 · D3.3 · D3.6
🇨🇦 British Columbia不列颠哥伦比亚Anatomy & Physiology 12: Big Idea 2 (gene expression, replication, biotechnology)解剖与生理 12：大概念 2（基因表达、复制、生物技术）
🇨🇦 Alberta阿尔伯塔Biology 30 Unit C GO3 · C3.2k · C3.3k · C3.6k
Full Syllabus Map lives in ../Study Guides/Unit_6_Molecular_Genetics.html. Note: NGSS assesses molecular genetics at a conceptual level (DNA structure, central dogma, mutations); enzyme-level replication detail and the operon model (Q8) carry the Honors flag.完整大纲对照表见 ../Study Guides/Unit_6_Molecular_Genetics.html。注：NGSS 在概念层面考查分子遗传学（DNA 结构、中心法则、突变）；酶级复制细节和操纵子模型（Q8）标有荣誉级标签。